Mensuration (Perimeter & Area of Plane Figures)

Given:

Length of the plot ($l$) = 135 m

Breadth of the plot ($b$) = 65 m

By formula,

The perimeter of a rectangle = $2(l + b)$

= 2(135 + 65) m

= 2 x 200 m

= 400 m

Cost of fencing the plot = Perimeter x rate of fencing

= 400 x ₹ 60

= ₹ 24,000

Hence, the perimeter of the plot is 400 m and the cost of fencing it is ₹ 24,000.

Given:

Ratio of length and breadth of the field = 7 : 4

Perimeter of the field = 440 m

Let the length = $7x$ m and breadth = $4x$ m

By formula,

Perimeter = $2(l + b)$

$$\begin{aligned}\Rightarrow 2(7x + 4x) = 440 \\[1em] \Rightarrow 2 \times 11x = 440 \\[1em] \Rightarrow 22x = 440 \\[1em] \Rightarrow x = \frac{440}{22} \\[1em] \Rightarrow x = 20\end{aligned}$$

∴ Length = $7x$ = 7 x 20 m = 140 m

And, breadth = $4x$ = 4 x 20 m = 80 m

Cost of fencing the field = Perimeter x rate of fencing

= 440 x ₹ 150

= ₹ 66,000

Hence, the length and breadth of the field are 140 m and 80 m respectively and the cost of fencing it is ₹ 66,000.

Given:

Length of the field ($l$) = 30 m

Diagonal of the field = 34 m

Let the breadth of the field be $b$ m.

Applying Pythagoras theorem in triangle ABC, we get :

AC² = AB² + BC²

$$\begin{aligned}\Rightarrow (34)^2 = (30)^2 + b^2 \\[1em] \Rightarrow 1156 = 900 + b^2 \\[1em] \Rightarrow b^2 = 1156 - 900 \\[1em] \Rightarrow b^2 = 256 \\[1em] \Rightarrow b = \sqrt{256} \text{ m} \\[1em] \Rightarrow b = 16 \text{ m}\end{aligned}$$

Now, perimeter of the field = $2(l + b)$

= 2(30 + 16) m

= 2 x 46 m

= 92 m

Hence, the breadth of the field is 16 m and its perimeter is 92 m.

Given:

Diagonal of the square = $12\sqrt{2}$ cm

By formula,

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

⇒ side x $\sqrt{2} = 12\sqrt{2}$

⇒ side = 12 cm

Now, perimeter of the square = 4 x side

= 4 x 12 cm

= 48 cm

Hence, the perimeter of the square is 48 cm.

Given:

Length of the rectangle ($l$) = 22.5 m

Breadth of the rectangle ($b$) = 16 dm

Since,

1 m = 10 dm

⇒ $b = \frac{16}{10}$ m = 1.6 m

Now, perimeter of the rectangle = $2(l + b)$

= 2(22.5 + 1.6) m

= 2 x 24.1 m

= 48.2 m

Hence, the perimeter of the rectangle is 48.2 m.

Given:

Length of the rectangle ($l$) = 24 cm

Diagonal of the rectangle = 25 cm

Let the breadth of the rectangle be $b$ cm.

Applying Pythagoras theorem in triangle ABC, we get :

AC² = AB² + BC²

$$\begin{aligned}\Rightarrow (25)^2 = (24)^2 + b^2 \\[1em] \Rightarrow 625 = 576 + b^2 \\[1em] \Rightarrow b^2 = 625 - 576 \\[1em] \Rightarrow b^2 = 49 \\[1em] \Rightarrow b = \sqrt{49} \text{ cm} \\[1em] \Rightarrow b = 7 \text{ cm}\end{aligned}$$

Now, perimeter of the rectangle = $2(l + b)$

= 2(24 + 7) cm

= 2 x 31 cm

= 62 cm

Hence, the perimeter of the rectangle is 62 cm.

Given:

Ratio of length and breadth of the land = 5 : 3

Total cost of fencing = ₹ 19,200

Rate of fencing = ₹ 48 per metre

Length of the fence x its rate = ₹ 19,200

⇒ Length of the fence = $\frac{19200}{48}$ m = 400 m

∴ Perimeter of the land = length of its fence = 400 m

Let the length = $5x$ m and breadth = $3x$ m

By formula,

$2(l + b)$ = perimeter

$$\begin{aligned}\Rightarrow 2(5x + 3x) = 400 \\[1em] \Rightarrow 2 \times 8x = 400 \\[1em] \Rightarrow 16x = 400 \\[1em] \Rightarrow x = \frac{400}{16} = 25\end{aligned}$$

∴ Length = $5x$ = 5 x 25 m = 125 m

And, breadth = $3x$ = 3 x 25 m = 75 m

Hence, the length and breadth of the land are 125 m and 75 m respectively.

Given:

Side of the square = 20 cm

Length of the rectangle ($l$) = 24 cm

Length of the wire = perimeter of the square

= 4 x side

= 4 x 20 cm

= 80 cm

Since the same wire is bent into a rectangle,

Perimeter of the rectangle (P) = length of the wire = 80 cm

By formula,

Breadth ($b$) = $\frac{P}{2} - l$

= $\frac{80}{2}$ cm - 24 cm

= 40 cm - 24 cm

= 16 cm

Hence, the breadth of the rectangle is 16 cm.

(i) Given:

$l$ = 38 cm and $b$ = 27 cm

By formula,

P = $2(l + b)$

= 2(38 cm + 27 cm)

= 2 x 65 cm

= 130 cm

Hence, P = 130 cm.

(ii) Given:

P = 88 cm and $l$ = 24 cm

By formula,

$b = \frac{P}{2} - l$

= $\frac{88}{2}$ cm - 24 cm

= 44 cm - 24 cm

= 20 cm

Hence, $\bm{b}$ = 20 cm.

(iii) Given:

P = 96 m and $b$ = 28 m

By formula,

$l = \frac{P}{2} - b$

= $\frac{96}{2}$ m - 28 m

= 48 m - 28 m

= 20 m

Hence, $\bm{l}$ = 20 m.

Given:

Rate of fencing = ₹ 75 per meter

Cost of fencing the square field = ₹ 67,500

∴ Length of the fence x its rate = ₹ 67,500

⇒ Length of the fence = $\frac{67500}{75}$ m = 900 m

∴ Perimeter of the square field = length of its fence = 900 m

Since, perimeter of a square = 4 x length of its side

⇒ Length of the side of the square = $\frac{\text{Perimeter}}{4}$

= $\frac{900}{4}$ m

= 225 m

Hence, the perimeter of the square field is 900 m and its side is 225 m.

