Perimeter & Area of Plane Figures

Given:

Sides of the triangle are 14 cm, 17 cm and 25 cm.

As we know, perimeter of a triangle = sum of the lengths of its three sides

∴ Perimeter of the triangle = (14 + 17 + 25) cm

= 56 cm

Hence, the perimeter of the triangle is 56 cm.

Given:

Perimeter of the triangle = 40 cm

Two sides of the triangle are 12.9 cm and 14.6 cm.

As we know, perimeter of a triangle = sum of the lengths of its three sides

Let the length of the third side be $x$ cm. Then,

$$\begin{aligned}\Rightarrow 12.9 + 14.6 + x = 40 \\[1em] \Rightarrow 27.5 + x = 40 \\[1em] \Rightarrow x = 40 - 27.5 \\[1em] \Rightarrow x = 12.5\end{aligned}$$

Hence, the length of the third side is 12.5 cm.

Given:

Length ($l$) = 25 cm and breadth ($b$) = 18 cm

Perimeter of the rectangle = $2(l + b)$

= 2 (25 + 18) cm

= 2 x 43 cm

= 86 cm

Area of the rectangle = $l \times b$

= 25 cm x 18 cm

= 450 cm²

Hence, the perimeter of the rectangle is 86 cm and its area is 450 cm².

Given:

Length ($l$) = 27 cm and breadth ($b$) = 17 cm

Perimeter of the rectangle = $2(l + b)$

= 2 (27 + 17) cm

= 2 x 44 cm

= 88 cm

Area of the rectangle = $l \times b$

= 27 cm x 17 cm

= 459 cm²

Hence, the perimeter of the rectangle is 88 cm and its area is 459 cm².

Given:

Length ($l$) = 7.2 m = 7.2 x 100 cm = 720 cm [1 m = 100 cm]

Breadth ($b$) = 80 cm

Perimeter of the rectangle = $2(l + b)$

= 2 (720 + 80) cm

= 2 x 800 cm

= 1600 cm

= $\frac{1600}{100}$ m

= 16 m

Area of the rectangle = $l \times b$

= 720 cm x 80 cm

= 57600 cm²

Hence, the perimeter of the rectangle is 16 m and its area is 57600 cm².

Given:

Side of the square field = 9 m

(i) Perimeter of the square field = 4 x side

= 4 x 9 m

= 36 m

Hence, the perimeter of the field is 36 m.

(ii) Area of the square field = (side)²

= 9 m x 9 m

= 81 m²

Hence, the area of the field is 81 m².

Given:

Area of the square = 256 cm²

(i) As we know, area of a square = (side)²

⇒ (side)² = 256 cm² = 16 cm x 16 cm

⇒ side = 16 cm

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

(ii) Perimeter of the square = 4 x side

= 4 x 16 cm

= 64 cm

Hence, the perimeter of the square is 64 cm.

Given:

Area of the field = 768 m²

Length of the field ($l$) = 32 m

Rate of fencing = ₹ 20 per metre

(i) Breadth of the field ($b$) = $\frac{\text{Area}}{\text{Length}}$

$= \frac{768}{32}$ m

= 24 m

Hence, the breadth of the field is 24 m.

(ii) Perimeter of the field = $2(l + b)$

= 2 (32 + 24) m

= 2 x 56 m

= 112 m

Cost of fencing the field = Perimeter x rate of fencing

= 112 x ₹ 20

= ₹ 2,240

Hence, the cost of fencing the field is ₹ 2,240.

Given:

Area of the field = 3400 m²

Length of the field ($l$) = 68 m

(i) Breadth of the field ($b$) = $\frac{\text{Area}}{\text{Length}}$

$= \frac{3400}{68}$ m

= 50 m

Hence, the breadth of the field is 50 m.

(ii) Perimeter of the field = $2(l + b)$

= 2 (68 + 50) m

= 2 x 118 m

= 236 m

Distance covered by the man in going 1 time around the field = 236 m

∴ Distance covered by the man in going 5 times around the field = 5 x 236 m = 1180 m

Hence, the distance covered by the man in going 5 times around the field is 1180 m.

Given:

Area of the park = 4100 m²

Breadth of the park ($b$) = 50 m

Rate of fencing = ₹ 30 per metre

Length of the park ($l$) = $\frac{\text{Area}}{\text{Breadth}}$

$= \frac{4100}{50}$ m

= 82 m

Perimeter of the park = $2(l + b)$

= 2 (82 + 50) m

= 2 x 132 m

= 264 m

Cost of fencing the park = Perimeter x rate of fencing

= 264 x ₹ 30

= ₹ 7,920

Hence, the cost of fencing the park is ₹ 7,920.

