Mensuration

(i) A triangle ABC having sides of lengths, AB = 7 cm, BC = 5 cm, AC = 8.5 cm.

Given:

Sides: AB = 7 cm, BC = 5 cm, AC = 8.5 cm

Perimeter of triangle = Sum of all sides

Perimeter of triangle = AB + BC + AC

= 7 cm + 5 cm + 8.5 cm

= 20.5 cm

Perimeter of triangle = 20.5 cm.

(ii) A rectangle of length 12 cm and breadth 7.5 cm.

Given:

Length = 12 cm, breadth = 7.5 cm

Perimeter of rectangle = 2(length + breadth)

= 2(12 cm + 7.5 cm)

= 2(19.5 cm)

= 39 cm

Perimeter of rectangle = 39 cm.

(iii) A square having each side of length 4.5 cm.

Given:

Side = 4.5 cm

Perimeter of square = 4 x side

= 4 x 4.5 cm $\quad$[Substituting the value]

= 18 cm

Perimeter of square = 18 cm.

(iv) A rhombus having each side of length 6.3 cm.

Given:

Side = 6.3 cm

Perimeter of rhombus = 4 x side

= 4 x 6.3 cm $\quad$[Substituting the value]

= 25.2 cm

Perimeter of rhombus = 25.2 cm.

Given:

Perimeter of rectangle = 186 m

Breadth = 35 m

Perimeter of rectangle = 2(length + breadth)

186 m = 2(length + 35 m)

186 m = (2 x length + 2 x 35 m)

186 m = 2 x length + 70 m

186 m - 70 m = 2 x length

116 m = 2 x length

⇒ length = $\frac{116}{2}$ m

⇒ length = 58 m

Length of the rectangular field = 58 m.

Given:

Perimeter = 172 m

Perimeter of square = 4 x side

172 m = 4 x side

⇒ side = $\frac{172}{4}$ m

⇒ side = 43 m

Length of each side of the square plot = 43 m.

Given:

Length = 125 m

Breadth = 72 m

Rate = ₹ 27 per metre

Fencing is done along the boundary, so we first find the perimeter.

Perimeter of rectangle = 2(length + breadth)

= 2(125 m + 72 m)

= 2(197 m)

= 394 m

Total cost = Perimeter x Rate

= 394 m x ₹ 27

= ₹ 10638

Total cost of fencing the rectangular plot = ₹ 10638.

Given:

Radius (r) = 17.5 cm

$\pi = \frac{22}{7}$

Circumference of a circle = 2$\pi$r

= 2 × $\frac{22}{7}$ × 17.5 cm

= 2 × 22 × 2.5 cm

= 110 cm

Circumference of the circle = 110 cm.

Given:

Circumference of the circular garden = 66 m

$\pi = \frac{22}{7}$

Circumference of a circle = $\pi d$ $\quad$[d is diameter]

66 m = $\frac{22}{7}$ × d

⇒ d = $\frac{66 \times 7}{22}$ m

⇒ d = $\frac{462}{22}$ m

⇒ d = 21 m

Diameter of the circular garden = 21 m.

Given:

Diameter = 1.12 m

Number of revolutions = 500

$\pi = \frac{22}{7}$

Distance covered in 1 revolution = Circumference of the wheel

Circumference of a circle = $\pi d$ $\quad$[d is diameter]

= $\frac{22}{7}$ x 1.12 m

= $\frac{24.64}{7}$ m

= 3.52 m

Total distance covered = Circumference x Number of Revolutions

= 3.52 m x 500

= 1760 m

Converting it into km:

1 km = 1000 m

∴ 1760 m = 1.76 km

The car covers a distance of 1.76 km.

Given:

Radius (r) = 70 cm

Time = 5 minutes

Number of rotations = 300

$\pi = \frac{22}{7}$

Distance covered in 1 rotation = Circumference of the wheel

Circumference of a circle = 2πr

= 2 × $\frac{22}{7}$ × 70 cm

= 2 × 22 × 10 cm

= 440 cm

Total distance = Circumference × number of rotations

= 440 cm × 300

= 132000 cm

Converting distance into km:

1 km = 100000 cm

∴ 132000 cm = $\frac{132000}{100000}$ km = 1.32 km

Converting time into hours:

1 hour = 60 minutes

∴ 5 minutes = $\frac{5}{60}$ hour = $\frac{1}{12}$ hour

Speed = $\frac{\text{Distance}}{\text{Time}}$

= $\frac{1.32}{\frac{1}{12}}$ km/hr

= 1.32 × 12 km/hr

= 15.84 km/hr

Speed of the cycle = 15.84 km/hr.

Given:

Diameter = 1.6 m

Total distance = 352 m

$\pi = \frac{22}{7}$

Distance covered in 1 revolution = Circumference of the wheel

Circumference of a circle = $\pi d \quad$ [d is diameter]

= $\frac{22}{7}$ × 1.6 m

= $\frac{35.2}{7}$ m

= 5.02 m

Number of revolutions = $\frac{\text{Total distance}}{\text{Circumference}}$

= $\frac{352}{\frac{35.2}{7}}$

= 352 × $\frac{7}{35.2}$

= 10 × 7

= 70

Number of revolutions made by the cycle wheel = 70.

Given:

Side of the square = 16.5 cm

$\pi = \frac{22}{7}$

When a wire is reshaped, its total length remains the same.

Perimeter of square = 4 × side

= 4 × 16.5 cm $\quad$[Substituting the value]

= 66 cm

So, circumference of the circle = 66 cm

Circumference of a circle = 2πr

66 cm = 2 × $\frac{22}{7}$ × r

66 cm = $\frac{44}{7}$ × r

⇒ r = $\frac{66}{\frac{44}{7}}$ cm

⇒ r = 66 × $\frac{7}{44}$ cm

⇒ r = $\frac{21}{2}$ cm

⇒ r = 10.5 cm

Radius of the circle = 10.5 cm.

Given:

Total distance = 1.76 km

Total rotations = 1000

Converting distance into cm:

1 km = 100000 cm

∴ 1.76 km = 1.76 × 100000 cm = 176000 cm

Distance covered in 1 rotation = Circumference of the wheel

= $\frac{\text{Total distance}}{\text{Total rotations}}$

= $\frac{176000}{1000}$ cm

= 176 cm

Circumference of a circle = 2πr

176 cm = 2 × $\frac{22}{7}$ × r

176 cm = $\frac{44}{7}$ × r

⇒ r = $\frac{176}{\frac{44}{7}}$ cm

⇒ r = 176 × $\frac{7}{44}$ cm

⇒ r = 28 cm

Radius of the wheel = 28 cm.

Full Squares: Counting the solid blue boxes in the main body and sleeves, we get approximately 48 full squares.

∴ 48 full squares = 48 cm²

Half or More Squares: The slanted parts of the sleeves and the curved neckline contribute about 12 squares that are half or more filled.

∴ 12 full squares = 12 cm²

Less than Half: We ignore the squares which are less than half.

Total Area: (48 + 12) cm² = 60 cm².

The area of T-shirt = 60 cm².

Full Squares: Counting the complete, solid blue squares within the neck and the main body of the bottle:

Neck: 4 full squares

Main Body: 47 full squares

Total Full Squares = 51

Area of Full Squares = 51 cm²

Half or More Squares: These are the squares along the slanted "shoulders" and the rounded base corners that are at least half-filled.

Shoulder Slants: 3 squares

Rounded Base: 2 squares

Total Half or More Squares = 5

Area of these Squares = 5 cm²

Less than Half Squares: These are the tiny slivers at the very edges of the rounded corners and the shoulder joints.

Ignore these squares and count them as 0 cm²

Total Area = (51 + 5) cm² = 56 cm²

The area of Bottle = 56 cm².

Dividing the shape into rectangles A, B, and C:

Rectangle A (Top Bar): 8 wide x 2 high = 16 cm²

Rectangle B (Middle): 2 wide x 5 high = 10 cm²

Rectangle C (Bottom Bar): 10 wide x 2 high = 20 cm²

Total Area: (16 + 10 + 20) cm² = 46 cm²

Total area = 46 cm².

Dividing the shape into rectangles A, B, C and D:

Rectangle A (Backbone): 2 wide x 12 high = 24 cm²

Rectangle B (Top Arm): 6 wide x 2 high = 12 cm²

Rectangle C (Middle Arm): 4 wide x 2 high = 8 cm²

Rectangle D (Bottom Arm): 6 wide x 2 high = 12 cm²

Total Area: (24 + 12 + 8 + 12) cm² = 56 cm²

Total area = 56 cm².

(i) Length = 16 cm, Breadth = 12 cm

Perimeter of a rectangle = 2(Length + Breadth)

= 2(16 cm + 12 cm)

= 2(28 cm)

= 56 cm

Area of a rectangle = (Length × Breadth)

= (16 cm × 12 cm)

= 192 cm²

Perimeter = 56 cm, Area = 192 cm².

(ii) Length = 9.6 cm, Breadth = 1.5 cm

Perimeter of a rectangle = 2(Length + Breadth)

= 2(9.6 cm + 1.5 cm)

= 2(11.1 cm)

= 22.2 cm

Area of a rectangle = (Length × Breadth)

= (9.6 cm × 1.5 cm)

= 14.4 cm²

Perimeter = 22.2 cm, Area = 14.4 cm².

(iii) Length = 8 m, Breadth = 9 dm

First, let us convert length into dm:

1 m = 10 dm

∴ 8 m = 80 dm

Perimeter of a rectangle = 2(Length + Breadth)

= 2(80 dm + 9 dm)

= 2(89 dm)

= 178 dm

Area of a rectangle = (Length × Breadth)

= (80 dm × 9 dm)

= 720 dm²

Perimeter = 178 dm, Area = 720 dm².

(iv) Length = 6 hm, Breadth = 8 dam

First, let us convert length and breadth into m:

1 hm = 100 m

∴ 6 hm = 600 m

1 dam = 10 m

∴ 8 dam = 80 m

Perimeter of a rectangle = 2(Length + Breadth)

= 2(600 m + 80 m)

= 2(680 m)

= 1360 m

Area of a rectangle = (Length × Breadth)

= (600 m × 80 m)

= 48000 m²

Perimeter = 1360 m, Area = 48000 m².

(v) Length = 1 m 25 cm, Breadth = 80 cm

First, let us convert length to cm:

1 m = 100 cm

∴ 1 m 25 cm = (100 + 25) cm = 125 cm

Perimeter of a rectangle = 2(Length + Breadth)

= 2(125 cm + 80 cm)

= 2(205 cm)

= 410 cm

Area of a rectangle = (Length × Breadth)

= (125 cm × 80 cm)

= 10000 cm²

Perimeter = 410 cm, Area = 10000 cm²

(i)

Given:

Side = 14 cm

Perimeter of a square = 4 × side

= 4 × 14 cm

= 56 cm

Area of a square = (side)²

= (14 cm)²

= 196 cm²

Perimeter = 56 cm, Area = 196 cm²

(ii)

Given:

Side = 3.5 cm

Perimeter of a square = 4 × side

= 4 × 3.5 cm

= 14 cm

Area of a square = (side)²

= (3.5 cm)²

= 12.25 cm²

Perimeter = 14 cm, Area = 12.25 cm²

(iii)

Given:

Side = 1 m 20 cm

First, let us convert side into m:

100 cm = 1 m

∴ 1 m 20 cm = (1 + 0.2) m = 1.2 m

Perimeter of a square = 4 × side

= 4 × 1.2 m

= 4.8 m

Area of a square = (side)²

= (1.2 m)²

= 1.44 m²

Perimeter = 4.8 m, Area = 1.44 m²

Given:

Perimeter = 240 m

Length = 63 m

We know the formula,

Perimeter of a rectangle = 2(Length + Breadth)

240 m = 2(63 m + Breadth)

240 m = (2 × 63 m + 2 × Breadth)

240 m = 126 m + (2 × Breadth)

240 m - 126 m = 2 × Breadth

114 m = 2 × Breadth

⇒ Breadth = $\frac{114}{2}$ m

⇒ Breadth = 57 m

Area of a rectangle = (Length × Breadth)

= (63 m × 57 m)

= 3591 m²

Breadth = 57 m, Area = 3591 m²

Given:

Perimeter = 189 m

Breadth = 10.5 m

We know the formula,

Perimeter of a rectangle = 2(Length + Breadth)

189 m = 2(Length + 10.5 m)

189 m = (2 × Length + 2 × 10.5 m)

189 m = (2 × Length) + 21 m

189 m - 21 m = 2 × Length

168 m = 2 × Length

⇒ Length = $\frac{168}{2}$ m

⇒ Length = 84 m

Area of a rectangle = (Length × Breadth)

= (84 m × 10.5 m)

= 882 m²

Length = 84 m, Area = 882 m²

This problem requires two different measurements: Perimeter for fencing and Area for ploughing.

Given:

Length = 175 m

Breadth = 96 m

Rate of fencing = ₹ 17.50 per metre

Rate of ploughing = ₹ 4.50 per square metre

We know the formula,

Perimeter of a rectangle = 2(Length + Breadth)

= 2(175 m + 96 m)

= 2(271 m)

= 542 m

Total cost of fencing = Perimeter × Rate of fencing

= 542 m × 17.50

= ₹ 9485

Area of a rectangle = (Length × Breadth)

= (175 m × 96 m)

= 16800 m²

Total cost of ploughing = Area × Rate of ploughing

= 16800 m² × ₹ 4.50

= ₹ 75600

Cost of fencing = ₹ 9485, Cost of ploughing = ₹ 75600.

To find the number of tiles, we divide the total area of the floor by the area of a single tile.

Given:

Rate of cost of putting the tiles = ₹ 131.40 per tile.

