Perimeter & Area of Plane Figures

Given:

Length of the rectangle = 15 cm

Breadth of the rectangle = 8 cm

By formula,

Perimeter of a rectangle = 2 x (length + breadth)

= 2 x (15 + 8) cm

= 2 x 23 cm

= 46 cm

By formula,

Area of a rectangle = length x breadth

= 15 x 8 cm²

= 120 cm²

By formula,

Diagonal of a rectangle = $\sqrt{\text{length}^2 + \text{breadth}^2}$

$$\begin{aligned}= \sqrt{(15)^2 + (8)^2} \text{ cm}\\[1em] = \sqrt{225 + 64} \text{ cm}\\[1em] = \sqrt{289} \text{ cm}\\[1em] = 17 \text{ cm}\end{aligned}$$

Hence, the perimeter, area and diagonal of the rectangle are 46 cm, 120 cm² and 17 cm respectively.

Given:

Length of the rectangle = 20 m

Breadth of the rectangle = 15 m

By formula,

Perimeter of a rectangle = 2 x (length + breadth)

= 2 x (20 + 15) m

= 2 x 35 m

= 70 m

By formula,

Area of a rectangle = length x breadth

= 20 x 15 m²

= 300 m²

By formula,

Diagonal of a rectangle = $\sqrt{\text{length}^2 + \text{breadth}^2}$

$$\begin{aligned}= \sqrt{(20)^2 + (15)^2} \text{ m}\\[1em] = \sqrt{400 + 225} \text{ m}\\[1em] = \sqrt{625} \text{ m}\\[1em] = 25 \text{ m}\end{aligned}$$

Hence, the perimeter, area and diagonal of the rectangle are 70 m, 300 m² and 25 m respectively.

Given:

Length of the rectangle = 3.2 m

Breadth of the rectangle = 2.4 m

By formula,

Perimeter of a rectangle = 2 x (length + breadth)

= 2 x (3.2 + 2.4) m

= 2 x 5.6 m

= 11.2 m

By formula,

Area of a rectangle = length x breadth

= 3.2 x 2.4 m²

= 7.68 m²

By formula,

Diagonal of a rectangle = $\sqrt{\text{length}^2 + \text{breadth}^2}$

$$\begin{aligned}= \sqrt{(3.2)^2 + (2.4)^2} \text{ m}\\[1em] = \sqrt{10.24 + 5.76} \text{ m}\\[1em] = \sqrt{16} \text{ m}\\[1em] = 4 \text{ m}\end{aligned}$$

Hence, the perimeter, area and diagonal of the rectangle are 11.2 m, 7.68 m² and 4 m respectively.

Given:

Perimeter of the rectangle = 51.8 m

Length of the rectangle = 16.5 m

Let the breadth of the rectangle be b m.

By formula,

Perimeter of a rectangle = 2 x (length + breadth)

⇒ 2 x (16.5 + b) = 51.8

⇒ 16.5 + b = $\frac{51.8}{2}$

⇒ 16.5 + b = 25.9

⇒ b = 25.9 - 16.5

⇒ b = 9.4 m

Now, area of the rectangle = length x breadth

= 16.5 x 9.4 m²

= 155.1 m²

Hence, the breadth of the rectangle is 9.4 m and its area is 155.1 m².

Given:

Perimeter of the rectangle = 42 m

Breadth of the rectangle = 7.4 m

Let the length of the rectangle be l m.

By formula,

Perimeter of a rectangle = 2 x (length + breadth)

⇒ 2 x (l + 7.4) = 42

⇒ l + 7.4 = $\frac{42}{2}$

⇒ l + 7.4 = 21

⇒ l = 21 - 7.4

⇒ l = 13.6 m

Now, area of the rectangle = length x breadth

= 13.6 x 7.4 m²

= 100.64 m²

Hence, the length of the rectangle is 13.6 m and its area is 100.64 m².

Given:

Perimeter of the rectangle = 68 m

Length of the rectangle = 24 m

Let the breadth of the rectangle be b m.

By formula,

Perimeter of a rectangle = 2 x (length + breadth)

⇒ 2 x (24 + b) = 68

⇒ 24 + b = $\frac{68}{2}$

⇒ 24 + b = 34

⇒ b = 34 - 24

⇒ b = 10 m

Now, area of the rectangle = length x breadth

= 24 x 10 m²

= 240 m²

By formula,

Diagonal of a rectangle = $\sqrt{\text{length}^2 + \text{breadth}^2}$

$$\begin{aligned}= \sqrt{(24)^2 + (10)^2} \text{ m}\\[1em] = \sqrt{576 + 100} \text{ m}\\[1em] = \sqrt{676} \text{ m}\\[1em] = 26 \text{ m}\end{aligned}$$

Hence, the breadth, area and diagonal of the rectangle are 10 m, 240 m² and 26 m respectively.

Given:

Length of the rectangle = 30 cm

Diagonal of the rectangle = 34 cm

Let the breadth of the rectangle be b cm.

Using Pythagoras theorem in the right triangle BAD formed by the length, breadth and diagonal,

⇒ length² + breadth² = diagonal²

⇒ (30)² + b² = (34)²

⇒ 900 + b² = 1156

⇒ b² = 1156 - 900

⇒ b² = 256

⇒ b = $\sqrt{256}$

⇒ b = 16 cm

Now, perimeter of the rectangle = 2 x (length + breadth)

= 2 x (30 + 16) cm

= 2 x 46 cm

= 92 cm

And, area of the rectangle = length x breadth

= 30 x 16 cm²

= 480 cm²

Hence, the breadth, perimeter and area of the rectangle are 16 cm, 92 cm and 480 cm² respectively.

Given:

Area of the rectangle = 19.6 m²

Length of the rectangle = 5.6 m

By formula,

Breadth of a rectangle = $\frac{\text{Area}}{\text{Length}}$

= $\frac{19.6}{5.6}$ m

= 3.5 m

Now, perimeter of the rectangle = 2 x (length + breadth)

= 2 x (5.6 + 3.5) m

= 2 x 9.1 m

= 18.2 m

Hence, the breadth of the rectangle is 3.5 m and its perimeter is 18.2 m.

Given:

Area of the rectangle = 52 m²

Breadth of the rectangle = 6.5 m

By formula,

Length of a rectangle = $\frac{\text{Area}}{\text{Breadth}}$

= $\frac{52}{6.5}$ m

= 8 m

Now, perimeter of the rectangle = 2 x (length + breadth)

= 2 x (8 + 6.5) m

= 2 x 14.5 m

= 29 m

Hence, the length of the rectangle is 8 m and its perimeter is 29 m.

Given:

Sides of the rectangular park are in the ratio 3 : 2.

Area of the park = 1536 m²

Rate of fencing = ₹ 23.50 per metre

Let the length of the park be 3x meters and its breadth be 2x meters.

By formula,

Area of a rectangle = length x breadth

⇒ 3x x 2x = 1536

⇒ 6x² = 1536

⇒ x² = $\frac{1536}{6}$

⇒ x² = 256

⇒ x = $\sqrt{256}$

⇒ x = 16

So, length of the park = 3x

= 3 x 16 m

= 48 m

And, breadth of the park = 2x

= 2 x 16 m

= 32 m

Now, perimeter of the park = 2 x (length + breadth)

= 2 x (48 + 32) m

= 2 x 80 m

= 160 m

Cost of fencing the park = 160 x ₹ 23.50

= ₹ 3,760

Hence, the cost of fencing the park is ₹ 3,760.

Given:

Length of the room = 12 m

Breadth of the room = 8 m

Breadth of the carpet = 75 cm = $\frac{75}{100}$ m = 0.75 m

Rate of carpeting = ₹ 116.50 per metre

Area of the room = length x breadth

= 12 x 8 m²

= 96 m²

Since the carpet covers the whole floor,

Area of the carpet = Area of the room = 96 m²

∴ Length of the carpet = $\frac{\text{Area of the carpet}}{\text{Breadth of the carpet}}$

= $\frac{96}{0.75}$ m

= 128 m

Now, cost of carpeting the room = 128 x ₹ 116.50

= ₹ 14,912

Hence, the cost of carpeting the room is ₹ 14,912.

Given:

Length of the verandah = 50 m

Breadth of the verandah = 12 m

Size of each tile = 6 dm by 5 dm

Area of the verandah = length x breadth

= 50 x 12 m²

= 600 m²

Length of 1 tile = 6 dm = $\frac{6}{10}$ m = 0.6 m

Breadth of 1 tile = 5 dm = $\frac{5}{10}$ m = 0.5 m

Area of 1 tile = length x breadth

= 0.6 x 0.5 m²

= 0.3 m²

∴ Number of tiles required = $\frac{\text{Area of the verandah}}{\text{Area of 1 tile}}$

= $\frac{600}{0.3}$

= 2000

Hence, 2000 tiles are needed.

Given:

Length and breadth of the rectangular field are in the ratio 7 : 5.

Perimeter of the field = 384

Rate of reaping = ₹ 1.25 per sq. metre

Let the length of the field be 7x metres and its breadth be 5x metres.

By formula,

Perimeter of a rectangle = 2 x (length + breadth)

⇒ 2 x (7x + 5x) = 384

⇒ 2 x 12x = 384

⇒ 24x = 384

⇒ x = $\frac{384}{24}$

⇒ x = 16

So, length of the field = 7x

= 7 x 16 m

= 112 m

And, breadth of the field = 5x

= 5 x 16 m

= 80 m

Now, area of the field = length x breadth

= 112 x 80 m²

= 8960 m²

Cost of reaping the field = 8960 x ₹ 1.25

= ₹ 11,200

Hence, the cost of reaping the field is ₹ 11,200.

Given:

Length of the room = 9.5 m

Width of the room = 6 m

Width of the verandah = 1.25 m

Rate of cementing = ₹ 28 per sq. metre

Area of the room = length x width

= 9.5 x 6 m²

= 57 m²

Length of the room including the verandah = [9.5 + (1.25 + 1.25)] m

= 12 m

Width of the room including the verandah = [6 + (1.25 + 1.25)] m

= 8.5 m

Area of the room including the verandah = length x width

= 12 x 8.5 m²

= 102 m²

∴ Area of the verandah = (102 - 57) m²

= 45 m²

Now, cost of cementing the floor of the verandah = 45 x ₹ 28

= ₹ 1,260

Hence, the cost of cementing the floor of the verandah is ₹ 1,260.

Given:

Length of the grassy plot = 125 m

Breadth of the grassy plot = 74 m

Width of the path = 2.5 m (all round it on the inside)

Rate of levelling = ₹ 6.80 per m²

Area of the whole plot = length x breadth

= 125 x 74 m²

= 9250 m²

Length of the plot excluding the path = [125 - (2.5 + 2.5)] m

= 120 m

Breadth of the plot excluding the path = [74 - (2.5 + 2.5)] m

= 69 m

Area of the plot excluding the path = length x breadth

= 120 x 69 m²

= 8280 m²

∴ Area of the path = (9250 - 8280) m²

= 970 m²

Now, cost of levelling the path = 970 x ₹ 6.80

= ₹ 6,596

Hence, the cost of levelling the path is ₹ 6,596.

Given:

Length of the plot = 95 m

Breadth of the plot = 72 m

Width of the path = 3.5 m (all around on the inside)

Rate of constructing the path = ₹ 46.50 per m²

Rate of laying the grass = ₹ 3.75 per m²

Area of the whole plot = length x breadth

= 95 x 72 m²

= 6840 m²

Length of the plot excluding the path = [95 - (3.5 + 3.5)] m

= 88 m

Breadth of the plot excluding the path = [72 - (3.5 + 3.5)] m

= 65 m

Area of the plot excluding the path = length x breadth

= 88 x 65 m²

= 5720 m²

∴ Area of the path = (6840 - 5720) m²

= 1120 m²

Cost of constructing the path = 1120 x ₹ 46.50

= ₹ 52,080

Cost of laying the grass = 5720 x ₹ 3.75

= ₹ 21,450

∴ Total expense = ₹ (52,080 + 21,450)

= ₹ 73,530

Hence, the total expense involved is ₹ 73,530.