Given:

Length of the field ($l$) = 35 m

Breadth of the field ($b$) = 18 m

Rate of cultivating = ₹ 20 per m²

Rate of fencing = ₹ 15 per metre

Area of the field = $l \times b$

= 35 m x 18 m

= 630 m²

Cost of cultivating the field = Area x rate of cultivating

= 630 x ₹ 20

= ₹ 12,600

Perimeter of the field = $2(l + b)$

= 2 (35 + 18) m

= 2 x 53 m

= 106 m

Cost of fencing the field = Perimeter x rate of fencing

= 106 x ₹ 15

= ₹ 1,590

Hence, the cost of cultivating the field is ₹ 12,600 and the cost of fencing the field is ₹ 1,590.

Given:

Perimeter of the plot = 174 m

Breadth of the plot ($b$) = 35 m

Let the length of the plot be $l$ m.

As we know, perimeter of a rectangle = $2(l + b)$

$$\begin{aligned}\Rightarrow 2(l + 35) = 174 \\[1em] \Rightarrow l + 35 = \frac{174}{2} \\[1em] \Rightarrow l + 35 = 87 \\[1em] \Rightarrow l = 87 - 35 = 52\end{aligned}$$

∴ Length of the plot = 52 m

Area of the plot = $l \times b$

= 52 m x 35 m

= 1820 m²

Hence, the length of the plot is 52 m and its area is 1820 m².

Given:

Side of the square = 18 cm

Width of the rectangle = 12 cm

Area of the square = (side)²

= 18 cm x 18 cm

= 324 cm²

As the area of the rectangle = area of the square

⇒ area of the rectangle = 324 cm²

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

$= \frac{324}{12}$ cm

= 27 cm

Hence, the length of the rectangle is 27 cm.

Given:

Length of the hall = 16 m = 16 x 100 cm = 1600 cm

Breadth of the hall = 12 m = 12 x 100 cm = 1200 cm

Length of each slab = 25 cm

Breadth of each slab = 15 cm

Area of the floor of the hall = length x breadth

= 1600 cm x 1200 cm

= 1920000 cm²

Area of each slab = 25 cm x 15 cm = 375 cm²

∴ Number of slabs required = $\frac{\text{Area of the floor of the hall}}{\text{Area of each slab}}$

$= \frac{1920000}{375}$

= 5120

Hence, 5120 marble slabs are needed to cover the floor of the hall.

Given:

Length of the room = 13 m

Breadth of the room = 9 m

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

Rate of the carpet = ₹ 50 per metre

Area of the floor of the room = length x breadth = 13 m x 9 m = 117 m²

Area of the carpet required = area of the floor = 117 m²

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

$= \frac{117}{0.75}$ m

= 156 m

Cost of carpeting the room = Length of the carpet x rate

= 156 x ₹ 50

= ₹ 7,800

Hence, the cost of carpeting the room is ₹ 7,800.

Given:

Length of the sheet = 384 cm

Breadth of the sheet = 172 cm

Size of paper required for each envelope = 16 cm by 12 cm

Area of the sheet = length x breadth

= 384 cm x 172 cm

= 66048 cm²

Area of paper required for each envelope = 16 cm x 12 cm = 192 cm²

∴ Number of envelopes = $\frac{\text{Ar. of sheet}}{\text{Paper Ar. for each envelope}}$

$= \frac{66048}{192}$

= 344

Hence, 344 envelopes can be made out of the sheet of paper.

Given:

Total cost of cultivating the field = ₹ 18,330

Rate of cultivating = ₹ 15 per m²

Breadth of the field ($b$) = 26 m

Rate of fencing = ₹ 20 per metre

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

$= \frac{18330}{15}$ m²

= 1222 m²

Length of the field ($l$) = $\frac{\text{Area}}{\text{Breadth}}$

$= \frac{1222}{26}$ m

= 47 m

Perimeter of the field = $2(l + b)$

= 2 (47 + 26) m

= 2 x 73 m

= 146 m

Cost of fencing the field = Perimeter x rate of fencing

= 146 x ₹ 20

= ₹ 2,920

Hence, the cost of fencing the field is ₹ 2,920.