Floor Length = 10 m = 1000 cm

Floor Breadth = 4.5 m = 450 cm

Side of square tile = 20 cm

Tile is in square shape,

∴ Area of tile = (Side)²

= (20 cm)²

= 400 cm²

Floor is in rectangular shape,

∴ Area of floor = Floor Length × Floor Breadth

= 1000 cm × 450 cm

= 450000 cm²

Number of tiles = $\frac{\text{Area of floor}}{\text{Area of tile}}$

= $\frac{450000}{400}$

= 1125

Cost of putting the tiles = Number of tiles × Rate

= 1125 × ₹ 131.40

= ₹ 147825

Number of tiles = 1125, Cost of putting the tiles = ₹ 147825.

Given:

Area = 1560 m²

Breadth = 24 m

Rate of cost of fencing = ₹ 22.50 per metre

We know the formula,

Area of a rectangle = Length × Breadth

1560 m² = Length × 24 m

⇒ Length = $\frac{1560}{24}$ m

⇒ Length = 65 m

Perimeter of a rectangle = 2(Length + Breadth)

= 2(65 m + 24 m)

= 2(89 m)

= 178 m

Cost of fencing = Perimeter × Rate

= 178 m × ₹ 22.50

= ₹ 4005

Length = 65 m, Perimeter = 178 m, Cost of fencing = ₹ 4005.

Given:

Area = 196 cm²

We know the formula,

Area of a square = (Side)²

196 cm² = (Side)²

⇒ Side = $\sqrt{196}$ cm

⇒ Side = 14 cm

Perimeter of a square = 4 × Side

= 4 × 14 cm

= 56 cm

The perimeter of a square = 56 cm

Given:

Area = 1 hectare

Converting area into m²:

1 hectare (ha) = 10000 m²

We know that,

Area of a square = (Side)²

10000 m² = (Side)²

⇒ Side = $\sqrt{10000}$ m

⇒ Side = 100 m

Perimeter of a square = 4 × Side

= 4 × 100 m

= 400 m

The perimeter of a square field = 400 m.

Given:

Total cost = ₹ 5400

Rate of fencing = ₹ 13.50 per metre

(i) the length of the side of the field

Perimeter of the field = $\frac{\text{Total cost of fencing}}{\text{Rate per metre}}$

= $\frac{5400}{13.50}$ m

= 400 m

We know the formula,

Perimeter = 4 × Side

⇒ Side = $\frac{\text{Perimeter}}{4}$

= $\frac{400}{4}$ m

= 100 m

The length of the side of the field = 100 m.

(ii) the area of the field

Area of a square = (Side)²

= (100 m)²

= 10000 m²

The area of the field = 10000 m².

Given:

Length = 50 m

Rate of cost of levelling = ₹ 12.50 per m²

Total cost = ₹ 20000

(i) the area of the plot

Area = $\frac{\text{Total cost}}{\text{Rate}}$

= $\frac{20000}{12.50}$ m²

= 1600 m²

The area of the plot = 1600 m².

(ii) the breadth of the plot

We know the formula,

Area of a rectangle = Length × Breadth

1600 m² = 50 m × Breadth

⇒ Breadth = $\frac{1600}{50}$ m

⇒ Breadth = 32 m

The breadth of the plot = 32 m.

(iii) the perimeter of the plot

Perimeter of a rectangle = 2(Length + Breadth)

= 2(50 m + 32 m)

= 2(82 m)

= 164 m

The perimeter of the plot = 164 m.

(i)

Let us partition the given figure into suitable rectangles by drawing the dotted lines, as shown below:

Thus, we obtain the rectangles ABKL, CDIJ and EFGH.

∴ Area of shaded figure = Area rectangle ABKL + Area rectangle CDIJ + Area rectangle EFGH

= (AB × AL) + (CD × DI) + (EF × FG)

= (9 m × 2 m) + (2.5 m × (1.5 m + 1 m + 4.5 m)) + (18 m × 1 m)

= 18 m² + (2.5 m × 7 m) + 18 m²

= 18 m² + 17.5 m² + 18 m²

= 53.5 m²

Perimeter of shaded figure = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA

= 9 m + 2.5 m + 2.5 m + 1.5 m + 18 m + 1 m + 18 m + 4.5 m + 2.5 m + 2.5 m + 9 m + 2 m

= 73 m

Area = 53.5 m², Perimeter = 73 m

(ii)

Let us partition the given figure into suitable rectangles by drawing the dotted lines, as shown below:

Thus, we obtain the rectangles ABKJ, BCDE and FGHI.

∴ Area of shaded figure = Area rectangle ABKJ + Area rectangle BCDE + Area rectangle FGHI

= (BJ × KJ) + (CD × BC) + (GH × IH)

= ((2 + 4 + 2 + 6) m × 2 m) + (2 m × 5 m) + (2 m × 4 m)

= (14 m × 2 m) + 10 m² + 8 m²

= 28 m² + 18 m²

= 46 m²

Perimeter of shaded figure = AK + KJ + JI + IH + HG + GF + FE + ED + DC + CA

= 14 m + 2 m + 6 m + 4 m + 2 m + 4 m + 4 m + 5 m + 2 m + 7 m

= 50 m

Area = 46 m², Perimeter = 50 m

(iii)

Let us partition the given figure into suitable rectangles by drawing the dotted lines, as shown below:

Thus, we obtain the rectangles ABQR, EFNO, IJKL, CDPQ and GHLM.

∴ Area of shaded figure = Area rectangle ABQR + Area rectangle EFNO + Area rectangle IJKL + Area rectangle CDPQ + Area rectangle GHLM

= (AR × AB) + (EO × EF) + (JK × IJ) + (CD × CQ) + (GH × GM)

= (9 m × 2 m) + ((9 + 2) m × 2 m) + (9 m × 2 m) + (2 m × 3 m) + (2 m × 3 m)

= 18 m² + (11 m × 2 m) + 18 m² + 6 m² + 6 m²

= 18 m² + 22 m² + 18 m² + 6 m² + 6 m²

= 70 m²

Perimeter of shaded figure = AR + RP + PO + ON + NM + MK + KJ + JI + IH + HG + GF + FE + ED + DC + CB + BA

= 9 m + 4 m + 2 m + 2 m + 2 m + 4 m + 9 m + 2 m + 6 m + 2 m + 6 m + 2 m + 6 m + 2 m + 6 m + 2 m

= 66 m

Area = 70 m², Perimeter = 66 m

(iv)

Let us partition the given figure into suitable rectangles by drawing the dotted lines, as shown below:

Thus, we obtain the rectangles ABCL, IFGH and KDEJ.

∴ Area of shaded figure = Area rectangle ABCL + Area rectangle IFGH + Area rectangle KDEJ

= (AB × BC) + (IF × FG) + (KJ × KD)

= (15 m × 3 m) + ((8 + 8 + 5) m × 2 m) + (8 m × 5 m)

= 45 m² + (21 m × 2 m) + 40 m²

= 85 m² + 42 m²

= 127 m²

Perimeter of shaded figure = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA

= 15 m + 3 m + 5 m + 8 m + 8 m + 2 m + 21 m + 2 m + 8 m + 8 m + 5 m + 3 m

= 88 m

Area = 127 m², Perimeter = 88 m

To find the area of the verandah, we calculate the area of the outer rectangle (room + verandah) and subtract the area of the inner rectangle (room).

Given:

Inner (Room): Length = 10 m, Breadth = 6 m

Outer (Room + Verandah): Since the verandah is 2 m wide on all sides, we add 2 + 2 = 4 m to both dimensions.

Outer Length = 10 + 4 = 14 m

Outer Breadth = 6 + 4 = 10 m

Rate of cost of flooring = ₹ 236 per m²

Area of Inner Room = Inner Length x Inner Breadth

= 10 x 6

= 60 m²

Area of Outer Rectangle = Inner Length x Inner Breadth

= 14 m x 10 m

= 140 m²

Area of Verandah = Outer Area - Inner Area

= 140 m² - 60 m²

= 80 m²

Cost of flooring the verandah:

Total cost = Area of Verandah x Rate

= ₹ 80 x 236

= ₹ 18880

The cost of flooring the verandah = ₹ 18880

In this case, the carpet is smaller than the room because a margin is left all around. We find the area of the carpeted portion.

Given:

Hall: Length = 16 m, Breadth = 12 m

Carpeted Area: A margin of 1 m is left on all sides, so we subtract 1 + 1 = 2 m from the hall's dimensions.

Carpet Length = 16 - 2 = 14 m

Carpet Breadth = 12 - 2 = 10 m

Rate of cost of carpeting = ₹ 615 per m²

Area of Carpet = Carpet Length x Carpet Breadth

= 14 m x 10 m

= 140 m²

Cost of carpeting:

Total cost = Area of Carpet x Rate

= ₹ 140 x 615

= ₹ 86100

The cost of carpeting = ₹ 86100

In this problem, the strip is removed from the inside of the original sheet.

Given:

Original Sheet: Length = 35 cm, Breadth = 25 cm

Remaining Sheet: A 5 cm strip is removed from all sides, so we subtract 5 + 5 = 10 cm from both dimensions.

Remaining Sheet Length = 35 - 10 = 25 cm

Remaining Sheet Breadth = 25 - 10 = 15 cm

Area of Remaining Sheet = Remaining Sheet Length x Remaining Sheet Breadth

= 25 cm x 15 cm

= 375 cm²

Area of Original Sheet = Original Sheet Length x Original Sheet Breadth

= 35 cm x 25 cm

= 875 cm²

Area of Cut-out Strip = Area of Original Sheet - Area of Remaining Sheet

= 875 cm² - 375 cm²

= 500 cm²

Area of Remaining Sheet = 375 cm², Area of Cut-out Strip = 500 cm²

Given:

Outer Field: Length = 116 m, Breadth = 76 m

Inner Field (excluding path): Subtract 2 x 3 m = 6 m from both dimensions.

Inner Length = 116 - 6 = 110 m

Inner Breadth = 76 - 6 = 70 m

Rate = ₹ 42.50 per m²

Area of Outer Field = Outer Length x Outer Breadth

= 116 m x 76 m

= 8816 m²

Area of Inner Field = Inner Length x Inner Breadth

= 110 m x 70 m = 7700 m²

Area of Path = Outer Area - Inner Area

= 8816 m² - 7700 m²

= 1116 m²

Cost needed to gravel the path:

Total cost = Area of Path x Rate

= ₹ 1116 x 42.50

= ₹ 47430

Money needed to gravel the path = ₹ 47430

Given:

Inner Plot: Length = 35 m, Breadth = 40 m

Outer Rectangle (Plot + Path): Add 2 x 2.5 m = 5 m to both dimensions.

Outer Length = 35 + 5 = 40 m

Outer Breadth = 40 + 5 = 45 m

Rate = ₹ 115.60 per m²

Area of Outer Rectangle = Outer Length x Outer Breadth

= 40 m x 45 m = 1800 m²

Area of Inner Plot = Inner Length x Inner Length

= 35 m x 40 m = 1400 m²

Area of Path = Area of Outer Rectangle - Area of Inner Plot

= 1800 m² - 1400 m²

= 400 m²

Money needed for tilling the path = Area of Path x Rate

= ₹ 400 x 115.60

= ₹ 46240

Area of Path = 400 m², Money needed for tilling the path = ₹ 46240

Let ABCD be one path:

Length = 115 m

Breadth = 2 m

Area of ABCD = Length x Breadth

= 115 m x 2 m

= 230 m²

Let EFGH be another path:

Length = 64 m

Breadth = 2.5 m

Area of EFGH = Length x Breadth

= 64 m x 2.5 m

= 160 m²

Then, JKLM is the area common to both:

Length = 2.5 m

Breadth = 2 m

Area of JKLM = Length x Breadth

= 2.5 m x 2 m

= 5 m²

∴ Area of cross-paths = Area ABCD + Area EFGH - Area JKLM

= 230 m² + 160 m² - 5 m²

= 390 m² - 5 m²

= 385 m²

Cost of gravelling the paths:

Rate of gravelling = ₹ 114 per m²

Total cost = Area of cross-paths x Rate

= ₹ 385 x 114

= ₹ 43890

The cost of gravelling the paths = ₹ 43890

Given:

Length of the hall = 22 m

Breadth of the hall = 15.5 m

Area of the hall = Length x Breadth

= 22 m x 15.5 m

= 341 m²

The carpet leaves a strip of 75 cm uncovered all around.

Converting strip width to m:

1 m = 100 cm

∴ 75 cm = 0.75 m

Since the strip is on all sides, we must subtract the width twice from both the length and the breadth of the hall:

Length of the carpet = 22 m - (0.75 m + 0.75 m)

= 22 m - 1.5 m

= 20.5 m

Breadth of the carpet = 15.5 m - (0.75 m + 0.75 m)

= 15.5 m - 1.5 m

= 14 m

Area of the carpet = Length of the carpet x Breadth of the carpet

= 20.5 m x 14 m

= 287 m²

Area of the Uncovered Strip: The strip is the area between the hall floor and the carpet.

Area of strip = Area of hall floor - Area of carpet

= 341 m² - 287 m²

= 54 m²

Area of the carpet = 287 m², Area of strip = 54 m²

(i)

From the figure,

Inner Rectangle Dimensions:

Inner Length = 12 m

Inner Breadth = 9 m

Area of Inner rectangle = Inner Length x Inner Breadth

= 12 m x 9 m

= 108 m²

Outer Rectangle Dimensions:

To find the outer dimensions, add the width of the shaded border to both sides:

Outer Length = 12 m + (3 m + 3 m )

= 12 m + 6 m

= 18 m

Outer Breadth = 9 m + (2 m + 2 m)

= 9 m + 4 m

= 13 m

Area of Outer rectangle = Outer Length x Outer Breadth

= 18 m x 13 m

= 234 m²

Area of shaded region = Area of Outer rectangle - Area of Inner rectangle

= 234 m² - 108 m²

= 126 m²

Area of shaded region = 126 m²

(ii)

From the figure,

Horizontal path dimensions:

Length = 150 m

Breadth = 2.5 m

Area 1 = Length x Breadth

= 150 m x 2.5 m

= 375 m²

Vertical path dimensions:

Length = 50 m

Breadth = 1.6 m

Area 2 = Length x Breadth

= 50 m x 1.6 m

= 80 m²

Overlapping area: The overlap is a rectangle where the widths of the two paths meet.

