Perimeter & Area (Mensuration)

The perimeter of a closed plane figure is the length of its boundary.

The given figure is a quadrilateral with sides 5 cm, 3 cm, 2 cm and 7 cm.

Perimeter = Sum of all sides

= 5 + 3 + 2 + 7 cm

= 17 cm

Hence, the perimeter of the figure is 17 cm.

The perimeter of a closed plane figure is the length of its boundary.

The given figure is a quadrilateral with sides 31 cm, 38 cm, 48 cm and 38 cm.

Perimeter = Sum of all sides

= 31 + 38 + 48 + 38 cm

= 155 cm

Hence, the perimeter of the figure is 155 cm.

The perimeter of a closed plane figure is the length of its boundary.

The given figure is a rhombus with each side measuring 19 cm.

Perimeter = 4 × side

= 4 × 19 cm

= 76 cm

Hence, the perimeter of the figure is 76 cm.

The perimeter of a closed plane figure is the length of its boundary.

The given figure is a regular pentagon with each side measuring 7 cm.

Perimeter = 5 × side

= 5 × 7 cm

= 35 cm

Hence, the perimeter of the figure is 35 cm.

(i) Perimeter of triangle = Sum of all sides

= 3 + 4 + 6 cm

= 13 cm

Hence, the perimeter of the triangle is 13 cm.

(ii) Perimeter of triangle = Sum of all sides

= 7 + 5.4 + 10.2 cm

= 22.6 cm

Hence, the perimeter of the triangle is 22.6 cm.

(iii) Perimeter of equilateral triangle = 3 × side

= 3 × 11 cm

= 33 cm

Hence, the perimeter of the equilateral triangle is 33 cm.

(iv) Perimeter of isosceles triangle = Sum of all sides

= 10 + 10 + 7 cm

= 27 cm

Hence, the perimeter of the isosceles triangle is 27 cm.

Given:

Length of the lid = 40 cm

Breadth of the lid = 10 cm

The length of the tape required will be equal to the perimeter of the rectangular lid.

Perimeter of rectangle = 2(length + breadth)

= 2(40 + 10) cm

= 2 × 50 cm

= 100 cm

= 1 m

Hence, the length of the tape required is 1 m.

Given:

Length of the table-top = 2 m 25 cm = 2.25 m

Breadth of the table-top = 1 m 50 cm = 1.50 m

Perimeter of rectangle = 2(length + breadth)

= 2(2.25 + 1.50) m

= 2 × 3.75 m

= 7.5 m

Hence, the perimeter of the tabletop is 7.5 m.

Given:

Perimeter of rectangle = 58 m

Length = 17 m

Let b be the breadth of the rectangle.

Perimeter of rectangle = 2(length + breadth)

⇒ 58 = 2(17 + b)

⇒ $\frac{58}{2}$ = 17 + b

⇒ 29 = 17 + b

⇒ b = 29 - 17

⇒ b = 12 m

Hence, the breadth of the rectangle is 12 m.

Given:

Length of the land = 0.7 km

Breadth of the land = 0.5 km

Perimeter of rectangular land = 2(length + breadth)

= 2(0.7 + 0.5) km

= 2 × 1.2 km

= 2.4 km

Since each side is to be fenced with 4 rows of wires,

Length of the wire needed = 4 × Perimeter

= 4 × 2.4 km

= 9.6 km

Hence, the length of the wire needed is 9.6 km.

Given:

Each side of the regular hexagon = 7.5 m

Perimeter of regular hexagon = 6 × side

= 6 × 7.5 m

= 45 m

Hence, the perimeter of the regular hexagon is 45 m.

Given:

Two sides of the triangle = 12 cm and 14 cm

Perimeter of the triangle = 36 cm

Let x be the third side of the triangle.

Perimeter of triangle = Sum of all sides

⇒ 36 = 12 + 14 + x

⇒ 36 = 26 + x

⇒ x = 36 - 26

⇒ x = 10 cm

Hence, the length of the third side is 10 cm.

Given:

Perimeter of regular pentagon = 100 cm

Perimeter of regular pentagon = 5 × side

⇒ 100 = 5 × side

⇒ side = $\frac{100}{5}$

⇒ side = 20 cm

Hence, each side of the regular pentagon is 20 cm.

The length of the string = 30 cm

(i) When the string is used to form a square, the perimeter of the square = 30 cm.

Perimeter of square = 4 × side

⇒ 30 = 4 × side

⇒ side = $\frac{30}{4}$

⇒ side = 7.5 cm

Hence, the length of each side of the square is 7.5 cm.