Draw a dotted line, as shown in the figure below, to divide the shaded region into two rectangles I and II.

From the figure,

Rectangle I (vertical) is 10 m long and 1 m wide.

Rectangle II (horizontal) is 8 m long and 2 m wide.

As we know, area of a rectangle = length x breadth

∴ Area of the shaded region = area of rectangle I + area of rectangle II

= (10 x 1 + 8 x 2) m²

= (10 + 16) m²

= 26 m²

Hence, the area of the shaded region is 26 m².

Draw dotted lines, as shown in the figure below, to divide the shaded region into three rectangles I, II and III.

From the figure,

Rectangle I (top) is 12 m long and 1 m wide.

Rectangle II (bottom) is 10 m long and 2 m wide.

Width of rectangle III (middle vertical) = 12 m - 10 m = 2 m

Height of rectangle III (middle vertical) = 13 m - 1 m - 2 m = 10 m

As we know, area of a rectangle = length x breadth

∴ Area of the shaded region = area of rectangle I + area of rectangle II + area of rectangle III

= (12 x 1 + 10 x 2 + 2 x 10) m²

= (12 + 20 + 20) m²

= 52 m²

Hence, the area of the shaded region is 52 m².

From the figure,

Length of the outer rectangle = 14 m and breadth = 12 m

Length of the inner (unshaded) rectangle = 14 m - 2 m = 12 m and breadth = 10.5 m

Area of the outer rectangle = 14 m x 12 m = 168 m²

Area of the inner rectangle = 12 m x 10.5 m = 126 m²

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

= 168 m² - 126 m²

= 42 m²

Hence, the area of the shaded region is 42 m².

From the figure,

Length of the outer rectangle = 16 m and breadth = 12 m

Width of the shaded border = 1.5 m

Length of the inner rectangle = 16 m - (1.5 + 1.5) m = 13 m

Breadth of the inner rectangle = 12 m - (1.5 + 1.5) m = 9 m

Area of the outer rectangle = 16 m x 12 m = 192 m²

Area of the inner rectangle = 13 m x 9 m = 117 m²

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

= 192 m² - 117 m²

= 75 m²

Hence, the area of the shaded region is 75 m².

Draw dotted lines, as shown in the figure below, to divide the shaded region into four rectangles I, II, III and IV placed one above the other.

From the figure, each rectangle is of height 1 m, and each upper rectangle is shorter than the one below it by 0.5 m on each side.

Length of rectangle I (bottom) = 4 m

Length of rectangle II = 4 m - (0.5 + 0.5) m = 3 m

Length of rectangle III = 3 m - (0.5 + 0.5) m = 2 m

Length of rectangle IV (top) = 2 m - (0.5 + 0.5) m = 1 m

As we know, area of a rectangle = length x breadth

∴ Area of the shaded region = area of rectangle I + area of rectangle II + area of rectangle III + area of rectangle IV

= (4 x 1 + 3 x 1 + 2 x 1 + 1 x 1) m²

= (4 + 3 + 2 + 1) m²

= 10 m²

Hence, the area of the shaded region is 10 m².

Given:

Ratio of the sides = 8 : 7

Perimeter of the rectangle = 90 cm

Let the length ($l$) be $8x$ cm and the breadth ($b$) be $7x$ cm.

Perimeter of the rectangle = $2(l + b)$

$$\begin{aligned}\Rightarrow 2(8x + 7x) = 90 \\[1em] \Rightarrow 2 \times 15x = 90 \\[1em] \Rightarrow 30x = 90 \\[1em] \Rightarrow x = \frac{90}{30} \\[1em] \Rightarrow x = 3\end{aligned}$$

∴ Length of the rectangle = 8 x 3 cm = 24 cm

Hence, option 2 is the correct option.

Given:

Area of the rectangle = 350 cm²

Breadth of the rectangle ($b$) = 14 cm

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

$= \frac{350}{14}$ cm

= 25 cm

Perimeter of the rectangle = $2(l + b)$

= 2 (25 + 14) cm

= 2 x 39 cm

= 78 cm

Hence, option 3 is the correct option.

Given:

Length of the field ($l$) = 35 m

Breadth of the field ($b$) = 18 m

Rate of fencing = ₹ 22.50 per metre

Perimeter of the field = $2(l + b)$

= 2 (35 + 18) m

= 2 x 53 m

= 106 m

Cost of fencing the field = Perimeter x rate of fencing

= 106 x ₹ 22.50

= ₹ 2,385

Hence, option 4 is the correct option.