Given:

Length of the hall = 22 m

Breadth of the hall = 15.5 m

Margin left all around = 75 cm = $\frac{75}{100}$ m = 0.75 m

Width of the carpet = 82 cm = $\frac{82}{100}$ m = 0.82 m

Rate of the carpet = ₹ 124 per metre

Length of the carpet = [22 - (2 x 0.75)] m

= (22 - 1.5) m

= 20.5 m

Breadth of the carpet = [15.5 - (2 x 0.75)] m

= (15.5 - 1.5) m

= 14 m

∴ Area of the carpet = length x breadth

= 20.5 x 14 m²

= 287 m²

Area of the hall = length x breadth

= 22 x 15.5 m²

= 341 m²

∴ Area of the strip left uncovered = (341 - 287) m²

= 54 m²

Now, length of the carpet needed = $\frac{\text{Area of the carpet}}{\text{Width of the carpet}}$

= $\frac{287}{0.82}$ m

= 350 m

Cost of the carpet = 350 x ₹ 124

= ₹ 43,400

Hence, the area of the carpet is 287 m², the area of the strip left uncovered is 54 m² and the cost of the carpet is ₹ 43,400.

Given:

Length of the lawn = 75 m

Breadth of the lawn = 60 m

Width of each road = 4 m

Rate of gravelling = ₹ 14.50 per sq. metre

Area of the road parallel to the length = length of the lawn x width of the road

= 75 x 4 m²

= 300 m²

Area of the road parallel to the breadth = breadth of the lawn x width of the road

= 60 x 4 m²

= 240 m²

The two roads cross each other in a square of side 4 m.

Area of the common portion = side x side

= 4 x 4 m²

= 16 m²

∴ Total area of the roads = (300 + 240 - 16) m²

= 524 m²

Now, cost of gravelling the roads = 524 x ₹ 14.50

= ₹ 7,598

Hence, the cost of gravelling the roads is ₹ 7,598.

Given:

Length and breadth of the rectangular park are in the ratio 5 : 2.

Width of the path = 2.5 metre (all around the outside)

Area of the path = 305 metre²

Let the length of the park be 5x metres and its breadth be 2x metres.

Then, length of the outer rectangle = (5x + 2 x 2.5) metre = (5x + 5) metre

And, breadth of the outer rectangle = (2x + 2 x 2.5) metre = (2x + 5) metre

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

⇒ (5x + 5)(2x + 5) - 5x x 2x = 305

⇒ 10x² + 25x + 10x + 25 - 10x² = 305

⇒ 35x + 25 = 305

⇒ 35x = 305 - 25

⇒ 35x = 280

⇒ x = $\frac{280}{35}$

⇒ x = 8

So, length of the park = 5x

= 5 x 8 metre

= 40 metre

And, breadth of the park = 2x

= 2 x 8 metre

= 16 metre

Hence, the dimensions of the park are 40 m by 16 m.

Given:

Side of the square = 16 cm

By formula,

Perimeter of a square = 4 x side

= 4 x 16 cm

= 64 cm

By formula,

Area of a square = side²

= (16)² cm²

= 256 cm²

By formula,

Diagonal of a square = $\sqrt{2}$ x side

= $16\sqrt{2}$ cm

= 16 x 1.414 cm

= 22.624 cm

Hence, the perimeter, area and diagonal of the square are 64 cm, 256 cm² and 22.624 cm respectively.

Given:

Side of the square = 8.5 m

By formula,

Perimeter of a square = 4 x side

= 4 x 8.5 m

= 34 m

By formula,

Area of a square = side²

= (8.5)² m²

= 72.25 m²

By formula,

Diagonal of a square = $\sqrt{2}$ x side

= $8.5\sqrt{2}$ m

= 8.5 x 1.414 m

= 12.019 m

= 12.02 m (approx.)

Hence, the perimeter, area and diagonal of the square are 34 m, 72.25 m² and 12.02 m respectively.

Given:

Side of the square = 2.5 dm

By formula,

Perimeter of a square = 4 x side

= 4 x 2.5 dm

= 10 dm

By formula,

Area of a square = side²

= (2.5)² dm²

= 6.25 dm²

By formula,

Diagonal of a square = $\sqrt{2}$ x side

= $2.5\sqrt{2}$ dm

= 2.5 x 1.414 dm

= 3.535 dm

Hence, the perimeter, area and diagonal of the square are 10 dm, 6.25 dm² and 3.535 dm respectively.

Given:

Perimeter of the square = 28 cm

By formula,

Perimeter of a square = 4 x side

⇒ 4 x side = 28

⇒ side = $\frac{28}{4}$

⇒ side = 7 cm

Now, area of the square = side²

= (7)² cm²

= 49 cm²

And, diagonal of the square = $\sqrt{2}$ x side

= $7\sqrt{2}$ cm

= 7 x 1.4142 cm (taking $\sqrt{2}$ = 1.4142)

= 9.8994 cm

= 9.899 cm (approx.)

Hence, the area of the square is 49 cm² and the length of its diagonal is 9.899 cm (approx.).

Given:

Diagonal of the square = $5\sqrt{2}$ m

By formula,

Area of a square = $\frac{1}{2}$ x (diagonal)²

= $\frac{1}{2}$ x $\left(5\sqrt{2}\right)^2$ m²

= $\frac{1}{2}$ x 50 m²

= 25 m²

Now, side of the square = $\sqrt{\text{Area}}$

= $\sqrt{25}$ m

= 5 m

∴ Perimeter of the square = 4 x side

= 4 x 5 m

= 20 m

Hence, the area of the square is 25 m² and its perimeter is 20 m.

Given:

Diagonal of the square = 12 cm

By formula,

Area of a square = $\frac{1}{2}$ x (diagonal)²

= $\frac{1}{2}$ x (12)² cm²

= $\frac{1}{2}$ x 144 cm²

= 72 cm²

Now, side of the square = $\sqrt{\text{Area}}$

$$\begin{aligned}= \sqrt{72} \text{ cm}\\[1em] = \sqrt{36 \times 2} \text{ cm}\\[1em] = 6\sqrt{2} \text{ cm}\\[1em] = 6 \times 1.414 \text{ cm}\\[1em] = 8.484 \text{ cm}\end{aligned}$$

∴ Perimeter of the square = 4 x side

= 4 x 8.484 cm

= 33.936 cm

Hence, the area of the square is 72 cm² and its perimeter is 33.936 cm.

Given:

Area of the square field = 32 m²

Let the diagonal of the square field be d m.

By formula,

Area of a square = $\frac{1}{2}$ x (diagonal)²

⇒ $\frac{1}{2}$ x d² = 32

⇒ d² = 32 x 2

⇒ d² = 64

⇒ d = $\sqrt{64}$

⇒ d = 8 m

Hence, the diagonal of the square field is 8 m.

Given:

Area of the square = 81 cm²

By formula,

Side of a square = $\sqrt{\text{Area}}$

= $\sqrt{81}$ cm

= 9 cm

Now, perimeter of the square = 4 x side

= 4 x 9 cm

= 36 cm

And, diagonal of the square = $\sqrt{2}$ x side

= $9\sqrt{2}$ cm

= 9 x 1.414 cm

= 12.726 cm

Hence, the perimeter of the square is 36 cm and the length of its diagonal is 12.726 cm.

Given:

Area of the square field = 6.25 ares

Rate of fencing = ₹ 32.50 per metre

As we know, 1 are = 100 m²

∴ Area of the square field = (6.25 x 100) m²

= 625 m²

Now, side of the square field = $\sqrt{\text{Area}}$

= $\sqrt{625}$ m

= 25 m

∴ Perimeter of the field = 4 x side

= 4 x 25 m

= 100 m

Cost of putting a fence round the field = 100 x ₹ 32.50

= ₹ 3,250

Hence, the cost of putting a fence round the field is ₹ 3,250.

Given:

Cost of ploughing the square field = ₹ 5,400

Rate of ploughing = ₹ 13.50 per square metre

Rate of fencing = ₹ 28.50 per metre

Area of the field = $\frac{\text{Total cost of ploughing}}{\text{Rate of ploughing}}$

= $\frac{5400}{13.50}$ m²

= 400 m²

Side of the square field = $\sqrt{\text{Area of the field}}$

= $\sqrt{400}$ m

= 20 m

∴ Perimeter of the field = 4 x side

= 4 x 20 m

= 80 m

Cost of fencing the field = 80 x ₹ 28.50

= ₹ 2,280

Hence, the cost of fencing the field is ₹ 2,280.

Given:

Base of the parallelogram = 65 cm

Height of the parallelogram = 7.8 cm

By formula,

Area of a parallelogram = base x height

= 65 x 7.8 cm²

= 507 cm²

Hence, the area of the parallelogram is 507 cm².

Given:

Base of the parallelogram = 6.4 m

Height of the parallelogram = 75 cm = $\frac{75}{100}$ m = 0.75 m

By formula,

Area of a parallelogram = base x height

= 6.4 x 0.75 m²

= 4.8 m²

Hence, the area of the parallelogram is 4.8 m².

Given:

Height of the parallelogram = $\frac{1}{3}$ x base

Area of the parallelogram = 192 cm²

Let the base of the parallelogram be b cm.

Then, height of the parallelogram = $\frac{b}{3}$ cm

By formula,

Area of a parallelogram = base x height

⇒ b x $\frac{b}{3}$ = 192

⇒ $\frac{b^2}{3}$ = 192

⇒ b² = 192 x 3

⇒ b² = 576

⇒ b = $\sqrt{576}$

⇒ b = 24 cm

So, height of the parallelogram = $\frac{24}{3}$ cm

= 8 cm

Hence, the height and the base of the parallelogram are 8 cm and 24 cm respectively.

Given:

PQ = 26 cm and QR = 20 cm

Distance between the longer sides = 12.5 cm

Since PQ = 26 cm > QR = 20 cm, so PQ and SR are the longer sides and QR and PS are the shorter sides.

(i) By formula,

Area of a parallelogram = base x height

Taking PQ as the base,

Area of parallelogram PQRS = 26 x 12.5 cm²

= 325 cm²

Hence, the area of the parallelogram is 325 cm².

(ii) Let the distance between the shorter sides be h cm.

Taking QR as the base,

Area of parallelogram PQRS = QR x h

⇒ 20 x h = 325

⇒ h = $\frac{325}{20}$

⇒ h = 16.25 cm

Hence, the distance between the shorter sides is 16.25 cm.

Given:

AB = 16 cm and BC = 14 cm

Area of parallelogram ABCD = 168 cm²

Since AB = 16 cm > BC = 14 cm, so AB and DC are the longer sides and BC and AD are the shorter sides.

By formula,

Area of a parallelogram = base x height

Let the distance between the longer sides be h₁ cm.

∴ AB x h₁ = 168

⇒ 16 x h₁ = 168

⇒ h₁ = $\frac{168}{16}$

⇒ h₁ = 10.5 cm

Let the distance between the shorter sides be h₂ cm.

∴ BC x h₂ = 168

⇒ 14 x h₂ = 168

⇒ h₂ = $\frac{168}{14}$

⇒ h₂ = 12 cm

Hence, the distance between the longer sides is 10.5 cm and that between the shorter sides is 12 cm.

Given:

AB = 28 cm, BC = 26 cm and diagonal AC = 30 cm

(i) We know that a diagonal of a parallelogram divides it into two triangles equal in area.

∴ Area of parallelogram ABCD = 2 x (area of ΔABC)

For ΔABC, let a = BC = 26 cm, b = AC = 30 cm and c = AB = 28 cm.

