Perimeter and Area

Given,

Side of the square ABCD = 24 cm

AE = 15 cm

∴ EB = AB - AE = 24 cm - 15 cm = 9 cm

Since EF is parallel to BC, AEFD and EBCF are rectangles with EF = AD = BC = 24 cm.

(i) As we know,

Perimeter of rectangle = 2(length + breadth)

Perimeter of rectangle AEFD = 2 (AE + EF)

= 2 (15 + 24) cm

= 2 x 39 cm

= 78 cm

Perimeter of rectangle EBCF = 2 (EB + BC)

= 2 (9 + 24) cm

= 2 x 33 cm

= 66 cm

∴ Perimeter of AEFD - perimeter of EBCF = 78 cm - 66 cm = 12 cm

Hence, the perimeter of AEFD exceeds the perimeter of EBCF by 12 cm.

(ii) As we know,

Area of rectangle = length x breadth

Area of rectangle AEFD = AE x EF

= 15 cm x 24 cm

= 360 cm²

Area of rectangle EBCF = EB x BC

= 9 cm x 24 cm

= 216 cm²

∴ Area of AEFD - area of EBCF = 360 cm² - 216 cm² = 144 cm²

Hence, the area of AEFD exceeds the area of EBCF by 144 cm².

Given,

Length of the park = 180 m

Breadth of the park = 120 m

Speed of Nagma = 7.5 km/hour

Distance covered by Nagma in one round of the park = perimeter of the park

= 2 (length + breadth)

= 2 (180 m + 120 m)

= 2 x 300 m

= 600 m

∴ Distance covered in five rounds = 5 x 600 m = 3000 m

Speed of running = 7.5 km/hour = 7.5 x 1000 m/hour = 7500 m/hour

∴ Time taken by Nagma in covering 3000 m = $\frac{\text{distance}}{\text{speed}}$

$$\begin{aligned}= \frac{3000 \text{ m}}{7500 \text{ m/hour}}\\[1em] = \frac{2}{5} \text{ hours}\\[1em] = \Big(\frac{2}{5} \times 60\Big) \text{ minutes}\\[1em] = 24 \text{ minutes}\end{aligned}$$

Hence, Nagma will complete five rounds in 24 minutes.

Given,

Area of the plot = 540 m²

Length of the plot ($l$) = 27 m

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

As we know,

Area of a rectangle = length x breadth

⇒ 27 x $b$ = 540

⇒ $b = \frac{540}{27}$ = 20

∴ Breadth of the plot = 20 m

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

= 2 (27 + 20) m

= 2 x 47 m

= 94 m

Hence, the breadth of the plot is 20 m and its perimeter is 94 m.

Given,

Perimeter of the field = 151 m

Breadth of the field ($b$) = 32 m

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

As we know,

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

$$\begin{aligned}\Rightarrow 2(l + 32) = 151\\[1em] \Rightarrow l + 32 = \frac{151}{2} = 75.5\\[1em] \Rightarrow l = 75.5 - 32 = 43.5\end{aligned}$$

∴ Length of the field = 43.5 m

Area of the field = length x breadth

= 43.5 m x 32 m

= 1392 m²

Hence, the length of the field is 43.5 m and its area is 1392 m².

Given,

Area of the plot = 340 m²

Breadth of the plot ($b$) = 17 m

Rate of fencing = ₹ 350 per metre

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

As we know,

Area of a rectangle = length x breadth

⇒ $l$ x 17 = 340

⇒ $l = \frac{340}{17}$ = 20

∴ Length of the plot = 20 m

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

= 2 (20 + 17) m

= 2 x 37 m

= 74 m

Cost of fencing the plot = Perimeter x rate of fencing

= 74 x ₹ 350

= ₹ 25,900

Hence, the cost of surrounding the plot with a fence is ₹ 25,900.

Given,

Side of the square park = 60 m

Length of the rectangular park = 90 m

Area of the square park = (side)²

= 60 m x 60 m

= 3600 m²

As the area of the rectangular park = area of the square park

⇒ area of the rectangular park = 3600 m²

Let the breadth of the rectangular park be $b$ m.

Area of the rectangular park = length x breadth

⇒ 90 x $b$ = 3600

⇒ $b = \frac{3600}{90}$ = 40

Hence, the breadth of the rectangular park is 40 m.

Given,

Length of the rectangle = 40 cm

Breadth of the rectangle = 22 cm

Length of the wire = perimeter of the rectangle

= 2 (length + breadth)

= 2 (40 + 22) cm

= 2 x 62 cm

= 124 cm

As the wire is rebent into the shape of a square,

perimeter of the square = length of the wire = 124 cm

As we know,

Perimeter of a square = 4 x side

⇒ 4 x side = 124

⇒ side = $\frac{124}{4}$ cm = 31 cm

Area of the rectangle = length x breadth = 40 cm x 22 cm = 880 cm²

Area of the square = (side)² = 31 cm x 31 cm = 961 cm²

∴ Area of the square - area of the rectangle = 961 cm² - 880 cm² = 81 cm²

Hence, each side of the square is 31 cm and the square encloses more area by 81 cm².

Given,

Length of the wall = 4.5 m

Height of the wall = 3.6 m

Breadth of the door = 1 m

Height of the door = 2 m

Area of the wall (including the door) = length x height = 4.5 m x 3.6 m = 16.2 m²

Area of the door = breadth x height = 1 m x 2 m = 2 m²

∴ Area of the wall to be white washed = 16.2 m² - 2 m² = 14.2 m²

Cost of white washing the wall = Area x rate of white washing

= 14.2 x ₹ 400

= ₹ 5,680

Hence, the cost of white washing the wall is ₹ 5,680.

Given,

Length of the park = 45 m

Breadth of the park = 30 m

Width of the path = 2.5 m

In the adjoining figure, the shaded portion shows the path.

Area of the park = length x breadth = 45 m x 30 m = 1350 m²

Length of the outer rectangle = 45 m + 2.5 m + 2.5 m = 50 m

Breadth of the outer rectangle = 30 m + 2.5 m + 2.5 m = 35 m

∴ Total area of the park and the path = length x breadth of the outer rectangle = 50 m x 35 m = 1750 m²

∴ Area of the path = total area of the park and the path - area of the park = 1750 m² - 1350 m² = 400 m²

Hence, the area of the path is 400 m².

Given,

Length of the carpet = 5 m

Breadth of the carpet = 2 m

Width of the red border = 25 cm = $\frac{25}{100}$ m = 0.25 m

Area of the carpet = length x breadth = 5 m x 2 m = 10 m²

Length of the blue (inner) portion = 5 m - 0.25 m - 0.25 m = 4.5 m

Breadth of the blue (inner) portion = 2 m - 0.25 m - 0.25 m = 1.5 m

∴ Area of the blue portion = length x breadth = 4.5 m x 1.5 m = 6.75 m²

Area of the red portion = Area of the carpet - area of the blue portion

= 10 m² - 6.75 m²

= 3.25 m²

∴ Ratio of the areas of red portion to blue portion = 3.25 : 6.75

= 325 : 675

= 13 : 27

Hence, the area of the blue portion is 6.75 m² and the ratio of the areas of red portion to blue portion is 13 : 27.

Given,

Length of the room = 5.5 m

Breadth of the room = 4 m

Width of the verandah = 2.25 m

(i) In the adjoining figure, the shaded portion shows the verandah.

Area of the room = length x breadth = 5.5 m x 4 m = 22 m²

Length of the outer rectangle = 5.5 m + 2.25 m + 2.25 m = 10 m

Breadth of the outer rectangle = 4 m + 2.25 m + 2.25 m = 8.5 m

∴ Total area of the room and the verandah = length x breadth of the outer rectangle = 10 m x 8.5 m = 85 m²

∴ Area of the verandah = total area of the room and the verandah - area of the room = 85 m² - 22 m² = 63 m²

Hence, the area of the verandah is 63 m².