Overlapping area = 2.5 m x 1.6 m

= 4 m²

Area of the shaded region = (Area 1 + Area 2) - Overlapping area

= (375 m² + 80 m²) - 4 m²

= 455 m² - 4 m²

= 451 m²

Area of shaded region = 451 m²

(i)

Given:

base = 16 cm, height = 7.5 cm

We have the formula,

Area of the triangle = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x 16 cm x 7.5 cm

= $\frac{1}{1}$ x 8 cm x 7.5 cm

= 60 cm²

Area of the triangle = 60 cm²

(ii)

Given:

base = 5.6 m, height = 3.5 m

We have the formula,

Area of the triangle = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x 5.6 m x 3.5 m

= $\frac{1}{1}$ x 2.8 m x 3.5 m

= 9.8 m²

Area of the triangle = 9.8 m²

(iii)

Given:

base = 6.4 m, height = 8 dm

Converting base into dm:

1 m = 10 dm

∴ 6.4 m = 64 dm

We have the formula,

Area of the triangle = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x 64 dm x 8 dm

= $\frac{1}{1}$ x 32 dm x 8 dm

= 256 dm²

Area of the triangle = 256 dm²

(iv)

Given:

base = 9.5 cm, height = 6 mm

Converting height into cm:

1 cm = 10 mm

∴ 6 mm = 0.6 cm

We have the formula,

Area of the triangle = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x 0.6 cm x 9.5 cm

= $\frac{1}{1}$ x 0.3 cm x 9.5 cm

= 2.85 cm²

Area of the triangle = 2.85 cm²

(i)

Given:

area = 28.9 m² base = 8.5 m

We have the formula,

Area of the triangle = $\frac{1}{2}$ x base x height

$$\begin{aligned}\phantom{=} \text{height} = \frac{2 \times \text{Area}}{\text{base}} \\[1em] = \frac{2 \times 28.9 \text{ m}^2}{8.5 \text{ m}} \\[1em] = \frac{57.8 \text{ m}^2}{8.5 \text{ m}} \\[1em] = 6.8 \text{ m}\end{aligned}$$

The height of the triangle = 6.8 m

(ii)

Given:

area = 56 dm² base = 2.8 m

Converting base into dm:

1 m = 10 dm

∴ 2.8 m = 28 dm

We have the formula,

Area of the triangle = $\frac{1}{2}$ x base x height

$$\begin{aligned}\phantom{=} \text{height} = \frac{2 \times \text{Area}}{\text{base}} \\[1em] = \frac{2 \times 56 \text{ dm}^2}{28 \text{ dm}} \\[1em] = \frac{2 \times 2 \text{ dm}^2}{1 \text{ dm}} \\[1em] = 4 \text{ dm}\end{aligned}$$

The height of the triangle = 4 dm

(i)

Given:

area = 4.2 m² height = 2.4 m

We have the formula,

Area of the triangle = $\frac{1}{2}$ x base x height

$$\begin{aligned}\phantom{=} \text{base} = \frac{2 \times \text{Area}}{\text{height}} \\[1em] = \frac{2 \times 4.2 \text{ m}^2}{2.4 \text{ m}} \\[1em] = \frac{1 \times 4.2 \text{ m}^2}{1.2 \text{ m}} \\[1em] = 3.5 \text{ m} \\[1em]\end{aligned}$$

= 3.5 m

The base of the triangle = 3.5 m

(ii)

Given:

area = 2.4 dm², height = 80 cm

Converting area into cm:

1 dm² = 100 cm²

∴ 2.4 dm² = 240 cm²

We have the formula,

Area of the triangle = $\frac{1}{2}$ x base x height

$$\begin{aligned}\phantom{=} \text{base} = \frac{2 \times \text{Area}}{\text{height}} \\[1em] = \frac{2 \times 240 \text{ cm}^2}{80 \text{ cm}} \\[1em] = \frac{2 \times 3 \text{ cm}^2}{1 \text{ cm}} \\[1em] = 6 \text{ cm}\end{aligned}$$

The base of the triangle = 6 cm

Given:

Sides: a = 13 cm, b = 20 cm, c = 21 cm

s = $\frac{1}{2}$ x (a + b + c)

= $\frac{1}{2}$ x (13 + 20 + 21) cm

= $\frac{1}{2}$ x 54 cm

= $\frac{1}{1}$ x 27 cm

= 27 cm

Area of the triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

$$\begin{aligned}= \sqrt{27(27 - 13)(27 - 20)(27 - 21)} \text{ cm}^2 \\[1em] = \sqrt{27 \times 14 \times 7 \times 6} \text{ cm}^2 \\[1em] = \sqrt{(3 \times 3 \times 3) \times (2 \times 7) \times 7 \times (2 \times 3)} \text{ cm}^2 \\[1em] = \sqrt{3^4 \times 2^2 \times 7^2} \text{ cm}^2 \\[1em] = 3^2 \times 2 \times 7 \text{ cm}^2 \\[1em] = 9 \times 14 \text{ cm}^2 \\[1em] = 126 \text{ cm}^2\end{aligned}$$

Largest side = 21 cm

Let the altitude corresponding to the largest side be h cm. Then

$$\begin{aligned}\phantom{=} \text{Area} = \frac{1}{2} \times \text{base} \times h \\[1em] 126 \text{ cm}^2 = \frac{1}{2} \times 21 \times h \\[1em] \Rightarrow h = \frac{126 \times 2}{21} \text{ cm} \\[1em] \Rightarrow h = 6 \times 2 \text{ cm} \\[1em] \Rightarrow h = 12 \text{ cm}\end{aligned}$$

Area = 126 cm², The required height = 12 cm

Given:

Sides: a = 50 cm, b = 48 cm, c = 14 cm

s = $\frac{1}{2}$ x (a + b + c)

= $\frac{1}{2}$ x (50 + 48 + 14) cm

= $\frac{1}{2}$ x 112 cm

= $\frac{1}{1}$ x 56 cm

= 56 cm

Area of the triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

$$\begin{aligned}= \sqrt{56(56 - 50)(56 - 48)(56 - 14)} \text{ cm}^2 \\[1em] = \sqrt{56 \times 6 \times 8 \times 42} \text{ cm}^2 \\[1em] = \sqrt{(8 \times 7) \times (6) \times 8 \times (6 \times 7)} \text{ cm}^2 \\[1em] = \sqrt{8^2 \times 7^2 \times 6^2} \text{ cm}^2 \\[1em] = 8 \times 7 \times 6 \text{ cm}^2 \\[1em] = 336 \text{ cm}^2 \\[1em]\end{aligned}$$

Let the altitude corresponding to the side of 48 cm be h cm. Then

$$\begin{aligned}\phantom{=} \text{Area} = \frac{1}{2} \times \text{base} \times h \\[1em] 336 \text{ cm}^2 = \frac{1}{2} \times 48 \times h \\[1em] \Rightarrow h = \frac{336 \times 2}{48} \text{ cm} \\[1em] \Rightarrow h = 7 \times 2 \text{ cm} \\[1em] \Rightarrow h = 14 \text{ cm}\end{aligned}$$

Area = 336 cm², The required height = 14 cm

Given:

Sides: a = 30 cm, b = 30 cm, c = 48 cm

s = $\frac{1}{2}$ x (a + b + c)

= $\frac{1}{2}$ x (30 + 30 + 48) cm

= $\frac{1}{2}$ x 108 cm

= $\frac{1}{1}$ x 54 cm

= 54 cm

Area of the triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

$$\begin{aligned}= \sqrt{54(54 - 30)(54 - 30)(54 - 48)} \text{ cm}^2 \\[1em] = \sqrt{54 \times 24 \times 24 \times 6} \text{ cm}^2 \\[1em] = \sqrt{(6 \times 9) \times 24^2 \times 6} \text{ cm}^2 \\[1em] = \sqrt{6^2 \times 3^2 \times 24^2} \text{ cm}^2 \\[1em] = 6 \times 3 \times 24 \text{ cm}^2 \\[1em] = 432 \text{ cm}^2 \\[1em]\end{aligned}$$

The area of an isosceles triangle = 432 cm²

Given:

The ratio of height and base of a triangle = 5 : 3

Area of triangle = 43.2 m²

Let the Base (b) = 5x

Let the Height (h) = 3x

Area of triangle = $\frac{1}{2}$ x base x height

43.2 m² = $\frac{1}{2}$ x (5x) x (3x)

43.2 m² = $\frac{1}{2}$ x (15x²)

43.2 m² = $\frac{15 x^2}{2}$

43.2 m² x 2 = 15x²

86.4 m² = 15x²

⇒ x² = $\frac{86.4}{15}$ m²

⇒ x² = 5.76 m²

⇒ x = $\sqrt{5.76}$ m²

⇒ x = 2.4 m

Base = 5x = 5 x 2.4 m = 12 m

Height = 3x = 3 x 2.4 m = 7.2 m

Base = 12 m, Height = 7.2 m

(i)

Given:

Side(a) = 12 cm

Area of the triangle = $\frac{\sqrt{3}}{4}a^2$

$$\begin{aligned}= \frac{\sqrt{3}}{4}(12^2) \text{ cm}^2 \\[1em] = \frac{\sqrt{3}}{4}(144) \text{ cm}^2 \\[1em] = \frac{\sqrt{3}}{1}(36) \text{ cm}^2 \\[1em] = 36\sqrt{3} \text{ cm}^2 \\[1em] = 36 \times 1.73 \text{ cm}^2 \\[1em] = 62.28 \text{ cm}^2\end{aligned}$$

Height of the triangle = $\frac{\sqrt{3}}{2}a$

$$\begin{aligned}= \frac{\sqrt{3}}{2} \times 12 \text{ cm} \\[1em] = \frac{\sqrt{3}}{1} \times 6 \text{ cm} \\[1em] = 6\sqrt{3} \text{ cm} \\[1em] = 6 \times 1.73 \text{ cm} \\[1em] = 10.38 \text{ cm}\end{aligned}$$

Area = 62.28 cm², Height = 10.38 cm

(ii)

Given:

Side(a) = 10 m

Area of the triangle = $\frac{\sqrt{3}}{4}a^2$

$$\begin{aligned}= \frac{\sqrt{3}}{4}(10^2) \text{ m}^2 \\[1em] = \frac{\sqrt{3}}{4}(100) \text{ m}^2 \\[1em] = \frac{\sqrt{3}}{1}(25) \text{ m}^2 \\[1em] = 25\sqrt{3} \text{ m}^2 \\[1em] = 25 \times 1.73 \text{ m}^2 \\[1em] = 43.25 \text{ m}^2\end{aligned}$$

Height of the triangle = $\frac{\sqrt{3}}{2}a$

$$\begin{aligned}= \frac{\sqrt{3}}{2} \times 10 \text{ m} \\[1em] = \frac{\sqrt{3}}{1} \times 5 \text{ m} \\[1em] = 5\sqrt{3} \text{ m} \\[1em] = 5 \times 1.73 \text{ m} \\[1em] = 8.65 \text{ m}\end{aligned}$$

Area = 43.25 m², Height = 8.65 m

(iii)

Given:

Side(a) = 6.4 m

Area of the triangle = $\frac{\sqrt{3}}{4}a^2$

$$\begin{aligned}= \frac{\sqrt{3}}{4}(6.4^2) \text{ m}^2 \\[1em] = \frac{\sqrt{3}}{4}(40.96) \text{ m}^2 \\[1em] = \frac{\sqrt{3}}{1}(10.24) \text{ m}^2 \\[1em] = 10.24\sqrt{3} \text{ m}^2 \\[1em] = 10.24 \times 1.73 \text{ m}^2 \\[1em] = 17.7152 \text{ m}^2\end{aligned}$$

Height of the triangle = $\frac{\sqrt{3}}{2}a$

$$\begin{aligned}= \frac{\sqrt{3}}{2} \times 6.4 \text{ m} \\[1em] = \frac{\sqrt{3}}{1} \times 3.2 \text{ m} \\[1em] = 3.2\sqrt{3} \text{ m} \\[1em] = 3.2 \times 1.73 \text{ m} \\[1em] =5.536 \text{ m}\end{aligned}$$

Area = 17.7152 m², Height = 5.536 m

Let ABC be the right angle triangle.

Hypotenuse AC = 26 cm

One side AB = 10 cm

BC² = AC² - AB²

BC² = 26² - 10²

BC² = 676 - 100

BC² = 576

BC = $\sqrt{576}$

BC = 24 cm

∴ Area of the triangle = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x BC x AB

= $\frac{1}{2}$ x (24 x 10) cm²

= $\frac{1}{2}$ x 240 cm²

= $\frac{1}{1}$ x 120 cm²

= 120 cm²

Area of the triangle = 120 cm²

Given:

Area = 240 cm²

One leg (AB) = 16 cm

Let the other leg (BC) be x.