Given:

Total cost of fencing = ₹ 2,000

Rate of fencing = ₹ 25 per metre

Length of the field ($l$) = 24 m

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

$= \frac{2000}{25}$ m

= 80 m

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

$$\begin{aligned}2(l + b) = \text{perimeter} \\[1em] \Rightarrow 2(24 + b) = 80 \\[1em] \Rightarrow (24 + b) = \frac{80}{2} \\[1em] \Rightarrow 24 + b = 40 \\[1em] \Rightarrow b = 40 - 24 \\[1em] \Rightarrow b = 16\end{aligned}$$

∴ Breadth of the field = 16 m

Hence, option 2 is the correct option.

Given:

$l : 2(l + b) = 1 : 3$

Let the length ($l$) = $x$.

Then,

$$\begin{aligned}2(l + b) = 3x \\[1em] \Rightarrow l + b = \frac{3x}{2} \\[1em] \Rightarrow b = \frac{3x}{2} - x = \frac{x}{2} \\[1em] \therefore l : b = x : \frac{x}{2} = 2 : 1\end{aligned}$$

Hence, option 4 is the correct option.

Given:

Total cost of fencing = ₹ 2,400

Rate of fencing = ₹ 30 per metre

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

$= \frac{2400}{30}$ m

= 80 m

Each side of the field = $\frac{\text{Perimeter}}{4}$

$= \frac{80}{4}$ m

= 20 m

Hence, option 2 is the correct option.

A regular hexagon has 6 equal sides.

∴ Perimeter of a regular hexagon = side + side + side + side + side + side = 6 x side

Hence, option 3 is the correct option.

1 m² = 1 m x 1 m

= 100 cm x 100 cm [1 m = 100 cm]

= 10000 cm²

Hence, option 3 is the correct option.

Given:

Area of the square = 100 m²

⇒ (side)² = 100 m² = 10 m x 10 m

⇒ side = 10 m

Perimeter of the square = 4 x side

= 4 x 10 m

= 40 m

Hence, option 4 is the correct option.

The four walls of a room are two pairs of rectangles: two walls of size length x height and two walls of size breadth x height.

∴ Area of 4 walls of a room = 2 x length x height + 2 x breadth x height

= 2 (length + breadth) x height

Hence, option 3 is the correct option.

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

$= \frac{120}{15}$ cm

= 8 cm

Hence, the breadth of the rectangle is 8 cm.

(ii) Let the length be $4x$ cm and the breadth be $3x$ cm.

Area of the rectangle = length x breadth

$$\begin{aligned}\Rightarrow 4x \times 3x = 108 \\[1em] \Rightarrow 12x^2 = 108 \\[1em] \Rightarrow x^2 = \frac{108}{12} \\[1em] \Rightarrow x^2 = 9 \\[1em] \Rightarrow x^2 = 3 \times 3 \\[1em] \Rightarrow x = \sqrt{3 \times 3} \\[1em] \Rightarrow x = 3\end{aligned}$$

∴ Breadth of the rectangle = 3 x 3 cm = 9 cm

Hence, the breadth of the rectangle is 9 cm.

(iii) A regular pentagon has 5 equal sides.

∴ Each side of the pentagon = $\frac{\text{Perimeter}}{5}$

$= \frac{16}{5}$ cm

= 3.2 cm

Hence, each side of the pentagon is 3.2 cm.

(iv) Let the side of the square be $a$ cm.

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

Given, area = perimeter (numerically)

$$\begin{aligned}\Rightarrow a^2 = 4a \\[1em] \Rightarrow a^2 - 4a = 0 \\[1em] \Rightarrow a(a - 4) = 0 \\[1em] \Rightarrow a = 0 \text{ or } a - 4 = 0 \\[1em] \Rightarrow a = 0 \text{ or } a = 4\end{aligned}$$

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

Hence, the side of the square is 4 cm.

(v) As we know, area of a rectangular room = length x breadth

∴ $\frac{\text{area}}{\text{breadth}} = \text{length}$

Hence, for a rectangular room, $\bm{\frac{\text{area}}{\text{breadth}} = \text{length}}$.

(i) Perimeter of a rectangle = 2 (length + breadth), i.e., twice the sum of its length and breadth, not the sum of its length and breadth.

Hence, the given statement is false (F).

(ii) 1 m² = 1 m x 1 m = 100 cm x 100 cm = 10000 cm², not 10 cm².