(ii) Cost of cementing the floor of the verandah = Area x rate of cementing

= 63 x ₹ 200

= ₹ 12,600

Hence, the cost of cementing the floor of the verandah is ₹ 12,600.

Given,

Length of the park = 70 m

Breadth of the park = 45 m

Width of each road = 5 m

The area covered by the roads is shown shaded in the adjoining diagram.

Area of the shaded portion parallel to the length = length of the park x width of the road = 70 m x 5 m = 350 m²

Area of the shaded portion parallel to the breadth = breadth of the park x width of the road = 45 m x 5 m = 225 m²

Area of the dark shaded portion (common to both roads) = width of one road x width of the other road = 5 m x 5 m = 25 m²

∴ The area covered by the cross roads = area of shaded portion

= 350 m² + 225 m² - 25 m²

= 550 m²

Note that we subtract the area of the dark shaded portion because it has occurred twice – firstly when considering area parallel to length and secondly when considering area parallel to breadth.

Cost of constructing the roads = Area x rate of constructing

= 550 x ₹ 105

= ₹ 57,750

Hence, the area of the roads is 550 m² and the cost of constructing the roads is ₹ 57,750.

Given,

Length of the room = 10 m

Breadth of the room = 7.5 m

Width of the carpet = 1.25 m

Rate of the carpet = ₹ 250 per metre

Area of the floor of the room = length x breadth = 10 m x 7.5 m = 75 m²

Area of the carpet required = area of the floor = 75 m²

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

= $\frac{75}{1.25}$ m

= 60 m

Cost of the carpet = Length of the carpet x rate

= 60 x ₹ 250

= ₹ 15,000

Hence, the cost of covering the floor with carpet is ₹ 15,000.

Given,

Length of the room = 6.5 m

Breadth of the room = 5 m

Side of each square tile = 25 cm

Rate = ₹ 94 per tile

Area of the floor of the room = length x breadth = 6.5 m x 5 m = 32.5 m²

= 32.5 x 10000 cm² = 325000 cm² [1 m² = 10000 cm²]

Area of one tile = (side)² = 25 cm x 25 cm = 625 cm²

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

= $\frac{325000}{625}$

= 520

Cost of flooring the room = Number of tiles x rate per tile

= 520 x ₹ 94

= ₹ 48,880

Hence, the cost of flooring the room is ₹ 48,880.

Given,

Side of the square floor = 4.8 m = 4.8 x 100 cm = 480 cm

Perimeter of each square tile = 1.2 m = 1.2 x 100 cm = 120 cm

Cost of each tile = ₹ 270

Side of each tile = $\frac{\text{Perimeter}}{4} = \frac{120}{4}$ cm = 30 cm

Area of the floor = (side)² = 480 cm x 480 cm = 230400 cm²

Area of one tile = (side)² = 30 cm x 30 cm = 900 cm²

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

= $\frac{230400}{900}$

= 256

Cost of covering the floor = Number of tiles x cost of each tile

= 256 x ₹ 270

= ₹ 69,120

Hence, the cost of covering the floor is ₹ 69,120.

Given,

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

Total cost of fencing = ₹ 46,800

Rate of fencing = ₹ 180 per metre

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

= $\frac{46800}{180}$ m

= 260 m

∴ Perimeter of the plot = 260 m

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

As we know,

$2(l + b)$ = perimeter

$$\begin{aligned}\Rightarrow 2(l + 50) = 260\\[1em] \Rightarrow l + 50 = \frac{260}{2} = 130\\[1em] \Rightarrow l = 130 - 50 = 80\end{aligned}$$

Hence, the length of the plot is 80 m.

(ii) Area of the plot = length x breadth

= 80 m x 50 m

= 4000 m²

Cost of leveling the plot = Area x rate of leveling

= 4000 x ₹ 7.60

= ₹ 30,400

Hence, the cost of leveling the plot is ₹ 30,400.

Given,

Area of the plot = 460 m²

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

Length of the plot = $b$ + 15% of $b$

$$\begin{aligned}= b + \frac{15}{100} \times b\\[1em] = \frac{115b}{100}\\[1em] = \frac{23b}{20} \text{ m}\end{aligned}$$

As we know,

Area of a rectangle = length x breadth

$$\begin{aligned}\Rightarrow \frac{23b}{20} \times b = 460\\[1em] \Rightarrow b^2 = 460 \times \frac{20}{23}\\[1em] \Rightarrow b^2 = 400\\[1em] \Rightarrow b = \sqrt{400} = 20\end{aligned}$$

∴ Breadth of the plot = 20 m

And, length of the plot = $\frac{23 \times 20}{20}$ m = 23 m

Perimeter of the plot = 2 (length + breadth)

= 2 (23 + 20) m

= 2 x 43 m

= 86 m

Hence, the perimeter of the plot is 86 m.

From the figure,

Base = 8 cm and height = 4.5 cm

Area of the parallelogram = base x height

= 8 cm x 4.5 cm

= 36 cm²

Hence, the area of the parallelogram is 36 cm².

From the figure,

Base = 2 cm and height = 4.4 cm

Area of the parallelogram = base x height

= 2 cm x 4.4 cm

= 8.8 cm²

Hence, the area of the parallelogram is 8.8 cm².

From the figure,

Base = 2.5 cm and height = 3.5 cm

Area of the parallelogram = base x height

= 2.5 cm x 3.5 cm

= 8.75 cm²

Hence, the area of the parallelogram is 8.75 cm².

From the figure,

Base = 6.4 cm and corresponding height = 6 cm

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

= $\frac{1}{2}$ x 6.4 cm x 6 cm

= 19.2 cm²

Hence, the area of the triangle is 19.2 cm².

From the figure,

Base = 5 cm and height = 6 cm

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

= $\frac{1}{2}$ x 5 cm x 6 cm

= 15 cm²

Hence, the area of the triangle is 15 cm².

From the figure,

Base = 4.5 cm and height = 6 cm

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

= $\frac{1}{2}$ x 4.5 cm x 6 cm

= 13.5 cm²

Hence, the area of the triangle is 13.5 cm².

As we know,

Area of a parallelogram = base x height

(i) Area of the parallelogram = 7.5 cm x 5.6 cm

= 42 cm²

Hence, the area of the parallelogram is 42 cm².

(ii) Let the height be $h$ cm.

$$\begin{aligned}\Rightarrow 20 \times h = 246\\[1em] \Rightarrow h = \frac{246}{20} = 12.3\end{aligned}$$

Hence, the height of the parallelogram is 12.3 cm.

(iii) Let the base be $b$ cm.

$$\begin{aligned}\Rightarrow b \times 15 = 154.5\\[1em] \Rightarrow b = \frac{154.5}{15} = 10.3\end{aligned}$$

Hence, the base of the parallelogram is 10.3 cm.

(iv) Let the base be $b$ cm.

$$\begin{aligned}\Rightarrow b \times 8.4 = 48.72\\[1em] \Rightarrow b = \frac{48.72}{8.4} = 5.8\end{aligned}$$

Hence, the base of the parallelogram is 5.8 cm.

As we know,

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

(i) Area of the triangle = $\frac{1}{2}$ x 23.2 m x 16.7 m

= 11.6 m x 16.7 m

= 193.72 m²

Hence, the area of the triangle is 193.72 m².

(ii) Let the height be $h$ cm.

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 15 \times h = 87\\[1em] \Rightarrow h = \frac{87 \times 2}{15} = 11.6\end{aligned}$$

Hence, the height of the triangle is 11.6 cm.

(iii) Let the height be $h$ mm.

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 22 \times h = 170.5\\[1em] \Rightarrow 11 \times h = 170.5\\[1em] \Rightarrow h = \frac{170.5}{11} = 15.5\end{aligned}$$

Hence, the height of the triangle is 15.5 mm.

(iv) Let the base be $b$ cm.

$$\begin{aligned}\Rightarrow \frac{1}{2} \times b \times 31.4 = 1256\\[1em] \Rightarrow b = \frac{1256 \times 2}{31.4} = 80\end{aligned}$$

Hence, the base of the triangle is 80 cm.