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{26 + 30 + 28}{2}\\[1em] = \frac{84}{2}\\[1em] = 42\end{aligned}$$

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

$$\begin{aligned}= \sqrt{42(42 - 26)(42 - 30)(42 - 28)} \text{ cm}^{2}\\[1em] = \sqrt{42 \times 16 \times 12 \times 14} \text{ cm}^{2}\\[1em] = \sqrt{112896} \text{ cm}^{2}\\[1em] = 336 \text{ cm}^{2}\end{aligned}$$

∴ Area of parallelogram ABCD = 2 x 336 cm²

= 672 cm²

Hence, the area of parallelogram ABCD is 672 cm².

(ii) Let the distance between AB and DC be h₁ cm.

By formula,

Area of a parallelogram = base x height

Taking AB as the base,

AB x h₁ = 672

⇒ 28 x h₁ = 672

⇒ h₁ = $\frac{672}{28}$

⇒ h₁ = 24 cm

Hence, the distance between AB and DC is 24 cm.

(iii) Let the distance between CB and DA be h₂ cm.

By formula,

Area of a parallelogram = base x height

Taking CB as the base,

CB x h₂ = 672

⇒ 26 x h₂ = 672

⇒ h₂ = $\frac{672}{26}$

⇒ h₂ = $\frac{336}{13}$ cm

⇒ h₂ = 25.84 cm (approx.)

Hence, the distance between CB and DA is $\mathbf{\frac{336}{13}}$ cm = 25.84 cm (approx.).

Given:

Perimeter of the rhombus = 48 cm

Altitude of the rhombus = 8.5 cm

By formula,

Perimeter of a rhombus = 4 x side

⇒ 4 x side = 48

⇒ side = $\frac{48}{4}$

⇒ side = 12 cm

Since a rhombus is a parallelogram,

Area of the rhombus = base x altitude

= 12 x 8.5 cm²

= 102 cm²

Hence, the area of the rhombus is 102 cm².

Given:

Area of the rhombus = 139.2 cm²

Altitude of the rhombus = 9.6 cm

By formula,

Area of a rhombus = base x altitude

⇒ base x 9.6 = 139.2

⇒ base = $\frac{139.2}{9.6}$

⇒ base = 14.5 cm

So, side of the rhombus = 14.5 cm

∴ Perimeter of the rhombus = 4 x side

= 4 x 14.5 cm

= 58 cm

Hence, the perimeter of the rhombus is 58 cm.

Given:

Area of the rhombus = 184 cm²

Perimeter of the rhombus = 64 cm

By formula,

Perimeter of a rhombus = 4 x side

⇒ 4 x side = 64

⇒ side = $\frac{64}{4}$

⇒ side = 16 cm

Now, area of the rhombus = base x altitude

⇒ 16 x altitude = 184

⇒ altitude = $\frac{184}{16}$

⇒ altitude = 11.5 cm

Hence, the altitude of the rhombus is 11.5 cm.

(i) Given:

Diagonals of the rhombus = 16 cm and 24 cm

By formula,

Area of a rhombus = $\frac{1}{2}$ x (product of diagonals)

= $\frac{1}{2}$ x 16 x 24 cm²

= $\frac{1}{2}$ x 384 cm²

= 192 cm²

Hence, the area of the rhombus is 192 cm².

(ii) Given:

Diagonals of the rhombus = 14.8 cm and 12.5 cm

By formula,

Area of a rhombus = $\frac{1}{2}$ x (product of diagonals)

= $\frac{1}{2}$ x 14.8 x 12.5 cm²

= $\frac{1}{2}$ x 185 cm²

= 92.5 cm²

Hence, the area of the rhombus is 92.5 cm².

Given:

Area of the rhombus = 138 cm²

One diagonal of the rhombus = 11.5 cm

Let the other diagonal of the rhombus be d cm.

By formula,

Area of a rhombus = $\frac{1}{2}$ x (product of diagonals)

⇒ $\frac{1}{2}$ x 11.5 x d = 138

⇒ 11.5 x d = 138 x 2

⇒ 11.5 x d = 276

⇒ d = $\frac{276}{11.5}$

⇒ d = 24 cm

Hence, the length of the other diagonal is 24 cm.

Given:

Each side of the rhombus = 20 cm

One diagonal of the rhombus = 24 cm

Let ABCD be the given rhombus in which AB = 20 cm and diagonal AC = 24 cm.

Let the diagonals AC and BD intersect at O.

We know that the diagonals of a rhombus bisect each other at right angles.

∴ AO = $\frac{1}{2}$ x AC

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

= 12 cm

and ∠AOB = 90°

Using Pythagoras theorem in right triangle AOB,

AB² = AO² + OB²

⇒ OB² = AB² - AO²

⇒ OB² = (20)² - (12)²

⇒ OB² = 400 - 144

⇒ OB² = 256

⇒ OB = $\sqrt{256}$

⇒ OB = 16 cm

So, BD = 2 x OB

= 2 x 16 cm

= 32 cm

Now, area of rhombus ABCD = $\frac{1}{2}$ x AC x BD

= $\frac{1}{2}$ x 24 x 32 cm²

= $\frac{1}{2}$ x 768 cm²

= 384 cm²

Hence, the area of the rhombus is 384 cm².

Given:

Parallel sides of the trapezium = 23.7 cm and 16.3 cm

Distance between the parallel sides = 11.4 cm

By formula,

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

$$\begin{aligned}= \frac{1}{2} \times (23.7 + 16.3) \times 11.4 \text{ cm}^{2}\\[1em] = \frac{1}{2} \times 40 \times 11.4 \text{ cm}^{2}\\[1em] = 20 \times 11.4 \text{ cm}^{2}\\[1em] = 228 \text{ cm}^{2}\end{aligned}$$

Hence, the area of the trapezium is 228 cm².

Given:

Parallel sides of the trapezium = 2.5 m and 1.3 m

Distance between the parallel sides = 80 cm = $\frac{80}{100}$ m = 0.8 m

By formula,

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

$$\begin{aligned}= \frac{1}{2} \times (2.5 + 1.3) \times 0.8 \text{ m}^{2}\\[1em] = \frac{1}{2} \times 3.8 \times 0.8 \text{ m}^{2}\\[1em] = \frac{1}{2} \times 3.04 \text{ m}^{2}\\[1em] = 1.52 \text{ m}^{2}\end{aligned}$$

Hence, the area of the trapezium is 1.52 m².

Given:

Lengths of the parallel sides of the trapezium are in the ratio 7 : 5.

Distance between the parallel sides = 14 cm

Area of the trapezium = 252 cm²

Let the lengths of the parallel sides be 7x centimetres and 5x centimetres.

By formula,

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

⇒ $\frac{1}{2}$ x (7x + 5x) x 14 = 252

⇒ $\frac{1}{2}$ x 12x x 14 = 252

⇒ 84x = 252

⇒ x = $\frac{252}{84}$

⇒ x = 3

So, one parallel side = 7x

= 7 x 3 cm

= 21 cm

And, the other parallel side = 5x

= 5 x 3 cm

= 15 cm

Hence, the lengths of the parallel sides are 21 cm and 15 cm.

Given:

Area of the trapezium = 162 cm²

Height of the trapezium = 6 cm

One of the bases = 23 cm

Let the other base of the trapezium be b cm.

By formula,

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

⇒ $\frac{1}{2}$ x (23 + b) x 6 = 162

⇒ 3 x (23 + b) = 162

⇒ 23 + b = $\frac{162}{3}$

⇒ 23 + b = 54

⇒ b = 54 - 23

⇒ b = 31 cm

Hence, the other base of the trapezium is 31 cm.

Given:

AB = 78 cm, DC = 52 cm, BC = 30 cm and AD = 28 cm

Draw CE ∥ DA and CF ⊥ AB.

Let CF = h cm. Then,

AE = DC = 52 cm, CE = DA = 28 cm

and EB = (AB - AE) = (78 - 52) cm = 26 cm

First we find the area of ΔCEB.

Let a = CE = 28 cm, b = EB = 26 cm and c = BC = 30 cm.

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{28 + 26 + 30}{2}\\[1em] = \frac{84}{2}\\[1em] = 42\end{aligned}$$

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

$$\begin{aligned}= \sqrt{42(42 - 28)(42 - 26)(42 - 30)} \text{ cm}^{2}\\[1em] = \sqrt{42 \times 14 \times 16 \times 12} \text{ cm}^{2}\\[1em] = \sqrt{112896} \text{ cm}^{2}\\[1em] = 336 \text{ cm}^{2}\end{aligned}$$

Also, area of ΔCEB = $\frac{1}{2}$ x EB x CF

= $\frac{1}{2}$ x 26 x h cm²

= 13h cm²

∴ 13h = 336

⇒ h = $\frac{336}{13}$ cm

Thus, the distance between the parallel sides = $\frac{336}{13}$ cm

Now, area of trapezium ABCD = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

$$\begin{aligned}= \frac{1}{2} \times (78 + 52) \times \frac{336}{13} \text{ cm}^{2}\\[1em] = \frac{1}{2} \times 130 \times \frac{336}{13} \text{ cm}^{2}\\[1em] = 65 \times \frac{336}{13} \text{ cm}^{2}\\[1em] = 5 \times 336 \text{ cm}^{2}\\[1em] = 1680 \text{ cm}^{2}\end{aligned}$$

Hence, the area of the trapezium is 1680 cm².

Given:

Parallel sides of the trapezium = 20 cm and 10 cm

Each non-parallel side = 13 cm

Let ABCD be the given trapezium in which AB = 20 cm, DC = 10 cm and AD = BC = 13 cm.

Then, FE = DC = 10 cm

Since the non-parallel sides are equal, AF = BE.

∴ AF = BE = $\frac{AB - FE}{2}$

= $\frac{20 - 10}{2}$ cm

= $\frac{10}{2}$ cm

= 5 cm

Using Pythagoras theorem in right triangle ADF,

AD² = AF² + DF²

⇒ DF² = AD² - AF²

⇒ DF² = (13)² - (5)²

⇒ DF² = 169 - 25

⇒ DF² = 144

⇒ DF = $\sqrt{144}$

⇒ DF = 12 cm

Now, area of trapezium ABCD = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

= $\frac{1}{2}$ x (20 + 10) x 12 cm²

= $\frac{1}{2}$ x 30 x 12 cm²

= 180 cm²

Hence, the area of the trapezium is 180 cm².

Given:

Area of the trapezium = 198 cm²

Height of the trapezium = 9 cm

One of the parallel sides is longer than the other by 8 cm.

Let the shorter parallel side be x cm.

Then, the longer parallel side = (x + 8) cm

By formula,

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

⇒ $\frac{1}{2}$ x (x + x + 8) x 9 = 198

⇒ (2x + 8) x 9 = 198 x 2

⇒ 2x + 8 = $\frac{396}{9}$

⇒ 2x + 8 = 44

⇒ 2x = 44 - 8

⇒ 2x = 36

⇒ x = 18

So, the shorter parallel side = 18 cm

And, the longer parallel side = (18 + 8) cm = 26 cm

Hence, the two parallel sides of the trapezium are 18 cm and 26 cm.

Given:

ABCD is a rectangle in which AB = 18 cm, BC = 8 cm and DE = 10 cm.

Since ABCD is a rectangle, AD = BC = 8 cm and DC = AB = 18 cm.

Using Pythagoras theorem in right triangle ADE,

DE² = AD² + AE²

⇒ AE² = DE² - AD²

⇒ AE² = (10)² - (8)²

⇒ AE² = 100 - 64

⇒ AE² = 36

⇒ AE = $\sqrt{36}$

⇒ AE = 6 cm

∴ EB = (AB - AE)

= (18 - 6) cm

= 12 cm

The shaded region EBCD is a trapezium in which DC ∥ EB and BC is the distance between them.