Given:

Length of the rectangle = 36 cm

Breadth of the rectangle = 28 cm

By formula,

Perimeter of the rectangle = 2(length + breadth)

= 2(36 + 28) cm

= 2 x 64 cm

= 128 cm

Given, perimeter of the square = perimeter of the rectangle = 128 cm

∴ Side of the square = $\frac{\text{Perimeter}}{4}$

= $\frac{128}{4}$ cm

= 32 cm

Hence, the side of the square is 32 cm.

Given:

Radius ($r$) = 21 cm and π = $3\frac{1}{7} = \frac{22}{7}$

By formula,

Circumference of the circle = $2\pi r$

= 2 x $\frac{22}{7}$ x 21 cm

= 2 x 22 x 3 cm

= 132 cm

Hence, the circumference of the circle is 132 cm.

Given:

Circumference of the circle = 440 cm

Let the radius of the circle be $r$ cm.

By formula,

Circumference of a circle = $2\pi r$

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r = 440\\[1em] \Rightarrow \frac{44}{7} \times r = 440\\[1em] \Rightarrow r = 440 \times \frac{7}{44}\\[1em] \Rightarrow r = 70 \text{ cm}\end{aligned}$$

And, diameter = 2 x radius

= 2 x 70 cm

= 140 cm

Hence, the radius and diameter of the circle are 70 cm and 140 cm respectively.

Given:

Diameter of the circular field = 56 m

∴ Radius of the field ($r$) = $\frac{56}{2}$ m = 28 m [Diameter = 2 x radius]

By formula,

Circumference of the field = $2\pi r$

= 2 x $\frac{22}{7}$ x 28 m

= 2 x 22 x 4 m

= 176 m

Cost of fencing the field = Circumference x rate of fencing

= 176 x ₹ 80

= ₹ 14,080

Hence, the circumference of the field is 176 m and the cost of fencing it is ₹ 14,080.

Given:

Radii of the two circles are 20 cm and 13 cm.

By formula,

Circumference of the circle with radius 20 cm = $2\pi r$

= 2 x $\frac{22}{7}$ x 20 cm

= $\frac{880}{7}$ cm

Circumference of the circle with radius 13 cm = $2\pi r$

= 2 x $\frac{22}{7}$ x 13 cm

= $\frac{572}{7}$ cm

∴ Difference between their circumferences

$$\begin{aligned}= \frac{880}{7} \text{ cm} - \frac{572}{7} \text{ cm}\\[1em] = \frac{308}{7} \text{ cm}\\[1em] = 44 \text{ cm}\end{aligned}$$

Hence, the difference between their circumferences is 44 cm.

Given:

Diameter of the circle = 42 cm

∴ Radius of the circle ($r$) = $\frac{42}{2}$ cm = 21 cm

By formula,

Perimeter (circumference) of the circle = $2\pi r$

= 2 x $\frac{22}{7}$ x 21 cm

= 2 x 22 x 3 cm

= 132 cm

Perimeter of the new circle = 2 x 132 cm = 264 cm

Let the radius of the new circle be $R$ cm.

$$\begin{aligned}\therefore 2\pi R = 264\\[1em] \Rightarrow 2 \times \frac{22}{7} \times R = 264\\[1em] \Rightarrow R = 264 \times \frac{7}{44}\\[1em] \Rightarrow R = 42 \text{ cm}\end{aligned}$$

Hence, the perimeter of the circle is 132 cm and the radius of the new circle is 42 cm.

Given:

Side of the square = 22 cm

(i) By formula,

Perimeter of the square = 4 x side

= 4 x 22 cm

= 88 cm

Hence, the perimeter of the square is 88 cm.

(ii) Given, circumference of the circle = perimeter of the square

= 88 cm

Hence, the circumference of the circle is 88 cm.

(iii) Let the radius of the circle be $r$ cm.

$$\begin{aligned}\therefore 2\pi r = 88\\[1em] \Rightarrow 2 \times \frac{22}{7} \times r = 88\\[1em] \Rightarrow r = 88 \times \frac{7}{44}\\[1em] \Rightarrow r = 14 \text{ cm}\end{aligned}$$

Hence, the radius of the circle is 14 cm.

Given:

Radii of the two circles are 15 cm and 8 cm.

Let the required radius be $R$ cm.

By formula,

Circumference of a circle = $2\pi r$

Circumference of the circle with radius $R$ cm = Sum of the circumferences of the circles with radii 15 cm and 8 cm

⇒ $2\pi R = 2\pi \times 15 + 2\pi \times 8$

Dividing each term by 2π, we get :

$R$ = 15 + 8 = 23 cm

Hence, the required radius is 23 cm.

Given:

Radii of the three circles are 10 cm, 12 cm and 18 cm.

Let the radius of the circle = $R$ cm

By formula,

Circumference of a circle = $2\pi r$

Circumference of the required circle = Sum of the circumferences of the circles with radii 10 cm, 12 cm and 18 cm

∴ $2\pi R = 2\pi \times 10 + 2\pi \times 12 + 2\pi \times 18$

On dividing each term by 2π, we get :

$R$ = 10 + 12 + 18 = 40 cm

∴ Radius of the circle obtained = 40 cm

And, its diameter = 2 x radius

= 2 x 40 cm

= 80 cm

Hence, the diameter of the circle is 80 cm.

Given:

Radius of the given circle = 12 cm

Let the radius of the required circle be $R$ cm.

By formula,

Circumference of a circle = $2\pi r$

Circumference of the required circle = 8 x circumference of the given circle

⇒ $2\pi R = 8 \times 2\pi \times 12$

Dividing both sides by 2π, we get :

$R$ = 8 x 12 = 96 cm

∴ Radius of the required circle = 96 cm

And, its diameter = 2 x radius

= 2 x 96 cm

= 192 cm

Hence, the diameter of the circle is 192 cm.

Given:

Ratio of the radii of the two circles = 3 : 5

Let the radii of the two circles be $3x$ and $5x$.

By formula,

Circumference of a circle = $2\pi r$

Circumference of the first circle = $2\pi \times 3x = 6\pi x$

Circumference of the second circle = $2\pi \times 5x = 10\pi x$

∴ Ratio between their circumferences = $6\pi x : 10\pi x$

= 6 : 10

= 3 : 5

Hence, the ratio between their circumferences is 3 : 5.

Given:

Ratio of the circumferences of the two circles = 5 : 7

Let the radii of the two circles be $r_1$ and $r_2$.

By formula,

Circumference of a circle = $2\pi r$

$$\begin{aligned}\Rightarrow \frac{2\pi r_1}{2\pi r_2} = \frac{5}{7}\\[1em] \Rightarrow \frac{r_1}{r_2} = \frac{5}{7}\\[1em] \Rightarrow r_1 : r_2 = 5 : 7\end{aligned}$$

Hence, the ratio between their radii is 5 : 7.