We know the formula:

∴ Area of the triangle = $\frac{1}{2}$ x base x height

240 cm² = $\frac{1}{2}$ x BC x AB

240 cm² = $\frac{1}{2}$ x (x x 16 cm)

240 cm² = $\frac{1}{1}$ x (x x 8 cm)

⇒ x = $\frac{240 \text{ cm}^2}{8 \text{ cm}}$

⇒ x = 30 cm

∴ BC = 30 cm

Length of the other leg = 30 cm

Given:

Area = 1014 cm²

Let the legs be 3x and 4x

Area of the triangle = $\frac{1}{2}$ x base x height

1014 cm² = $\frac{1}{2}$ x (3x) x (4x)

1014 cm² = $\frac{1}{2}$ x (12x²)

1014 cm² = $\frac{1}{1}$ x 6x²

⇒ x² = $\frac{1014}{6}$ cm²

⇒ x² = 169 cm²

⇒ x = $\sqrt{169}$ cm²

⇒ x = 13 cm

AB = 3x = 3 x 13 = 39 cm

BC = 4x = 4 x 13 = 52 cm

Hypotenuse² = AB² + BC²

AC² = (39² + 52²) cm²

AC² = (1521 + 2704) cm²

AC² = 4225 cm²

AC = $\sqrt{4225}$ cm²

AC = 65 cm

Hypotenuse = 65 cm

Let the sides be 13x, 14x and 15x

Perimeter of the triangle = 84 cm

Perimeter of the triangle = AB + BC + AC

84 cm = 13x + 14x + 15x

84 cm = 42x

⇒ x = $\frac{84}{42}$ cm

⇒ x = 2 cm

∴ AB = 13x = 13 x 2 = 26 cm

BC = 14x = 14 x 2 = 28 cm

AC = 15x = 15 x 2 = 30 cm

s = $\frac{AB + BC + AC}{2}$

s = $\frac{(26 + 28 + 30)}{2}$ cm

s = $\frac{84}{2}$ cm

s = 42 cm

Area of the triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

$$\begin{aligned}= \sqrt{42(42 - 26)(42 - 28)(42 - 30)} \text{ cm}^2 \\[1em] = \sqrt{42 \times 16 \times 14 \times 12} \text{ cm}^2 \\[1em] = \sqrt{(14 \times 3) \times 16 \times 14 \times (4 \times 3)} \text{ cm}^2 \\[1em] = \sqrt{14^2 \times 3^2 \times 4^2 \times 4^2} \text{ cm}^2 \\[1em] = 14 \times 3 \times 4 \times 4 \text{ cm}^2 \\[1em] = 336 \text{ cm}^2\end{aligned}$$

Area of the triangle = 336 cm²

In an isosceles triangle, two sides are equal.

Let the equal sides be a and the base be b = 12 cm.

Perimeter = 32 cm

Perimeter of the triangle = AB + AC + BC

32 cm = a + a + 12

(32 - 12) cm = 2a

20 cm = 2a

⇒ a = $\frac{20}{2}$ cm

⇒ a = 10 cm

∴ AB = AC = 10 cm

The altitude to the base of an isosceles triangle bisects the base. Using the Pythagorean theorem on one of the right-angled halves:

$$\begin{aligned}\phantom{=} h^2 + \left(\frac{b}{2}\right)^2 = a^2 \\[1em] h^2 + 6^2 = 10^2 \\[1em] h^2 + 36 = 100 \\[1em] h^2 = 100 - 36 \\[1em] h^2 = 64 \\[1em] h = \sqrt{64} \\[1em] h = 8\end{aligned}$$

Area of the triangle = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x 12 x 8 cm²

= $\frac{1}{2}$ x 96 cm²

= $\frac{1}{1}$ x 48 cm²

= 48 cm²

Area of the triangle = 48 cm²

Given:

Height = 24.5 m

Total Cost = ₹ 176.40

Rate = 80 paise per m²

Converting rate into rupees:

1 rupee = 100 paise

∴ 80 paise = ₹ 0.80 per m²

Area = $\frac{\text{Total Cost}}{\text{Rate}}$

= $\frac{176.40}{0.80}$ m²

= 220.5 m²

We know the formula:

Area of the triangle = $\frac{1}{2}$ x base x height

220.5 m² = $\frac{1}{2}$ x base x 24.5

220.5 m² = $\frac{1}{1}$ x base x 12.25

220.5 m² = 12.25 x base

⇒ base = $\frac{220.5 \text{ m}^2}{12.25 \text{ m}}$

base = 18 m

To find the area of the quadrilateral ABCD, we can split it into two triangles: the equilateral triangle △ABD and the right-angled triangle △BCD.

Given:

AB = BD = AD = 10 cm

∠BCD = 90°

CD = 8 cm

Since AB = BD = AD = 10 cm, △ABD is an equilateral triangle.

The formula for the area of an equilateral triangle is:

Area (△ABD) = $\frac{\sqrt{3}}{4} \times (\text{side})^2$

$$\begin{aligned}= \frac{1.732}{4} \times (10)^2 \text{ cm}^2 \\[1em] = \frac{1.732}{4} \times 100 \text{ cm}^2 \\[1em] = \frac{1.732}{1} \times 25 \text{ cm}^2 \\[1em] = 43.25 \text{ cm}^2\end{aligned}$$

Area of △BCD:

△BCD is a right-angled triangle at ∠C = 90°.

We know the hypotenuse BD = 10 cm and one side CD = 8 cm.

Let's find the base (BC) using Pythagoras' theorem:

BC² + CD² = BD²

BC² + (8 cm)² = (10 cm)²

BC² + 64 cm² = 100 cm²

BC² = (100 - 64) cm²

BC² = 36 cm²

BC = $\sqrt{36}$ cm²

BC = 6 cm

Area(△BCD) = $\frac{1}{2}$ x base x height

$$\begin{aligned}= \frac{1}{2} \times 6 \times 8 \text{ cm}^2 \\[1em] = \frac{1}{2} \times 48 \text{ cm}^2 \\[1em] = \frac{1}{1} \times 24 \text{ cm}^2 \\[1em] = 24 \text{ cm}^2\end{aligned}$$

Let's find total area of Quadrilateral ABCD:

Area of Quadrilateral ABCD = Area(△ABD) + Area(△BCD)

= 43.25 cm² + 24 cm²

= 67.25 cm²

The total area of the quadrilateral = 67.25 cm²

To calculate the area of the quadrilateral PQRS, we can split it into two right-angled triangles: △ PRS and △ PRQ.

Given:

PQ = 8 cm

RQ = 17 cm

∠RPQ = 90°

RS = 6 cm

∠PRS = 90°

Let's find PR which acts as the height for one triangle and the base for the other.

In the right-angled triangle △ PRQ, the right angle is at P (∠RPQ = 90°).

Using Pythagoras' theorem:

PR² + PQ² = RQ²

PR² + (8 cm)² = (17 cm)²

PR² + 64 cm² = 289 cm²

PR² = 289 cm² - 64 cm²

PR² = 225 cm²

PR = $\sqrt{225}$ cm²

PR = 15 cm

Let's find area of △PRQ:

Area of △PRQ = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x PQ x PR

= $\frac{1}{2}$ x 8 cm x 15 cm

= $\frac{1}{2}$ x 120 cm²

= $\frac{1}{1}$ x 60 cm²

= 60 cm²

Now, let's find area of △PRS:

Area of △PRS = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x RS x PR

= $\frac{1}{2}$ x 6 cm x 15 cm

= $\frac{1}{2}$ x 90 cm²

= $\frac{1}{1}$ x 45 cm²

= 45 cm²

Let's find total Area of Quadrilateral PQRS:

Area of Quadrilateral PQRS = Area of △ PRQ + Area of △ PRS

= 60 cm² + 45 cm²

= 105 cm²

The total area of the quadrilateral = 105 cm²

Given:

AC = 25 cm

BE = 3.6 cm

DF = 2.4 cm

To find the area of this quadrilateral, we treat it as two triangles sharing a common base (the diagonal AC).

Area of △ABC = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x AC x BE

= $\frac{1}{2}$ x 25 cm x 3.6 cm

= $\frac{1}{2}$ x 90 cm²

= $\frac{1}{1}$ x 45 cm²

= 45 cm²

Area of △ACD = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x AC x DF

= $\frac{1}{2}$ x 25 cm x 2.4 cm

= $\frac{1}{2}$ x 60 cm²

= $\frac{1}{1}$ x 30 cm²

= 30 cm²

Let's find the total area of quadrilateral ABCD:

Area of quadrilateral ABCD = Area of △ABC + Area of △ACD

= 45 cm² + 30 cm²

= 75 cm²

The area of the quadrilateral ABCD = 75 cm²

Given:

To find the area of the quadrilateral ABCD, we can split it into two triangles △ABD and △DBC.

Area of △ABD:

The sides of the triangle are a = 48 cm, b = 14 cm, c = 50 cm

Semi perimeter (s1) = $\frac{a + b + c}{2} = \frac{48 + 14 + 50}{2} \text{ cm} = \frac{112}{2} \text{ cm} = 56 \text{ cm}$

Area of △ABD = $\sqrt{s1(s1 - a)(s1 - b)(s1 - c)}$

$$\begin{aligned}= \sqrt{56(56 - 48)(56 - 14)(56 - 50)} \text{ cm}^2 \\[1em] = \sqrt{56 \times 8 \times 42 \times 6} \text{ cm}^2 \\[1em] = \sqrt{(7 \times 8) \times 8 \times (7 \times 6) \times 6} \text{ cm}^2 \\[1em] = \sqrt{7^2 \times 8^2 \times 6^2} \text{ cm}^2 \\[1em] = 7 \times 8 \times 6 \text{ cm}^2 \\[1em] = 336 \text{ cm}^2\end{aligned}$$

Area of △DBC:

The sides of the triangle are a = 50 cm, b = 78 cm, c = 112 cm

Semi perimeter (s2) = $\frac{a + b + c}{2} = \frac{50 + 78 + 112}{2} \text{ cm} = \frac{240}{2} \text{ cm} = 120 \text{ cm}$

Area of △DBC = $\sqrt{s2(s2 - a)(s2 - b)(s2 - c)}$

$$\begin{aligned}= \sqrt{120(120-50)(120-78)(120-112)} \text{ cm}^2 \\[1em] = \sqrt{120 \times 70 \times 42 \times 8} \text{ cm}^2 \\[1em] = \sqrt{(10 \times 12) \times (10 \times 7) \times (6 \times 7) \times 8} \text{ cm}^2 \\[1em] = \sqrt{10^2 \times 7^2 \times (12 \times 6 \times 8)} \text{ cm}^2 \\[1em] = \sqrt{10^2 \times 7^2 \times 576} \text{ cm}^2 \\[1em] = 10 \times 7 \times 24 \text{ cm}^2 \\[1em] = 1680\text{ cm}^2\end{aligned}$$

Let's find the total area of quadrilateral ABCD:

Area of quadrilateral ABCD = Area of △ABD + Area of △DBC

= 336 cm² + 1680 cm²

= 2016 cm²

The area of the quadrilateral ABCD = 2016 cm²

(i)

Given:

base = 14.5 cm, height = 2.4 cm

Area of parallelogram = base x height

= 14.5 cm x 2.4 cm

= 34.8 cm²

Area of parallelogram = 34.8 cm²

(ii)

Given:

base = 3.8 m, height = 1.25 m

Area of parallelogram = base x height

= 3.8 m x 1.25 m

= 4.75 m²

Area of parallelogram = 4.75 m²

(iii) base = 75 cm, height = 1.4 m

Given:

base = 75 cm, height = 1.4 m

Converting base into m:

1 m = 100 cm

∴ 75 cm = 0.75 m

Area of parallelogram = base x height

= 0.75 m x 1.4 m

= 1.05 m²

Area of parallelogram = 1.05 m²

Given:

Area = 96 cm²

height is three-eighths of its base.

Let the base of the parallelogram be x cm.

∴ height = $\frac{3}{8}x$ cm

We know the formula:

Area of parallelogram = base x height

96 cm² = x cm x $\frac{3}{8}x$ cm

96 cm² = $\frac{3}{8}x^2$ cm²

⇒ x² = $\frac{96 \times 8}{3}$

⇒ x² = $\frac{32 \times 8}{1}$

⇒ x² = 256

⇒ x = $\sqrt{256}$

⇒ x = 16

∴ base = 16 cm

height = $\frac{3}{8}x \text{ cm} = \frac{3}{8}(16) \text{ cm} = \frac{3}{1}(2) \text{ cm} = 6 \text{ cm}$

height = 6 cm, base = 16 cm

Given:

AB = 35 cm, BC = 28 cm

Distance between its longer sides = 8 cm

Since opposite sides of a parallelogram are equal,

∴ CD = 35 cm

AD = 28 cm

The "longer sides" are AB and CD, which both measure 35 cm. The distance between these sides is the height which is given as:

h1 = 8 cm.

(i) the area of the parallelogram

We know the formula:

Area of the parallelogram = base x height

= CD x h1

= 35 cm x 8 cm

= 280 cm²

∴ Area of parallelogram = 280 cm²

(ii) the distance between its shorter sides.

The "shorter sides" are BC and AD, which both measure 28 cm.

Let the distance between these sides be h2.

Since the area of the parallelogram is constant 280 cm², we can use the shorter side as the base:

Area of the parallelogram = base x height

280 cm² = BC x h2

280 cm² = 28 cm x h2

⇒ h2 = $\frac{280 \text{ cm}^2}{28 \text{ cm}}$

⇒ h2 = $\frac{10}{1}$ cm

⇒ h2 = 10 cm

∴ The distance between its shorter sides = 10 cm.

(i)

Given:

d1 = 12 cm

d2 = 21 cm

Area of rhombus = $\frac{1}{2}$ x d1 x d2

= $\frac{1}{2}$ x 12 cm x 21 cm

= $\frac{1}{2}$ x 252 cm²

= $\frac{1}{1}$ x 126 cm²

= 126 cm²

∴ Area of rhombus = 126 cm².