∴ Area of the shaded region EBCD = $\frac{1}{2}$ x (DC + EB) x BC

= $\frac{1}{2}$ x (18 + 12) x 8 cm²

= $\frac{1}{2}$ x 30 x 8 cm²

= 120 cm²

Hence, the area of the shaded region EBCD is 120 cm².

Given:

AC = 17 m, BC = 8 m, EF = 9 m, GD = 6 m, GL ⊥ EF and GL = 3.6 m

ABCH is a rectangle, so ∠ABC = 90°.

Using Pythagoras theorem in right triangle ABC,

AC² = AB² + BC²

⇒ AB² = AC² - BC²

⇒ AB² = (17)² - (8)²

⇒ AB² = 289 - 64

⇒ AB² = 225

⇒ AB = $\sqrt{225}$

⇒ AB = 15 m

Area of rectangle ABCH = AB x BC

= 15 x 8 m²

= 120 m²

EFGD is a trapezium in which EF ∥ GD and GL is the distance between them.

Area of trapezium EFGD = $\frac{1}{2}$ x (EF + GD) x GL

= $\frac{1}{2}$ x (9 + 6) x 3.6 m²

= $\frac{1}{2}$ x 15 x 3.6 m²

= 27 m²

∴ Area of the figure ABCDEFGH = Area of rectangle ABCH + Area of trapezium EFGD

= (120 + 27) m²

= 147 m²

Hence, the area of the figure ABCDEFGH is 147 m².

The given figure can be divided into three parts — a rectangle at the top, a trapezium in the middle and a rectangle at the bottom.

Upper rectangle:

Length = 3.5 m and breadth = 3 m

Area of the upper rectangle = length x breadth

= 3.5 x 3 m²

= 10.5 m²

Middle trapezium:

Parallel sides = 3.5 m and 6 m

Distance between the parallel sides = (4.6 - 3) m = 1.6 m

Area of the trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

= $\frac{1}{2}$ x (3.5 + 6) x 1.6 m²

= $\frac{1}{2}$ x 9.5 x 1.6 m²

= $\frac{1}{2}$ x 15.2 m²

= 7.6 m²

Lower rectangle:

Length = 6 m and breadth = 2.5 m

Area of the lower rectangle = length x breadth

= 6 x 2.5 m²

= 15 m²

∴ Area of the shaded region = (10.5 + 7.6 + 15) m²

= 33.1 m²

Hence, the area of the shaded region is 33.1 m².

The given figure can be divided into three parts — a trapezium at the top, a parallelogram in the middle and a rectangle at the bottom.

Upper trapezium:

Parallel sides = 18 m and 12 m

Distance between the parallel sides = 3 m

Area of the upper trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

= $\frac{1}{2}$ x (18 + 12) x 3 m²

= $\frac{1}{2}$ x 30 x 3 m²

= 45 m²

Middle parallelogram:

Base = 10 m and height = 3 m

Area of the middle parallelogram = base x height

= 10 x 3 m²

= 30 m²

Lower rectangle:

Length = 18 m and breadth = 3 m

Area of the lower rectangle = length x breadth

= 18 x 3 m²

= 54 m²

∴ Area of the shaded region = (45 + 30 + 54) m²

= 129 m²

Hence, the area of the shaded region is 129 m².

Given:

AC = 14 cm, BL ⊥ AC, DM ⊥ AC, BL = 8 cm and DM = 6 cm

By formula,

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

Area of quad. ABCD = area (ΔADC) + area (ΔABC)

Substituting values we get:

Area of quad. ABCD = ($\frac{1}{2}$ x 14 x 6 + $\frac{1}{2}$ x 14 x 8) cm²

= (42 + 56) cm²

= 98 cm²

Hence, the area of quad. ABCD is 98 cm².

Given:

AF = 9 cm, AG = 13 cm, AH = 19 cm, AD = 24 cm, BF = 6 cm, CH = 8 cm and EG = 9 cm

From the figure,

FH = AH - AF = (19 - 9) cm = 10 cm

HD = AD - AH = (24 - 19) cm = 5 cm

Area of the pentagon ABCDE

= area (ΔABF) + area (trap. FBCH) + area (ΔHCD) + area (ΔAED)

Area of ΔABF = $\frac{1}{2}$ x AF x BF

= $\frac{1}{2}$ x 9 x 6 cm²

= 27 cm²

Area of trap. FBCH = $\frac{1}{2}$ x (BF + CH) x FH

= $\frac{1}{2}$ x (6 + 8) x 10 cm²

= $\frac{1}{2}$ x 14 x 10 cm²

= 70 cm²

Area of ΔHCD = $\frac{1}{2}$ x HD x CH

= $\frac{1}{2}$ x 5 x 8 cm²

= 20 cm²

Area of ΔAED = $\frac{1}{2}$ x AD x EG

= $\frac{1}{2}$ x 24 x 9 cm²

= 108 cm²

∴ Area of the pentagon ABCDE = (27 + 70 + 20 + 108) cm²

= 225 cm²

Hence, the area of the pentagon ABCDE is 225 cm².

Given:

PR = 10 cm, PS = 12 cm, QD = 3 cm, RE = 7 cm and TF = 5 cm

Area of the pentagon PQRST

= area (ΔPQR) + area (ΔPRS) + area (ΔPTS)

Area of ΔPQR = $\frac{1}{2}$ x PR x QD

= $\frac{1}{2}$ x 10 x 3 cm²

= 15 cm²

Area of ΔPRS = $\frac{1}{2}$ x PS x RE

= $\frac{1}{2}$ x 12 x 7 cm²

= 42 cm²

Area of ΔPTS = $\frac{1}{2}$ x PS x TF

= $\frac{1}{2}$ x 12 x 5 cm²

= 30 cm²

∴ Area of the pentagon PQRST = (15 + 42 + 30) cm²

= 87 cm²

Hence, the area of the pentagon PQRST is 87 cm².

Given:

AJ = 6 cm, AK = 10 cm, AL = 18 cm, AM = 21 cm, AD = 27 cm, BJ = 5 cm, CL = 6 cm, EM = 4 cm and FK = 6 cm

From the figure,

JL = AL - AJ = (18 - 6) cm = 12 cm

LD = AD - AL = (27 - 18) cm = 9 cm

KM = AM - AK = (21 - 10) cm = 11 cm

MD = AD - AM = (27 - 21) cm = 6 cm

Area of the hexagon ABCDEF

= area (ΔABJ) + area (trap. JBCL) + area (ΔLCD) + area (ΔAFK) + area (trap. KFEM) + area (ΔMED)

Area of ΔABJ = $\frac{1}{2}$ x AJ x BJ

= $\frac{1}{2}$ x 6 x 5 cm²

= 15 cm²

Area of trap. JBCL = $\frac{1}{2}$ x (BJ + CL) x JL

= $\frac{1}{2}$ x (5 + 6) x 12 cm²

= $\frac{1}{2}$ x 11 x 12 cm²

= 66 cm²

Area of ΔLCD = $\frac{1}{2}$ x LD x CL

= $\frac{1}{2}$ x 9 x 6 cm²

= 27 cm²

Area of ΔAFK = $\frac{1}{2}$ x AK x FK

= $\frac{1}{2}$ x 10 x 6 cm²

= 30 cm²

Area of trap. KFEM = $\frac{1}{2}$ x (FK + EM) x KM

= $\frac{1}{2}$ x (6 + 4) x 11 cm²

= $\frac{1}{2}$ x 10 x 11 cm²

= 55 cm²

Area of ΔMED = $\frac{1}{2}$ x MD x EM

= $\frac{1}{2}$ x 6 x 4 cm²

= 12 cm²

∴ Area of the hexagon ABCDEF = (15 + 66 + 27 + 30 + 55 + 12) cm²

= 205 cm²

Hence, the area of the hexagon ABCDEF is 205 cm².

Given:

Each side of the hexagon LMNPQR = 5 cm

Height MQ = 11 cm and width RP = 8 cm

The diagonal MQ divides the hexagon LMNPQR into two equal trapeziums MNPQ and MQRL.

∴ Area of the hexagon LMNPQR = 2 x (area of trapezium MNPQ)

In trapezium MNPQ, the parallel sides are MQ = 11 cm and NP = 5 cm.

Since RP ⊥ MQ and MQ passes through the mid-point of RP,

Distance between MQ and NP = $\frac{1}{2}$ x RP

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

= 4 cm

Area of trapezium MNPQ = $\frac{1}{2}$ x (MQ + NP) x 4

= $\frac{1}{2}$ x (11 + 5) x 4 cm²

= $\frac{1}{2}$ x 16 x 4 cm²

= 32 cm²

∴ Area of the hexagon LMNPQR = 2 x 32 cm²

= 64 cm²

Hence, the area of the hexagon LMNPQR is 64 cm².

Given:

Each side of the octagon ABCDEFGH = 5 cm

HC = 11 cm, AP ⊥ HC and AP = 4 cm

The octagon ABCDEFGH is divided by HC and GD into two equal trapeziums ABCH and DEFG and a rectangle CDGH.

Trapezium ABCH:

Parallel sides are AB = 5 cm and HC = 11 cm and the distance between them is AP = 4 cm.

Area of trapezium ABCH = $\frac{1}{2}$ x (AB + HC) x AP

= $\frac{1}{2}$ x (5 + 11) x 4 cm²

= $\frac{1}{2}$ x 16 x 4 cm²

= 32 cm²

∴ Area of trapezium DEFG = Area of trapezium ABCH = 32 cm²

Rectangle CDGH:

Length HC = 11 cm and breadth CD = 5 cm

Area of rectangle CDGH = HC x CD

= 11 x 5 cm²

= 55 cm²

∴ Area of the octagon ABCDEFGH = 2 x (area of trapezium DEFG) + area of rectangle CDGH

= (2 x 32 + 55) cm²

= (64 + 55) cm²

= 119 cm²

Hence, the area of the octagon ABCDEFGH is 119 cm².

Given:

AE ∥ BD, AF ⊥ BD, CG ⊥ BD, AE = 12 cm, BD = 16 cm, AF = 6.5 cm and CG = 8.5 cm

Area of the figure ABCDE = area (trap. ABDE) + area (ΔBCD)

Trapezium ABDE:

Parallel sides are AE = 12 cm and BD = 16 cm and the distance between them is AF = 6.5 cm.

Area of trap. ABDE = $\frac{1}{2}$ x (AE + BD) x AF

$$\begin{aligned}= \frac{1}{2} \times (12 + 16) \times 6.5 \text{ cm}^{2}\\[1em] = \frac{1}{2} \times 28 \times 6.5 \text{ cm}^{2}\\[1em] = 14 \times 6.5 \text{ cm}^{2}\\[1em] = 91 \text{ cm}^{2}\end{aligned}$$

Triangle BCD:

Base BD = 16 cm and height CG = 8.5 cm

Area of ΔBCD = $\frac{1}{2}$ x BD x CG

= $\frac{1}{2}$ x 16 x 8.5 cm²

= 8 x 8.5 cm²

= 68 cm²

∴ Area of the figure ABCDE = (91 + 68) cm²

= 159 cm²

Hence, the area of the figure ABCDE is 159 cm².