(ii) When the string is used to form an equilateral triangle, the perimeter of the triangle = 30 cm.

Perimeter of equilateral triangle = 3 × side

⇒ 30 = 3 × side

⇒ side = $\frac{30}{3}$

⇒ side = 10 cm

Hence, the length of each side of the equilateral triangle is 10 cm.

(iii) When the string is used to form a regular hexagon, the perimeter of the hexagon = 30 cm.

Perimeter of regular hexagon = 6 × side

⇒ 30 = 6 × side

⇒ side = $\frac{30}{6}$

⇒ side = 5 cm

Hence, the length of each side of the regular hexagon is 5 cm.

Given:

Length of the park = 225 m

Breadth of the park = 115 m

Rate of fencing = ₹ 13 per metre

Perimeter of rectangular park = 2(length + breadth)

= 2(225 + 115) m

= 2 × 340 m

= 680 m

Cost of fencing = Perimeter × Rate

= 680 × 13

= ₹ 8840

Hence, the cost of fencing the rectangular park is ₹ 8840.

Given:

Length of the park = 140 m

Breadth of the park = 90 m

Number of rounds = 5

Perimeter of rectangular park = 2(length + breadth)

= 2(140 + 90) m

= 2 × 230 m

= 460 m

Distance covered in 5 rounds = 5 × Perimeter

= 5 × 460 m

= 2300 m

Hence, the distance covered by Meera is 2300 m.

For Pinky:

Perimeter of rectangular park = 2(length + breadth)

= 2(80 + 55) m

= 2 × 135 m

= 270 m

Distance covered by Pinky in 8 rounds = 8 × 270 m = 2160 m

For Pankaj:

Perimeter of square park = 4 × side

= 4 × 75 m

= 300 m

Distance covered by Pankaj in 7 rounds = 7 × 300 m = 2100 m

Difference in distances covered = 2160 m - 2100 m = 60 m

Hence, Pinky covers more distance by 60 m.

Given:

Length of the park = 104 m

Breadth of the park = 56 m

Speed of Anjali = 4 km/h

Number of rounds = 5

Perimeter of rectangular park = 2(length + breadth)

= 2(104 + 56) m

= 2 × 160 m

= 320 m

Total distance covered in 5 rounds = 5 × 320 m = 1600 m

Speed of Anjali = 4 km/h = 4 × 1000 m/h = 4000 m/h

So, Anjali covers 4000 m in 60 minutes.

∴ Anjali covers 1 m in $\frac{60}{4000}$ minutes.

∴ Anjali covers 1600 m in $\Big(\frac{60}{4000} \times 1600\Big)$ minutes

$$\begin{aligned}= \frac{60 \times 1600}{4000}\\[1em] = \frac{96000}{4000}\\[1em] = 24 \text{ minutes}\end{aligned}$$

Hence, Anjali takes 24 minutes in making 5 rounds of the park.

Given:

Side of regular pentagon = 12 cm

Length of rectangle = $\frac{3}{2}$ × breadth

Length of the wire = Perimeter of regular pentagon

= 5 × side

= 5 × 12 cm

= 60 cm

Let the breadth of the rectangle be b cm.

Then, length of the rectangle = $\frac{3}{2}$ b cm

Since the wire is rebent into the shape of a rectangle,

Perimeter of rectangle = Length of wire

⇒ 2(length + breadth) = 60

⇒ 2 $\Big(\frac{3}{2}b + b\Big)$ = 60

⇒ 2 × $\frac{3b + 2b}{2}$ = 60

⇒ 5b = 60

⇒ b = $\frac{60}{5}$

⇒ b = 12 cm

Length of rectangle = $\frac{3}{2}$ × 12 = 18 cm

Hence, the length of the rectangle is 18 cm and the breadth is 12 cm.

To estimate the area of a closed plane figure using a square paper, the following conventions are adopted:

1. The area of one complete square is taken as 1 sq. unit.
2. If more than half the square is covered by the region, then count it as 1 sq. unit.
3. If less than half the square is covered by the region, then ignore this portion.
4. If exactly half the square is covered by the region, then take its area as $\frac{1}{2}$ sq. unit.

(i) By counting the unit squares enclosed by the figure:

Number of complete covered squares = 9

Area of the region enclosed by the figure = 9 × 1 sq. unit = 9 sq. units

Hence, the area of the region enclosed by figure (i) is 9 sq. units.

(ii) By counting the unit squares enclosed by the figure:

Area of the region enclosed by the figure = 5 sq. units

Hence, the area of the region enclosed by figure (ii) is 5 sq. units.