(ii) Given:

d1 = 17.8 cm

d2 = 25 cm

Area of rhombus = $\frac{1}{2}$ x d1 x d2

= $\frac{1}{2}$ x 17.8 cm x 25 cm

= $\frac{1}{2}$ x 445 cm²

= $\frac{1}{1}$ x 222.5 cm²

= 222.5 cm²

∴ Area of rhombus = 222.5 cm².

Let ABCD be the rhombus.

AB = BC = CD = DA = 20 cm

Diagonal (d1) AC = 24 cm

In a rhombus, the diagonals bisect each other at right angles 90°.

∴ AO = $\frac{1}{2}$(AC) = $\frac{1}{2}$(24) = 12 cm

∠AOB = 90°

Using Pythagorean Theorem in right angled △AOB, we have:

AB² = AO² + OB²

(20 cm)² = (12 cm)² + OB²

400 cm² = 144 cm² + OB²

⇒ OB² = (400 - 144) cm²

⇒ OB² = 256 cm²

⇒ OB = $\sqrt{256}$ cm²

⇒ OB = 16 cm

Let's find the length of second diagonal d2:

d2 = BD = (2 x OB) = (2 x 16 cm) = 32 cm

∴ Area of a rhombus ABCD = $\frac{1}{2}$ x d1 x d2

= $\frac{1}{2}$ x AC x BD

= $\frac{1}{2}$ x 24 cm x 32 cm

= $\frac{1}{2}$ x 768 cm²

= 384 cm²

∴ Area of rhombus ABCD = 384 cm².

Given:

Area = 234 cm²

d1 = 19.5 cm

d2 = ?

We know the formula:

Area of a rhombus = $\frac{1}{2}$ x d1 x d2

234 cm² = $\frac{1}{2}$ x 19.5 cm x d2

234 cm² = $\frac{19.5 \text{ cm}}{2}$ x d2

234 cm² x 2 = 19.5 cm x d2

468 cm² = 19.5 cm x d2

d2 = $\frac{468 \text{ cm}^2}{19.5 \text{ cm}}$

d2 = 24

∴ The length of the other diagonal is 24 cm.

Given:

PQ = 20 cm

QR = 21 cm

PR = 13 cm

(i) the area of the parallelogram PQRS.

Since opposite sides of a parallelogram are equal,

∴ RS = 20 cm, SP = 21 cm

In △PQR we have:

a = PQ = 20 cm

b = QR = 21 cm

c = PR = 13 cm

semi-perimeter (s) = $\frac{a + b + c}{2}$ = $\frac{20 + 21 + 13}{2} \text{ cm} = \frac{54}{2} \text{ cm} = 27 \text{ cm}$

Now, let's find the area of △PQR:

Area of △PQR = $\sqrt{s(s - a)(s - b)(s - c)}$

$$\begin{aligned}= \sqrt{27(27 - 20)(27 - 21)(27 - 13)} \text{ cm}^2 \\[1em] = \sqrt{27 \times 7 \times 6 \times 14} \text{ cm}^2 \\[1em] = \sqrt{(3 \times 3 \times 3) \times 7 \times (2 \times 3) \times (2 \times 7)} \text{ cm}^2 \\[1em] = \sqrt{3^4 \times 7^2 \times 2^2} \text{ cm}^2 \\[1em] = 3^2 \times 7 \times 2 \text{ cm}^2 \\[1em] = 9 \times 14 \text{ cm}^2 \\[1em] = 126 \text{ cm}^2\end{aligned}$$

Area of parallelogram PQRS = 2 x Area of △PQR

= 2 x 126 cm²

= 252 cm²

∴ Area of parallelogram = 252 cm².

(ii) the distance between the longer sides.

The "longer sides" are QR and PS, which both measure 21 cm.

Let the distance between these sides be h1.

Area of parallelogram PQRS = base x height

= 252 cm² = QR x h1

= 252 cm² = 21 cm x h1

⇒ h1 = $\frac{252 \text{ cm}^2}{21 \text{ cm}}$

⇒ h1 = 12 cm

∴ The distance between the longer sides = 12 cm.

(iii) the distance between the shorter sides.

The "shorter sides" are PQ and RS, which both measure 20 cm.

Let the distance between these sides be h2

Area of parallelogram = base x height

= 252 cm² = RS x h2

= 252 cm² = 20 cm x h2

⇒ h2 = $\frac{252 \text{ cm}^2}{20 \text{ cm}}$

⇒ h2 = 12.6 cm

∴ The distance between the shorter sides = 12.6 cm.

To find the area of the shaded region, we split the figure into two parts: a rectangle (ABCF) and a trapezium (FCDE).

Let's find area of Rectangle ABCF:

AB = 15 cm, AF = 9 cm, ∠FAB = ∠CBA = 90°

Since ED || AB || FC, the figure ABCF has parallel opposite sides and right angles, making it a rectangle.

Length (AB) = 15 cm

Breadth (AF) = 9 cm

Area of rectangle ABCF = Length x Breadth

= 15 cm x 9 cm

= 135 cm²

Now let's find area of Trapezium FCDE:

ED = 8 cm

Since ABCF is a rectangle, the side FC is equal to AB:

Base 1 (FC) = 15 cm

Base 2 (ED) = 8 cm

The total distance between AB and ED is given as 13 cm.

Let's find the height of the trapezium (h):

h = Distance between AB and ED - AF

h = 13 cm - 9 cm

h = 4 cm

Area of trapezium FCDE = $\frac{1}{2}$ x (sum of parallel sides) x height

= $\frac{1}{2}$ x (FC + ED) x h

= $\frac{1}{2}$ x (15 + 8) cm x 4 cm

= $\frac{1}{2}$ x 23 cm x 4 cm

= $\frac{1}{2}$ x 92 cm²

= $\frac{1}{1}$ x 46 cm²

= 46 cm²

∴ Total Area of the Shaded Region = Area of Rectangle ABCF + Area of trapezium FCDE

= 135 cm² + 46 cm²

= 181 cm²

∴ Area of the Shaded Region = 181 cm².

To find the area of the shaded region, subtract the areas of the three unshaded triangles from the total area of the square ABCD.

Let's find area of Square ABCD:

Side of the square = 12 cm $\quad$[Given]

Area of the square = (Side)²

= (12)²

= (12 cm x 12 cm)

= 144 cm²

Now let's find areas of the Unshaded Triangles:

The area of a right-angled triangle is $\frac{1}{2}$ x base x height

(i) Area of △AFG:

Base AG = 7 cm, Height FA = 5 cm

Area of △AFG = $\frac{1}{2}$ x 7 cm x 5 cm

= $\frac{1}{2}$ x 35 cm²

= 17.5 cm²

(ii) Area of △BGC:

Base BG = 5 cm, Height BC = 12 cm

Area of △BGC = $\frac{1}{2}$ x 5 cm x 12 cm

= $\frac{1}{2}$ x 60 cm²

= 30 cm²

(iii) Area of △DEF:

Base DE = (DC - CE) = (12 - 4) cm = 8 cm

Height DF = (AD - FA) = (12 - 5) cm = 7 cm

Area of △DEF = $\frac{1}{2}$ x 8 cm x 7 cm

= $\frac{1}{2}$ x 56 cm²

= 28 cm²

∴ Total Area of the Shaded Region = Area of square ABCD - (Sum areas of the Unshaded Triangles)

= Area of square ABCD - (Area of △AFG + Area of △BGC + Area of △DEF)

= 144 cm² - (17.5 + 30 + 28) cm²

= 144 cm² - 75.5 cm²

= 68.5 cm²

∴ Area of the Shaded Region = 68.5 cm².

To find the total area of this Z-shaped shaded region, we divide it into three separate parallelograms: the top horizontal bar ABCD, the bottom horizontal bar FHIJ, and the middle slanted connector EDGF as shown below:

The area of parallelogram = base x height

(i) Area of the Top Parallelogram ABCD:

Base: CD = (CE + ED) = (15 + 3) m = 18 m

Height: 2.5 m

Area = (18 m x 2.5 m) = 45 m²

(ii) Area of the Middle Slanted Parallelogram EDGF:

Base: GD = (GB - DB) = (27 - 3) m = 24 m

Height: 2 m

Area = (24 m x 2 m) = 48 m²

(iii) Area of the Bottom Parallelogram FHIJ:

Base: FH = (FG + GH) = (3 + 24) m = 27 m

Height = 3 m

Area = (27 m x 3 m) = 81 m²

∴ Area of the Shaded Region = (Area of ABCD + Area of EDGF + Area of FHIJ)

= 45 m² + 48 m² + 81 m²

= 174 m²

Area of the Shaded Region = 174 m².

(i)

To find the area of the shaded region, we break the figure into rectangle (ABDE) and a right-angled triangle (BDC).

In right-angled triangle BDC:

height = CD = (CE - DE) = (17 - 8) cm = 9 cm

BC = 15 cm $\quad$[Given]

Using Pythagoras theorem:

BD² + CD² = BC²

BD² + (9 cm)² = (15 cm)²

BD² + 81 cm ² = 225 cm²

BD² = (225 - 81) cm²

BD² = 144 cm²

BD = $\sqrt{144}$ cm²

BD = 12 cm

Area of triangle BDC = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x 12 cm x 9 cm

= $\frac{1}{2}$ x 108 cm²

= $\frac{1}{1}$ x 54 cm²

= 54 cm²

In rectangle ABDE:

Length = BD = 12 cm

Breadth = AB = 8 cm $\quad$[Given]

Area of rectangle ABDE = Length x Breadth

= 12 cm x 8 cm

= 96 cm²

∴ Area of the Shaded Region = (Area of triangle BDC + Area of rectangle ABDE)

= 54 cm² + 96 cm²

Area of shaded region = 150 cm²

(ii)

To find the area of the shaded region, we break the figure into rectangle (ABEF) and two identical right-angled triangles (AGH and BCD).

In right-angled triangle AGH:

AH = 5 cm

GH = 3 cm

Using Pythagoras Theorem,

AG² + GH² = AH²

AG² + (3 cm)² = (5 cm)²

AG² + 9 cm² = 25 cm²

AG² = (25 - 9) cm²

AG² = 16 cm²

AG = $\sqrt{16}$ cm²

AG = 4 cm

Since the triangles AGH and BCD are identical, their corresponding sides are equal.

∴ BD = 4 cm

Area of Triangles AGH and BCD:

Both triangles are identical with base 3 cm and height 4 cm.

Total Area of 2 Triangles = 2 x ($\frac{1}{2}$ x base x height)

= 2 x ($\frac{1}{2}$ x 3 cm x 4 cm)

= 2 x ($\frac{1}{2}$ x 12 cm²)

= 2 x ($\frac{1}{1}$ x 6 cm²)

= 2 x 6 cm²

= 12 cm²

In rectangle ABEF:

Length = BE = (BD + DE) = (4 + 6) cm = 10 cm

Breadth = EF = 6 cm

Area of rectangle ABEF = Length x Breadth

= 10 cm x 6 cm

= 60 cm²

∴ Area of the Shaded Region = (Area of 2 Triangles AGH and BCD + Area of rectangle ABEF)

= 12 cm² + 60 cm²

Area of shaded region = 72 cm²

(iii)

To find the area of the shaded region, we break the figure into the triangle on the left (ABC), the main central rectangle (BCDI), the smaller rectangle (EFHI), and the triangle on the right (FGH).

In △ABC:

Base: BC = 9 cm

Height: AJ = 6 cm

Area of △ABC = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x 9 cm x 6 cm

= $\frac{1}{2}$ x 54 cm²

= $\frac{1}{1}$ x 27 cm²

= 27 cm²

In rectangle BCDI:

BH = 26 cm

Length of the small section: EF = 8 cm

Since the opposite sides of a triangle are equal, therefore IH = 8 cm

Length BI = (BH - IH) = (26 - 8) cm = 18 cm

Breadth: BC = 9 cm

Area of rectangle BCDI = length x breadth

= 18 cm x 9 cm

= 162 cm²

In △FGH:

FG = GH = FH = 5 cm

Area of △FGH =

$$\begin{aligned}\frac{\sqrt{3}}{4} \times 5^2 \\[1em] = \frac{1.73}{4} \times 25 \\[1em] = 0.4325 \times 25 \\[1em] = 10.8125 \text{ cm}^2\end{aligned}$$

In rectangle EFHI:

Length: EF = 8 cm

Breadth: FH = 5 cm

Area of rectangle EFHI = length x breadth

= 8 cm x 5 cm

= 40 cm²

∴ Area of the shaded region = Area of △ABC + Area of rectangle BCDI + Area of △FGH + Area of rectangle EFHI

= 27 cm² + 162 cm² + 10.8125 cm² + 40 cm²

= 239.8125 cm²

Area of shaded region = 239.8125 cm²

(iv)

To find the area of the shaded region, subtract the area of the unshaded triangle from the area of the rectangle.

In rectangle ABCD:

Length: AB = 25 cm

Breadth: BC = 14 cm

Area of rectangle ABCD = Length x Breadth

= 25 cm x 14 cm

= 350 cm²

In △ABE:

Triangle ABE share the same base as the rectangle (AB) and its vertex E lies on the opposite side CD.

This means the height of the triangle is the same as the width of the rectangle.

Base: AB = 25 cm

Height: EF = 14 cm

Area of △ABE = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x 25 cm x 14 cm

= $\frac{1}{2}$ x 350 cm²

= $\frac{1}{1}$ x 175 cm²

= 175 cm²

∴ Area of shaded region = Area of rectangle ABCD - Area of △ABE

= 350 cm² - 175 cm²

Area of shaded region = 175 cm²

(v)

To find the area of the shaded region, we split the shape into two rectangles (ABCD and EFID) and two identical triangles (EFG and BCH).