Given:

Ratio of the perimeters of the two squares = 8 : 15

Let the sides of the two squares be $s_1$ and $s_2$.

By formula,

Perimeter of a square = 4 x side

$$\begin{aligned}\Rightarrow \frac{4s_1}{4s_2} = \frac{8}{15}\\[1em] \Rightarrow \frac{s_1}{s_2} = \frac{8}{15}\\[1em] \Rightarrow s_1 : s_2 = 8 : 15\end{aligned}$$

Hence, the ratio between the lengths of their sides is 8 : 15.

Given:

Ratio of the sides of the two squares = 8 : 15

Let the sides of the two squares be $8x$ and $15x$.

By formula,

Perimeter of a square = 4 x side

Perimeter of the first square = $4 \times 8x = 32x$

Perimeter of the second square = $4 \times 15x = 60x$

∴ Ratio between their perimeters = $32x : 60x$

= 32 : 60

= 8 : 15

Hence, the ratio between their perimeters is 8 : 15.

Given:

Side of the square = 44 cm

By formula,

Perimeter of the square = 4 x side

= 4 x 44 cm

= 176 cm

Given, circumference of the circle = perimeter of the square = 176 cm

Let the radius of the circle be $r$ cm.

$$\begin{aligned}\therefore 2\pi r = 176\\[1em] \Rightarrow 2 \times \frac{22}{7} \times r = 176\\[1em] \Rightarrow r = 176 \times \frac{7}{44}\\[1em] \Rightarrow r = 28 \text{ cm}\end{aligned}$$

Hence, the perimeter of the square is 176 cm and the radius of the circle is 28 cm.

Given:

Length of the rectangle = 25 cm

Breadth of the rectangle = 16 cm

By formula,

Area of a rectangle = length x breadth

= 25 cm x 16 cm

= 400 cm²

Hence, the area of the rectangle is 400 cm².

Given:

Diagonal of the board = 1 m = 100 cm

Length of the board = 96 cm

In right-angled triangle ABC,

AB² + BC² = AC²

⇒ (96)² + BC² = (100)²

⇒ 9216 + BC² = 10000

⇒ BC² = 10000 - 9216 = 784

∴ BC = $\sqrt{784}$ cm = 28 cm

Area of the board = length x breadth

= AB x BC

= 96 cm x 28 cm

= 2688 cm²

Hence, the area of the board is 2688 cm².

Given:

Ratio of the sides of the park = 4 : 3

Area of the park = 1728 m²

Let length = $4x$ m and breadth = $3x$ m

By formula,

Area of a rectangle = length x breadth

$$\begin{aligned}\Rightarrow 4x \times 3x = 1728 \\[1em] \Rightarrow 12x^2 = 1728 \\[1em] \Rightarrow x^2 = \frac{1728}{12} = 144 \\[1em] \Rightarrow x = \sqrt{144} = 12\end{aligned}$$

∴ Length = $4x$ = 4 x 12 m = 48 m

And, breadth = $3x$ = 3 x 12 m = 36 m

(i) Perimeter of the park = 2(length + breadth)

= 2(48 + 36) m

= 2 x 84 m

= 168 m

Hence, the perimeter of the park is 168 m.

(ii) Cost of fencing the park = Perimeter x rate of fencing

= 168 x ₹ 40

= ₹ 6,720

Hence, the cost of fencing the park is ₹ 6,720.

Given:

Length of the floor = 40 m = 40 x 100 cm = 4000 cm

Breadth of the floor = 15 m = 15 x 100 cm = 1500 cm

By formula,

Area of the floor = length x breadth

= 4000 cm x 1500 cm

= 60,00,000 cm²

Area of one tile = length x breadth

= 60 cm x 50 cm

= 3000 cm²

∴ Number of tiles required = $\frac{\text{Area of the floor}}{\text{Area of one tile}}$

= $\frac{6000000}{3000}$

= 2000

Hence, 2000 tiles are required to cover the floor.

Given:

Ratio of length and breadth of the land = 5 : 3

Total cost of fencing = ₹ 9600

Rate of fencing = ₹ 24 per metre

(i) Length of the fence = $\frac{\text{Cost of fencing}}{\text{Rate of fencing}}$

= $\frac{9600}{24}$ m

= 400 m

∴ Perimeter of the land = 400 m

Let length = $5x$ m and breadth = $3x$ m

By formula,

2(length + breadth) = perimeter

$$\begin{aligned}\Rightarrow 2 \times (5x + 3x) = 400 \\[1em] \Rightarrow 2 \times 8x = 400 \\[1em] \Rightarrow x = \frac{400}{16} = 25\end{aligned}$$

∴ Length = $5x$ = 5 x 25 m = 125 m

And, breadth = $3x$ = 3 x 25 m = 75 m

Hence, the length and breadth of the land are 125 m and 75 m respectively.

(ii) Area of the land = length x breadth

= 125 m x 75 m

= 9375 m²

Hence, the area of the land is 9375 m².

(iii) Cost of levelling the land = Area of the land x rate of levelling

= 9375 x ₹ 60

= ₹ 5,62,500

Hence, the cost of levelling the land is ₹ 5,62,500.

Given:

Perimeter of the square = 56 cm

Since, perimeter of a square = 4 x side

⇒ side of the square = $\frac{56}{4}$ cm = 14 cm

Area of the square = (side)²

= (14 cm)²

= 196 cm²

Hence, the area of the square is 196 cm².

Given:

Width of the path = 2.5 m

Area of the path = 165 m²

Let each side of the square lawn be $x$ m.

∴ Side of the outer square (lawn with path) = $(x + 2 \times 2.5)$ m = $(x + 5)$ m

Area of the path = Area of outer square - area of the lawn

$$\begin{aligned}\Rightarrow (x + 5)^2 - x^2 = 165 \\[1em] \Rightarrow x^2 + 10x + 25 - x^2 = 165 \\[1em] \Rightarrow 10x + 25 = 165 \\[1em] \Rightarrow 10x = 165 - 25 = 140 \\[1em] \Rightarrow x = \frac{140}{10} = 14 \text{ m}\end{aligned}$$

∴ Side of the lawn = 14 m

Area of the lawn = (side)²

= (14 m)²

= 196 m²

Hence, the area of the lawn is 196 m².

Given:

From the figure,

Length of the outer rectangle = 20 cm and its breadth = 16 cm

Length of the inner (unshaded) rectangle = 15 cm and its breadth = 10 cm

By formula,

Area of a rectangle = length x breadth

∴ Area of the outer rectangle = 20 cm x 16 cm

= 320 cm²

And, area of the inner (unshaded) rectangle = 15 cm x 10 cm

= 150 cm²

∴ Area of the shaded portion = Outer area - inner area

= 320 cm² - 150 cm²

= 170 cm²

Hence, the area of the shaded region is 170 cm².