(iii) By counting the unit squares enclosed by the figure:

Area of the region enclosed by the figure = 10 sq. units

Hence, the area of the region enclosed by figure (iii) is 10 sq. units.

(iv) By counting the unit squares enclosed by the figure:

Area of the region enclosed by the figure = 6 sq. units

Hence, the area of the region enclosed by figure (iv) is 6 sq. units.

(v) By counting the unit squares enclosed by the figure:

Area of the region enclosed by the figure = 4 sq. units

Hence, the area of the region enclosed by figure (v) is 4 sq. units.

(vi) By counting the unit squares enclosed by the figure:

Area of the region enclosed by the figure = 6 sq. units

Hence, the area of the region enclosed by figure (vi) is 6 sq. units.

Area of rectangle = length × breadth

(i) Length = 9 m, Breadth = 6 m

Area = 9 × 6 = 54 sq. m

(ii) Length = 17 m, Breadth = 3 m

Area = 17 × 3 = 51 sq. m

(iii) Length = 14 m, Breadth = 4 m

Area = 14 × 4 = 56 sq. m

On comparing the areas: 56 sq. m > 54 sq. m > 51 sq. m

Hence, rectangle (iii) has the largest area (56 sq. m) and rectangle (ii) has the smallest area (51 sq. m).

Area of rectangle = length × breadth

(i) Length = 23 cm, Breadth = 14 cm

Area = 23 × 14 = 322 sq. cm

Hence, the area of the rectangle is 322 sq. cm.

(ii) Length = 4 km, Breadth = 3 km

Area = 4 × 3 = 12 sq. km

Hence, the area of the rectangle is 12 sq. km.

(iii) Length = 2 m = 200 cm, Breadth = 90 cm

Area = 200 × 90 = 18000 sq. cm

Converting to sq. m:

Area = $\frac{18000}{10000}$ sq. m = 1.8 sq. m

Hence, the area of the rectangle is 1.8 sq. m.

Area of square = side × side

(i) Side = 8 cm

Area = 8 × 8 = 64 sq. cm

Hence, the area of the square is 64 sq. cm.

(ii) Side = 14 m

Area = 14 × 14 = 196 sq. m

Hence, the area of the square is 196 sq. m.

(iii) Side = 2 m 50 cm = 2.5 m

Area = 2.5 × 2.5 = 6.25 sq. m

Hence, the area of the square is 6.25 sq. m.

Given:

Length of the room = 4 m

Breadth of the room = 3 m 25 cm = 3.25 m

Area of carpet needed = Area of the floor of the room

= length × breadth

= 4 × 3.25 sq. m

= 13 sq. m

Hence, 13 sq. m of carpet is needed to cover the floor of the room.

Given:

Length of the field = 50 m

Breadth of the field = 30 m

Rate of tiling = ₹ 20 per sq. m

Area of rectangular field = length × breadth

= 50 × 30 sq. m

= 1500 sq. m

Cost of tiling = Area × Rate

= 1500 × 20

= ₹ 30000

Hence, the cost of tiling the rectangular field is ₹ 30000.

Given:

Length of the floor = 5 m

Breadth of the floor = 4 m

Side of the square carpet = 3 m

Area of the floor = length × breadth

= 5 × 4 sq. m

= 20 sq. m

Area of the square carpet = side × side

= 3 × 3 sq. m

= 9 sq. m

Area of the floor not carpeted = Area of the floor - Area of the carpet

= 20 - 9

= 11 sq. m

Hence, the area of the floor that is not carpeted is 11 sq. m.

Given:

Length of the outer rectangle = 100 m

Breadth of the outer rectangle = 60 m

Width of the path = 2 m

Area of the outer rectangle = length × breadth

= 100 × 60 sq. m

= 6000 sq. m

Length of the inner rectangle = 100 - (2 + 2) = 96 m

Breadth of the inner rectangle = 60 - (2 + 2) = 56 m

Area of the inner rectangle = 96 × 56 sq. m

= 5376 sq. m

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

= 6000 - 5376

= 624 sq. m

Hence, the area of the path is 624 sq. m.

Given:

Length of the land = 8 m

Breadth of the land = 6 m 50 cm = 6.5 m

Side of each flower bed = 1 m 50 cm = 1.5 m

Number of flower beds = 4

Area of the rectangular land = length × breadth

= 8 × 6.5 sq. m

= 52 sq. m

Area of one square flower bed = side × side

= 1.5 × 1.5 sq. m

= 2.25 sq. m

Area of 4 flower beds = 4 × 2.25 = 9 sq. m

Area of the remaining part of land = Area of land - Area of 4 flower beds

= 52 - 9

= 43 sq. m

Hence, the area of the remaining part of land is 43 sq. m.