Given,

AB = 6 cm and AD = 4 cm

Height corresponding to the base AB = 3 cm

(i) Taking AB as the base of the parallelogram, its height is 3 cm.

Area of parallelogram ABCD = base x height

= 6 cm x 3 cm

= 18 cm²

Hence, the area of parallelogram ABCD is 18 cm².

(ii) Let the height corresponding to the base AD be $h$ cm.

Area of parallelogram ABCD = base x height = AD x $h$

$$\begin{aligned}\Rightarrow 4 \times h = 18\\[1em] \Rightarrow h = \frac{18}{4} = 4.5\end{aligned}$$

Hence, the height corresponding to the base AD is 4.5 cm.

Given,

AB = AC = 7.5 cm, BC = 9 cm and AD = 6 cm

(i) Area of ΔABC = $\frac{1}{2}$ x base x height

$$\begin{aligned}= \frac{1}{2} \times BC \times AD\\[1em] = \frac{1}{2} \times 9 \text{ cm} \times 6 \text{ cm}\\[1em] = 27 \text{ cm}^2\end{aligned}$$

Hence, the area of ΔABC is 27 cm².

(ii) In ΔABC, AB = base = 7.5 cm and height = CE

Area of ΔABC = $\frac{1}{2}$ x base x height = $\frac{1}{2}$ x AB x CE

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 7.5 \times CE = 27\\[1em] \Rightarrow CE = \frac{27 \times 2}{7.5}\\[1em] \Rightarrow CE = \frac{54}{7.5} = 7.2\end{aligned}$$

Hence, the height CE from C to AB is 7.2 cm.

Let ABC be a right-angled triangle at B i.e. ∠B = 90° and its base BC = 8 cm and hypotenuse AC = 17 cm.

From ΔABC, by Pythagoras theorem, we get

(Hypotenuse)² = (Base)² + (Perpendicular)²

$$\begin{aligned}\Rightarrow AC^2 = BC^2 + AB^2\\[1em] \Rightarrow 17^2 = 8^2 + AB^2\\[1em] \Rightarrow AB^2 = 17^2 - 8^2\\[1em] \Rightarrow AB^2 = 289 - 64 = 225\\[1em] \Rightarrow AB = \sqrt{225} \text{ cm} = 15 \text{ cm}\end{aligned}$$

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

$= \frac{1}{2}$ × 8 cm × 15 cm

= 60 cm²

Hence, the area of the triangle is 60 cm².

Given,

AB = 6 cm and BC = 8 cm, ∠B = 90°

As ΔABC is right angled at B, by Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

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

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

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

In ΔABC, AC = base = 10 cm and height = BN

Area of ΔABC = $\frac{1}{2}$ x base x height = $\frac{1}{2}$ x AC x BN

⇒ $\frac{1}{2}$ x 10 x BN = 24

⇒ 5 x BN = 24

⇒ BN = $\frac{24}{5}$ = 4.8

Hence, the length of perpendicular BN on the side AC is 4.8 cm.

Given,

AB = 10 cm and EF = 16 cm

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

$$\begin{aligned}= \frac{1}{2} \times AB \times EF\\[1em] = \frac{1}{2} \times 10 \text{ cm} \times 16 \text{ cm}\\[1em] = 80 \text{ cm}^2\end{aligned}$$

∴ Area of parallelogram ABCD = area of ΔABE = 80 cm²

Let the altitude of the parallelogram to the base AB be $h$ cm.

Area of parallelogram ABCD = base x height

⇒ 10 x $h$ = 80

⇒ $h = \frac{80}{10}$ = 8

∴ Altitude of the parallelogram to the base AB = 8 cm

Since ABCD is a parallelogram, DC = AB = 10 cm

As M is the mid-point of DC,

DM = $\frac{1}{2}$ x DC = $\frac{1}{2}$ x 10 cm = 5 cm

In ΔAMD, taking DM as base, the corresponding height is the distance between AB and DC, i.e., 8 cm.

∴ Area of ΔAMD = $\frac{1}{2}$ x DM x height

= $\frac{1}{2}$ x 5 cm x 8 cm

= 20 cm²

Hence, the length of the altitude of the parallelogram is 8 cm and the area of ΔAMD is 20 cm².

Given,

AB = 18 cm and BC = 10 cm

In ΔBEC, ∠E = 90° and EC = 8 cm

As ΔBEC is right angled at E, by Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

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

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

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

Area of rectangle ABCD = length x breadth

= 18 cm x 10 cm

= 180 cm²

∴ Area of the shaded region = area of rectangle ABCD - area of ΔBEC

= 180 cm² - 24 cm²

= 156 cm²

Hence, the area of the shaded region is 156 cm².

As ABCD is a rectangle, AB = DC = 18 cm and AD = BC = 10 cm.

From figure,

DE = 10 cm, DF = 6 cm

EC = DC - DE = 18 cm - 10 cm = 8 cm

As we know,

Area of a rectangle = length x breadth

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

∴ Area of the shaded region

= area of rect. ABCD - area of ΔFDE - area of ΔECB

= (18 x 10 - $\frac{1}{2}$ x 10 x 6 - $\frac{1}{2}$ x 8 x 10) cm²

= (180 - 30 - 40) cm²

= 110 cm²

Hence, the area of the shaded region is 110 cm².

From figure,

AB = AE + EB = 10 cm + 10 cm = 20 cm

BC = 20 cm, so ABCD is a square of side 20 cm.

AF = 10 cm

FD = AD - AF = 20 cm - 10 cm = 10 cm

As we know,

Area of a square = (side)²

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

∴ Area of the shaded region

= area of square ABCD - area of ΔFAE - area of ΔEBC - area of ΔFDC

= (20 x 20 - $\frac{1}{2}$ x 10 x 10 - $\frac{1}{2}$ x 10 x 20 - $\frac{1}{2}$ x 10 x 20) cm²

= (400 - 50 - 100 - 100) cm²

= 150 cm²

Hence, the area of the shaded region is 150 cm².

As we know,

Circumference of a circle = $2\pi r$

Given,

Radius ($r$) = 7 cm

Circumference of the circle = 2 x $\frac{22}{7}$ x 7 cm

= 2 x 22 cm

= 44 cm

Hence, the circumference of the circle is 44 cm.

As we know,

Circumference of a circle = $2\pi r$

Given,

Radius ($r$) = 21 cm

Circumference of the circle = 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.

As we know,

Circumference of a circle = $2\pi r$

Given,

Radius ($r$) = 28 mm

Circumference of the circle = 2 x $\frac{22}{7}$ x 28 mm

= 2 x 22 x 4 mm

= 176 mm

Hence, the circumference of the circle is 176 mm.

As we know,

Circumference of a circle = $2\pi r$

Given,

Radius ($r$) = 3.5 cm

Circumference of the circle = 2 x $\frac{22}{7}$ x 3.5 cm

= 2 x 22 x 0.5 cm

= 22 cm

Hence, the circumference of the circle is 22 cm.

As we know,

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

Given,

Radius ($r$) = 14 mm

Area of the circle = $\frac{22}{7}$ x 14 x 14 mm²

= 22 x 2 x 14 mm²

= 616 mm²

Hence, the area of the circle is 616 mm².

As we know,

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

Given,

Diameter = 49 m

∴ Radius ($r$) = $\frac{49}{2}$ m

$$\begin{aligned}\text{ Area of the circle }= \frac{22}{7} \times \frac{49}{2} \times \frac{49}{2} \text{ m}^2\\[1em] = \frac{11 \times 7 \times 49}{2} \text{ m}^2\\[1em] = \frac{3773}{2} \text{ m}^2\\[1em] = 1886.5 \text{ m}^2\end{aligned}$$

Hence, the area of the circle is 1886.5 m².

As we know,

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

Given,

Diameter = 9.8 m

∴ Radius ($r$) = $\frac{9.8}{2}$ m = 4.9 m

Area of the circle = $\frac{22}{7}$ x 4.9 x 4.9 m²

= 22 x 0.7 x 4.9 m²

= 75.46 m²

Hence, the area of the circle is 75.46 m².