Given:

From the figure,

Outer figure = 30 cm x 20 cm

Inner (unshaded) figure = 12 cm x 12 cm

By formula,

Area of a rectangle = length x breadth

∴ Area of the shaded portion = Area of outer figure - area of inner figure

= 30 x 20 cm² - 12 x 12 cm²

= 600 cm² - 144 cm²

= 456 cm²

Hence, the area of the shaded region is 456 cm².

Given:

From the figure,

Outer figure = 40 cm x 40 cm

Inner (unshaded) figure = 32 cm x 15 cm

By formula,

Area of a rectangle = length x breadth

∴ Area of the shaded portion = Area of outer figure - area of inner figure

= 40 x 40 cm² - 32 x 15 cm²

= 1600 cm² - 480 cm²

= 1120 cm²

Hence, the area of the shaded region is 1120 cm².

Given:

From the figure,

Outer figure = 40 cm x 40 cm

Inner (unshaded) figure = 15 cm x 15 cm

By formula,

Area of a rectangle = length x breadth

∴ Area of the shaded portion = Area of outer figure - area of inner figure

= 40 x 40 cm² - 15 x 15 cm²

= 1600 cm² - 225 cm²

= 1375 cm²

Hence, the area of the shaded region is 1375 cm².

Given:

From the figure, the shaded region can be divided into three rectangles : top, left and bottom.

Width of the top rectangle = 2 cm and its length = 20 cm

Width of the left rectangle = (20 - 18) cm = 2 cm and its length = 8 cm

Width of the bottom rectangle = (12 - 2 - 8) cm = 2 cm and its length = (2 + 12) cm = 14 cm

By formula,

Area of a rectangle = length x breadth

Area of the top rectangle = 20 x 2 cm² = 40 cm²

Area of the left rectangle = 8 x 2 cm² = 16 cm²

Area of the bottom rectangle = 14 x 2 cm² = 28 cm²

∴ Area of the shaded portion = (40 + 16 + 28) cm²

= 84 cm²

Hence, the area of the shaded region is 84 cm².

Let ABCD be the //gm.

Base of the parallelogram (AB) = 20 cm

Height (DE) = 16 cm

By formula,

Area of the parallelogram = base x height

= AB x DE

= 20 cm x 16 cm

= 320 cm²

Hence, the area of the parallelogram is 320 cm².

Given:

Area of the parallelogram = 768 cm²

Let the height of the parallelogram = $x$ cm

∴ Its base = $3x$ cm

By formula,

Area of the parallelogram = base x height

$$\begin{aligned}\Rightarrow 3x \times x = 768 \\[1em] \Rightarrow 3x^2 = 768 \\[1em] \Rightarrow x^2 = \frac{768}{3} = 256 \\[1em] \Rightarrow x = \sqrt{256} = 16 \text{ cm}\end{aligned}$$

∴ Height = $x$ cm = 16 cm

And, base = $3x$ cm = 3 x 16 cm = 48 cm

Hence, the base and the height of the parallelogram are 48 cm and 16 cm respectively.

Given:

Diagonals of the rhombus are $d_1$ = 30 cm and $d_2$ = 24 cm

By formula,

Area of the rhombus = $\frac{1}{2}$ x product of its diagonals

= $\frac{1}{2} \times d_1 \times d_2$

= $\frac{1}{2}$ x 30 cm x 24 cm

= 360 cm²

Hence, the area of the rhombus is 360 cm².

Given:

Area of the rhombus = 112 cm²

One diagonal ($d_1$) = 14 cm

Let the other diagonal be $d_2$ cm.

By formula,

Area of the rhombus = $\frac{1}{2} \times d_1 \times d_2$

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 14 \times d_2 = 112 \\[1em] \Rightarrow 7 \times d_2 = 112 \\[1em] \Rightarrow d_2 = \frac{112}{7} = 16 \text{ cm}\end{aligned}$$

Hence, the other diagonal of the rhombus is 16 cm.

Given:

Base of the parallelogram = 18 cm

Area of the parallelogram = 153 cm²

Let the distance of the given side from its opposite side (height) be $h$ cm.

By formula,

Area of the parallelogram = base x height

$$\begin{aligned}\Rightarrow 18 \times h = 153 \\[1em] \Rightarrow h = \frac{153}{18} = 8.5\text{ cm}\end{aligned}$$

Hence, the distance of the given side from its opposite side is 8.5 cm.

Given:

Longer side of the parallelogram = 15 cm

Shorter side of the parallelogram = 10 cm

Distance between the longer sides = 6 cm

By formula,

Area of a parallelogram = base x height

Taking the longer side as base,

Area of the parallelogram = 15 cm x 6 cm

= 90 cm²

Let the distance between the shorter sides be $h$ cm.

Taking the shorter side as base,

Area of the parallelogram = $10 \times h$

$$\begin{aligned}\Rightarrow 10 \times h = 90 \\[1em] \Rightarrow h = \frac{90}{10} = 9 \text{ cm}\end{aligned}$$

Hence, the distance between the shorter sides is 9 cm.

Given:

Area of the rhombus = 84 cm²

Perimeter of the rhombus = 56 cm

As all the sides of a rhombus are equal,

⇒ side of the rhombus = $\frac{56}{4}$ cm = 14 cm

Let the height of the rhombus be $h$ cm.

Since, a rhombus is also a parallelogram,

By formula,

Area of the rhombus = base x height

$$\begin{aligned}\Rightarrow 14 \times h = 84 \\[1em] \Rightarrow h = \frac{84}{14} = 6 \text{ cm}\end{aligned}$$

Hence, the height of the rhombus is 6 cm.

Given:

Base of the triangle = 30 cm

Height of the triangle = 18 cm

By formula,

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

= $\frac{1}{2}$ x 30 cm x 18 cm

= 270 cm²

Hence, the area of the triangle is 270 cm².

Given:

Base of the triangle = 18 cm

Area of the triangle = 270 cm²

Let the height of the triangle be $h$ cm.

By formula,

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

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 18 \times h = 270 \\[1em] \Rightarrow 9 \times h = 270 \\[1em] \Rightarrow h = \frac{270}{9} = 30 \text{ cm}\end{aligned}$$

Hence, the height of the triangle is 30 cm.

Given:

Area of the right-angled triangle = 160 cm²

One leg of the triangle = 16 cm

Let the length of the other leg be $x$ cm.

By formula,

Area of a right-angled triangle = $\frac{1}{2}$ x product of legs

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 16 \times x = 160 \\[1em] \Rightarrow 8x = 160 \\[1em] \Rightarrow x = \frac{160}{8} = 20 \text{ cm}\end{aligned}$$

Hence, the length of the other leg is 20 cm.

Given:

Consider a right-angled triangle ABC in which ∠B = 90°, AC = hypotenuse = 13 cm and AB = 12 cm.