Area of one tile = length × breadth

= 12 × 5 sq. cm

= 60 sq. cm

(i) Length of the region = 70 cm

Breadth of the region = 36 cm

Area of the rectangular region = 70 × 36 sq. cm

= 2520 sq. cm

Number of tiles needed = $\frac{\text{Area of region}}{\text{Area of one tile}}$

= $\frac{2520}{60}$

= 42

Hence, 42 tiles are needed to cover the rectangular region.

(ii) Length of the region = 144 cm

Breadth of the region = 1 m = 100 cm

Area of the rectangular region = 144 × 100 sq. cm

= 14400 sq. cm

Number of tiles needed = $\frac{\text{Area of region}}{\text{Area of one tile}}$

= $\frac{14400}{60}$

= 240

Hence, 240 tiles are needed to cover the rectangular region.

Given:

Area of rectangular plot = 340 sq. m

Breadth = 17 m

Area of rectangle = length × breadth

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

= $\frac{340}{17}$

= 20 m

Perimeter of rectangle = 2(length + breadth)

= 2(20 + 17) m

= 2 × 37 m

= 74 m

Hence, the length of the plot is 20 m and the perimeter is 74 m.

Given:

Area of rectangular plot = 144 sq. m

Length = 16 m

Rate of fencing = ₹ 60 per metre

Area of rectangle = length × breadth

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

= $\frac{144}{16}$

= 9 m

Perimeter of rectangle = 2(length + breadth)

= 2(16 + 9) m

= 2 × 25 m

= 50 m

Cost of fencing = Perimeter × Rate

= 50 × 60

= ₹ 3000

Hence, the breadth of the plot is 9 m and the cost of fencing is ₹ 3000.

Given:

Length of rectangle = 17 m

Breadth of rectangle = 11 m

Perimeter of square = Perimeter of rectangle

Perimeter of rectangle = 2(length + breadth)

= 2(17 + 11) m

= 2 × 28 m

= 56 m

Perimeter of square = 4 × side

⇒ 56 = 4 × side

⇒ side = $\frac{56}{4}$

⇒ side = 14 m

Area of square = side × side

= 14 × 14 sq. m

= 196 sq. m

Hence, the area of the square is 196 sq. m.

Given:

Side of square = 10 m

Breadth of rectangle = 8 m

Area of square = side × side

= 10 × 10 sq. m

= 100 sq. m

Area of rectangle = Area of square = 100 sq. m

Area of rectangle = length × breadth

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

= $\frac{100}{8}$

= 12.5 m

Perimeter of rectangle = 2(length + breadth)

= 2(12.5 + 8) m

= 2 × 20.5 m

= 41 m

Hence, the perimeter of the rectangle is 41 m.

(i) Splitting the L-shape into two rectangles:

Rectangle 1 (vertical part): length = 12 cm, breadth = 2 cm

Area of Rectangle 1 = 12 × 2 = 24 sq. cm

Rectangle 2 (horizontal part at the bottom): length = 8 cm, breadth = 2 cm

Area of Rectangle 2 = 8 × 2 = 16 sq. cm

Total area = Area of Rectangle 1 + Area of Rectangle 2

= 24 + 16

= 40 sq. cm

Hence, the area of figure (i) = 40 sq. cm.

(ii) Splitting the plus-shaped figure into three rectangles:

Central horizontal rectangle: length = 7 + 7 + 7 = 21 cm, breadth = 7 cm

Area = 21 × 7 = 147 sq. cm

Top square : side = 7 cm

Area = 7 × 7 = 49 sq. cm

Bottom square : side = 7 cm

Area = 7 × 7 = 49 sq. cm

Total area = 147 + 49 + 49

= 245 sq. cm.

Hence, the area of figure (ii) is 245 sq. cm.

(iii) Splitting the T-shape into two rectangles:

Top bar of T: length = 5 cm, breadth = 1 cm

Area of top bar = 5 × 1 = 5 sq. cm

Stem of T (vertical part): length = 4 cm, breadth = 1 cm

Area of stem = 4 × 1 = 4 sq. cm

Total area = 5 + 4

= 9 sq. cm.

Hence, the area of figure (iii) is 9 sq. cm.

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

(ii) The unit of measurement of perimeter is same as that of length.