As we know,

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

Given,

Radius ($r$) = 5 cm

$$\begin{aligned}\text{ Area of the circle }= \frac{22}{7} \times 5 \times 5 \text{ cm}^2\\[1em] = \frac{550}{7} \text{ cm}^2\\[1em] = 78\frac{4}{7} \text{ cm}^2\end{aligned}$$

Hence, the area of the circle is $\bm{78\frac{4}{7}}$ cm².

Given,

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

Circumference of the circle = $2\pi r$

= 2 x 3.14 x 20 cm

= 125.6 cm

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

= 3.14 x 20 x 20 cm²

= 1256 cm²

Hence, the circumference and the area of the circle are 125.6 cm and 1256 cm² respectively.

Given,

Length of the minute hand = 1.4 m

In 1 hour, the minute hand completes one full round, so its tip moves along a circle of radius equal to the length of the minute hand.

∴ Radius of the circle ($r$) = 1.4 m

Distance moved by the tip in 1 hour = circumference of the circle

= $2\pi r$

= 2 x $\frac{22}{7}$ x 1.4 m

= 2 x 22 x 0.2 m

= 8.8 m

Hence, the tip of the minute hand moves 8.8 m in 1 hour.

Given,

Diameter of the garden = 21 m

∴ Radius of the garden ($r$) = $\frac{21}{2}$ m

Rate of the rope = ₹ 40 per metre

Circumference of the garden = $2\pi r$

= 2 x $\frac{22}{7} \times \frac{21}{2}$ m

= 22 x 3 m

= 66 m

Length of the rope needed for 2 rounds = 2 x 66 m = 132 m

Cost of the rope = Length of the rope x rate

= 132 x ₹ 40

= ₹ 5,280

Hence, the length of the rope needed is 132 m and its cost is ₹ 5,280.

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

∴ Diameter = $2r$ cm

Given,

Circumference of a circle exceeds its diameter by 30 cm.

As we know,

Circumference of a circle = $2\pi r$

$$\begin{aligned}\therefore 2\pi r - 2r = 30\\[1em] \Rightarrow 2r(\pi - 1) = 30\\[1em] \Rightarrow 2r\Big(\frac{22}{7} - 1\Big) = 30\\[1em] \Rightarrow 2r\Big(\frac{22 - 7}{7}\Big) = 30\\[1em] \Rightarrow 2r \times \frac{15}{7} = 30\\[1em] \Rightarrow \frac{30r}{7} = 30\\[1em] \Rightarrow r = 30 \times \frac{7}{30}\\[1em] \Rightarrow r = 7\end{aligned}$$

Hence, the radius of the circle is 7 cm.

Given,

Circumference of the circle = 44 cm

Let the diameter of the circle be $d$ cm.

As we know,

Circumference of a circle = $2\pi r = \pi d$

$$\begin{aligned}\Rightarrow \frac{22}{7} \times d = 44\\[1em] \Rightarrow d = 44 \times \frac{7}{22}\\[1em] \Rightarrow d = 14\end{aligned}$$

Hence, the length of the diameter of the circle is 14 cm.

Given,

Circumference of the circle = 31.4 cm and π = 3.14

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

As we know,

Circumference of a circle = $2\pi r$

$$\begin{aligned}\Rightarrow 2 \times 3.14 \times r = 31.4\\[1em] \Rightarrow 6.28 \times r = 31.4\\[1em] \Rightarrow r = \frac{31.4}{6.28}\\[1em] \Rightarrow r = 5\end{aligned}$$

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

= 3.14 x 5 x 5 cm²

= 78.5 cm²

Hence, the radius of the circle is 5 cm and its area is 78.5 cm².

Given,

Area of the circle = 144π cm²

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

As we know,

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

$$\begin{aligned}\Rightarrow \pi r^2 = 144\pi\\[1em] \Rightarrow r^2 = 144\\[1em] \Rightarrow r = \sqrt{144} = 12\end{aligned}$$

∴ Radius of the circle = 12 cm

Circumference of the circle = $2\pi r$

= 2 x π x 12 cm

= 24π cm

Hence, the radius of the circle is 12 cm and its circumference is 24π cm.

Given,

Diameter of the wheel = 56 cm

∴ Radius of the wheel ($r$) = $\frac{56}{2}$ cm = 28 cm

Circumference of the wheel = $2\pi r$

$$\begin{aligned}= 2 \times \frac{22}{7} \times 28 \text{ cm}\\[1em] = 2 \times 22 \times 4 \text{ cm}\\[1em] = 176 \text{ cm}\end{aligned}$$

∴ The distance covered by the wheel in one revolution = 176 cm

Since the distance to be covered by the wheel = 88 km

= 88 x 1000 x 100 cm

= 88,00,000 cm

∴ The number of times the wheel would rotate = $\frac{\text{distance to be covered}}{\text{distance covered in one revolution}}$

= $\frac{8800000}{176}$

= 50000

Hence, the wheel will rotate 50000 times.

Given,

Side of the square cardboard = 21 cm

The diameter of the circle of maximum area that can be cut out of the square cardboard is equal to the side of the square.

∴ Diameter of the circle = 21 cm

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

Area of the square cardboard = (side)²

= 21 cm x 21 cm

= 441 cm²

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

$$\begin{aligned}= \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \text{ cm}^2\\[1em] = \frac{693}{2} \text{ cm}^2\\[1em] = 346.5 \text{ cm}^2\end{aligned}$$

∴ Area of the cardboard left = Area of the square cardboard - area of the circle

= 441 cm² - 346.5 cm²

= 94.5 cm²

Hence, the area of the cardboard left is 94.5 cm².

Given,

Side of the equilateral triangle = 4.4 cm

Length of the wire = perimeter of the equilateral triangle

= 3 x side

= 3 x 4.4 cm

= 13.2 cm

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

As we know,

Circumference of a circle = $2\pi r$

As the wire is rebent into the shape of a circle,

circumference of the circle = length of the wire

$$\begin{aligned}\Rightarrow 2\pi r = 13.2\\[1em] \Rightarrow 2 \times \frac{22}{7} \times r = 13.2\\[1em] \Rightarrow r = 13.2 \times \frac{7}{44}\\[1em] \Rightarrow r = 2.1\end{aligned}$$

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

= $\frac{22}{7}$ x 2.1 x 2.1 cm²

= 22 x 0.3 x 2.1 cm²

= 13.86 cm²

Hence, the radius of the circle is 2.1 cm and its area is 13.86 cm².

Given,

Side of the square = 27.5 cm

Length of the wire = perimeter of the square

= 4 x side

= 4 x 27.5 cm

= 110 cm

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

As we know,

Circumference of a circle = $2\pi r$

As the wire is bent into the shape of a circle,

circumference of the circle = length of the wire

$$\begin{aligned}\Rightarrow 2\pi r = 110\\[1em] \Rightarrow 2 \times \frac{22}{7} \times r = 110\\[1em] \Rightarrow r = 110 \times \frac{7}{44}\\[1em] \Rightarrow r = 17.5\end{aligned}$$

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

= $\frac{22}{7}$ x 17.5 x 17.5 cm²

= 22 x 2.5 x 17.5 cm²

= 962.5 cm²

Hence, the area of the circle is 962.5 cm².

Given,

Length of the rectangle = 18.7 cm

Breadth of the rectangle = 14.3 cm

Length of the wire = perimeter of the rectangle

= 2 (length + breadth)

= 2 (18.7 + 14.3) cm

= 2 x 33 cm

= 66 cm

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

As we know,

Circumference of a circle = $2\pi r$

As the wire is bent into the shape of a circle,

circumference of the circle = length of the wire

$$\begin{aligned}\Rightarrow 2\pi r = 66\\[1em] \Rightarrow 2 \times \frac{22}{7} \times r = 66\\[1em] \Rightarrow r = 66 \times \frac{7}{44}\\[1em] \Rightarrow r = 10.5\end{aligned}$$

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

= $\frac{22}{7}$ x 10.5 x 10.5 cm²

= 22 x 1.5 x 10.5 cm²

= 346.5 cm²

Hence, the radius of the circle is 10.5 cm and its area is 346.5 cm².