In rectangle EFID:

Length: IF = 9 cm

Breadth: EF = 5 cm

Area of rectangle EFID = Length x Breadth

= 9 cm x 5 cm

= 45 cm²

In rectangle ABCD:

Length: CD = (CI + ID) = (9 + 5) cm = 14 cm

Breadth: BC = 5 cm

Area of rectangle ABCD = Length x Breadth

= 14 cm x 5 cm

= 70 cm²

In right-angled △BCH:

Base: BH = (AH - AB) = (17 - 14) cm = 3 cm $\quad$[AB = CD]

Height: BC = 5 cm $\quad$[Given]

Area of △BCH = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x 3 cm x 5 cm

= $\frac{1}{2}$ x 15 cm²

= $\frac{1}{1}$ x 7.5 cm²

= 7.5 cm²

Since △BCH and △EFG are identical, their areas are equal.

∴ Area of △EFG = 7.5 cm²

Area of Shaded Region = Area of rectangle ABCD + Area of rectangle EFID + Sum of area of △BCH and △EFG

= 70 cm² + 45 cm² + (7.5 cm² + 7.5 cm²)

= 115 cm² + 15 cm²

Area of shaded region = 130 cm²

(i)

Given:

r = 2.8 m

Area of circle = πr²

$$\begin{aligned}= \frac{22}{7} \times (2.8 \text{ m})^2 \\[1em] = \frac{22}{7} \times (2.8 \text{ m} \times 2.8 \text{ m}) \\[1em] = \frac{22}{7} \times 7.84 \text{ m}^2 \\[1em] = 22 \times 1.12 \text{ m}^2 \\[1em] = 24.64 \text{ m}^2\end{aligned}$$

Area of circle = 24.64 m²

(ii)

Given:

r = 8.4 m

Area of circle = πr²

$$\begin{aligned}= \frac{22}{7} \times (8.4 \text{ m})^2 \\[1em] = \frac{22}{7} \times (8.4 \text{ m} \times 8.4 \text{ m}) \\[1em] = \frac{22}{7} \times 70.56 \text{ m}^2 \\[1em] = 22 \times 10.08 \text{ m}^2 \\[1em] = 221.76 \text{ m}^2\end{aligned}$$

Area of circle = 221.76 m²

(i)

Given:

d = 18.2 cm

Radius (r) = $\frac{\text{Diameter}}{2}$

r = $\frac{18.2}{2}$ cm = 9.1 cm

Area of circle = πr²

$$\begin{aligned}= \frac{22}{7} \times (9.1 \text{ cm})^2 \\[1em] = \frac{22}{7} \times (9.1 \text{ cm} \times 9.1 \text{ cm}) \\[1em] = \frac{22}{7} \times 82.81 \text{ cm}^2 \\[1em] = 22 \times 11.83 \text{ cm}^2 \\[1em] = 260.26 \text{ cm}^2\end{aligned}$$

Area of circle = 260.26 cm²

(ii)

Given:

d = 30.8 cm

Radius (r) = $\frac{\text{Diameter}}{2}$

r = $\frac{30.8}{2}$ cm = 15.4 cm

Area of circle = πr²

$$\begin{aligned}= \frac{22}{7} \times (15.4 \text{ cm})^2 \\[1em] = \frac{22}{7} \times (15.4 \text{ cm} \times 15.4 \text{ cm}) \\[1em] = \frac{22}{1} \times (2.2 \text{ cm} \times 15.4 \text{ cm}) \\[1em] = (22 \times 2.2 \times 15.4) \text{ cm}^2 \\[1em] = 745.36 \text{ cm}^2\end{aligned}$$

Area of circle = 745.36 cm²

(i)

Given:

r = 10 cm

Area of circle = πr²

= 3.14 x (10 cm)²

= 3.14 x 10 cm x 10 cm

= 3.14 x 100 cm²

= 314 cm²

Area of circle = 314 cm²

(ii)

Given:

r = 15 m

Area of circle = πr²

= 3.14 x (15 m)²

= 3.14 x 15 m x 15 m

= 3.14 x 225 m²

= 706.5 m²

Area of circle = 706.5 m²

(i)

Given:

Area = 55.44 m²

Area of circle = πr²

⇒ r = $\sqrt{\frac{\text{Area}}{\pi}}$

$$\begin{aligned}\Rightarrow r = \sqrt{\frac{55.44}{\frac{22}{7}}} \text{ m}^2 \\[1em] \Rightarrow r = \sqrt{55.44 \times \frac{7}{22}} \text{ m}^2 \\[1em] \Rightarrow r = \sqrt{2.52 \times \frac{7}{1}} \text{ m}^2 \\[1em] \Rightarrow r = \sqrt{17.64} \text{ m}^2 \\[1em] \Rightarrow r = 4.2 \text{ m}\end{aligned}$$

Circumference of circle = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 4.2 \text{ m} \\[1em] = 2 \times \frac{22}{1} \times 0.6 \text{ m} \\[1em] = 2 \times 22 \times 0.6 \text{ m} \\[1em] = 26.4 \text{ m}\end{aligned}$$

radius = 4.2 m, Circumference = 26.4 m

(ii)

Given:

Area = 186.34 cm²

Area of circle = πr²

⇒ r = $\sqrt{\frac{\text{Area}}{\pi}}$

$$\begin{aligned}\Rightarrow r = \sqrt{\frac{186.34}{\frac{22}{7}}} \text{ cm}^2 \\[1em] \Rightarrow r = \sqrt{186.34 \times \frac{7}{22}} \text{ cm}^2 \\[1em] \Rightarrow r = \sqrt{8.47 \times \frac{7}{1}} \text{ cm}^2 \\[1em] \Rightarrow r = \sqrt{59.29} \text{ cm}^2 \\[1em] \Rightarrow r = 7.7 \text{ cm}\end{aligned}$$

Circumference of circle = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 7.7 \text{ cm} \\[1em] = 2 \times \frac{22}{1} \times 1.1 \text{ cm} \\[1em] = 2 \times 22 \times 1.1 \text{ cm} \\[1em] = 48.4 \text{ cm}\end{aligned}$$

radius = 7.7 cm, Circumference = 48.4 cm

(i)

Given:

Area = 200.96 cm²

Area of circle = πr²

⇒ r = $\sqrt{\frac{\text{Area}}{\pi}}$

$$\begin{aligned}\Rightarrow r = \sqrt{\frac{200.96}{3.14}} \text{ cm}^2 \\[1em] \Rightarrow r = \sqrt{64} \text{ cm}^2 \\[1em] \Rightarrow r = 8 \text{ cm}\end{aligned}$$

Circumference of circle = 2πr

= 2 x 3.14 x 8 cm

= 50.24 cm

radius = 8 cm, Circumference = 50.24 cm

(ii)

Given:

Area = 379.94 m²

Area of circle = πr²

⇒ r = $\sqrt{\frac{\text{Area}}{\pi}}$

$$\begin{aligned}\Rightarrow r = \sqrt{\frac{379.94}{3.14}} \text{ m}^2 \\[1em] \Rightarrow r = \sqrt{121} \text{ m}^2 \\[1em] \Rightarrow r = 11 \text{ m}\end{aligned}$$

Circumference of circle = 2πr

= 2 x 3.14 x 11 m

= 69.08 m

radius = 11 m, Circumference = 69.08 m

To find the cost of turfing the remaining portion, we first calculate the areas of the rectangular plot and the circular garden.

In rectangle:

Length = 70 m $\quad$[Given]

Breadth = 40 m $\quad$[Given]

Area of rectangle = Length x Breadth

= 70 m x 40 m

= 2800 m²

Area of circular garden = πr²

radius (r) = 17.5 m $\quad$[Given]

$$\begin{aligned}= \frac{22}{7} \times (17.5 \text{ m})^2 \\[1em] = \frac{22}{7} \times (17.5 \text{ m} \times 17.5 \text{ m}) \\[1em] = \frac{22}{1} \times (2.5 \text{ m} \times 17.5 \text{ m}) \\[1em] = (22 \times 2.5 \times 17.5) \text{ m}^2 \\[1em] = 962.5 \text{ m}^2\end{aligned}$$

Area of the Remaining portion = Area of rectangle - Area of circular garden

= 2800 m² - 962.5 m²

= 1837.5 m²

Cost of Turfing = Remaining portion x Rate

Rate = ₹ 28.50 per sq. metre $\quad$[Given]

Substituting the values in above, we get:

Cost of Turfing = 1837.5 m² x ₹ 28.50

= ₹ 52368.75

The cost of turfing the remaining portion is ₹ 52368.75.

Given:

Area of square = 146.41 cm²

We know the formula,

Area of square = (Side)²

146.41 cm² = (Side)²

⇒ Side = $\sqrt{146.41}$ cm²

⇒ Side = 12.1 cm

Let's find the perimeter (length of wire):

We know the formula,

Perimeter of square = 4 x Side

= 4 x 12.1 cm

= 48.4 cm

∴ Length of wire = 48.4 cm

Let's find the radius of the Circle:

When the wire is bent into a circle, its circumference is equal to the length of the wire (48.4 cm).

We know the formula,

Circumference of circle = 2πr

$$\begin{aligned}48.4 \text{ cm} = 2 \times \frac{22}{7} \times r \\[1em] 48.4 \text{ cm} = \frac{44}{7} \times r \\[1em] r = \frac{48.4 \times 7}{44}\text{ cm} \\[1em] r = \frac{1.1 \times 7}{1}\text{ cm} \\[1em] r = 7.7 \text{ cm} \\[1em]\end{aligned}$$

Area of circle = πr²

$$\begin{aligned}= \frac{22}{7} \times (7.7 \text{ m})^2 \\[1em] = \frac{22}{7} \times (7.7 \text{ m} \times 7.7 \text{ m}) \\[1em] = \frac{22}{1} \times (1.1 \text{ m} \times 7.7 \text{ m}) \\[1em] = (22 \times 1.1 \times 7.7) \text{ m}^2 \\[1em] = 186.34 \text{ m}^2\end{aligned}$$

The area of the circle formed is 186.34 cm².

To find the area of the remaining portion, we subtract the total area of all the circular plates from the initial area of the rectangular sheet.

In rectangle:

Length = 145 cm $\quad$[Given]

Breadth = 32 cm $\quad$[Given]

Area of Sheet = Length x Breadth

= 145 cm x 32 cm

= 4640 cm²

Let's find area of One Circular Plate:

Area of one plate = πr²

Diameter = 8 cm $\quad$[Given]

radius = $\frac{\text{Diameter}}{2} = \frac{8}{2} \text{ cm} = 4 \text{ cm}$

Substituting the values in above, we get:

$$\begin{aligned}= \frac{22}{7} \times (4 \text{ cm})^2 \\[1em] = \frac{22}{7} \times (4 \text{ cm} \times 4 \text{ cm}) \\[1em] = \frac{22}{7} \times (16 \text{ cm}^2) \\[1em] = \frac{352}{7} \text{ cm}^2 \\[1em]\end{aligned}$$

Total Area of all circular Plates = Number of plates x Area of one plate

$$\begin{aligned}= 42 \times \frac{352}{7} \text{ cm}^2 \\[1em] = 6 \times \frac{352}{1} \text{ cm}^2 \\[1em] = 2112 \text{ cm}^2\end{aligned}$$

Area of the Remaining Portion = Area of Sheet - Total Area of all circular Plates

= 4640 cm² - 2112 cm²

= 2528 cm²

The area of the remaining portion of the sheet is 2528 cm².

To find the area of the circle inscribed in the square, we first need to determine the side of the square, which will be equal to the diameter of the circle.

Let's find the Side of the Square:

Area of the square = 784 cm²

We know the formula,

Area of the square = (Side)²

784 cm² = (Side)²

⇒ Side = $\sqrt{784}$ cm²

⇒ Side = 28 cm

∴ Diameter of circle = 28 cm

radius = $\frac{\text{Diameter}}{2} = \frac{28}{2} \text{ cm} = 14 \text{ cm}$

Area of circle = πr²

$$\begin{aligned}= \frac{22}{7} \times (14 \text{ cm})^2 \\[1em] = \frac{22}{7} \times (14 \text{ cm} \times 14 \text{ cm}) \\[1em] = \frac{22}{1} \times (2 \text{ cm} \times 14 \text{ cm}) \\[1em] = (22 \times 28) \text{ cm}^2 \\[1em] = 616 \text{ cm}^2\end{aligned}$$

The area of the inscribed circle is 616 cm².

To find the area of the space enclosed between two concentric circles, we subtract the area of the smaller inner circle from the area of the larger outer circle.

Given:

Radius of the outer circle (R) = 25 cm

Radius of the inner circle (r) = 17 cm

The area of the space between the circles is given by:

Area = Area of Outer circle - Area of inner circle

Area = πR² - πr²

$$\begin{aligned}= \Big(\frac{22}{7} \times (25^2) - \frac{22}{7} \times (17^2)\Big) \text{ cm}^2 \\[1em] = \frac{22}{7} \times (25^2 - 17^2) \text{ cm}^2 \\[1em] = \frac{22}{7} \times (625 - 289) \text{ cm}^2 \\[1em] = \frac{22}{7} \times (336) \text{ cm}^2 \\[1em] = \frac{22}{1} \times (48) \text{ cm}^2 \\[1em] = 1056 \text{ cm}^2\end{aligned}$$

The area of the space enclosed by the two concentric circles is 1056 cm².

To find the area of a path around a circular boundary, we calculate the area of the outer circle and subtract the area of the inner circle.