In right-angled triangle ABC,

AB² + BC² = AC²

⇒ (12)² + BC² = (13)²

⇒ 144 + BC² = 169

⇒ BC² = 169 - 144 = 25

∴ BC = $\sqrt{25}$ cm = 5 cm

Area of the triangle = $\frac{1}{2}$ x product of legs

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

= $\frac{1}{2}$ x 12 cm x 5 cm

= 30 cm²

Hence, the area of the right-angled triangle is 30 cm².

Given:

Side of the equilateral triangle = 16 cm

By formula,

Area of an equilateral triangle = $\frac{\sqrt{3}}{4}$ x (side)²

= $\frac{1.73}{4}$ x 16 x 16 cm²

= 1.73 x 64 cm²

= 110.72 cm²

Hence, the area of the equilateral triangle is 110.72 cm².

Given:

Sides of the triangle are 21 cm, 17 cm and 10 cm.

Let $a, b$ and $c$ represent sides of the triangle so that $a$ = 21 cm, $b$ = 17 cm and $c$ = 10 cm

∴ $a + b + c$ = 21 cm + 17 cm + 10 cm = 48 cm

and $s = \frac{a + b + c}{2} = \frac{48}{2}$ cm = 24 cm

By formula,

Area of the triangle

$$\begin{aligned}= \sqrt{s(s - a)(s - b)(s - c)}\\[1em] = \sqrt{24(24 - 21)(24 - 17)(24 - 10)} \text{ cm}^2\\[1em] = \sqrt{24 \times 3 \times 7 \times 14} \text{ cm}^2\\[1em] = \sqrt{2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 2 \times 7} \text{ cm}^2\\[1em] = 2 \times 2 \times 3 \times 7 \text{ cm}^2\\[1em] = 84 \text{ cm}^2\end{aligned}$$

Hence, the area of the triangle is 84 cm².

Given:

Each of the equal sides, $a$ = 10 cm

Base, $b$ = 16 cm

By formula,

Area of an isosceles triangle = $\frac{1}{4} \times b \times \sqrt{4a^2 - b^2}$

$$\begin{aligned}= \frac{1}{4} \times 16 \times \sqrt{4 \times 10^2 - 16^2} \text{ cm}^2\\[1em] = 4 \times \sqrt{400 - 256} \text{ cm}^2\\[1em] = 4 \times \sqrt{144} \text{ cm}^2\\[1em] = 4 \times 12 \text{ cm}^2\\[1em] = 48 \text{ cm}^2\end{aligned}$$

Hence, the area of the isosceles triangle is 48 cm².

Given:

Area of the triangle = 360 cm²

Height of the triangle = 24 cm

Let the base of the triangle be $b$ cm.

By formula,

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

$$\begin{aligned}\Rightarrow \frac{1}{2} \times b \times 24 = 360 \\[1em] \Rightarrow 12b = 360 \\[1em] \Rightarrow b = \frac{360}{12} \\[1em] \Rightarrow b = 30 \text{ cm}\end{aligned}$$

Hence, the base of the triangle is 30 cm.

Given:

Ratio of the legs = 4 : 3

Area of the triangle = 4056 cm²

Let the legs of the triangle be $4x$ cm and $3x$ cm.

By formula,

Area of a right-angled triangle = $\frac{1}{2}$ x product of legs

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 4x \times 3x = 4056 \\[1em] \Rightarrow 6x^2 = 4056 \\[1em] \Rightarrow x^2 = \frac{4056}{6} = 676 \\[1em] \Rightarrow x = \sqrt{676} = 26\end{aligned}$$

∴ One leg = $4x$ = 4 x 26 cm = 104 cm

And, other leg = $3x$ = 3 x 26 cm = 78 cm

Hence, the lengths of the legs are 104 cm and 78 cm.

Given:

Area of the equilateral triangle = $64\sqrt{3}$ cm²

Let each side of the triangle be $s$ cm.

By formula,

Area of an equilateral triangle = $\frac{\sqrt{3}}{4} \times s^2$

$$\begin{aligned}\Rightarrow \frac{\sqrt{3}}{4} \times s^2 = 64\sqrt{3}\\[1em] \Rightarrow s^2 = 64\sqrt{3} \times \frac{4}{\sqrt{3}}\\[1em] \Rightarrow s^2 = 256\\[1em] \Rightarrow s = \sqrt{256} = 16 \text{ cm}\end{aligned}$$

Hence, the length of each side of the triangle is 16 cm.

Given:

Ratio of the sides of the triangle = 15 : 13 : 14

Perimeter of the triangle = 168 cm

Let the sides of the triangle be $15x$ cm, $13x$ cm and $14x$ cm.

$$\begin{aligned}\therefore 15x + 13x + 14x = 168 \\[1em] \Rightarrow 42x = 168 \\[1em] \Rightarrow x = \frac{168}{42} \\[1em] \Rightarrow x = 4\end{aligned}$$

∴ Sides of the triangle are :

$a$ = 15 x 4 cm = 60 cm, $b$ = 13 x 4 cm = 52 cm and $c$ = 14 x 4 cm = 56 cm

and $s = \frac{a + b + c}{2} = \frac{168}{2}$ cm = 84 cm

By formula,

Area of the triangle

$$\begin{aligned}= \sqrt{s(s - a)(s - b)(s - c)}\\[1em] = \sqrt{84(84 - 60)(84 - 52)(84 - 56)} \text{ cm}^2\\[1em] = \sqrt{84 \times 24 \times 32 \times 28} \text{ cm}^2\\[1em] = \sqrt{(2 \times 2 \times 3 \times 7) \times (2 \times 2 \times 2 \times 3) \times (2 \times 2 \times 2 \times 2 \times 2) \times (2 \times 2 \times 7)} \text{ cm}^2\\[1em] = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 7 \text{ cm}^2\\[1em] = 1344 \text{ cm}^2\end{aligned}$$

Hence, the area of the triangle is 1344 cm².

Given:

Diameter of the circle = 20 cm

∴ Radius of the circle $(r) = \frac{20}{2}$ cm = 10 cm

By formula,

Circumference of the circle = $2\pi r$

= 2 x 3.14 x 10 cm

= 62.8 cm

Area of the circle = $\pi r^2$

= 3.14 x 10 x 10 cm²

= 314 cm²

Hence, the circumference and the area of the circle are 62.8 cm and 314 cm² respectively.

Given:

Circumference of the circle - diameter of the circle = 18 cm

Let the radius of the circle be $r$ cm.