Given,

Diameter of the circular park = 84 m

∴ Radius of the park ($r$) = $\frac{84}{2}$ m = 42 m

Width of the road = 3.5 m

Let $R$ be the radius of the outer circle.

Then $R$ = 42 m + 3.5 m = 45.5 m = $\frac{91}{2}$ m

Area of the road = $\pi R^2 - \pi r^2 = \pi(R^2 - r^2)$

$$\begin{aligned}= \frac{22}{7}\Big(\frac{91}{2} \times \frac{91}{2} - 42 \times 42\Big) \text{ m}^2\\[1em] = \frac{22}{7}\Big(\frac{8281}{4} - 1764\Big) \text{ m}^2\\[1em] = \frac{22}{7}\Big(\frac{8281 - 7056}{4}\Big) \text{ m}^2\\[1em] = \frac{22}{7} \times \frac{1225}{4} \text{ m}^2\\[1em] = \frac{22 \times 175}{4} \text{ m}^2\\[1em] = 962.5 \text{ m}^2\end{aligned}$$

Cost of constructing the road = Area x rate

= 962.5 x ₹ 240

= ₹ 2,31,000

Hence, the cost of constructing the road is ₹ 2,31,000.

Given,

Width of the circular path = 2 m

Outer circumference of the circular path = 44 m

Let the radius of the outer circle be $R$ m.

Outer circumference = $2\pi R$

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

∴ Radius of the inner circle ($r$) = Radius of the outer circle - width of the path

= 7 m - 2 m

= 5 m

Inner circumference of the circular path = $2\pi r$

$$\begin{aligned}= 2 \times \frac{22}{7} \times 5 \text{ m}\\[1em] = \frac{220}{7} \text{ m}\\[1em] = 31\frac{3}{7} \text{ m}\end{aligned}$$

Area of the path = $\pi R^2 - \pi r^2 = \pi(R^2 - r^2)$

$$\begin{aligned}= \frac{22}{7}(7 \times 7 - 5 \times 5) \text{ m}^2\\[1em] = \frac{22}{7}(49 - 25) \text{ m}^2\\[1em] = \frac{22}{7} \times 24 \text{ m}^2\\[1em] = \frac{528}{7} \text{ m}^2\\[1em] = 75\frac{3}{7} \text{ m}^2\end{aligned}$$

Hence, the inner circumference of the circular path is $\bm{31\frac{3}{7}}$ m and the area of the path is $\bm{75\frac{3}{7}}$ m².

Given,

Area enclosed between the concentric circles = 770 cm²

Radius of the outer circle ($R$) = 21 cm

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

Area enclosed between the circles = $\pi R^2 - \pi r^2 = \pi(R^2 - r^2)$

$$\begin{aligned}\Rightarrow \frac{22}{7}(21^2 - r^2) = 770\\[1em] \Rightarrow 441 - r^2 = 770 \times \frac{7}{22}\\[1em] \Rightarrow 441 - r^2 = 245\\[1em] \Rightarrow r^2 = 441 - 245 = 196\\[1em] \Rightarrow r = \sqrt{196} = 14\end{aligned}$$

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

Given,

Radius of the circular card sheet ($R$) = 14 cm

Radius of each small circle ($r$) = 3.5 cm

Length of the rectangle = 3 cm and breadth = 1 cm

Area of the circular card sheet = $\pi R^2$

= $\frac{22}{7}$ x 14 x 14 cm²

= 616 cm²

Area of two small circles = $2 \times \pi r^2$

= 2 x $\frac{22}{7}$ x 3.5 x 3.5 cm²

= 2 x 38.5 cm²

= 77 cm²

Area of the rectangle = length x breadth = 3 cm x 1 cm = 3 cm²

∴ Area of the remaining sheet = Area of the card sheet - area of two circles - area of the rectangle

= 616 cm² - 77 cm² - 3 cm²

= 536 cm²

Hence, the area of the remaining sheet is 536 cm².

From the figure, the rectangle is 7 cm wide and 10 cm long, and the unshaded semicircle is drawn on the top side of 7 cm as diameter.

∴ Radius of the semicircle ($r$) = $\frac{7}{2}$ cm = 3.5 cm

Length of the semicircular arc = $\frac{1}{2} \times 2\pi r = \pi r$

$= \frac{22}{7}$ × 3.5 cm

= 11 cm

The boundary of the shaded region consists of two sides of 10 cm each, the bottom side of 7 cm and the semicircular arc.

∴ Length of the boundary = (10 + 10 + 7 + 11) cm = 38 cm

Area of the rectangle = length x breadth = 10 cm x 7 cm = 70 cm²

Area of the semicircle = $\frac{1}{2} \times \pi r^2$

= $\frac{1}{2} \times \frac{22}{7}$ x 3.5 x 3.5 cm²

= 19.25 cm²

∴ Area of the shaded region = Area of the rectangle - area of the semicircle

= 70 cm² - 19.25 cm²

= 50.75 cm²

Hence, the length of the boundary is 38 cm and the area of the shaded region is 50.75 cm².

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

The diameter of each semicircle = side of the square = $a$ cm

∴ Radius of each semicircle = $\frac{a}{2}$ cm

From the figure,

radius of top semicircle + side of the square + radius of bottom semicircle = 28

$$\begin{aligned}\Rightarrow \frac{a}{2} + a + \frac{a}{2} = 28\\[1em] \Rightarrow 2a = 28\\[1em] \Rightarrow a = 14\end{aligned}$$

∴ Side of the square = 14 cm and radius of each semicircle ($r$) = 7 cm

The boundary of the figure is made up of 4 semicircles of diameter 14 cm.

Length of each semicircular arc = $\frac{1}{2} \times \pi d$

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

= 22 cm

∴ Length of the boundary = 4 x 22 cm = 88 cm

Area of the square = (side)² = 14 cm x 14 cm = 196 cm²

Area of 4 semicircles = $4 \times \frac{1}{2} \times \pi r^2$

= 2 x $\frac{22}{7}$ x 7 x 7 cm²

= 308 cm²

∴ Area of the shaded region = Area of the square + area of 4 semicircles

= 196 cm² + 308 cm²

= 504 cm²

Hence, the length of the boundary is 88 cm and the area of the shaded region is 504 cm².

From the figure, the shaded region consists of a rectangle 10.5 cm long and 4 cm wide and a semicircle drawn on the top side.

Diameter of the semicircle = 10.5 cm - 3.5 cm = 7 cm

∴ Radius of the semicircle ($r$) = $\frac{7}{2}$ cm = 3.5 cm

Length of the semicircular arc = $\pi r$

$= \frac{22}{7}$ x 3.5 cm

= 11 cm

∴ Perimeter of the shaded region = (4 + 3.5 + 11 + 4 + 10.5) cm

= 33 cm

Area of the rectangle = length x breadth = 10.5 cm x 4 cm = 42 cm²

Area of the semicircle = $\frac{1}{2} \times \pi r^2$

= $\frac{1}{2} \times \frac{22}{7}$ x 3.5 x 3.5 cm²

= 19.25 cm²

∴ Area of the shaded region = Area of the rectangle + area of the semicircle

= 42 cm² + 19.25 cm²

= 61.25 cm²

Hence, the perimeter of the shaded region is 33 cm and its area is 61.25 cm².

From the figure, the shaded region consists of a triangle and a semicircle drawn on its base as diameter.

Let ABC be the triangle with AB = 10 cm and AD = 8 cm perpendicular to BC, where D is the centre of the semicircle.