Given:

AP = 20 m, AQ = 35 m, AR = 58 m, AS = 65 m, AD = 75 m, BP = 15 m, CR = 20 m, ES = 15 m and FQ = 10 m

From the figure,

QS = AS - AQ = (65 - 35) m = 30 m

SD = AD - AS = (75 - 65) m = 10 m

PR = AR - AP = (58 - 20) m = 38 m

RD = AD - AR = (75 - 58) m = 17 m

Area of the field ABCDEFA

= area (ΔAFQ) + area (trap. QFES) + area (ΔSED) + area (ΔABP) + area (trap. PBCR) + area (ΔRCD)

Area of ΔAFQ = $\frac{1}{2}$ x AQ x FQ

= $\frac{1}{2}$ x 35 x 10 m²

= 175 m²

Area of trap. QFES = $\frac{1}{2}$ x (FQ + ES) x QS

= $\frac{1}{2}$ x (10 + 15) x 30 m²

= $\frac{1}{2}$ x 25 x 30 m²

= 375 m²

Area of ΔSED = $\frac{1}{2}$ x SD x ES

= $\frac{1}{2}$ x 10 x 15 m²

= 75 m²

Area of ΔABP = $\frac{1}{2}$ x AP x BP

= $\frac{1}{2}$ x 20 x 15 m²

= 150 m²

Area of trap. PBCR = $\frac{1}{2}$ x (BP + CR) x PR

= $\frac{1}{2}$ x (15 + 20) x 38 m²

= $\frac{1}{2}$ x 35 x 38 m²

= 665 m²

Area of ΔRCD = $\frac{1}{2}$ x RD x CR

= $\frac{1}{2}$ x 17 x 20 m²

= 170 m²

∴ Area of the field ABCDEFA = (175 + 375 + 75 + 150 + 665 + 170) m²

= 1610 m²

Hence, the area of the field ABCDEFA is 1610 m².

Given:

Radius of the semi-circle (r) = 35 cm

By formula,

Perimeter of a semi-circle = r(π + 2)

$$\begin{aligned}= 35 \times \left(\frac{22}{7} + 2\right) \text{ cm}\\[1em] = 35 \times \frac{36}{7} \text{ cm}\\[1em] = 5 \times 36 \text{ cm}\\[1em] = 180 \text{ cm}\end{aligned}$$

By formula,

Area of a semi-circle = $\frac{1}{2}$ πr²

= $\frac{1}{2} \times \frac{22}{7} \times 35 \times 35$ cm²

= 11 x 5 x 35 cm²

= 1925 cm²

Hence, the perimeter and area of the semi-circle are 180 cm and 1925 cm² respectively.

Given:

Radius of the semi-circle (r) = 4.2 cm

By formula,

Perimeter of a semi-circle = r(π + 2)

$$\begin{aligned}= 4.2 \times \left(\frac{22}{7} + 2\right) \text{ cm}\\[1em] = 4.2 \times \frac{36}{7} \text{ cm}\\[1em] = 0.6 \times 36 \text{ cm}\\[1em] = 21.6 \text{ cm}\end{aligned}$$

By formula,

Area of a semi-circle = $\frac{1}{2}$ πr²

= $\frac{1}{2} \times \frac{22}{7} \times 4.2 \times 4.2$ cm²

= 11 x 0.6 x 4.2 cm²

= 27.72 cm²

Hence, the perimeter and area of the semi-circle are 21.6 cm and 27.72 cm² respectively.

Given:

Radius of the semi-circle (r) = 15.4 cm

By formula,

Perimeter of a semi-circle = r(π + 2)

$$\begin{aligned}= 15.4 \times \left(\frac{22}{7} + 2\right) \text{ cm}\\[1em] = 15.4 \times \frac{36}{7} \text{ cm}\\[1em] = 2.2 \times 36 \text{ cm}\\[1em] = 79.2 \text{ cm}\end{aligned}$$

By formula,

Area of a semi-circle = $\frac{1}{2}$ πr²

= $\frac{1}{2} \times \frac{22}{7} \times 15.4 \times 15.4$ cm²

= 11 x 2.2 x 15.4 cm²

= 372.68 cm²

Hence, the perimeter and area of the semi-circle are 79.2 cm and 372.68 cm² respectively.

Given:

Diameter of the semi-circle = 18.2 cm

∴ Radius of the semi-circle (r) = $\frac{18.2}{2}$ cm = 9.1 cm

By formula,

Perimeter of a semi-circle = r(π + 2)

$$\begin{aligned}= 9.1 \times \left(\frac{22}{7} + 2\right) \text{ cm}\\[1em] = 9.1 \times \frac{36}{7} \text{ cm}\\[1em] = 1.3 \times 36 \text{ cm}\\[1em] = 46.8 \text{ cm}\end{aligned}$$

By formula,

Area of a semi-circle = $\frac{1}{2}$ πr²

= $\frac{1}{2} \times \frac{22}{7} \times 9.1 \times 9.1$ cm²

= 11 x 1.3 x 9.1 cm²

= 130.13 cm²

Hence, the perimeter and area of the semi-circle are 46.8 cm and 130.13 cm² respectively.

Given:

Diameter of the semi-circle = 5.6 cm

∴ Radius of the semi-circle (r) = $\frac{5.6}{2}$ cm = 2.8 cm

By formula,

Perimeter of a semi-circle = r(π + 2)

$$\begin{aligned}= 2.8 \times \left(\frac{22}{7} + 2\right) \text{ cm}\\[1em] = 2.8 \times \frac{36}{7} \text{ cm}\\[1em] = 0.4 \times 36 \text{ cm}\\[1em] = 14.4 \text{ cm}\end{aligned}$$

By formula,

Area of a semi-circle = $\frac{1}{2}$ πr²

= $\frac{1}{2} \times \frac{22}{7} \times 2.8 \times 2.8$ cm²

= 11 x 0.4 x 2.8 cm²

= 12.32 cm²

Hence, the perimeter and area of the semi-circle are 14.4 cm and 12.32 cm² respectively.

Given:

Radius of the semi-circle (r) = 8 cm and π = 3.14

By formula,

Perimeter of a semi-circle = r (π + 2)

= 8 x (3.14 + 2) cm

= 8 x 5.14 cm

= 41.12 cm

By formula,

Area of a semi-circle = $\frac{1}{2}$ πr²

= $\frac{1}{2}$ x 3.14 x 8 x 8 cm²

= 1.57 x 64 cm²

= 100.48 cm²

Hence, the perimeter and area of the semi-circle are 41.12 cm and 100.48 cm² respectively.

Given:

Radius of the semi-circle (r) = 15 cm and π = 3.14

By formula,

Perimeter of a semi-circle = r (π + 2)

= 15 x (3.14 + 2) cm

= 15 x 5.14 cm

= 77.1 cm

By formula,

Area of a semi-circle = $\frac{1}{2}$ πr²

= $\frac{1}{2}$ x 3.14 x 15 x 15 cm²

= 1.57 x 225 cm²

= 353.25 cm²

Hence, the perimeter and area of the semi-circle are 77.1 cm and 353.25 cm² respectively.

Given:

Radius of the semi-circle (r) = 20 cm and π = 3.14

By formula,

Perimeter of a semi-circle = r (π + 2)

= 20 x (3.14 + 2) cm

= 20 x 5.14 cm

= 102.8 cm

By formula,

Area of a semi-circle = $\frac{1}{2}$ πr²

= $\frac{1}{2}$ x 3.14 x 20 x 20 cm²

= 1.57 x 400 cm²

= 628 cm²

Hence, the perimeter and area of the semi-circle are 102.8 cm and 628 cm² respectively.

Given:

Perimeter of the semi-circle = 54 cm

Let the radius of the semi-circle be r cm.

By formula,

Perimeter of a semi-circle = r(π + 2)

$$\begin{aligned}\Rightarrow r \times \left(\frac{22}{7} + 2\right) = 54\\[1em] \Rightarrow r \times \frac{36}{7} = 54\\[1em] \Rightarrow r = 54 \times \frac{7}{36} \\[1em] \Rightarrow r = 10.5 \text{ cm}\end{aligned}$$

Now,

Area of the semi-circle = $\frac{1}{2}$ πr²

= $\frac{1}{2} \times \frac{22}{7} \times 10.5 \times 10.5$ cm²

= 11 x 1.5 x 10.5 cm²

= 173.25 cm²

Hence, the area of the semi-circle is 173.25 cm².

Given:

Perimeter of the semi-circle = 144 cm

Let the radius of the semi-circle be r cm.

By formula,

Perimeter of a semi-circle = r(π + 2)

$$\begin{aligned}\Rightarrow r \times \left(\frac{22}{7} + 2\right) = 144\\[1em] \Rightarrow r \times \frac{36}{7} = 144\\[1em] \Rightarrow r = 144 \times \frac{7}{36}\\[1em] \Rightarrow r = 28 \text{ cm }\end{aligned}$$

Now,

Area of the semi-circle = $\frac{1}{2}$ πr²

= $\frac{1}{2} \times \frac{22}{7} \times 28 \times 28$ cm²

= 11 x 4 x 28 cm²

= 1232 cm²

Hence, the area of the semi-circle is 1232 cm².

Given:

Area of the semi-circle = 77 cm²

Let the radius of the semi-circle be r cm.

By formula,

Area of a semi-circle = $\frac{1}{2}$ πr²

$$\begin{aligned}\Rightarrow \frac{1}{2} \times \frac{22}{7} \times r^2 = 77\\[1em] \Rightarrow \frac{11}{7} \times r^2 = 77\\[1em] \Rightarrow r^2 = 77 \times \frac{7}{11}\\[1em] \Rightarrow r^{2} = 49\\[1em] \Rightarrow r = \sqrt{49}\\[1em] \Rightarrow r = 7 \text{ cm}\end{aligned}$$

Now,

Perimeter of the semi-circle = r(π + 2)

= 7 x $\left(\frac{22}{7} + 2\right)$ cm

= 7 x $\frac{36}{7}$ cm

= 36 cm

Hence, the radius of the semi-circle is 7 cm and its perimeter is 36 cm.

Given:

Area of the semi-circle = 12.32 m²

Let the radius of the semi-circle be r m.

By formula,

Area of a semi-circle = $\frac{1}{2}$ πr²

$$\begin{aligned}\Rightarrow \frac{1}{2} \times \frac{22}{7} \times r^2 = 12.32\\[1em] \Rightarrow \frac{11}{7} \times r^2 = 12.32\\[1em] \Rightarrow r^2 = 12.32 \times \frac{7}{11}\\[1em] \Rightarrow r^{2} = 7.84\\[1em] \Rightarrow r = \sqrt{7.84}\\[1em] \Rightarrow r = 2.8 \text{ m}\end{aligned}$$

Now,

Perimeter of the semi-circle = r(π + 2)

$$\begin{aligned}= 2.8 \times \left(\frac{22}{7} + 2\right) \text{ m}\\[1em] = 2.8 \times \frac{36}{7} \text{ m}\\[1em] = 0.4 \times 36 \text{ m}\\[1em] = 14.4 \text{ m}\end{aligned}$$

Hence, the radius of the semi-circle is 2.8 m and its perimeter is 14.4 m.

Given:

Radius of the semi-circle (r) = 28 cm

Since the wire is in the form of a semi-circular arc,

Length of the wire = Length of the semi-circular arc

= πr

= $\frac{22}{7}$ x 28 cm

= 88 cm

When this wire is bent into the form of a square, the perimeter of the square = 88 cm.

By formula,

Perimeter of a square = 4 x side

⇒ 4 x side = 88

⇒ side = $\frac{88}{4}$

⇒ side = 22 cm

Hence, the length of the side of the square is 22 cm.

Given:

Side of the square = 49.5 cm

Length of the wire = Perimeter of the square

= 4 x side

= 4 x 49.5 cm

= 198 cm

When this wire is bent into a semi-circle, it forms the semi-circular arc, so the length of the semi-circular arc = 198 cm.

Let the radius of the semi-circle be r cm.

By formula,

Length of a semi-circular arc = πr

⇒ $\frac{22}{7}$ x r = 198

⇒ r = 198 x $\frac{7}{22}$

⇒ r = 63 cm

Hence, the radius of the semi-circle so formed is 63 cm.

Given:

Area of the square = 756.25 cm²

By formula,

Side of a square = $\sqrt{\text{Area}}$

= $\sqrt{756.25}$ cm

= 27.5 cm

∴ Length of the wire = Perimeter of the square

= 4 x side

= 4 x 27.5 cm

= 110 cm

When this wire is bent to form a semi-circle, it forms the semi-circular arc, so the length of the semi-circular arc = 110 cm.