∴ Diameter = $2r$ cm

By formula,

Circumference of a circle = $2\pi r$

∴ Circumference - diameter = 18

$$\begin{aligned}\Rightarrow 2\pi r - 2r = 18\\[1em] \Rightarrow 2r(\pi - 1) = 18\\[1em] \Rightarrow 2r\left(\frac{22}{7} - 1\right) = 18\\[1em] \Rightarrow 2r \times \frac{15}{7} = 18\\[1em] \Rightarrow r = 18 \times \frac{7}{30}\\[1em] \Rightarrow r = 4.2 \text{ cm}\end{aligned}$$

Hence, the radius of the circle is 4.2 cm.

Given:

Ratio of the radii of the two circles = 5 : 7

Let the radii of the two circles be $5x$ and $7x$.

By formula,

Circumference of a circle = $2\pi r$ and area of a circle = $\pi r^2$

(i) Circumference of the first circle = $2\pi \times 5x = 10\pi x$

Circumference of the second circle = $2\pi \times 7x = 14\pi x$

∴ Ratio between their circumferences = $10\pi x : 14\pi x$

= 5 : 7

Hence, the ratio between their circumferences is 5 : 7.

(ii) Area of the first circle = $\pi \times (5x)^2 = 25\pi x^2$

Area of the second circle = $\pi \times (7x)^2 = 49\pi x^2$

∴ Ratio between their areas = $25\pi x^2 : 49\pi x^2$

= 25 : 49

Hence, the ratio between their areas is 25 : 49.

Given:

Ratio of the areas of the two circles = 16 : 9

Let the radii of the two circles be $r_1$ and $r_2$.

By formula,

Area of a circle = $\pi r^2$

(i) According to question:

$$\begin{aligned}\Rightarrow \frac{\pi r_1^2}{\pi r_2^2} = \frac{16}{9}\\[1em] \Rightarrow \frac{r_1^2}{r_2^2} = \frac{16}{9}\\[1em] \Rightarrow \frac{r_1}{r_2} = \sqrt{\frac{16}{9}} = \frac{4}{3}\end{aligned}$$

Hence, the ratio between their radii is 4 : 3.

(ii) Ratio between their diameters = $2r_1 : 2r_2$

= $r_1 : r_2$

= 4 : 3

Hence, the ratio between their diameters is 4 : 3.

(iii) Ratio between their circumferences = $2\pi r_1 : 2\pi r_2$

= $r_1 : r_2$

= 4 : 3

Hence, the ratio between their circumferences is 4 : 3.

Given:

Inner circumference = 528 m

Outer circumference = 616 m

Let the inner radius be $r$ m and the outer radius be $R$ m.

By formula,

Inner circumference = $2\pi r$

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r = 528\\[1em] \Rightarrow r = 528 \times \frac{7}{44}\\[1em] \Rightarrow r = 84 \text{ m}\end{aligned}$$

Outer circumference = $2\pi R$

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times R = 616\\[1em] \Rightarrow R = 616 \times \frac{7}{44}\\[1em] \Rightarrow R = 98 \text{ m}\end{aligned}$$

∴ Width of the track = $R - r$

= 98 m - 84 m

= 14 m

Hence, the width of the track is 14 m.

Given:

Inner circumference = 264 m

Width of the track = 7 m

By formula,

Circumference of a circle = $2\pi r$

(i) Let the radius of the inner track be $r$ m.

$$\begin{aligned}\therefore 2\pi r = 264\\[1em] \Rightarrow 2 \times \frac{22}{7} \times r = 264\\[1em] \Rightarrow r = 264 \times \frac{7}{44}\\[1em] \Rightarrow r = 42 \text{ m}\end{aligned}$$

Hence, the radius of the inner track is 42 m.

(ii) Radius of the outer circumference ($R$) = inner radius + width of the track

= 42 m + 7 m

= 49 m

Hence, the radius of the outer circumference is 49 m.

(iii) Length of the outer circumference = $2\pi R$

= 2 x $\frac{22}{7}$ x 49 m

= 2 x 22 x 7 m

= 308 m

Hence, the length of the outer circumference is 308 m.

(iv) Cost of fencing the outer circumference = Length of the outer circumference x rate of fencing

= 308 x ₹ 50

= ₹ 15,400

Hence, the cost of fencing the outer circumference is ₹ 15,400.

Given:

Diameter of the wheel = 63 cm

∴ Radius of the wheel ($r$) = $\frac{63}{2}$ cm

In one revolution of the wheel, car moves = circumference of the wheel

= $2\pi r$

= 2 x $\frac{22}{7}$ x $\frac{63}{2}$ cm

= 22 x 9 cm

= 198 cm

In 2000 revolutions, car will move = $2\pi r \times n$

= 198 x 2000 cm

= 3,96,000 cm

Since,

1 km = 1000 m = 1000 x 100 cm = 1,00,000 cm

∴ Distance moved by the car = $\frac{396000}{100000}$ km = 3.96 km

Hence, the car will move 3.96 km during 2000 revolutions of its wheel.

Given:

Diameter of the wheel = 70 cm

∴ Radius of the wheel ($r$) = $\frac{70}{2}$ cm = 35 cm

In one revolution of the wheel, car moves = circumference of the wheel

= $2\pi r$

= 2 x $\frac{22}{7}$ x 35 cm

= 220 cm

Distance to be travelled = 1 km = 1,00,000 cm

Let the number of revolutions be $n$.

In $n$ revolutions, car will move = $2\pi r \times n$

$$\begin{aligned}\Rightarrow 220 \times n = 100000\\[1em] \Rightarrow n = \frac{100000}{220}\\[1em] \Rightarrow n = \frac{5000}{11}\\[1em] \Rightarrow n = 454\frac{6}{11}\end{aligned}$$

Hence, the wheel will make $454\frac{6}{11}$ revolutions to travel one kilometre.

Given:

Area of the square = 484 cm²

By formula,

Area of a square = (side)²

⇒ (side)² = 484

⇒ side = $\sqrt{484}$ cm = 22 cm

Length of the wire = perimeter of the square

= 4 x side

= 4 x 22 cm

= 88 cm

Hence, the length of the wire is 88 cm.

(i) When a wire is bent to form the largest circle of radius $r$,

The length of the wire = circumference of the circle formed = $2\pi r$

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r = 88\\[1em] \Rightarrow r = 88 \times \frac{7}{44}\\[1em] \Rightarrow r = 14 \text{ cm}\end{aligned}$$

Hence, the radius of the circle formed is 14 cm.

(ii) Area of the circle = $\pi r^2$

= $\frac{22}{7}$ x 14 x 14 cm²

= 22 x 2 x 14 cm²

= 616 cm²

Hence, the area of the circle is 616 cm².