∴ BD = DC (radii of the semicircle)

In right angled triangle ADB, by Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

⇒ AB² = AD² + BD²

⇒ 10² = 8² + BD²

⇒ BD² = 100 - 64 = 36

⇒ BD = $\sqrt{36}$ cm = 6 cm

∴ DC = BD = 6 cm

∴ BC = BD + DC = 6 cm + 6 cm = 12 cm

∴ Diameter of the semicircle = 12 cm and its radius ($r$) = 6 cm

In right angled triangle ADC, by Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

⇒ AC² = AD² + DC²

⇒ AC² = 8² + 6² = 64 + 36 = 100

⇒ AC = $\sqrt{100}$ cm = 10 cm

Length of the semicircular arc = $\pi r$

$$\begin{aligned}= \frac{22}{7} \times 6 \text{ cm}\\[1em] = \frac{132}{7} \text{ cm}\end{aligned}$$

∴ Perimeter of the shaded region = AB + AC + semicircular arc

$$\begin{aligned}= \Big(10 + 10 + \frac{132}{7}\Big) \text{ cm}\\[1em] = \frac{272}{7} \text{ cm}\\[1em] = 38\frac{6}{7} \text{ cm}\end{aligned}$$

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

$$\begin{aligned}= \frac{1}{2} \times BC \times AD\\[1em] = \frac{1}{2} \times 12 \text{ cm} \times 8 \text{ cm}\\[1em] = 48 \text{ cm}^2\end{aligned}$$

Area of the semicircle = $\frac{1}{2} \times \pi r^2$

$$\begin{aligned}= \frac{1}{2} \times \frac{22}{7} \times 6 \times 6 \text{ cm}^2\\[1em] = \frac{396}{7} \text{ cm}^2\end{aligned}$$

∴ Area of the shaded region = Area of ΔABC + area of the semicircle

$$\begin{aligned}= \Big(48 + \frac{396}{7}\Big) \text{ cm}^2\\[1em] = \frac{732}{7} \text{ cm}^2\\[1em] = 104\frac{4}{7} \text{ cm}^2\end{aligned}$$

Hence, the perimeter of the shaded region is $\bm{38\frac{6}{7}}$ cm and its area is $\bm{104\frac{4}{7}}$ cm².

Given,

Area of the square = 81 cm²

As we know,

Area of a square = (side)²

⇒ (side)² = 81

⇒ side = $\sqrt{81}$ cm = 9 cm

Length of the wire = perimeter of the square

= 4 x side

= 4 x 9 cm

= 36 cm

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

Perimeter of the semicircle (including its diameter) = $\pi r + 2r$

As the same wire is bent into the semicircle,

$$\begin{aligned}\Rightarrow \pi r + 2r = 36\\[1em] \Rightarrow r\Big(\frac{22}{7} + 2\Big) = 36\\[1em] \Rightarrow r \times \frac{36}{7} = 36\\[1em] \Rightarrow r = 36 \times \frac{7}{36}\\[1em] \Rightarrow r = 7\end{aligned}$$

Area of the semicircle = $\frac{1}{2} \times \pi r^2$

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

= 77 cm²

Hence, the area of the shape so formed is 77 cm².

(i) All the sides of a regular polygon are equal.

Hence, perimeter of a regular polygon = number of sides × length of a side.

(ii) Area is measured by comparing it with a unit square.

Hence, the unit of measurement of area is square units.

(iii) All the four sides of a rhombus are equal.

Hence, the perimeter of a rhombus is = 4 × length of a side.

(iv) 1 km² = 1000 m x 1000 m = 1000000 m² and 1 hectare = 10000 m²

∴ 1 km² = $\frac{1000000}{10000}$ hectare = 100 hectare

Hence, an area of 1 km² = 100 hectare.

(v) Let the length of the adjacent side be $x$ cm.

Perimeter of the parallelogram = 2 (sum of two adjacent sides) = $2(12 + x)$

$$\begin{aligned}\Rightarrow 2(12 + x) = 40\\[1em] \Rightarrow 12 + x = 20\\[1em] \Rightarrow x = 8\end{aligned}$$

Hence, the length of the adjacent side is 8 cm.

(vi) Polishing is done over the whole surface of the table-top.

Hence, to find the cost of polishing a table-top, we need to find the area of the table-top.

(vii) As we know, $\frac{\text{circumference}}{\text{diameter}}$ = π

Hence, the ratio of circumference to the diameter of a circle is π.

(viii) Let the length of the altitude be $h$ cm.

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

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 20 \times h = 90\\[1em] \Rightarrow 10 \times h = 90\\[1em] \Rightarrow h = 9\end{aligned}$$

Hence, the length of the altitude corresponding to 20 cm long base is 9 cm.

(i) A diagonal of a rectangle divides it into two congruent right angled triangles, so their areas are equal.

Hence, the given statement is true (T).

(ii) A diagonal of a parallelogram divides it into two congruent triangles, so their areas are equal.

Hence, the given statement is true (T).

(iii) Two parallelograms with sides 6 cm, 4 cm and heights (on the 6 cm side) 3 cm and 2 cm have equal perimeters (20 cm each), but their areas are 18 cm² and 12 cm² respectively.

Hence, the given statement is false (F).

(iv) A rectangle 6 cm x 4 cm and a rectangle 12 cm x 2 cm both have area 24 cm², but their perimeters are 20 cm and 28 cm respectively.

Hence, the given statement is false (F).

(v) Radius = $\frac{d}{2}$

∴ Area of the circle = $\pi \times \Big(\frac{d}{2}\Big)^2 = \frac{\pi d^2}{4}$, not $\pi d^2$.

Hence, the given statement is false (F).

(vi) Area of a parallelogram = base x height, not the product of its two adjacent sides.

Hence, the given statement is false (F).

Given,

Perimeter of the square = 24 cm

⇒ Side of the square = $\frac{\text{Perimeter}}{4} = \frac{24}{4}$ cm = 6 cm

∴ Area of the square = (side)²

= (6 cm)²

= 36 cm²

Hence, option 3 is the correct option.

Given,

Area of the parallelogram = 54 cm²

Base = 7.5 cm

Let the corresponding height be $h$ cm.

As we know,

Area of a parallelogram = base x height

⇒ 7.5 x $h$ = 54

⇒ $h = \frac{54}{7.5}$ = 7.2

Hence, option 1 is the correct option.

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

As we know,

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 remains same.

Hence, option 4 is the correct option.

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

As we know,

Area of a parallelogram = base x height

∴ Original area of the parallelogram = $b \times h$

New base = $3b$ and new height = $2h$

∴ New area of the parallelogram = $3b \times 2h = 6bh$

= 6 x original area

Hence, option 3 is the correct option.

Given,

Diameter of the circle ($d$) = 28 cm

Circumference of the circle = $\pi d$

$$\begin{aligned}= \frac{22}{7} \times 28 \text{ cm}\\[1em] = 22 \times 4 \text{ cm}\\[1em] = 88 \text{ cm}\end{aligned}$$

Hence, option 2 is the correct option.

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

$\Rightarrow \frac{\text{Circumference}}{\text{Area}} = \frac{2\pi r}{\pi r^2} = \frac{2}{r}$

∴ Ratio of circumference to the area = $2 : r$

Hence, option 1 is the correct option.

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

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

Given,

Area = circumference (numerically)

$$\begin{aligned}\Rightarrow \pi r^2 = 2\pi r\\[1em] \Rightarrow \pi \times r \times r = 2 \times \pi \times r\end{aligned}$$

Dividing both sides by $\pi r$, we get

⇒ $r$ = 2

∴ Radius of the circle = 2 units

Hence, option 2 is the correct option.

Given,

Diameter of the circle = $d$

∴ Radius of the circle $(r) = \frac{d}{2}$

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

$$\begin{aligned}= \pi \times \Big(\frac{d}{2}\Big)^2\\[1em] = \frac{\pi d^2}{4}\end{aligned}$$

Hence, option 4 is the correct option.

Let the radii of the two circles be $2x$ and $3x$.

As we know,

Circumference of a circle = $2\pi r$

Circumference of the first circle = $2\pi \times 2x = 4\pi x$

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

∴ Ratio of their circumferences = $4\pi x : 6\pi x$

= 2 : 3

Hence, option 1 is the correct option.