(iii) Since a rhombus has 4 equal sides, perimeter = 4 × side = 4 × 7 = 28 cm.

If the side of a rhombus is 7 cm then its perimeter is 28 cm.

(iv) The area of a closed plane figure is measured in square units (sq. units).

(i) False. Centimetre is the unit of length, not area. Area is measured in square centimetres (sq. cm).

(ii) False. The sum of lengths of the sides of a polygon is called its perimeter, not its area.

(iii) True. If side = s, then original perimeter = 4s. If side is doubled (i.e., 2s), then new perimeter = 4(2s) = 8s = 2 × 4s. So, perimeter is also doubled.

(iv) False. If side = s, then original area = s². If side is doubled (i.e., 2s), then new area = (2s)² = 4s². So, area becomes four times, not double.

(v) True. Construction of a road covers its surface, so we find its area.

(vi) True. Fencing is done along the boundary, so we find the perimeter.

Given:

Perimeter of square = 50 cm

Perimeter of square = 4 × side

⇒ 50 = 4 × side

⇒ side = $\frac{50}{4}$

⇒ side = 12.5 cm

Hence, option 4 is the correct option.

Given:

Length = 25 cm

Breadth = 12 cm

Area of rectangle = length × breadth

= 25 × 12

= 300 sq. cm

Hence, option 3 is the correct option.

Given:

Perimeter of square = 36 cm

Perimeter of square = 4 × side

⇒ 36 = 4 × side

⇒ side = $\frac{36}{4}$

⇒ side = 9 cm

Area of square = side × side

= 9 × 9

= 81 sq. cm

Hence, option 4 is the correct option.

Given:

Area of rectangular plot = 180 sq. m

Length = 15 m

Area of rectangle = length × breadth

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

= $\frac{180}{15}$

= 12 m

Hence, option 1 is the correct option.

Let the length = l and the breadth = b.

Original perimeter = 2(l + b)

If length and breadth are doubled, new length = 2l and new breadth = 2b.

New perimeter = 2(2l + 2b)

= 2 × 2(l + b)

= 2 × Original perimeter

Thus, the new perimeter is twice the original perimeter.

Hence, option 2 is the correct option.

Let the length = l and the breadth = b.

Original area = l × b

If length and breadth are doubled, new length = 2l and new breadth = 2b.

New area = 2l × 2b

= 4 × (l × b)

= 4 × Original area

Thus, the new area becomes four times the original area.

Hence, option 4 is the correct option.

Let the side of the square = s.

Original area = s × s = s²

If the side is halved, new side = $\frac{\text s}{2}$

New area = $\frac{\text s}{2} \times \frac{\text s}{2} = \frac{\text s^2}{4}$

= $\frac{1}{4}$ × Original area

Thus, the new area becomes one-fourth of the original area.

Hence, option 3 is the correct option.

When the square of side 1 cm is joined to the square of side 3 cm, one side of the smaller square (1 cm) coincides with a part of one side of the bigger square.

Perimeter of bigger square = 4 × 3 = 12 cm

Perimeter of smaller square = 4 × 1 = 4 cm

When joined, the side of the smaller square (1 cm) is no longer on the boundary, and 1 cm of the side of the bigger square is also no longer on the boundary.

Perimeter of new figure = (Perimeter of bigger square - 1) + (Perimeter of smaller square - 1)

= (12 - 1) + (4 - 1)

= 11 + 3

= 14 cm

Hence, option 2 is the correct option.

Given:

Perimeter of each regular hexagon = 30 cm

Side of each hexagon = $\frac{30}{6}$ = 5 cm

When two hexagons are joined along one side, one side of each hexagon (5 cm) coincides.

Sum of perimeters of both hexagons = 30 + 30 = 60 cm

Perimeter of new figure = 60 - 2 × 5

= 60 - 10

= 50 cm

Hence, option 4 is the correct option.

Let the side of the square = s.

Area of square = s²

Perimeter of square = 4s

Given, Area = Perimeter

⇒ s² = 4s

⇒ s² - 4s = 0

⇒ s(s - 4) = 0

⇒ s = 4 (since s ≠ 0)

Thus, the length of each side is 4 units.

Hence, option 4 is the correct option.

Given:

Side of each square table = 1.3 m

Number of tables = 4

When four square tables are placed end to end, the length of the rectangular table = 4 × 1.3 = 5.2 m

The breadth of the rectangular table = 1.3 m

Perimeter of rectangle = 2(length + breadth)

= 2(5.2 + 1.3) m

= 2 × 6.5 m

= 13 m

Hence, option 3 is the correct option.