Given:

Radius of the circle ($r$) = 28 cm

Length of the wire = circumference of the circle

= $2\pi r$

= 2 x $\frac{22}{7}$ x 28 cm

= 2 x 22 x 4 cm

= 176 cm

Since the same wire is bent in the form of a square,

Perimeter of the square = 176 cm

⇒ 4 x side = 176

⇒ side = $\frac{176}{4}$ cm = 44 cm

∴ Area of the square = (side)²

= (44 cm)²

= 1936 cm²

Hence, the area of the square formed is 1936 cm².

Given:

Length of the rectangular paper = 35 cm

Breadth of the rectangular paper = 22 cm

The diameter of the largest circle which can be cut out of this paper is equal to the breadth of the paper.

∴ Diameter of the circle = 22 cm

⇒ Radius of the circle ($r$) = $\frac{22}{2}$ cm = 11 cm

Area of the circle = $\pi r^2$

$$\begin{aligned}= \frac{22}{7} \times 11 \times 11 \text{ cm}^2\\[1em] = \frac{2662}{7} \text{ cm}^2\\[1em] = 380\frac{2}{7} \text{ cm}^2 = 380.29 \text{ cm}^2\end{aligned}$$

Hence, the area of the largest circle is $380\frac{2}{7}$ cm² = 380.29 cm².

Given:

Length of the rectangular paper = 30 cm

Breadth of the rectangular paper = 20 cm

Radius of each quadrant ($r$) = 7 cm

Area of the rectangular paper = length x breadth

= 30 cm x 20 cm

= 600 cm²

Since a quadrant is one-fourth of a circle, four quadrants of the same radius together make one complete circle.

∴ Area of 4 quadrants = area of a circle of radius 7 cm

= $\pi r^2$

= $\frac{22}{7}$ x 7 x 7 cm²

= 154 cm²

∴ Area of the remaining paper = Area of the rectangular paper - area of 4 quadrants

= 600 cm² - 154 cm²

= 446 cm²

Hence, the area of the remaining paper is 446 cm².

Given:

Perimeter of the square = 32 cm

⇒ Side of the square = $\frac{\text{Perimeter}}{4} = \frac{32}{4}$ cm = 8 cm

∴ Area of the square = (side)²

= (8 cm)²

= 64 cm²

Hence, option 3 is the correct option.

Given:

Side of the square = 20 cm

Length of the rectangle ($l$) = 25 cm

Length of the wire = perimeter of the square

= 4 x side

= 4 x 20 cm

= 80 cm

∴ Perimeter of the rectangle ($P$) = 80 cm

By formula,

Breadth of the rectangle ($b$) = $\frac{P}{2} - l$

= $\frac{80}{2}$ cm - 25 cm

= 40 cm - 25 cm

= 15 cm

Since 15 cm is not given in the options, the correct answer is none of these.

Hence, option 4 is the correct option.

Given:

Area of △ABC = 18 cm²

Height AD = 3 cm

By formula,

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

⇒ $\frac{1}{2}$ x BC x AD = 18

⇒ $\frac{1}{2}$ x BC x 3 = 18

⇒ BC = $\frac{18 \times 2}{3}$

⇒ BC = 12 cm

Hence, option 3 is the correct option.

Given:

From the figure, the given triangle is right-angled with hypotenuse = 10 cm and one leg (base) = 8 cm.

Let the other leg (height) be $h$ cm.

By Pythagoras theorem,

$$\begin{aligned}\Rightarrow h^2 + 8^2 = 10^2 \\[1em] \Rightarrow h^2 + 64 = 100 \\[1em] \Rightarrow h^2 = 100 - 64 = 36 \\[1em] \Rightarrow h = \sqrt{36} = 6 \text{ cm}\end{aligned}$$

∴ Area of the triangle = $\frac{1}{2}$ x product of legs

= $\frac{1}{2}$ x 8 cm x 6 cm

= 24 cm²

Hence, option 3 is the correct option.

Given:

AB = 8 cm, BC = 12 cm, AC = 10 cm and altitude AD = 6 cm

By formula,

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

Taking BC as base and AD as the corresponding altitude,

Area of △ABC = $\frac{1}{2}$ x BC x AD

= $\frac{1}{2}$ x 12 cm x 6 cm

= 36 cm²

Taking AB as base and CE as the corresponding altitude,

Area of △ABC = $\frac{1}{2}$ x AB x CE

⇒ $\frac{1}{2}$ x 8 x CE = 36

⇒ 4 x CE = 36

⇒ CE = $\frac{36}{4}$ = 9 cm

Hence, option 2 is the correct option.

Given:

Area of the square = perimeter of the square (numerically)

Let the side of the square be $s$ cm.

By formula,

Area of the square = $s^2$ and perimeter of the square = $4s$

$$\begin{aligned}\Rightarrow s^2 = 4s \\[1em] \Rightarrow s^2 - 4s = 0 \\[1em] \Rightarrow s(s - 4) = 0 \\[1em] \Rightarrow s - 4 = 0 \quad (\text{as } s \neq 0) \\[1em] \Rightarrow s = 4 \text{ cm}\end{aligned}$$

Hence, option 2 is the correct option.

Given:

Radius of the outer circle ($R$) = 2 cm

Radius of the inner circle ($r$) = 1 cm

Area of the shaded portion = Area of outer circle - area of inner circle

= $\pi R^2 - \pi r^2$

= π x 2² - π x 1² cm²

= 4π - π cm²

= 3π cm²

Since 3π cm² is not given in the options, the correct answer is none of these.

Hence, option 4 is the correct option.

Given:

Base of the triangle is doubled and its height is halved.

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

By formula,

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

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

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

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

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

Since the new area is equal to the original area, the area is not changed.

Hence, option 4 is the correct option.

Given:

Circumferences of the two circles are 7 cm and 10 cm.

Let the radii of the two circles be $r_1$ and $r_2$.

By formula,

Circumference of a circle = $2\pi r$

$$\begin{aligned}\therefore \frac{2\pi r_1}{2\pi r_2} = \frac{7}{10}\\[1em] \Rightarrow \frac{r_1}{r_2} = \frac{7}{10}\\[1em] \Rightarrow r_1 : r_2 = 7 : 10\end{aligned}$$

Hence, option 1 is the correct option.

Given:

Length of the room = 3 m = 300 cm

Breadth of the room = 2 m = 200 cm

Side of each square tile = 20 cm

By formula,

Area of the floor = length x breadth

= 300 cm x 200 cm

= 60000 cm²

Area of one tile = (side)²

= (20 cm)²

= 400 cm²

∴ Number of tiles required = $\frac{\text{Area of the floor}}{\text{Area of one tile}}$

= $\frac{60000}{400}$

= 150

Hence, option 3 is the correct option.

Given:

Area of the circle = circumference of the circle (numerically)

Let the radius of the circle be $r$ units.