Let the radius of the semi-circle be r cm.

By formula,

Length of a semi-circular arc = πr

⇒ $\frac{22}{7}$ x r = 110

⇒ r = 110 x $\frac{7}{22}$

⇒ r = 35 cm

Hence, the radius of the semi-circle so formed is 35 cm.

Given:

Radius of each semi-circle (r) = 17.5 m

Length of the rectangle = 40 m

Since the semi-circles are described on the breadth of the rectangle,

Breadth of the rectangle = 2 x 17.5 m = 35 m

Area of the rectangle = length x breadth

= 40 x 35 m²

= 1400 m²

Area of the two semi-circles = $2 \times \frac{1}{2}$ πr²

= $\frac{22}{7} \times 17.5 \times 17.5$ m²

= 22 x 2.5 x 17.5 m²

= 962.5 m²

∴ Area of the park = (1400 + 962.5) m²

= 2362.5 m²

Perimeter of the park = 2 x length + (circumference of the two semi-circles)

= 2 x length + 2πr

= (2 x 40) m + $\left(2 \times \frac{22}{7} \times 17.5\right)$ m

= 80 m + 110 m

= 190 m

Hence, the area of the park is 2362.5 m² and its perimeter is 190 m.

Given:

OAB is an equilateral triangle of side 28 cm.

A semi-circle is described on AB as diameter.

∴ Radius of the semi-circle (r) = $\frac{28}{2}$ cm = 14 cm

Area of the equilateral triangle OAB = $\frac{\sqrt{3}}{4} \times \text{side}^2$

$$\begin{aligned}= \frac{\sqrt{3}}{4} \times (28)^2 \text{ cm}^{2}\\[1em] = \frac{1.73}{4} \times 784 \text{ cm}^{2}\\[1em] = 1.73 \times 196 \text{ cm}^{2}\\[1em] = 339.08 \text{ cm}^{2}\end{aligned}$$

Area of the semi-circle = $\frac{1}{2}$ πr²

= $\frac{1}{2} \times \frac{22}{7} \times 14 \times 14$ cm²

= 11 x 2 x 14 cm²

= 308 cm²

∴ Area of the shaded region = (339.08 + 308) cm²

= 647.08 cm²

Perimeter of the shaded region = OA + OB + (length of the semi-circular arc)

= (28 + 28) cm + πr

= 56 cm + $\left(\frac{22}{7} \times 14\right)$ cm

= 56 cm + 44 cm

= 100 cm

Hence, the perimeter of the shaded region is 100 cm and its area is 647.08 cm².

Given:

AB ∥ CD, AB = 12 cm, CD = 15 cm, BD = 4 cm and π = 3.14

A semi-circle is described on BD as diameter.

Draw BE ∥ AC.

Then, CE = AB = 12 cm

∴ ED = (CD - CE)

= (15 - 12) cm

= 3 cm

Using Pythagoras theorem in right triangle BDE,

BE² = ED² + BD²

⇒ BE² = (3)² + (4)²

⇒ BE² = 9 + 16

⇒ BE² = 25

⇒ BE = $\sqrt{25}$

⇒ BE = 5 cm

Since ACEB is a parallelogram, AC = BE = 5 cm.

Radius of the semi-circle (r) = $\frac{BD}{2}$ = $\frac{4}{2}$ cm = 2 cm

Area of trapezium ABDC = $\frac{1}{2}$ x (AB + CD) x BD

= $\frac{1}{2}$ x (12 + 15) x 4 cm²

= $\frac{1}{2}$ x 27 x 4 cm²

= 54 cm²

Area of the semi-circle = $\frac{1}{2}$ πr²

= $\frac{1}{2}$ x 3.14 x 2 x 2 cm²

= 6.28 cm²

∴ Area of the figure = (54 + 6.28) cm²

= 60.28 cm²

Perimeter of the figure = AB + AC + CD + (length of the semi-circular arc)

= (12 + 5 + 15) cm + πr

= 32 cm + (3.14 x 2) cm

= 32 cm + 6.28 cm

= 38.28 cm

Hence, the perimeter of the figure is 38.28 cm and its area is 60.28 cm².

Given:

Length of the rectangle = 20 m

Breadth of the rectangle = 7 m

Two semi-circles are cut off from the rectangle, each having the breadth of the rectangle as its diameter.

∴ Radius of each semi-circle (r) = $\frac{7}{2}$ m = 3.5 m

Area of the rectangle = length x breadth

= 20 x 7 m²

= 140 m²

Area of the two semi-circles = $2 \times \frac{1}{2}$ πr²

= $\frac{22}{7} \times 3.5 \times 3.5$ m²

= 22 x 0.5 x 3.5 m²

= 38.5 m²

∴ Area of the shaded region = (140 - 38.5) m²

= 101.5 m²

Hence, the area of the shaded region is 101.5 m².

The shaded region consists of two rectangles, each 3.5 m by 12 m, and a semi-circular ring.

Area of the two rectangles = 2 x (length x breadth)

= 2 x (3.5 x 12) m²

= 2 x 42 m²

= 84 m²

For the semi-circular ring,

Inner radius (r) = $\frac{7}{2}$ m = 3.5 m

Outer radius (R) = $\frac{(3.5 + 7 + 3.5)}{2}$ m = $\frac{14}{2}$ m = 7 m

Area of the semi-circular ring = $\frac{1}{2}$ π (R² - r²)

$$\begin{aligned}= \frac{1}{2} \times \frac{22}{7} \times \left[(7)^2 - (3.5)^2\right] \text{ m}^{2}\\[1em] = \frac{11}{7} \times (49 - 12.25) \text{ m}^{2}\\[1em] = \frac{11}{7} \times 36.75 \text{ m}^{2}\\[1em] = 11 \times 5.25 \text{ m}^{2}\\[1em] = 57.75 \text{ m}^{2}\end{aligned}$$

∴ Area of the shaded region = (84 + 57.75) m²

= 141.75 m²

Hence, the area of the shaded region is 141.75 m².

The inner white region is a rectangle 50 m by 21 m with a semi-circle at each end, and the shaded region is a border of uniform width 3.5 m around it.

For the inner region,

Radius of each inner semi-circle (r) = $\frac{21}{2}$ m = 10.5 m

Area of the inner region = (length x breadth) + $2 \times \frac{1}{2}$ πr²

= (50 x 21) m² + $\left(2 \times \frac{1}{2} \times \frac{22}{7} \times 10.5 \times 10.5\right)$ m²

= 1050 m² + 346.5 m²

= 1396.5 m²

For the outer region,

Radius of each outer semi-circle (R) = (10.5 + 3.5) m = 14 m

Breadth of the outer rectangle = (21 + 3.5 + 3.5) m = 28 m

Area of the outer region = (length x breadth) + $2 \times \frac{1}{2}$ πR²

= (50 x 28) m² + $\left(2 \times \frac{1}{2} \times \frac{22}{7} \times 14 \times 14\right)$ m²

= 1400 m² + 616 m²

= 2016 m²

∴ Area of the shaded region = (2016 - 1396.5) m²

= 619.5 m²

Hence, the area of the shaded region is 619.5 m².

Diameter of the bigger circle = 21 m

∴ Radius of the bigger circle (R) = $\frac{21}{2}$ m = 10.5 m

Area of the bigger circle = πR²

= $\frac{22}{7} \times 10.5 \times 10.5$ m²

= 22 x 1.5 x 10.5 m²

= 346.5 m²

Diameter of each of the two unshaded semi-circles = 10.5 m

∴ Radius of each unshaded semi-circle (r) = $\frac{10.5}{2}$ m = 5.25 m

Area of the two unshaded semi-circles = $2 \times \frac{1}{2}$ πr²

= $\frac{22}{7} \times 5.25 \times 5.25$ m²

= 22 x 0.75 x 5.25 m²

= 86.625 m²

∴ Area of the shaded region = (346.5 - 86.625) m²

= 259.875 m²

Hence, the area of the shaded region is 259.875 m².

Diameter of the bigger semi-circle = 21 m

∴ Radius of the bigger semi-circle (R) = $\frac{21}{2}$ m = 10.5 m

Area of the bigger semi-circle = $\frac{1}{2}$ πR²

= $\frac{1}{2} \times \frac{22}{7} \times 10.5 \times 10.5$ m²

= 11 x 1.5 x 10.5 m²

= 173.25 m²

Diameter of each of the two smaller semi-circles = 7 m

∴ Radius of each smaller semi-circle (r) = $\frac{7}{2}$ m = 3.5 m

Area of the two smaller semi-circles = $2 \times \frac{1}{2}$ πr²

= $\frac{22}{7} \times 3.5 \times 3.5$ m²

= 22 x 0.5 x 3.5 m²

= 38.5 m²

∴ Area of the shaded region = (173.25 - 38.5) m²

= 134.75 m²

Hence, the area of the shaded region is 134.75 m².

Let the side of the square be a cm.

A semi-circle is described on each side of the square, so the radius of each semi-circle = $\frac{a}{2}$ cm.

From the figure,

$a + \frac{a}{2} + \frac{a}{2}$ = 42

⇒ 2a = 42

⇒ a = 21

So, side of the square = 21 cm

∴ Radius of each semi-circle (r) = $\frac{21}{2}$ cm = 10.5 cm

Area of the square = side²

= (21)² cm²

= 441 cm²

Area of the four semi-circles = $4 \times \frac{1}{2}$ πr²

= $2 \times \frac{22}{7} \times 10.5 \times 10.5$ cm²

= 2 x 22 x 1.5 x 10.5 cm²

= 693 cm²

∴ Area of the shaded region = (441 + 693) cm²

= 1134 cm²

Hence, the area of the shaded region is 1134 cm².

The figure consists of a bigger semi-circle of diameter 42 cm and two smaller semi-circles, each of diameter 21 cm.

Radius of the bigger semi-circle (R) = $\frac{42}{2}$ cm = 21 cm

Area of the bigger semi-circle = $\frac{1}{2}$ πR²

= $\frac{1}{2} \times \frac{22}{7} \times 21 \times 21$ cm²

= 11 x 3 x 21 cm²

= 693 cm²

Radius of each smaller semi-circle (r) = $\frac{21}{2}$ cm = 10.5 cm

Area of the two smaller semi-circles = $2 \times \frac{1}{2}$ πr²

= $\frac{22}{7} \times 10.5 \times 10.5$ cm²

= 22 x 1.5 x 10.5 cm²

= 346.5 cm²

∴ Area of the shaded region = (693 + 346.5) cm²

= 1039.5 cm²

Hence, the area of the shaded region is 1039.5 cm².

Let the breadth of the rectangle be b cm.

Then, length of the rectangle = (b + 2) cm

By formula,

Perimeter of a rectangle = 2 x (length + breadth)

⇒ 2 x (b + 2 + b) = 48

⇒ 2b + 2 = $\frac{48}{2}$

⇒ 2b + 2 = 24

⇒ 2b = 22

⇒ b = 11 cm

So, length of the rectangle = (11 + 2) cm = 13 cm

Now, area of the rectangle = length x breadth

= 13 x 11 cm²

= 143 cm²

Hence, option 3 is the correct option.

Let the breadth of the hall be b m.

Then, length of the hall = $\frac{4}{3}$b m

By formula,

Area of a rectangle = length x breadth

⇒ $\frac{4}{3}$b x b = 300

⇒ $\frac{4b^2}{3}$ = 300

⇒ 4b² = 900

⇒ b² = 225

⇒ b = $\sqrt{225}$

⇒ b = 15 m

So, length of the hall = $\frac{4}{3}$ x 15 m = 20 m

∴ Difference between the length and the breadth = (20 - 15) m

= 5 m

Hence, option 3 is the correct option.