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

As we know,

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

Area of the first circle = $\pi \times (3x)^2 = 9\pi x^2$

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

∴ Ratio of their areas = $9\pi x^2 : 25\pi x^2$

= 9 : 25

Hence, option 4 is the correct option.

Given,

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

Perimeter of the semicircle = Length of semicircular arc + 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.

Area of the rectangle = length x breadth = 14 cm x 11 cm = 154 cm²

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

As we know,

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

Given,

Area of the circle = area of the rectangle

$$\begin{aligned}\Rightarrow \pi r^2 = 154\\[1em] \Rightarrow \frac{22}{7} \times r^2 = 154\\[1em] \Rightarrow r^2 = 154 \times \frac{7}{22}\\[1em] \Rightarrow r^2 = 49\\[1em] \Rightarrow r = \sqrt{49} = 7\end{aligned}$$

Hence, option 4 is the correct option.

Length of each wire = 88 cm

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

As we know,

Circumference of a circle = $2\pi r$

Circumference of the circle = length of the wire

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

Area enclosed by the circular wire = $\pi r^2$

= $\frac{22}{7}$ x 14 x 14 cm²

= 616 cm²

Side of the square = $\frac{\text{Perimeter}}{4} = \frac{88}{4}$ cm = 22 cm

Area enclosed by the square = (side)² = 22 cm x 22 cm = 484 cm²

Since 616 cm² > 484 cm², the area enclosed by the circular wire is greater than the square.

So, Statement I is true.

Radius of the circle = $\frac{d}{2}$

∴ Area of the circle = π x $\Big(\frac{d}{2}\Big)^2 = \frac{\pi d^2}{4}$

So, Statement II is true.

∴ Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

Given,

Length : breadth = 2 : 1

∴ Length = $2b$, where $b$ is the breadth.

Perimeter of the rectangle = 2 (length + breadth)

$$\begin{aligned}= 2(2b + b)\\[1em] = 2 \times 3b\\[1em] = 6b\end{aligned}$$

So, Statement I is true.

The perimeter of a closed plane figure is the length of its boundary. So, the perimeter of a rectangle is the sum of all its 4 sides.

So, Statement II is true.

∴ Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

A diagonal of a parallelogram divides it into two congruent triangles, so the two triangles have equal area.

So, Statement I is true.

Area of a parallelogram = base x height, not the product of its diagonals.

So, Statement II is false.

∴ Statement I is true but Statement II is false.

Hence, option 1 is the correct option.

Let the other two sides be $3x$ cm and $4x$ cm.

By Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

$$\begin{aligned}\Rightarrow (3x)^2 + (4x)^2 = 10^2\\[1em] \Rightarrow 9x^2 + 16x^2 = 100\\[1em] \Rightarrow 25x^2 = 100\\[1em] \Rightarrow x^2 = 4\\[1em] \Rightarrow x = 2\end{aligned}$$

∴ The other two sides are 3 x 2 cm = 6 cm and 4 x 2 cm = 8 cm.

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

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

= 24 cm²

The area of the triangle is 24 cm² and not 12 cm².

So, Statement I is false.

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

So, Statement II is true.

∴ Statement I is false but Statement II is true.

Hence, option 2 is the correct option.

Given,

Length of the park = 125 m

Breadth of the park = 65 m

Width of the path = 3 m

In the adjoining figure, the shaded portion shows the path.

Area of the park = length x breadth = 125 m x 65 m = 8125 m²

Length of the outer rectangle = 125 m + 3 m + 3 m = 131 m

Breadth of the outer rectangle = 65 m + 3 m + 3 m = 71 m

∴ Total area of the park and the path = length x breadth of the outer rectangle = 131 m x 71 m = 9301 m²

∴ Area of the path = total area of the park and the path - area of the park = 9301 m² - 8125 m² = 1176 m²

Hence, the area of the path is 1176 m².

From the figure,

Length of the rectangle = 22 m and breadth = 14 m

Width of the shaded strip parallel to the length = 3 m

Width of the shaded strip parallel to the breadth = 2 m

(i) Area of the shaded portion parallel to the length = length of the rectangle x width of the strip = 22 m x 3 m = 66 m²

Area of the shaded portion parallel to the breadth = breadth of the rectangle x width of the strip = 14 m x 2 m = 28 m²

Area of the dark shaded portion (common to both strips) = width of one strip x width of the other strip = 3 m x 2 m = 6 m²

∴ Area of the shaded region = 66 m² + 28 m² - 6 m²

= 88 m²

Note that we subtract the area of the dark shaded portion because it has occurred twice.

Hence, the area of the shaded region is 88 m².

(ii) Area of the rectangle = length x breadth = 22 m x 14 m = 308 m²

∴ Area of the unshaded region = Area of the rectangle - area of the shaded region

= 308 m² - 88 m²

= 220 m²

Hence, the area of the unshaded region is 220 m².

As we know,

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

(i) Given,

Base = 2 m and height = 1.5 m

Area of the triangle = $\frac{1}{2}$ x 2 m x 1.5 m

= 1.5 m²

Hence, the area of the triangle is 1.5 m².

(ii) Given,

Base = 3.4 m and height = 90 cm = $\frac{90}{100}$ m = 0.9 m

Area of the triangle = $\frac{1}{2}$ x 3.4 m x 0.9 m

= 1.7 m x 0.9 m

= 1.53 m²

Hence, the area of the triangle is 1.53 m².

Given,

SR = 12 cm, PS = 8 cm and QM = 7.6 cm

(i) Taking SR as the base of the parallelogram, its height is QM.

Area of the parallelogram PQRS = base x height

= SR x QM

= 12 cm x 7.6 cm

= 91.2 cm²

Hence, the area of the parallelogram PQRS is 91.2 cm².

(ii) Taking PS as the base of the parallelogram, its height is QN.

Area of the parallelogram PQRS = base x height = PS x QN

⇒ 8 x QN = 91.2

⇒ QN = $\frac{91.2}{8}$ = 11.4

Hence, the length of QN is 11.4 cm.

From the figure,

ΔABC is right angled at A, AC = 4 cm and AB = 3 cm.

AD is the altitude from A to BC.

(i) Area of ΔABC = $\frac{1}{2}$ x base x height

$$\begin{aligned}= \frac{1}{2} \times AB \times AC\\[1em] = \frac{1}{2} \times 3 \text{ cm} \times 4 \text{ cm}\\[1em] = 6 \text{ cm}^2\end{aligned}$$

Hence, the area of ΔABC is 6 cm².

(ii) As ΔABC is right angled at A, by Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

BC² = AB² + AC²

⇒ BC² = 3² + 4² = 9 + 16 = 25

⇒ BC = $\sqrt{25}$ cm = 5 cm

Hence, the length of BC is 5 cm.

(iii) In ΔABC, BC = base = 5 cm and height = AD

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

$$\begin{aligned}= \frac{1}{2} \times BC \times AD\\[1em] \Rightarrow \frac{1}{2} \times 5 \times AD = 6\\[1em] \Rightarrow AD = \frac{6 \times 2}{5}\\[1em] \Rightarrow AD = \frac{12}{5} = 2.4\end{aligned}$$

Hence, the length of altitude from A to BC is 2.4 cm.

Given,

Area of the right angled triangle = 54 cm²

One leg = 12 cm

Let the other leg be $x$ cm.

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

$$\begin{aligned}\Rightarrow \frac{1}{2} \times 12 \times x = 54 \\[1em] \Rightarrow 6x = 54 \\[1em] \Rightarrow x = \frac{54}{6} \\[1em] \Rightarrow x = 9\end{aligned}$$

∴ Other leg = 9 cm

Let the hypotenuse be $h$ cm.

By Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

$$\begin{aligned}\Rightarrow h^2 = 12^2 + 9^2\\[1em] \Rightarrow h^2 = 144 + 81 = 225\\[1em] \Rightarrow h = \sqrt{225} = 15\end{aligned}$$

∴ Hypotenuse = 15 cm

Perimeter of the triangle = sum of its three sides = (12 + 9 + 15) cm = 36 cm

Hence, the perimeter of the triangle is 36 cm.