Hence, the given statement is false (F).

(iii) The four walls of a room consist of two walls of size length x height and two walls of size breadth x height.

∴ Area of 4 walls of a room = 2 x length x height + 2 x breadth x height = 2 x (Length + Breadth) x Height

Hence, the given statement is true (T).

(iv) Area of the square = (side)² = 2 m x 2 m = 4 m²

Hence, the given statement is true (T).

(v) 1 m² = 1 m x 1 m = 100 cm x 100 cm = 10000 cm²

Hence, the given statement is true (T).

1. Given:

Total cost of cultivating the field = ₹ 24,440

Rate of cultivating = ₹ 20 per m²

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

$= \frac{24440}{20} \text{ m}^2$

= 1222 m²

Hence, option (c) is the correct option.

2. Given:

Area of the field = 1222 m² (from part 1)

Breadth of the field = 26 m

Length of the field ($l$) = $\frac{\text{Area}}{\text{Breadth}}$

$= \frac{1222}{26}$ m

= 47 m

Hence, option (b) is the correct option.

3. Given:

Length of the field ($l$) = 47 m (from part 2)

Breadth of the field ($b$) = 26 m

Perimeter of the field = $2(l + b)$

= 2 (47 + 26) m

= 2 x 73 m

= 146 m

Hence, option (c) is the correct option.

4. Given:

Perimeter of the field = 146 m (from part 3)

Rate of fencing = ₹ 25 per m

Cost of fencing the field = Perimeter x rate of fencing

= 146 x ₹ 25

= ₹ 3,650

Hence, option (a) is the correct option.

The frame of a picture runs along its boundary, i.e., along all its sides. So, to find the cost of the frame, we need the total length of its boundary, which is its perimeter.

So, Assertion (A) is true.

The perimeter of a plane figure is the sum of the lengths of all its sides.

So, Reason (R) is true.

Since the length of the frame is the sum of the lengths of all the sides of the picture, Reason (R) correctly explains Assertion (A).

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

Hence, option 1 is the correct option.

Let the length of the rectangle be $l$ and its breadth be $b$.

∴ Original area of the rectangle = $l \times b$

New length = $\frac{l}{2}$ and new breadth = $2b$

∴ New area of the rectangle = $\frac{l}{2} \times 2b = l \times b$

Since the new area is equal to the original area, the area remains the same.

So, Assertion (A) is true.

Area of a rectangle = length x breadth, while 2 (length + breadth) is the perimeter of a rectangle.

So, Reason (R) is false.

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

Hence, option 3 is the correct option.

1 km² = 1 km x 1 km

= 1000 m x 1000 m [1 km = 1000 m]

= 1000000 m²

So, Assertion (A) is true.

1 m² = 1 m x 1 m

= 100 cm x 100 cm [1 m = 100 cm]

= 10000 cm²

So, Reason (R) is true.

Assertion (A) is based on the conversion 1 km = 1000 m, whereas Reason (R) is about converting m² into cm². So, Reason (R) does not explain Assertion (A).

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

Hence, option 2 is the correct option.

Given:

Side of the square = $x$

Area of the square = (side)² = $x^2$

Perimeter of the square = 4 x side = $4x$

∴ Length of the rectangle = perimeter of the square = $4x$

And, area of the rectangle = area of the square = $x^2$

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

$$\begin{aligned}= \frac{x^2}{4x}\\[1em] = \frac{x}{4}\\[1em] = x \div 4\end{aligned}$$

Hence, option 3 is the correct option.

From the figure, the base of the shaded triangle covers 6 squares and its height covers 4 squares.

The shaded triangle lies in a rectangle of 6 x 4 = 24 squares and covers exactly half of it.

∴ Area of the rectangle = 24 x 1 cm² = 24 cm²

∴ Area of the shaded figure = $\frac{1}{2}$ x 24 cm² = 12 cm²

Hence, option 3 is the correct option.

Given:

Side of the equilateral triangle = 14 cm

From the figure, the equilateral triangle is drawn on the length of the rectangle.

∴ Length of the rectangle = side of the equilateral triangle = 14 cm

Width of the rectangle = $\frac{1}{2}$ x length = $\frac{1}{2}$ x 14 cm = 7 cm

The boundary of the figure consists of the top side of the rectangle, its two widths and the two outer sides of the equilateral triangle.