Given:

Perimeter of the rectangular field = 200 m

Breadth of the field = 40 m

By formula,

Perimeter of a rectangle = 2 x (length + breadth)

⇒ 2 x (length + 40) = 200

⇒ length + 40 = 100

⇒ length = 60 m

Now, area of the field = length x breadth

= 60 x 40 m²

= 2400 m²

Hence, option 2 is the correct option.

The longest pole that can be placed in the park will be along its diagonal.

By formula,

Diagonal of a rectangle = $\sqrt{\text{length}^2 + \text{breadth}^2}$

$$\begin{aligned}= \sqrt{(10)^2 + (8)^2} \text{ m}\\[1em] = \sqrt{100 + 64} \text{ m}\\[1em] = \sqrt{164} \text{ m}\\[1em] = 12.8 \text{ m (approx.)}\end{aligned}$$

Hence, option 2 is the correct option.

Given:

Length of the garden = 10 m and width of the garden = 5 m

Width of the road = 1 m (all around the outside)

Rate of cementing = ₹ 200 per m²

Area of the garden = length x width

= 10 x 5 m²

= 50 m²

Length of the garden including the road = (10 + 1 + 1) m = 12 m

Width of the garden including the road = (5 + 1 + 1) m = 7 m

Area of the garden including the road = length x width

= 12 x 7 m²

= 84 m²

∴ Area of the road = (84 - 50) m²

= 34 m²

Cost of cementing the road = 34 x ₹ 200

= ₹ 6,800

Hence, option 3 is the correct option.

Area of the hall = length x breadth

= 40 x 15 m²

= 600 m²

Length of each stone = 6 dm = 0.6 m

Breadth of each stone = 5 dm = 0.5 m

Area of 1 stone = length x breadth

= 0.6 x 0.5 m²

= 0.3 m²

∴ Number of stones required = $\frac{\text{Area of the hall}}{\text{Area of 1 stone}}$

= $\frac{600}{0.3}$

= 2000

Hence, option 2 is the correct option.

Area of the road parallel to the length = length of the lawn x width of the road

= 60 x 5 m²

= 300 m²

Area of the road parallel to the breadth = breadth of the lawn x width of the road

= 40 x 5 m²

= 200 m²

Area of the common portion = side x side

= 5 x 5 m²

= 25 m²

∴ Total area of the roads = (300 + 200 - 25) m²

= 475 m²

Cost of gravelling the roads = 475 x ₹ 60

= ₹ 28,500

Hence, option 3 is the correct option.

Area of the plot = length x breadth

= 90 x 40 m²

= 3600 m²

Length of the plot including the lawn = (90 + 5 + 5) m = 100 m

Breadth of the plot including the lawn = (40 + 5 + 5) m = 50 m

Area of the plot including the lawn = length x breadth

= 100 x 50 m²

= 5000 m²

∴ Total area of the lawn = (5000 - 3600) m²

= 1400 m²

Hence, option 4 is the correct option.

Breadth of the rectangle = $\frac{\text{Area}}{\text{Length}}$

= $\frac{150}{15}$ cm

= 10 cm

New area of the rectangle = $1\frac{1}{3}$ x 150 cm²

= $\frac{4}{3}$ x 150 cm²

= 200 cm²

Since only the length is increased, the breadth remains 10 cm.

∴ New length = $\frac{200}{10}$ cm

= 20 cm

Now, new perimeter = 2 x (length + breadth)

= 2 x (20 + 10) cm

= 2 x 30 cm

= 60 cm

Hence, option 2 is the correct option.

Dimensions of the floor = 4 m x 3 m = 400 cm x 300 cm

Area of the floor = length x breadth

= 400 x 300 cm²

= 120000 cm²

Area of 1 tile = length x breadth

= 8 x 6 cm²

= 48 cm²

∴ Number of tiles required = $\frac{120000}{48}$

= 2500

Number of tiles along the length = $\frac{400}{8}$ = 50

Number of tiles along the breadth = $\frac{300}{6}$ = 50

Since both divisions are exact, 50 x 50 = 2500 tiles cover the floor fully without breaking any tile.

Hence, option 2 is the correct option.

Area of the field = length x breadth

= 120 x 80 m²

= 9600 m²

Since the park of width 12 m is made all around inside the field,

Length of the field excluding the park = [120 - (12 + 12)] m = 96 m

Breadth of the field excluding the park = [80 - (12 + 12)] m = 56 m

Area of the field excluding the park = length x breadth

= 96 x 56 m²

= 5376 m²

∴ Area of the park = (9600 - 5376) m²

= 4224 m²

Hence, option 3 is the correct option.

Area of the square = side²

= (84)² m²

= 7056 m²

Since the area of the rectangle is the same as the area of the square,

Area of the rectangle = 7056 m²

∴ Width of the rectangle = $\frac{\text{Area}}{\text{Length}}$

= $\frac{7056}{144}$ m

= 49 m

Hence, option 3 is the correct option.

Let the side of the given square be a.

Then, the side of the resulting square = 2a

Area of the given square = a²

Area of the resulting square = (2a)² = 4a²

∴ Required ratio = $\frac{4a^2}{a^2}$

= $\frac{4}{1}$

= 4 : 1

Hence, option 4 is the correct option.

Length of the cloth = 1.75 m = 175 cm

Width of the cloth = 105 cm

Area of the cloth = length x width

= 175 x 105 cm²

= 18375 cm²

Area of 1 handkerchief = side²

= (35)² cm²

= 1225 cm²

∴ Number of handkerchiefs made = $\frac{18375}{1225}$

= 15

Hence, option 2 is the correct option.

Let the sides of the two squares be a and b.

Then, $\frac{a^2}{b^2} = \frac{9}{1}$

⇒ $\frac{a}{b} = \sqrt{\frac{9}{1}}$

⇒ $\frac{a}{b} = \frac{3}{1}$

Now, ratio of their perimeters = $\frac{4a}{4b}$

= $\frac{a}{b}$

= 3 : 1

Hence, option 2 is the correct option.

Let the radius of the circle be r m.

By formula,

Circumference of a circle = 2πr

⇒ 2 x $\frac{22}{7}$ x r = 352

⇒ r = 352 x $\frac{7}{44}$

⇒ r = 56 m

Now, area of the circle = πr²

= $\frac{22}{7} \times 56 \times 56$ m²

= 22 x 8 x 56 m²

= 9856 m²

Hence, option 4 is the correct option.

Let the radius of the circle be r cm.

By formula,

Area of a circle = πr²

$$\begin{aligned}\Rightarrow \frac{22}{7} \times r^2 = 38.5\\[1em] \Rightarrow r^2 = 38.5 \times \frac{7}{22}\\[1em] \Rightarrow r^{2} = 12.25\\[1em] \Rightarrow r = \sqrt{12.25}\\[1em] \Rightarrow r = 3.5 \text{ cm}\end{aligned}$$

Now, circumference of the circle = 2πr

= $2 \times \frac{22}{7} \times 3.5$ cm

= 22 cm

Hence, option 3 is the correct option.

For the largest circle drawn inside a square,

Diameter of the circle = Side of the square = 14 cm

∴ Radius of the circle (r) = $\frac{14}{2}$ cm = 7 cm

Now, area of the circle = πr²

= $\frac{22}{7} \times 7 \times 7$ cm²

= 154 cm²

Hence, option 2 is the correct option.

For the largest circle drawn inside a rectangle,

Diameter of the circle = Breadth of the rectangle = 6 m

∴ Radius of the circle (r) = $\frac{6}{2}$ m = 3 m

Now, area of the circle = πr²

= $\frac{22}{7}$ x 3 x 3 m²

= $\frac{198}{7}$ m²

= $28\frac{2}{7}$ m²

Hence, option 1 is the correct option.

Length of the wire = Circumference of the circle

= 2πr

= $2 \times \frac{22}{7} \times 42$ cm

= 264 cm

When the wire is bent in the form of a rectangle, its perimeter = 264 cm.

Let the sides of the rectangle be 6x centimetres and 5x centimetres.

By formula,

Perimeter of a rectangle = 2 x (length + breadth)

∴ 2 x (6x + 5x) = 264

⇒ 2 x 11x = 264

⇒ 22x = 264

⇒ x = 12

So, the smaller side of the rectangle = 5x

= 5 x 12 cm

= 60 cm

Hence, option 2 is the correct option.

The cow grazes a circular region whose radius is the length of the rope.

Let the length of the rope be r m.

By formula,

Area of a circle = πr²

$$\begin{aligned}\Rightarrow \frac{22}{7} \times r^2 = 9856\\[1em] \Rightarrow r^2 = 9856 \times \frac{7}{22}\\[1em] \Rightarrow r^{2} = 3136\\[1em] \Rightarrow r = \sqrt{3136}\\[1em] \Rightarrow r = 56 \text{ m}\end{aligned}$$

Hence, option 1 is the correct option.

The shaded region consists of a bigger semi-circle of diameter 28 cm and two smaller semi-circles, each of diameter 14 cm.

Radius of the bigger semi-circle (R) = $\frac{28}{2}$ cm = 14 cm

Area of the bigger semi-circle = $\frac{1}{2}$ πR²

= $\frac{1}{2} \times \frac{22}{7} \times 14 \times 14$ cm²

= 308 cm²

Radius of each smaller semi-circle (r) = $\frac{14}{2}$ cm = 7 cm

Area of the two smaller semi-circles = $2 \times \frac{1}{2}$ πr²

= $\frac{22}{7} \times 7 \times 7$ cm²

= 154 cm²

∴ Area of the shaded region = (308 + 154) cm²

= 462 cm²

Hence, option 3 is the correct option.

Let ABCD be the rhombus with AB = 5 m and diagonal AC = 8 m, and let its diagonals intersect at O.

We know that the diagonals of a rhombus bisect each other at right angles.

∴ AO = $\frac{1}{2}$ x 8 m = 4 m and ∠AOB = 90°

Using Pythagoras theorem in right triangle AOB,

AB² = AO² + OB²

⇒ OB² = AB² - AO²

⇒ OB² = (5)² - (4)²

⇒ OB² = 25 - 16

⇒ OB² = 9

⇒ OB = 3 m

∴ BD = 2 x OB = 2 x 3 m = 6 m

Hence, option 2 is the correct option.

Let the shorter diagonal of the rhombus be d cm.

Then, the longer diagonal = 2d cm

By formula,

Area of a rhombus = $\frac{1}{2}$ x (product of diagonals)

⇒ $\frac{1}{2}$ x d x 2d = 144

⇒ d² = 144

⇒ d = $\sqrt{144}$

⇒ d = 12 cm

So, the longer diagonal = 2 x 12 cm = 24 cm

Hence, option 2 is the correct option.

Let the longer diagonal of the rhombus be d.

Then, the other diagonal = 80% of d = $\frac{80}{100}$d = $\frac{4}{5}$d

By formula,

Area of a rhombus = $\frac{1}{2}$ x (product of diagonals)

= $\frac{1}{2}$ x d x $\frac{4}{5}$d

= $\frac{2}{5}$d²

∴ Area of the rhombus = $\frac{2}{5}$ x (square of the length of the longer diagonal)

Hence, option 3 is the correct option.

By formula,

Area of a parallelogram = base x height

Here, base = 14 cm and height = 16 cm

∴ Area of the parallelogram = 14 x 16 cm²

= 224 cm²

Hence, option 2 is the correct option.

By formula,

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

= $\frac{1}{2}$ x (20 + 30) x 6 cm²

= $\frac{1}{2}$ x 50 x 6 cm²

= 150 cm²

Hence, option 2 is the correct option.

Let the parallel sides of the trapezium be 3x centimetres and 5x centimetres.