Let's find the radius of the pool (r):

Area of pool = πr²

$$\begin{aligned}962.5 \text{ m}^2 = \frac{22}{7} \times r^2 \\[1em] r^2 = \frac{962.5 \times 7}{22} \text{ m}^2 \\[1em] r^2 = 43.75 \times 7 \text{ m}^2 \\[1em] r^2 = 306.25 \text{ m}^2 \\[1em] r = \sqrt{306.25} \text{ m}^2 \\[1em] r = 17.5 \text{ m}\end{aligned}$$

Let's find the outer radius (R):

The path is around the pool, so we add the width to the inner radius.

R = (r + width) = (17.5 + 3.5) m = 21 m

Let's find the area of the path:

Area of the path = Area of outer circle - Area of pool

Area = πR² - πr²

$$\begin{aligned}= \Big(\frac{22}{7} \times (21^2) - \frac{22}{7} \times (17.5^2)\Big) \text{ m}^2 \\[1em] = \frac{22}{7} \times (21^2 - 17.5^2) \text{ m}^2 \\[1em] = \frac{22}{7} \times (441 - 306.25) \text{ m}^2 \\[1em] = \frac{22}{7} \times (134.75) \text{ m}^2 \\[1em] = \frac{22}{1} \times (19.25) \text{ m}^2 \\[1em] = 423.5 \text{ m}^2\end{aligned}$$

∴ The area of the path is 423.5 m².

The path is inside the field, so the field radius is the outer radius (R).

Given:

Width of the path = 5 m

Rate of cost of paving the path = ₹ 25 per sq. metre

Outer radius (R) = 41 m

Inner radius (r) = (Outer radius - width) = (41 - 5) m = 36 m

Area of the path = Area of outer circle - Area of inner circle

Area of the path = πR² - πr²

$$\begin{aligned}= \Big(\frac{22}{7} \times (41^2) - \frac{22}{7} \times (36^2)\Big) \text{ m}^2 \\[1em] = \frac{22}{7} \times (41^2 - 36^2) \text{ m}^2 \\[1em] = \frac{22}{7} \times (1681 - 1296) \text{ m}^2 \\[1em] = \frac{22}{7} \times (385) \text{ m}^2 \\[1em] = \frac{22}{1} \times (55) \text{ m}^2 \\[1em] = 1210 \text{ m}^2\end{aligned}$$

Cost of paving the path = Area of the path x Rate

= 1210 m² x ₹ 25

= ₹ 30250

∴ The cost of paving the path is ₹ 30250.

Given:

Area of the ring = 528 cm²

Radius of the outer circle (R) = 17 cm

(i) Let's find the radius of the smaller circle:

Area of ring = Area of outer circle - Area of inner circle

Area of the path = πR² - πr²

$$\begin{aligned}528 \text{ cm}^2 = \Big(\frac{22}{7} \times (17^2) - \frac{22}{7} \times r^2 \Big) \\[1em] 528 \text{ cm}^2 = \frac{22}{7} \times (17^2 - r^2) \\[1em] \frac{528 \times 7}{22} \text{ cm}^2 = (289 - r^2) \\[1em] (24 \times 7) \text{ cm}^2 = (289 - r^2) \\[1em] 168 \text{ cm}^2 = (289 - r^2) \\[1em] r^2 = (289 - 168) \text{ cm}^2 \\[1em] r^2 = 121 \text{ cm}^2 \\[1em] r = \sqrt{121} \text{ cm}^2 \\[1em] r = 11 \text{ cm}\end{aligned}$$

∴ The radius of the smaller circle is 11 cm.

(ii)

Let's find the width of the ring:

The width of the ring is the difference between the outer radius and the inner radius.

Width = (Outer radius - Inner radius) = (R - r) = (17 - 11) cm = 6 cm

∴ The width of the ring is 6 cm.

To solve this, we use the area of the garden to find its radius, and then the combined area of the garden and path to find the outer radius.

(i) the radius of the circular garden.

Let the radius of the garden be r.

Area = πr²

$$\begin{aligned}5544 \text{ m}^2 = \frac{22}{7} \times r^2 \\[1em] r^2 = \frac{5544 \times 7}{22} \text{ m}^2 \\[1em] r^2 = \frac{252 \times 7}{1} \text{ m}^2 \\[1em] r^2 = 1764 \text{ m}^2 \\[1em] r = \sqrt{1764} \text{ m}^2 \\[1em] r = 42 \text{ m}\end{aligned}$$

∴ The radius of the circular garden is 42 m.

(ii) the width of the path.

To find the width, we first need the outer radius (R) of the path.

Total area = Area of garden + Area of path

= 5544 m² + 2002 m²

= 7546 m²

Now, let's solve for the outer radius R:

Total area = πR²

$$\begin{aligned}7546 \text{ m}^2 = \frac{22}{7} \times R^2 \\[1em] R^2 = \frac{7546 \times 7}{22} \text{ m}^2 \\[1em] R^2 = \frac{343 \times 7}{1} \text{ m}^2 \\[1em] R^2 = 2401 \text{ m}^2 \\[1em] R = \sqrt{2401} \text{ m}^2 \\[1em] R = 49 \text{ m}\end{aligned}$$

Now, calculate the width of the path:

Width = (R - r) = (49 - 42) m = 7 m

∴ The width of the path is 7 m.

If we draw a diagonal in a rectangle, it forms a right-angled triangle.

By the Pythagorean theorem:

Diagonal² = Length² + Breadth².

To find the diagonal, we take the square root.

∴ Diagonal² = l² + b²

Diagonal = $\sqrt{l^2 + b^2}$

Hence, option 4 is the correct option.

First, let's find the breadth using the diagonal:

Diagonal = $\sqrt{l^2 + b^2}$

⇒ Breadth = $\sqrt{\text{diagonal}^2 - \text{length}^2}$

⇒ Breadth = $\sqrt{15^2 - 12^2}$ cm²

⇒ Breadth = $\sqrt{225 - 144}$ cm²

⇒ Breadth = $\sqrt{81}$ cm²

⇒ Breadth = 9 cm

Now, Area = Length x Breadth

= 12 cm x 9 cm

= 108 cm²

Hence, option 3 is the correct option.

Let breadth = x

Then length = 3x

Perimeter = 560 m $\quad$[Given]

We know the formula,

Perimeter = 2(Length + Breadth)

560 m = 2(3x + x)

560 m = 2(4x)

560 m = 8x

⇒ x = $\frac{560}{8}$ m

⇒ x = 70 m

∴ length = 3x = (3 x 70) m = 210 m

Hence, option 3 is the correct option.

Like a square, all four sides of a rhombus are equal.

So, Perimeter = a + a + a + a = 4a units

Hence, option 2 is the correct option.

When we draw a diagonal, it splits the square into two identical right-angled triangles.

In any right-angled triangle, the square of the longest side (the hypotenuse) is equal to the sum of the squares of the other two sides.

Hypotenuse² = Side1² + Side2²

For our square, the diagonal (d) is the hypotenuse, and the two sides of the square are the other two sides of the triangle.

∴ d² = a² + a² units

d² = 2a² units

d = $\sqrt{2a}^2$ units

d = a$\sqrt{2}$ units

Hence, option 1 is the correct option.

Area of original square = a²

The side of the new square is the diagonal = a$\sqrt{2}$

Area of new square = (a$\sqrt{2}$)²

= 2a²

Ratio = a² : 2a² = 1 : 2

Hence, option 2 is the correct option.

1 are is the area of a square whose sides are exactly 10 metres long.

To find the area of such a square, we multiply the length by the width:

Area = Side x Side

= 10 m x 10 m

= 100 m²

Hence, option 3 is the correct option.

Given:

Area = A sq. units

Length = l units

We know the formula,

Area of rectangle = Length x Breadth

⇒ Breadth = $\frac{\text{Area}}{\text{Length}}$ units

⇒ Breadth = $\frac{\text{A}}{\text{l}}$ units

Looking at the provided options, no option correctly match the answer. So, none of the options is correct.

Area of square = (Side)²

A sq. units = (Side)²

⇒ Side = $\sqrt{A}$ units

Hence, option 3 is the correct option.

To find the height of an equilateral triangle, draw a perpendicular line (height, h) from the top vertex to the base. This divides the equilateral triangle into two identical right-angled triangles.

The hypotenuse is the side of the original triangle (a).

The base of this smaller right triangle is exactly half of the original base $\Big(\frac{a}{2}\Big)$.

Using Pythagoras' theorem (a² + b² = c²)

$$\begin{aligned}h^2 + \left(\frac{a}{2}\right)^2 = a^2 \\[1em] h^2 + \frac{a^2}{4} = a^2 \\[1em] h^2 = a^2 - \frac{a^2}{4} \\[1em] h^2 = \frac{3a^2}{4} \\[1em] h = \sqrt\frac{3a^2}{4} \\[1em] h = \frac{\sqrt{3}}{2} \text{ a}\end{aligned}$$

Looking at the provided options, no option correctly match the answer. So, none of the options is correct.

$\frac{\sqrt3}{4} a^2$ sq. units

This is the standard formula for the area of an equilateral triangle.

Hence, option 2 is the correct option.

Given:

Sides: a = 13 cm, b = 14 cm, c = 15 cm

Semi-perimeter: s = $\frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} \text{ cm} = \frac{42}{2} \text{ cm} = 21 \text{ cm}$

Area = $\sqrt{s(s - a)(s - b)(s - c)}$

$$\begin{aligned}= \sqrt{21(21 - 13)(21 - 14)(21 - 15)} \text{ cm}^2 \\[1em] = \sqrt{21 \times 8 \times 7 \times 6} \text{ cm}^2 \\[1em] = \sqrt{(3 \times 7) \times (2 \times 2 \times 2) \times 7 \times (2 \times 3)} \text{ cm}^2 \\[1em] = \sqrt{2^4 \times 3^2 \times 7^2} \text{ cm}^2 \\[1em] = 2^2 \times 3 \times 7 \text{ cm}^2 \\[1em] = 4 \times 21 \text{ cm}^2 \\[1em] = 84 \text{ cm}^2\end{aligned}$$

Hence, option 1 is the correct option.

For any parallelogram:

Area = base x height

From the figure,

base = a

height = h

So, Area = a x h

Hence, option 2 is the correct option.

Given:

d1 = 14 cm, d2 = 24 cm

Area of rhombus = $\frac{1}{2}$ x d1 x d2

= $\frac{1}{2}$ x 14 cm x 24 cm

= $\frac{1}{2}$ x 336 cm²

= $\frac{1}{1}$ x 168 cm²

= 168 cm²

Hence, option 2 is the correct option.

We find the area of the big circle (πR²) and subtract the area of the small circle (πr²).

i.e., πR² - πr²

= π(R² - r²)

Hence, option 3 is the correct option.

(i) The perimeter of a simple closed figure is the length of its boundary.

(ii) The perimeter of a circle is called its circumference.

(iii) Pie (π) is the ratio of the circumference of any circle to its diameter.

(iv) The area of a simple closed figure is the measure of the surface enclosed by it.

(v) An area of 1 hectare is equal to 10000 m².

(vi) A unit square grid sheet is a sheet containing horizontal and vertical lines forming squares, each of dimensions 1 cm x 1 cm.

(vii) In a triangle having sides of lengths a units, b units and c units respectively, if s = $\frac{a + b + c}{2}$ units, then s is called the semi perimeter of the triangle.

(viii) The area of a △ABC having BE ⊥ AC, is given by $\frac{1}{2}$ x AC x BE.

(ix) Area of a right triangle is equal to $\frac{1}{2}$ x base x height.

(x) The area of a triangle having sides of lengths a units, b units and c units respectively and semiperimeter s units is $\sqrt{s(s - a)(s - b)(s - c)}$ sq. units.

(xi) The area of a parallelogram is equal to the product of the length of the side and the corresponding height.

(xii) Area of a rhombus is equal to $\frac{1}{2}$ x (product of diagonals).

(xiii) The region enclosed between two concentric circles of different radii is called a ring.

Explanation

(i)

The perimeter is the total distance around the outside of a 2D shape. If we imagine walking along the edge of a park until we return to our starting point, the distance we covered is the length of its boundary.

(ii)

While we use the word "perimeter" for shapes with straight sides (like squares or pentagons), a circle is a special case because it is perfectly curved. Therefore, the perimeter of a circle has its own specific name: the circumference.

(iii)

π is a mathematical constant that represents a fixed ratio found in every circle. No matter how large or small the circle is, if we divide the circumference by the diameter (the distance across the center), we will always get approximately 3.14159...

π = $\frac{\text{Circumference}}{\text{Diameter}}$

(iv)

While perimeter measures the "fence" around a shape, area measures the actual "field" or region inside that fence. It tells us how much flat surface the shape covers.

(v)

A hectare is a metric unit used primarily for measuring large plots of land. It is defined as the area of a square with sides of 100 meters.

Since Area = Side x Side:

= 100 m x 100 m

= 10000 m²

(vi)

A grid sheet is used to measure area by counting squares. A "unit" grid means each square has a side length of 1. Therefore, the dimensions of each small square are 1 unit x 1 unit (for example, 1 cm x 1 cm).

(vii)

In the study of triangles (specifically for Heron's Formula), we often need to use half of the total perimeter.

Perimeter = a + b + c

s = $\frac{a + b + c}{2}$

The term "s" stands for semi-perimeter (where "semi" means half).

(viii)

The standard formula for the area of a triangle is $\frac{1}{2}$ x base x height

In the diagram, AC is the base because the altitude BE drops onto it at a 90° angle.

∴ The area of a △ABC = $\frac{1}{2}$ x AC x BE

(ix)

In a right-angled triangle, the two sides that meet at the 90° angle serve as the base and the height. To find the area, we simply multiply these two sides and divide by 2.

(x)

$\sqrt{s(s - a)(s - b)(s - c)}$

This specific formula is known as Heron’s Formula. It is used to calculate the area of any triangle when we know the lengths of all three sides but don't have the height. The value s is the semi-perimeter.