Given:

Area of rectangle = 180 cm²

Length = 15 cm

Area = length × breadth

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

= $\frac{180}{15}$

= 12 cm

Ratio of breadth to length = 12 : 15

= $\frac{12}{15}$

= $\frac{4}{5}$

= 4 : 5

Hence, option 4 is the correct option.

Given:

Width of picture = 60 cm

Length of picture = 1.8 m = 180 cm

Perimeter of picture = 2(length + width)

= 2(180 + 60) cm

= 2 × 240 cm

= 480 cm

Ratio of width to perimeter = 60 : 480

= $\frac{60}{480}$

= $\frac{1}{8}$

= 1 : 8

Hence, option 4 is the correct option.

Statement I: The perimeter of a closed shape made from a wire is equal to the length of the wire. Since both wires are 27 m each, the perimeter of both the circle and the equilateral triangle will be 27 m. So, both shapes have the same perimeter. Statement I is true.

Statement II: The perimeter of a closed shape is the length of its boundary. This is the correct definition of perimeter. Statement II is true.

Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

Statement I: When two rectangles of length 20 m and width 10 m are placed edge to edge along their longer sides (20 m), the resulting figure is a rectangle of length 20 m and width 20 m (i.e., a square).

Perimeter = 2(20 + 20) = 2 × 40 = 80 m. Statement I is true.

Statement II: The perimeter of a pentagon is five times the length of any side only if it is a regular pentagon (all sides equal). In general, a pentagon has 5 sides which may not all be equal. Statement II is false.

Statement I is true but Statement II is false.

Hence, option 1 is the correct option.

Statement I:

Area of the front (or back) of the mobile phone = length × breadth = 15 × 8 = 120 cm²

Sum of the areas of the front and the back = 2 × 120 = 240 cm². Statement I is true.

Statement II: The area of a square with side a is given by side × side = a × a = a². Statement II is true.

Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

Statement I:

Sum of widths of two equal doors = 3 m

So, width of each door = $\frac{3}{2}$ = 1.5 m

Perimeter of each door = 7 m

Perimeter = 2(length + width)

⇒ 7 = 2(height + 1.5)

⇒ $\frac{7}{2}$ = height + 1.5

⇒ 3.5 = height + 1.5

⇒ height = 3.5 - 1.5

⇒ height = 2 m. Statement I is true.

Statement II:

Perimeter of rectangle = 2(length + width)

⇒ $\frac{1}{2}$ × Perimeter = length + width

⇒ Length = $\frac{1}{2}$ Perimeter - width

So, the correct formula is Length = $\frac{1}{2}$ perimeter - width, not + width. Statement II is false.

Statement I is true but Statement II is false.

Hence, option 1 is the correct option.

Statement I: Perimeter is measured in units of length (cm, m, km, etc.), while area is measured in square units (sq. cm, sq. m, sq. km, etc.). So, area and perimeter do not have the same units. Statement I is false.

Statement II: The area of a closed plane figure is indeed the measurement of the region (surface) enclosed by its boundary. Statement II is true.

Statement I is false but Statement II is true.

Hence, option 2 is the correct option.

Given:

Triangle PQR has sides PQ = 6 cm, QR = 7 cm and PR = 5 cm.

Perimeter of △PQR = Sum of all sides

= 6 + 7 + 5 cm

= 18 cm

Perimeter of square ABCD = 2 × Perimeter of △PQR

= 2 × 18 cm

= 36 cm

Perimeter of square = 4 × side

⇒ 36 = 4 × side

⇒ side = $\frac{36}{4}$

⇒ side = 9 cm

Area of square ABCD = side × side

= 9 × 9

= 81 sq. cm

Hence, the area of the square ABCD is 81 sq. cm.

Given:

Perimeter of equilateral triangle = 42 cm

Perimeter of equilateral triangle = 3 × side

⇒ 42 = 3 × side

⇒ side = $\frac{42}{3}$

⇒ side of triangle = 14 cm

Side of square = 2 × side of triangle

= 2 × 14

= 28 cm

Perimeter of square = 4 × side

= 4 × 28

= 112 cm

Hence, the perimeter of the square is 112 cm.

Given:

Total length of wire = 60 cm

Length of rectangle = 10 cm

Width of rectangle = 8 cm

Perimeter of rectangle = 2(length + width)

= 2(10 + 8) cm

= 2 × 18 cm

= 36 cm

Length of wire used for rectangle = 36 cm

Length of wire used for hexagon = 60 - 36 = 24 cm

Perimeter of regular hexagon = 6 × side

⇒ 24 = 6 × side

⇒ side = $\frac{24}{6}$

⇒ side = 4 cm

Hence, the length of each side of the hexagon is 4 cm.