∴ Perimeter of the figure = (14 + 7 + 7 + 14 + 14) cm

= 56 cm

Hence, option 2 is the correct option.

Given:

Length of the photograph ($l$) = 39.5 cm

Breadth of the photograph ($b$) = 31 cm

Length of the wooden strip required = perimeter of the photograph

= $2(l + b)$

= 2 (39.5 + 31) cm

= 2 x 70.5 cm

= 141 cm

Hence, option 3 is the correct option.

From the figure, the given figure is made up of two squares of side 10 cm each, overlapping each other.

Width of the overlapping part = 10 cm - 8 cm = 2 cm

Height of the overlapping part = 1.5 cm

Sides of the upper square on the boundary of the figure:

top side = 10 cm, left side = 10 cm, bottom side = 10 cm - 2 cm = 8 cm and right side = 10 cm - 1.5 cm = 8.5 cm

Sides of the lower square on the boundary of the figure:

right side = 10 cm, bottom side = 10 cm, top side = 10 cm - 2 cm = 8 cm and left side = 10 cm - 1.5 cm = 8.5 cm

∴ Perimeter of the figure = (10 + 10 + 8 + 8.5 + 10 + 10 + 8 + 8.5) cm

= 73 cm

Hence, option 4 is the correct option.

Given:

Length of the plot ($l$) = 43.15 m ≈ 43 m

Breadth of the plot ($b$) = 30.72 m ≈ 31 m

Estimated perimeter of the plot = $2(l + b)$

= 2 (43 + 31) m

= 2 x 74 m

= 148 m

Hence, option 2 is the correct option.

Given:

Side of the square = 18 cm

Width of the rectangle ($b$) = 12 cm

Area of the square = (side)² = 18 cm x 18 cm = 324 cm²

∴ Area of the rectangle = area of the square = 324 cm²

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

$= \frac{324}{12}$ cm

= 27 cm

Perimeter of the rectangle = $2(l + b)$

= 2 (27 + 12) cm

= 2 x 39 cm

= 78 cm

Hence, option 4 is the correct option.

Given:

Length of the land ($l$) = 3 m

Breadth of the land ($b$) = 2 m

Perimeter of the land = $2(l + b)$

= 2 (3 + 2) m

= 2 x 5 m

= 10 m

Length of wire needed for one row = 10 m

∴ Length of wire needed for three rows = 3 x 10 m = 30 m

Hence, option 3 is the correct option.

Given:

Length of the sheet = 384 cm

Breadth of the sheet = 168 cm

Size of paper required for each envelope = 16 cm by 12 cm

Area of the sheet = length x breadth

= 384 cm x 168 cm

= 64512 cm²

Area of paper required for each envelope = 16 cm x 12 cm = 192 cm²

∴ Number of envelopes = $\frac{\text{Area of sheet}}{\text{Paper Ar.for each envelope}}$

$= \frac{64512}{192}$

= 336

Hence, option 3 is the correct option.

From the figure,

Length of the floor = 30 m and breadth = 24 m

Unshaded rectangle at the top-right corner = 7 m x 6 m

Unshaded rectangle at the bottom-left corner = 6 m x 8 m

Area of the floor = 30 m x 24 m = 720 m²

Area of the unshaded rectangle at the top-right corner = 7 m x 6 m = 42 m²

Area of the unshaded rectangle at the bottom-left corner = 6 m x 8 m = 48 m²

∴ Area of the shaded region = (720 - 42 - 48) m²

= 630 m²

Cost of tiling = Area x rate of tiling

= 630 x ₹ 15

= ₹ 9,450

Hence, option 4 is the correct option.

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

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

Given, area = perimeter (numerically)

$$\begin{aligned}\Rightarrow a^2 = 4a \\[1em] \Rightarrow a^2 - 4a = 0 \\[1em] \Rightarrow a(a - 4) = 0 \\[1em] \Rightarrow a = 0 \text{ or } a - 4 = 0 \\[1em] \Rightarrow a = 0 \text{ or } a = 4\end{aligned}$$

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

∴ Side of the square = 4 units

Hence, option 3 is the correct option.

From the figure,

Length of the field = 56 m and width = 42 m

Area of the field = 56 m x 42 m = 2352 m²

New length of the field = 56 m - 7 m = 49 m

As the area remains unchanged,

New width of the field = $\frac{\text{Area}}{\text{New length}}$

$= \frac{2352}{49}$ m

= 48 m

Hence, option 1 is the correct option.