Given,

Area of the circle = 78.5 cm² and π = 3.14

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

As we know,

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

$$\begin{aligned}\Rightarrow 3.14 \times r^2 = 78.5\\[1em] \Rightarrow r^2 = \frac{78.5}{3.14}\\[1em] \Rightarrow r^2 = 25\\[1em] \Rightarrow r = \sqrt{25} = 5\end{aligned}$$

Circumference of the circle = $2\pi r$

= 2 x 3.14 x 5 cm

= 31.4 cm

Hence, the circumference of the circle is 31.4 cm.

Given,

Diameter of the given circle = 7 cm

∴ Radius of the given circle = $\frac{7}{2}$ cm = 3.5 cm

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

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

∴ Area of the required circle = 16 x 38.5 cm² = 616 cm²

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

Area of the required circle = $\pi R^2$

$$\begin{aligned}\Rightarrow \pi R^2 = 616\\[1em] \Rightarrow \frac{22}{7} \times R^2 = 616\\[1em] \Rightarrow R^2 = 616 \times \frac{7}{22}\\[1em] \Rightarrow R^2 = 196\\[1em] \Rightarrow R = \sqrt{196} = 14\end{aligned}$$

Circumference of the required circle = $2\pi R$

= 2 x $\frac{22}{7}$ x 14 cm

= 88 cm

Hence, the circumference of the circle is 88 cm.

Given,

Area of the circle = 154 cm²

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

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

$$\begin{aligned}\Rightarrow \pi r^2 = 154\\[1em] \Rightarrow \frac{22}{7} \times r^2 = 154\\[1em] \Rightarrow r^2 = 154 \times \frac{7}{22}\\[1em] \Rightarrow r^2 = 49\\[1em] \Rightarrow r = \sqrt{49} = 7\end{aligned}$$

∴ Diameter of the circle = 2 x 7 cm = 14 cm

Since the circle of biggest area is cut out of the square cardboard,

side of the square cardboard = diameter of the circle = 14 cm

∴ Original area of the cardboard = (side)²

= 14 cm x 14 cm

= 196 cm²

Hence, the original area of the cardboard is 196 cm².

Given,

Circumference of the circular park = 88 m

Width of the road = 3.5 m

Rate of paving = ₹ 60 per m²

Let the radius of the park be $r$ m.

Circumference of the park = $2\pi r$

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

Let $R$ be the radius of the outer circle.

Then $R$ = 14 m + 3.5 m = 17.5 m

Area of the road = $\pi R^2 - \pi r^2 = \pi(R^2 - r^2)$

$$\begin{aligned}= \frac{22}{7}(17.5 \times 17.5 - 14 \times 14) \text{ m}^2\\[1em] = \frac{22}{7}(306.25 - 196) \text{ m}^2\\[1em] = \frac{22}{7} \times 110.25 \text{ m}^2\\[1em] = 22 \times 15.75 \text{ m}^2\\[1em] = 346.5 \text{ m}^2\end{aligned}$$

Cost of paving the road = Area x rate

= 346.5 x ₹ 60

= ₹ 20,790

Hence, the cost of paving the road is ₹ 20,790.

Given,

Side of the square ABCD = 14 cm

From the figure, four equal circles are placed in the square, two along each side.

∴ Diameter of each circle = $\frac{14}{2}$ cm = 7 cm

⇒ Radius of each circle $(r) = \frac{7}{2}$ cm = 3.5 cm

Area of the square ABCD = (side)² = 14 cm x 14 cm = 196 cm²

Area of 4 circles = $4 \times \pi r^2$

= 4 x $\frac{22}{7}$ x 3.5 x 3.5 cm²

= 4 x 38.5 cm²

= 154 cm²

∴ Area of the shaded region = Area of the square - area of 4 circles

= 196 cm² - 154 cm²

= 42 cm²

Hence, the area of the shaded region is 42 cm².

Given,

Diameter of the larger semicircle = 28 cm

∴ Radius of the larger semicircle ($R$) = 14 cm

Since the two smaller semicircles are equal and together make up the diameter of the larger semicircle,

diameter of each smaller semicircle = $\frac{28}{2}$ cm = 14 cm

∴ Radius of each smaller semicircle ($r$) = 7 cm

(i) As we know,

Length of a semicircular arc = $\pi r$

Length of the boundary = arc of the larger semicircle + arcs of the two smaller semicircles

= $\pi R + \pi r + \pi r$

= $\frac{22}{7}$ x 14 cm + 2 x $\frac{22}{7}$ x 7 cm

= 44 cm + 44 cm

= 88 cm

Hence, the length of the boundary is 88 cm.

(ii) From the figure, one smaller semicircle is cut out from the larger semicircle and the other smaller semicircle is added to it. As the two smaller semicircles are equal, the area removed is equal to the area added.

∴ Area of the shaded region = area of the larger semicircle

= $\frac{1}{2} \times \pi R^2$

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

= 308 cm²

Hence, the area of the shaded region is 308 cm².

From the figure,

ED = 7 m, AC = 13 m and height of the trapezium = 6.5 m

Since ABDE is a parallelogram, AB = ED = 7 m

∴ BC = AC - AB = 13 m - 7 m = 6 m

Area of parallelogram ABDE = base x height

= 7 m x 6.5 m

= 45.5 m²

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

= $\frac{1}{2}$ x 6 m x 6.5 m

= 19.5 m²

∴ Area of the trapezium ACDE = Area of parallelogram ABDE + area of ΔBCD

= 45.5 m² + 19.5 m²

= 65 m²

Hence, the area of the trapezium ACDE is 65 m².

Given,

Length of the uncovered side = 20 m

Area of the plot = 320 m²

Let the other side of the plot be $x$ m.

Area of the plot = length x breadth

⇒ 20 x $x$ = 320

⇒ $x = \frac{320}{20}$ = 16

∴ Other side of the plot = 16 m

The three sides to be fenced are 20 m, 16 m and 16 m.

∴ Length of fencing required = (20 + 16 + 16) m = 52 m

Hence, 52 m of fencing is required.

Given,

Length of the rectangle = 28 cm

From the figure, two equal circles touching each other are drawn in the rectangle, each touching the sides of the rectangle.

∴ Diameter of each circle = $\frac{28}{2}$ cm = 14 cm

⇒ Radius of each circle ($r$) = 7 cm

And, breadth of the rectangle = diameter of a circle = 14 cm

Each circle is inscribed in a square of side 14 cm.

Area of one square = (side)² = 14 cm x 14 cm = 196 cm²

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

= $\frac{22}{7}$ x 7 x 7 cm²

= 154 cm²

∴ Area of the square not covered by the circle = 196 cm² - 154 cm² = 42 cm²

This area is made up of 4 equal corner pieces, and the shaded region is one such corner piece.

∴ Area of the shaded region = $\frac{42}{4}$ cm² = 10.5 cm²

Hence, the area of the shaded region is 10.5 cm².

Given,

Side of the square ABCD = 14 cm

Since the four arcs are of equal radius and adjacent arcs meet on the sides of the square,

radius of each arc ($r$) = $\frac{14}{2}$ cm = 7 cm

Each arc is a quadrant (one-fourth of a circle) as the angle at each corner of the square is 90°.

Perimeter of the shaded region = 4 x length of each quadrant arc

= 4 x $\frac{1}{4} \times 2\pi r$

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

= 44 cm

Four quadrants of the same radius together make one complete circle.

∴ Area of 4 quadrants = $\pi r^2$

= $\frac{22}{7}$ x 7 x 7 cm²

= 154 cm²

Area of the square = (side)² = 14 cm x 14 cm = 196 cm²

∴ Area of the shaded region = Area of the square - area of 4 quadrants

= 196 cm² - 154 cm²

= 42 cm²

Hence, the perimeter of the shaded region is 44 cm and its area is 42 cm².