Given:

Side of square = 10 cm

Length of rectangle = 12 cm

Length of wire = Perimeter of square

= 4 × side

= 4 × 10

= 40 cm

Since the wire is rebent into a rectangle,

Perimeter of rectangle = Length of wire = 40 cm

Perimeter of rectangle = 2(length + breadth)

⇒ 40 = 2(12 + breadth)

⇒ $\frac{40}{2}$ = 12 + breadth

⇒ 20 = 12 + breadth

⇒ breadth = 20 - 12

⇒ breadth = 8 cm

Area of square = side × side

= 10 × 10

= 100 sq. cm

Area of rectangle = length × breadth

= 12 × 8

= 96 sq. cm

Since 100 sq. cm > 96 sq. cm, the square encloses more area.

Difference in areas = 100 - 96 = 4 sq. cm

Hence, the breadth of the rectangle is 8 cm and the square encloses more area than the rectangle by 4 sq. cm.

Given:

Length of room = 9 m

Breadth of room = 6 m

Width of carpet = 2 m

Rate of carpet = ₹ 350 per metre

Area of the floor = length × breadth

= 9 × 6 sq. m

= 54 sq. m

Area of carpet = Area of floor (since the carpet exactly covers the floor) = 54 sq. m

Length of carpet = $\frac{\text{Area of carpet}}{\text{Width of carpet}}$

= $\frac{54}{2}$

= 27 m

Cost of carpet = Length × Rate

= 27 × 350

= ₹ 9450

Hence, the cost of covering the floor with carpet is ₹ 9450.

Given:

Total cost of fencing = ₹ 2400

Rate of fencing = ₹ 30 per metre

Perimeter of square plot = $\frac{\text{Total cost}}{\text{Rate}}$

= $\frac{2400}{30}$

= 80 m

Perimeter of square = 4 × side

⇒ 80 = 4 × side

⇒ side = $\frac{80}{4}$

⇒ side = 20 m

Hence, the length of each side of the field is 20 m.

Given:

Total cost of fencing = ₹ 2400

Rate of fencing = ₹ 30 per metre

Length of the park = 24 m

Perimeter of rectangular park = $\frac{\text{Total cost}}{\text{Rate}}$

= $\frac{2400}{30}$

= 80 m

Perimeter of rectangle = 2(length + breadth)

⇒ 80 = 2(24 + breadth)

⇒ $\frac{80}{2}$ = 24 + breadth

⇒ 40 = 24 + breadth

⇒ breadth = 40 - 24

⇒ breadth = 16 m

Hence, the breadth of the park is 16 m.

(i) Splitting the U-shape figure into three rectangles:

Rectangle DEFG: length = 5 cm, breadth = 1 cm

Area = 5 × 1 = 5 sq. cm

Rectangle JIHG: length = 2 cm, breadth = 1 cm

Area = 2 × 1 = 2 sq. cm

Rectangle ABCD: length = 2 cm, breadth = 1 cm

Area = 2 × 1 = 2 sq. cm

Total area of figure (i) = 5 + 2 + 2

= 9 sq. cm

Hence, the area of figure (i) is 9 sq. cm.

(ii) Splitting the figure:

As, LC = 4 and EM = 3 = FD = LF + LD

IF + FL = JK = 3

HG = IF = 2

$\Rightarrow$ 2 + FL = 3

Therefore, FL = 1

Now, LC = 4, FL + LD = 3

$\Rightarrow$ LD = 2

LD + DC = 4

2 + DC = 4

$\Rightarrow$ DC = 2

Rectangle ABCD: length = 2 cm, breadth = 4 cm

Area = 2 × 4 = 8 sq. cm

Rectangle HIGF: length = 2 cm, breadth = 1 cm

Area = 2 × 1 = 2 sq. cm

Square MEFD: length = 3 cm, breadth = 3 cm

Area = 3 × 3 = 9 sq. cm

Rectangle IJKL: length = 3 cm, breadth = 3 cm

Area = 3 × 3 = 9 sq. cm

Total area of figure (ii) = 8 + 2 + 9 + 9

= 28 sq. cm

Hence, the area of figure (ii) is 28 sq. cm.