By formula,

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

$$\begin{aligned}\Rightarrow \frac{1}{2} \times (3x + 5x) \times 12 = 384\\[1em] \Rightarrow \frac{1}{2} \times 8x \times 12 = 384\\[1em] \Rightarrow 48x = 384\\[1em] \Rightarrow x = \frac{384}{48}\\[1em] \Rightarrow x = 8\end{aligned}$$

So, the smaller of the parallel sides = 3x

= 3 x 8 cm

= 24 cm

Hence, option 2 is the correct option.

The cross-section of the canal is a trapezium whose parallel sides are 10 m and 6 m.

Let the required length be l m.

By formula,

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

⇒ $\frac{1}{2}$ x (10 + 6) x l = 640

⇒ $\frac{1}{2}$ x 16 x l = 640

⇒ 8l = 640

⇒ l = $\frac{640}{8}$

⇒ l = 80 m

Hence, option 2 is the correct option.

The rectangle is inscribed in the circle, so the diagonal of the rectangle is the diameter of the circle.

By formula,

Diagonal of a rectangle = $\sqrt{\text{length}^2 + \text{breadth}^2}$

∴ Diameter of the circle = $\sqrt{(8)^2 + (6)^2}$ m

= $\sqrt{64 + 36}$ m

= $\sqrt{100}$ m

= 10 m

∴ Radius of the circle (r) = $\frac{10}{2}$ m = 5 m

Area of the circle = πr²

= $\frac{22}{7}$ x 5 x 5 m²

= $\frac{550}{7}$ m²

= 78.57 m²

Area of the rectangle = length x breadth

= 8 x 6 m²

= 48 m²

∴ Area of the shaded portion = (78.57 - 48) m²

= 30.57 m²

= 30.6 m² (approx.)

Hence, option 1 is the correct option.

Given:

Side of the square = 21 m

A semi-circle is described on each side of the square.

∴ Radius of each semi-circle (r) = $\frac{21}{2}$ m = 10.5 m

Area of the square = side²

= (21)² m²

= 441 m²

Area of the four semi-circles = $4 \times \frac{1}{2}$ πr²

= $2 \times \frac{22}{7} \times 10.5 \times 10.5$ m²

= 2 x 22 x 1.5 x 10.5 m²

= 693 m²

∴ Total area of the bed of roses = (441 + 693) m²

= 1134 m²

Since each rose plant needs 6 m² of space,

Number of plants = $\frac{1134}{6}$

= 189

Hence, option 4 is the correct option.

Since the four horses are tethered at the four corners of the square plot so that they just cannot reach one another,

Radius of each grazed region (r) = $\frac{1}{2}$ x side of the square

= $\frac{63}{2}$ m

= 31.5 m

Area of the square plot = side²

= (63)² m²

= 3969 m²

Each horse grazes a quadrant of a circle of radius 31.5 m.

∴ Total grazed area = $4 \times \frac{1}{4}$ πr²

= $\frac{22}{7} \times 31.5 \times 31.5$ m²

= 22 x 4.5 x 31.5 m²

= 3118.5 m²

∴ Area left ungrazed = (3969 - 3118.5) m²

= 850.5 m²

Hence, option 4 is the correct option.

(i) As we know, 1 m = 100 cm

∴ 1 m² = (100 x 100) cm² = 10000 cm²

So, 100 m² = (100 x 10000) cm²

= 1000000 cm²

Hence, 100 m² = 1000000 cm².

(ii) As we know, 1 dm = 10 cm

∴ 1 dm² = (10 x 10) cm² = 100 cm²

So, 5000 cm² = $\frac{5000}{100}$ dm²

= 50 dm²

Hence, 5000 cm² = 50 dm².

(iii) As we know, 1 hectare = 10000 m²

Hence, 1 hectare = 10000 m².

(iv) The area of a rhombus = $\frac{1}{2}$ x (product of its diagonals)

Hence, the area of a rhombus is half the product of its diagonals.

(v) Let the side of the square be a units.

Perimeter of the square = 4a units and area of the square = a² sq. units

Since the perimeter and the area are numerically equal,

⇒ a² = 4a

⇒ a² - 4a = 0

⇒ a(a - 4) = 0

⇒ a = 0 or a = 4

Since the side of a square cannot be 0, a = 4.

Hence, the length of its side is 4 units.

(vi) Let the length of the rectangle be 3x centimetres and its breadth be 2x centimetres.

By formula,

Area of a rectangle = length x breadth

⇒ 3x x 2x = 96

⇒ 6x² = 96

⇒ x² = 16

⇒ x = 4

So, length of the rectangle = 3x

= 3 x 4 cm

= 12 cm

Hence, its length is 12 cm.

Let the trapezium be ABCD in which AB = 2a units and DC = AD = BC = a units.

Join the vertices D and C to the mid-point M of AB.

Then, AM = MB = a

Since AM = AD = a and DC = AM with DC ∥ AM, AMCD is a rhombus, so DM = a.

Similarly, MB = BC = a and DC = MB with DC ∥ MB, so CM = a.

∴ ΔADM, ΔDMC and ΔCMB all have each side equal to a, that is, the trapezium is divided into three equilateral triangles of equal area.

So, Assertion (A) is true.

The area of a trapezium = $\frac{1}{2}$ (sum of parallel sides) x distance between them.

So, Reason (R) is true.

But Reason (R) is a formula for the area of a trapezium and it does not explain how the trapezium is divided into three equilateral triangles. So, Reason (R) is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

The perimeter of a semi-circle of radius r is r (π + 2).

So, Reason (R) is true.

Let the radius of the semi-circle be r cm.

Then, perimeter = r (π + 2)

$$\begin{aligned}\Rightarrow r \times \left(\frac{22}{7} + 2\right) = 18\\[1em] \Rightarrow r \times \frac{36}{7} = 18\\[1em] \Rightarrow r = 18 \times \frac{7}{36}\\[1em] \Rightarrow r = 3.5 \text{ cm}\end{aligned}$$

So, Assertion (A) is true and Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

The area of a rhombus = $\frac{1}{2}$ x (product of its diagonals) and not the product of its diagonals.

So, Reason (R) is false.

Let ABCD be the rhombus with AB = 10 cm and diagonal AC = 12 cm, and let its diagonals intersect at O.

We know that the diagonals of a rhombus bisect each other at right angles.

∴ AO = $\frac{1}{2}$ x 12 cm = 6 cm and ∠AOB = 90°

Using Pythagoras theorem in right triangle AOB,

AB² = AO² + OB²

⇒ OB² = AB² - AO²

⇒ OB² = (10)² - (6)²

⇒ OB² = 100 - 36

⇒ OB² = 64

⇒ OB = 8 cm

∴ BD = 2 x OB = 2 x 8 cm = 16 cm

Now, area of the rhombus = $\frac{1}{2}$ x AC x BD

= $\frac{1}{2}$ x 12 x 16 cm²

= 96 cm²

So, Assertion (A) is true.

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

Hence, option 3 is the correct option.

Let the breadth of the hall be b m.

Then, length of the hall = $\frac{4}{3}$b m

By formula,

Area of a rectangle = length x breadth

⇒ $\frac{4}{3}$b x b = 300

⇒ $\frac{4b^2}{3}$ = 300

⇒ 4b² = 900

⇒ b² = 225

⇒ b = $\sqrt{225}$

⇒ b = 15 m

So, length of the hall = $\frac{4}{3}$ x 15 m = 20 m

∴ Difference between length and breadth = (20 - 15) m

= 5 m

Hence, option 2 is the correct option.

Let the base of the triangle be b and its height be h.

∴ Original area of the triangle = $\frac{1}{2}$ x b x h

Now, the base is doubled and the height is halved.

New base = 2b and new height = $\frac{h}{2}$

∴ New area of the triangle = $\frac{1}{2} \times 2b \times \frac{h}{2}$

= $\frac{1}{2}$ x b x h

Since the new area is equal to the original area, there is no change in the area.

Hence, option 4 is the correct option.

Let the diagonal of the square be d.

By formula,

Area of a square = $\frac{1}{2}$ x (diagonal)²

∴ Original area = $\frac{1}{2}$d²

When the diagonal is doubled, the new diagonal = 2d.

∴ New area = $\frac{1}{2}$ x (2d)²

= $\frac{1}{2}$ x 4d²

= 4 x $\frac{1}{2}$d²

So, the area becomes 4 times the original area.

Hence, option 3 is the correct option.

Let the longer diagonal of the rhombus be d.

Then, the other diagonal = 80% of d = $\frac{80}{100}$d = $\frac{4}{5}$d

By formula,

Area of a rhombus = $\frac{1}{2}$ x (product of diagonals)

= $\frac{1}{2}$ x d x $\frac{4}{5}$d

= $\frac{2}{5}$d²

∴ Area of the rhombus = $\frac{2}{5}$ x (square of the length of the longer diagonal)

Hence, option 4 is the correct option.

From the figure, the diagonals of the rhombus ABCD meet at O, where OA = 4 cm and OD = 3 cm.

We know that the diagonals of a rhombus bisect each other.

∴ AC = 2 x OA = 2 x 4 cm = 8 cm

And, BD = 2 x OD = 2 x 3 cm = 6 cm

Now, area of rhombus ABCD = $\frac{1}{2}$ x (product of diagonals)

= $\frac{1}{2}$ x 8 x 6 cm²

= 24 cm²

Hence, option 2 is the correct option.

From the figure, ABCD is a trapezium in which AB ∥ DC.

The parallel sides are AB = a and DC = c.

Since ∠B = ∠C = 90°, BC = b is the distance between the parallel sides.

By formula,

Area of a trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

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

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

Hence, option 3 is the correct option.

From the figure, the diagonals of the kite ABCD are DB = 12 cm and CA = 8 cm, and they cut each other at right angles.

By formula,

Area of a kite = $\frac{1}{2}$ x (product of diagonals)

= $\frac{1}{2}$ x 12 x 8 cm²

= 48 cm²

Hence, option 1 is the correct option.

Since the sides of a mirror are of the same length, each mirror is a rhombus.

From the figure, the diagonals of each rhombus are 10 cm and 5 cm.

Area of 1 mirror = $\frac{1}{2}$ x (product of diagonals)

= $\frac{1}{2}$ x 10 x 5 cm²

= 25 cm²

Since the decoration is made of four congruent mirrors,

Area occupied by the decoration = 4 x 25 cm²

= 100 cm²

Hence, option 3 is the correct option.

Since AB = CD and BC = AD, ABCD is a parallelogram.

From the figure, AC = 12 cm is a diagonal and the distance of D from AC is 3 cm.

By formula,

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

Area of ΔACD = $\frac{1}{2}$ x AC x 3

= $\frac{1}{2}$ x 12 x 3 cm²

= 18 cm²

We know that a diagonal of a parallelogram divides it into two triangles equal in area.

∴ Area of quadrilateral ABCD = 2 x (area of ΔACD)

= 2 x 18 cm²

= 36 cm²

Hence, option 4 is the correct option.

The given figure can be divided into two trapeziums.

Upper trapezium:

Since the total height of the left side is 12 cm and the lower part is 3 cm high,

Length of the left parallel side = (12 - 3) cm = 9 cm

Length of the right parallel side = 6 cm

Distance between them = 4 cm

Area of the upper trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

= $\frac{1}{2}$ x (9 + 6) x 4 cm²

= $\frac{1}{2}$ x 15 x 4 cm²

= 30 cm²

Lower trapezium:

Length of the lower parallel side = 16 cm

Length of the upper parallel side = (4 + 8) cm = 12 cm

Distance between them = 3 cm

Area of the lower trapezium = $\frac{1}{2}$ x (sum of parallel sides) x (distance between them)

= $\frac{1}{2}$ x (16 + 12) x 3 cm²

= $\frac{1}{2}$ x 28 x 3 cm²

= 42 cm²

∴ Area of the given figure = (30 + 42) cm²

= 72 cm²

Hence, option 3 is the correct option.