By formula,

Area of a circle = $\pi r^2$ and circumference of a circle = $2\pi r$

$$\begin{aligned}\Rightarrow \pi r^2 = 2\pi r \\[1em] \Rightarrow r^2 = 2r \quad (\text{dividing both sides by } \pi) \\[1em] \Rightarrow r^2 - 2r = 0 \\[1em] \Rightarrow r(r - 2) = 0 \\[1em] \Rightarrow r - 2 = 0 \quad (\text{as } r \neq 0) \\[1em] \Rightarrow r = 2 \text{ units}\end{aligned}$$

Hence, option 2 is the correct option.

Given:

Radius of the semicircle ($r$) = 7 cm

By formula,

Perimeter of the semicircle = Half of circumference + diameter

= $\pi r + 2r$

= $\frac{22}{7}$ x 7 cm + 2 x 7 cm

= 22 cm + 14 cm

= 36 cm

Hence, option 3 is the correct option.

Given:

Ratio of the areas of the two circles = 16 : 25

Let the radii of the two circles be $r_1$ and $r_2$.

By formula,

Area of a circle = $\pi r^2$

$$\begin{aligned}\therefore \frac{\pi r_1^2}{\pi r_2^2} = \frac{16}{25}\\[1em] \Rightarrow \frac{r_1}{r_2} = \sqrt{\frac{16}{25}} = \frac{4}{5}\end{aligned}$$

Ratio between their circumferences = $2\pi r_1 : 2\pi r_2$

= $r_1 : r_2$

= 4 : 5

Hence, option 1 is the correct option.

Given:

From the figure, the outer rectangle is 10 cm long and 8 cm broad, and the unshaded square has side 4 cm.

By formula,

Area of a rectangle = length x breadth and area of a square = (side)²

∴ Area of the shaded region = Area of outer figure - area of inner figure

= 10 x 8 cm² - 4 x 4 cm²

= 80 cm² - 16 cm²

= 64 cm²

Hence, option 4 is the correct option.

Given:

BD = 20 cm, AP = 8 cm and CQ = 12 cm

By formula,

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

The diagonal BD divides the quadrilateral ABCD into two triangles ABD and CBD, with AP and CQ as their altitudes on BD.

Area of △ABD = $\frac{1}{2}$ x BD x AP

= $\frac{1}{2}$ x 20 cm x 8 cm

= 80 cm²

Area of △CBD = $\frac{1}{2}$ x BD x CQ

= $\frac{1}{2}$ x 20 cm x 12 cm

= 120 cm²

∴ Area of quadrilateral ABCD = Area of △ABD + area of △CBD

= 80 cm² + 120 cm²

= 200 cm²

Hence, option 3 is the correct option.

Given:

Height of the equilateral triangle = $\frac{5\sqrt{3}}{2}$ cm

By formula,

Height of an equilateral triangle = $\frac{\sqrt{3}}{2}$ x side

⇒ $\frac{\sqrt{3}}{2}$ x side = $\frac{5\sqrt{3}}{2}$

⇒ side = 5 cm

∴ Area of equilateral triangle = $\frac{\sqrt{3}}{4}$ x (side)²

= $\frac{\sqrt{3}}{4}$ x 25 cm²

= $\frac{1.732}{4}$ x 25 cm²

= 10.825 cm²

So, Statement 1 is false.

Height ($h$) of the equilateral triangle = $\frac{\sqrt{3}}{2}$ x side

⇒ side = $\frac{2h}{\sqrt{3}}$

∴ Area of equilateral triangle = $\frac{\sqrt{3}}{4}$ x (side)²

$$\begin{aligned}= \frac{\sqrt{3}}{4} \times \frac{4h^2}{3}\\[1em] = \frac{1}{\sqrt{3}} \times h^2\\[1em] = \frac{1}{\sqrt{3}} \times (\text{height})^2 \text{ square unit}\end{aligned}$$

So, Statement 2 is true.

∴ Statement 1 is false, and Statement 2 is true.

Hence, option 4 is the correct option.

The distance around the edge of a circle is its circumference, which is equal to $2\pi r$ units.

So, Statement 1 is true.

Given:

Ratio of the areas of the two circles = 64 : 169

Let the radii of the two circles be $r_1$ and $r_2$.

By formula,

Area of a circle = $\pi r^2$

$$\begin{aligned}\therefore \frac{\pi r_1^2}{\pi r_2^2} = \frac{64}{169}\\[1em] \Rightarrow \frac{r_1}{r_2} = \sqrt{\frac{64}{169}} = \frac{8}{13}\end{aligned}$$

Ratio of their circumferences = $2\pi r_1 : 2\pi r_2$

= $r_1 : r_2$

= 8 : 13

So, Statement 2 is true.

∴ Both the Statements are true.

Hence, option 1 is the correct option.

Given:

Area of the square = 2.56 m²

⇒ (side)² = 2.56

⇒ side = $\sqrt{2.56}$ m = 1.6 m

∴ Perimeter of the square = 4 x side

= 4 x 1.6 m

= 6.4 m

The perimeter of the square is 6.4 m and not 6.4 cm.

So, Assertion (A) is false.

The perimeter of a square = (4 x side) unit.

So, Reason (R) is true.

∴ A is false, R is true.

Hence, option 2 is the correct option.

Given:

Base of the triangle = 15 cm

Area of the triangle = 0.9 m² = 0.9 x 10000 cm² = 9000 cm² [1 m² = 10⁴ cm²]

Let the altitude of the triangle be $h$ cm.

By formula,

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

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 15 \times h = 9000 \\[1em] \Rightarrow h = \frac{9000 \times 2}{15} \\[1em] \Rightarrow h = 1200 \text{ cm} = \frac{1200}{100} \text{ m} = 12 \text{ m}\end{aligned}$$

So, Assertion (A) is true.

Area of a triangle = ($\frac{1}{2}$ x base x corresponding height) unit².

So, Reason (R) is true.

∴ Both A and R are true.

Hence, option 3 is the correct option.

Given:

Shorter side of the parallelogram = 8 cm

Longer side of the parallelogram = 12 cm

Let the distance between the shorter sides be $h_1$ cm and the distance between the longer sides be $h_2$ cm.

By formula,

Area of a parallelogram = base x corresponding height

Taking the shorter side as base, area = $8 \times h_1$

Taking the longer side as base, area = $12 \times h_2$

$$\begin{aligned}\therefore 8 \times h_1 = 12 \times h_2\\[1em] \Rightarrow \frac{h_1}{h_2} = \frac{12}{8} = \frac{3}{2}\\[1em] \Rightarrow h_1 : h_2 = 3 : 2\end{aligned}$$

The ratio of the distance between its shorter sides to the distance between its longer sides is 3 : 2 and not 2 : 3.

So, Assertion (A) is false.

Area of a parallelogram = (base x corresponding height) unit².

So, Reason (R) is true.

∴ A is false, R is true.

Hence, option 2 is the correct option.