Given:

Length of sheet = 4 m = 400 cm

Breadth of sheet = 1.2 m = 120 cm

Size of each envelope = 25 cm × 15 cm

Area of the rectangular sheet = length × breadth

= 400 × 120 sq. cm

= 48000 sq. cm

To make one envelope, paper of double the size of the envelope is needed (because an envelope is formed by folding paper to make front and back).

Paper needed for one envelope = 2 × (25 × 15) sq. cm

= 2 × 375 sq. cm

= 750 sq. cm

Number of envelopes that can be made = $\frac{\text{Area of sheet}}{\text{Paper needed for one envelope}}$

= $\frac{48000}{750}$

= 64

Hence, 64 envelopes can be made from the rectangular sheet.

Given:

Perimeter of rectangle = 36 cm

Perimeter of rectangle = 2(length + breadth)

⇒ 36 = 2(length + breadth)

⇒ length + breadth = $\frac{36}{2}$

⇒ length + breadth = 18 cm

The possible pairs of (length, breadth) in natural numbers with sum 18 are:

Length	Breadth	Area (length × breadth)
17	1	17 sq. cm
16	2	32 sq. cm
15	3	45 sq. cm
14	4	56 sq. cm
13	5	65 sq. cm
12	6	72 sq. cm
11	7	77 sq. cm
10	8	80 sq. cm
9	9	81 sq. cm

(i) The maximum area is 81 sq. cm when length = 9 cm and breadth = 9 cm.

Hence, for maximum area, length = 9 cm and breadth = 9 cm.

(ii) The minimum area is 17 sq. cm when length = 17 cm and breadth = 1 cm.

Hence, for minimum area, length = 17 cm and breadth = 1 cm.

Given:

Area of square A = 25 sq. cm

Perimeter of square B = 12 cm

For square A:

Area of square = side × side

⇒ 25 = side × side

⇒ side² = 25

⇒ side = $\sqrt{25}$

⇒ side of A = 5 cm

For square B:

Perimeter of square = 4 × side

⇒ 12 = 4 × side

⇒ side = $\frac{12}{4}$

⇒ side of B = 3 cm

From the figure, side of square C = side of A + side of B

= 5 + 3

= 8 cm

Perimeter of square C = 4 × side

= 4 × 8

= 32 cm

Area of square C = side × side

= 8 × 8

= 64 sq. cm

Hence, the perimeter of square C is 32 cm and the area is 64 sq. cm.

(i) Finding missing measurements:

The total width of the house = 44 ft.

The top portion of the house contains: Master Bedroom, Toilet, Kitchen and Utility (placed side by side).

For Utility:

Area of Utility = 77 sq. ft, Height = 11 ft (same as Kitchen)

Width of Utility = $\frac{\text{Area}}{\text{Height}}$ = $\frac{77}{11}$ = 7 ft

So, Utility dimensions = 7 ft × 11 ft.

For Toilet:

Width of Toilet = Total width - (Master Bedroom width + Kitchen width + Utility width)

= 44 - (13 + 18 + 7)

= 44 - 38

= 6 ft

Height of Toilet = 11 ft (same as Kitchen and Utility)

So, Toilet dimensions = 6 ft × 11 ft.

Area of Toilet = 6 × 11 = 66 sq. ft.

For Hall:

Width of Hall = 24 ft

Height of Hall = 16 ft (same as Master Bedroom)

So, Hall dimensions = 24 ft × 16 ft.

Area of Hall = 24 × 16 = 384 sq. ft.

For Garden:

Width of Garden = Total width - (Small Bedroom width + Hall width)

= 44 - (13 + 24)

= 44 - 37

= 7 ft

Height of Garden = 16 ft (same as Hall)

So, Garden dimensions = 7 ft × 16 ft.

Area of Garden = 7 × 16 = 112 sq. ft.

For Small Bedroom:

Small Bedroom dimensions = 13 ft × 11 ft.

Area of Small Bedroom = 13 × 11 = 143 sq. ft.

The missing measurements are:

• Toilet: 6 ft × 11 ft, Area = 66 sq. ft
• Utility: 7 ft × 11 ft, Area = 77 sq. ft
• Hall: 24 ft × 16 ft, Area = 384 sq. ft
• Garden: 7 ft × 16 ft, Area = 112 sq. ft
• Small Bedroom: 13 ft x 11 ft, Area = 143 sq. ft

(ii) Total area of the house:

Total length of the house = Master Bedroom height + Small Bedroom height = 16 + 11 = 27 ft

Total width of the house = 44 ft

Total area of the house = length × width

= 44 × 27

= 1188 sq. ft

Hence, the total area of the house (including the Garden) is 1188 sq. ft.