(xi)

For a parallelogram, the area is simply base x height. However, the height must be "corresponding," meaning it must be the perpendicular distance between the chosen base and the side opposite to it.

(xii)

A rhombus has a unique property where its diagonals bisect each other at right angles. Because of this symmetry, we can find its area by multiplying the lengths of the two diagonals (d1 and d2) and dividing by 2.

(xiii)

Concentric circles are circles that share the same center point but have different radii (like a bulls eye). The "doughnut" shaped space caught between the inner circle and the outer circle is mathematically called a ring.

(i) False
Reason — The ratio of the circumference to the diameter is π $\Big(π = \frac{\text{Circumference}}{\text{diameter}}\Big)$. The ratio $\frac{\text{diameter}}{\text{circumference}}$ is actually the reciprocal, $\frac{1}{\pi}$.

(ii) True
Reason — Since 1 m = 100 cm, then 1 m² = (100 cm)² = 10000 cm².

(iii) False
Reason — The area of a parallelogram is simply base × height. The fraction $\frac{1}{2}$ is used for triangles, not parallelograms.

(iv) False
Reason — The area is half the product of its legs ($\frac{1}{2}$ × leg1 × leg2). The product of the legs alone would give the area of a rectangle.

(v) True
Reason — Using the standard formula $\frac{1}{2}$ × base × height , if base is × and height is y,

the area is $\frac{1}{2}$ × x × y = $\frac{1}{2}xy$.

(vi) False
Reason — Area of circle = π × (radius)².

If the radius is 2r,

then Area = π × (radius)²

= π × (2r)²

= π × (4r²)

= 4πr²

(vii) False
Reason — If diameter = x, then radius = $\frac{x}{2}$.

Area = π x (radius)²

= π x $\Big(\frac{x}{2}\Big)^2$

= $\frac{\pi x^2}{4}$

(1)

Length of the string = 66 cm

The total length of the string represents the perimeter of the shape.

An equilateral triangle has 3 equal sides.

Side = $\frac{\text{Perimeter}}{3} = \frac{66}{3} \text{ cm} = 22 \text{ cm}$

Hence, option 3 is the correct option.

(2)

First, let's find the side of the square:

Side = $\frac{\text{Perimeter}}{4} = \frac{66}{4} \text{ cm} = 16.5\text{ cm}$

Now, let's calculate the area:

Area = Side x Side

= 16.5 cm x 16.5 cm

= 272.25 cm²

Hence, option 2 is the correct option.

(3)

The string length is the circumference:

C = 2πr

66 cm = 2 x $\frac{22}{7}$ x r

66 cm = $\frac{44}{7}$ x r

r = $\frac{66 \times 7}{44}$ cm

r = $\frac{3 \times 7}{2}$ cm

r = $\frac{21}{2}$ cm

r = 10.5 cm

Hence, option 4 is the correct option.

(4)

Area of circle = πr²

Using the radius r = 10.5 cm

= $\frac{22}{7}$ x (10.5 cm)²

= $\frac{22}{7}$ x (110.25) cm²

= $\frac{22}{1}$ x (15.75) cm²

= 346.5 cm²

Hence, option 2 is the correct option.

Given:

AB = BC = CD = AD = 250 m

DE = 75 m, BD = 125 m

(1)

The area of ABCD = Base x Height

Here, the side AB acts as the base, and DE is the perpendicular height.

Area = AB x DE

= 250 m x 75 m

= 18750 m²

Hence, option 3 is the correct option.

(2)

We know that,

Area of rhombus ABCD = $\frac{1}{2}$ x d1 x d2

From the image, one diagonal BD = 125 m

Substituting the values in above, we get:

18750 m² = $\frac{1}{2}$ x 125 m x d2

18750 m² = $\frac{1}{1}$ x 62.5 m x d2

d2 = $\frac{18750 \text{ m}^2}{62.5 \text{ m}}$

d2 = 300 m

Hence, option 2 is the correct option.

(3)

The lentils are grown in △DBE.

In this right-angled triangle, we need the base EB.

Using Pythagoras' theorem in △DBE:

EB² = BD² - DE²

= (125 m)² - (75 m)²

= (15625 - 5625) m²

= 10000 m²

EB = $\sqrt{10000}$ m²

EB = 100 m

Area of △DBC = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x EB x DE

= $\frac{1}{2}$ x 100 m x 75 m

= $\frac{1}{1}$ x 50 m x 75 m

= 3750 m²

Hence, option 1 is the correct option.

(4)

A diagonal (BD) divides a rhombus into two triangles of equal area.

Area of △DBC = $\frac{1}{2}$ x Area of rhombus ABCD

= $\frac{1}{2}$ x 18750 m²

= $\frac{1}{1}$ x 9375 m²

= 9375 m²

Hence, option 2 is the correct option.

Assertion (A) is false but Reason (R) is true.

Explanation

In a parallelogram ABCD, AB and AD are two adjacent sides. The area is not simply the product of two sides unless the figure is a rectangle (where the sides meet at 90°).

For a general parallelogram, the area must be the product of a side and its perpendicular height, not another slanted side.

So, Assertion is false.

The fundamental formula for the area of any parallelogram is indeed Base x Height.

The height (altitude) must be the perpendicular distance between the base and the opposite side.

So, Reason is true.

Hence, option 4 is the correct option.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Explanation

In the provided figure, △ABC and △ACD share the same vertex A and their bases BC and CD lie on the same straight line BD. This means both triangles have the exact same perpendicular height (AC).

When triangles share the same height, the ratio of their areas is strictly determined by the ratio of their bases.

$\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle ACD} = \frac{BC}{CD}$

The Reason provides the mathematical foundation for this calculation.

The area of any triangle is $\frac{1}{2}$ x base x height

So, Reason is true.

Hence, option 1 is the correct option.

Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Explanation

The circumference (C) of a circle is calculated using the formula C = 2π r, where r is the radius. If we write the ratio of the circumference to the radius, it looks like this:

Ratio = $\frac{\text{Circumference}}{\text{Radius}} = \frac{2\pi r}{r} = 2\pi$

Ratio = 2π : 1

So, Assertion is true.

For practical calculations, we often take:

π ≈ $\frac{22}{7}$ or 3.14.

So, Reason is true.

But, the reason does not explain the assertion.

Hence, option 2 is the correct option.

Assertion (A) is true but Reason (R) is false.

Explanation

It is entirely possible for two different shapes to cover the same amount of surface (Area) while having different boundary lengths (Perimeter).

For example, consider two rectangles:

Rectangle 1: Sides 4 cm and 9 cm. Area = 36 cm², Perimeter = 2(4 + 9) = 26 cm.

Rectangle 2: Sides 6 cm and 6 cm (a square). Area = 36 cm², Perimeter = 4(6) = 24 cm.

Both have the same area, but different perimeters. So, Assertion is true.

Congruent figures must have the same shape and same size. Same area alone does not guarantee congruency.

So, Reason is false.

Hence, option 3 is the correct option.

Original dimensions:

Length = 6 cm

Breadth = 3 cm

New dimensions:

Length increases by 3 cm → 9 cm

Breadth decreases by 1 cm → 2 cm

Original perimeter:

2(6 + 3) = 18 cm

New perimeter:

2(9 + 2) = 22 cm

Increase in perimeter:

22 − 18 = 4 cm

Original area:

6 × 3 = 18 cm²

New area:

9 × 2 = 18 cm²

Area remains the same.

Hence, option 1 is the correct option

Outer rectangle dimensions:

Length = 250 m, Breadth = 150 m

Area of park:

250 m × 150 m = 37500 m²

Each flower bed:

Length = 120.5 m, Breadth = 70.5 m

Area of one flower bed:

120.5 m × 70.5 m = 8495.25 m²

There are 4 flower beds,

Area of 4 flower beds = 4 x Area of one flower bed

= 4 × 8495.25 m²

= 33981 m²

Area of running/jogging path = Area of park - Area of 4 flower beds

= 37500 m² − 33981 m²

= 3519 m²

Hence, option 1 is the correct option

Area of parallelogram = Base x Height

CD x AM = BC x AN

Since opposite sides are equal, BC = AD = 10.2 cm. Substituting the known values:

CD x 9 cm = 10.2 cm x 7.5 cm

CD = $\frac{76.5}{9}$ cm

CD = 8.5 cm

The diagonal AC cuts the parallelogram into two triangles of equal area. For △ACD, the side CD acts as the base, and AM is its corresponding perpendicular height:

Area of △ACD = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x CD x AM

= $\frac{1}{2}$ (8.5 x 9) cm²

Hence, option 1 is the correct option

Given:

The wire is made up of 5 semicircles.

π = 3.14

Reading the diameters of the semicircles from the figure:

d1 = 4 cm, d2 = 8 cm, d3 = 12 cm, d4 = 8 cm, d5 = 4 cm

The length of the wire is the sum of the curved (arc) length of each semicircle.

Length of one semicircular arc = πr = $\frac{\pi d}{2}$

So, the radii of the semicircles are:

r1 = 2 cm, r2 = 4 cm, r3 = 6 cm, r4 = 4 cm, r5 = 2 cm

Length of wire = πr1 + πr2 + πr3 + πr4 + πr5

= π(r1 + r2 + r3 + r4 + r5) $\quad$[Taking π common]

= π(2 cm + 4 cm + 6 cm + 4 cm + 2 cm) $\quad$[Substituting the values]

= π x 18 cm

= 3.14 x 18 cm

= 56.52 cm

Length of the wire = 56.52 cm

Hence, option 2 is the correct option.

Given:

Total land is a rectangle of length 25 m and breadth 15 m.

Cow's enclosure is a rectangle of length 6 m and breadth 3 m.

Ox's enclosure is a circle of diameter 7 m.

Plants' enclosure is a rectangle of length 8 m and breadth 2 m.

π = $\frac{22}{7}$

Total area of the land = length x breadth

= 25 m x 15 m $\quad$[Substituting the values]

= 375 m²

Area kept for cow = length x breadth

= 6 m x 3 m

= 18 m²

Area kept for ox = πr²
[Circle of diameter 7 m, so radius r = $\frac{7}{2}$ = 3.5 m]

= $\frac{22}{7}$ x (3.5 m)²

= $\frac{22}{7}$ x 12.25 m²

= 38.5 m²

Area kept for plants = length x breadth

= 8 m x 2 m

= 16 m²

Total area for animals and plants = Area of cow + Area of ox + Area of plants

= 18 m² + 38.5 m² + 16 m²

= 72.5 m²

Required ratio = Area for animals and plants : Total area of land

= 72.5 : 375

= 145 : 750 $\quad$[Multiplying both terms by 2]

= 29 : 150 $\quad$[Dividing both terms by 5]

Ratio of areas for animals and plants to the total area = 29 : 150

Hence, option 3 is the correct option.

Given:

Length of cardboard = 1 m 45 cm = 145 cm

Breadth of cardboard = 80 cm

Width of border = 20 cm

The two borders overlap at the centre, forming a cross.

Total area of the cardboard = length x breadth

= 145 cm x 80 cm $\quad$[Substituting the values]

= 11600 cm²

Area of the horizontal border = length x width

= 145 cm x 20 cm

= 2900 cm²

Area of the vertical border = breadth x width

= 80 cm x 20 cm

= 1600 cm²

Area of the overlapping square at the centre = width x width

= 20 cm x 20 cm

= 400 cm²

Total area of the border (cross) = Horizontal border + Vertical border - Overlap

= 2900 cm² + 1600 cm² - 400 cm²
[The overlap is counted twice, so subtract it once]

= 4100 cm²

Painted area = Total area of cardboard - Area of the border

= 11600 cm² - 4100 cm²

= 7500 cm²

Area painted by Aditi = 7500 cm²

Hence, option 3 is the correct option.

Given:

The plot is a square in which the area is equal to the perimeter.

Let the side of the square be 'a' units.

Area of square = a²

Perimeter of square = 4 x a

Since Ravi's area and Manoj's perimeter are equal,

a² = 4 x a $\quad$[Substituting the values]

⇒ $\frac{a^2}{a}$ = 4 $\quad$[Dividing both sides by a]

⇒ a = 4

Length of the side of the plot = 4 units

Hence, option 4 is the correct option.

Given:

ABCD is a rectangle of dimensions 8 cm x 3 cm.

So, BC = DA = 8 cm and AB = CD = 3 cm.

DEFG is a square whose side = 2 x CD.

GHIJ is a rectangle in which GH = 4 x AB and HI = $\frac{5}{4}$ x BC.

Side of square DEFG = 2 x CD

= 2 x 3 cm $\quad$[Substituting the value]

= 6 cm

So, DE = EF = FG = GD = 6 cm

For rectangle GHIJ:

GH = 4 x AB = 4 x 3 cm = 12 cm

HI = $\frac{5}{4}$ x BC = $\frac{5}{4}$ x 8 cm = 10 cm

So, IJ = GH = 12 cm and JG = HI = 10 cm

The perimeter is the sum of all the outer sides of the staircase figure.

Perimeter = IJ + IH + HF + FE + EC + CB + BA + AJ

where,

HF = GH - FG $\quad$[Vertical step from big rectangle down to square]

= 12 cm - 6 cm

= 6 cm

EC = DE - CD $\quad$[Vertical step from square down to small rectangle]

= 6 cm - 3 cm

= 3 cm

AJ = DA + GD + JG $\quad$[Full bottom edge]

= 8 cm + 6 cm + 10 cm

= 24 cm

Substituting all the values:

Perimeter = 12 cm + 10 cm + 6 cm + 6 cm + 3 cm + 8 cm + 3 cm + 24 cm

= 72 cm

Perimeter of the figure = 72 cm

Hence, option 1 is the correct option.