Area & Perimeter of Plane Figures

The area of a triangle is given by:

Area = $\frac{1}{2}$ x length x altitude

When the length of the side and the corresponding altitude of the triangle are doubled:

Area₁ = $\frac{1}{2}$ x 2 x length x 2 x altitude

= $4 \times \Big( \frac{1}{2} \times \text{length} \times \text{altitude} \Big)$

= 4 x Area

Area₁ = 4 x Area

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

Hence, option 3 is the correct option.

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

If each side of the equilateral triangle is halved, then the new area will be:

New area = $\frac{\sqrt{3}}{4} \times \Big(\frac{side}{2}\Big)^2$

= $\frac{1}{4}\Big(\frac{\sqrt{3}}{4} \times side^2\Big)$

= $\frac{1}{4}$ x Original area

Therefore, the new area of the equilateral triangle will be one-fourth of the original area.

Hence, option 4 is the correct option.

Given:

Area = 37.5 cm²

Base = 12.5 cm

Let h be the altitude of triangle.

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

⇒ 37.5 = $\frac{1}{2}$ x 12.5 x h

⇒ h = $\frac{37.5 \times 2}{12.5}$

⇒ h = $\frac{75}{12.5}$

⇒ h = 6 cm

Hence, option 3 is the correct option.

Given:

AB = AC = 12 cm

∠A = 90°

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

= $\frac{1}{2}$ x 12 x 12 cm²

= $\frac{1}{2}$ x 144 cm²

= 72 cm²

Hence, option 3 is the correct option.

Let the sides of the triangle be:

a = 9 cm, b = 12 cm and c = 15 cm.

The semi-perimeter s:

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{9 + 12 + 15}{2}\\[1em] = \frac{36}{2}\\[1em] = 18\end{aligned}$$

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

= $\sqrt{18(18 - 9)(18 - 12)(18 - 15)}$ cm²

= $\sqrt{18 \times 9 \times 6 \times 3}$ cm²

= $\sqrt{2,916}$ cm²

= 54 cm²

Hence, option 1 is the correct option.

Let the sides of the triangle be:

a = 18 cm, b = 24 cm and c = 30 cm.

The semi-perimeter s:

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{18 + 24 + 30}{2}\\[1em] = \frac{72}{2}\\[1em] = 36\end{aligned}$$

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

= $\sqrt{36(36 - 18)(36 - 24)(36 - 30)}$ cm²

= $\sqrt{36 \times 18 \times 12 \times 6}$ cm²

= $\sqrt{46,656}$ cm²

= 216 cm²

Using the area formula to find the altitude corresponding to the largest side (base = 30 cm):

Area = $\frac{1}{2}$ x base x altitude

Let h be the altitude:

⇒ 216 = $\frac{1}{2}$ x 30 x h

⇒ 216 = 15 x h

⇒ h = $\frac{216}{15}$

⇒ h = 14.4 cm

Hence, the area of the triangle is 216 cm² and the length of altitude corresponding to the largest side is 14.4 cm.

It is given that the lengths of the sides of a triangle are in the ratio 3 : 4 : 5.

Let the lengths of the sides be 3a, 4a and 5a.

The perimeter of the triangle is 144 cm.

Perimeter = sum of all sides of triangle

⇒ 144 = 3a + 4a + 5a

⇒ 144 = 12a

⇒ a = $\frac{144}{12}$

⇒ a = 12

So, the sides of triangle = 3a, 4a and 5a

= 3 x 12, 4 x 12 and 5 x 12

= 36, 48 and 60

Let a = 36 cm, b = 48 cm and c = 60 cm.

The semi-perimeter s:

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{36 + 48 + 60}{2}\\[1em] = \frac{144}{2}\\[1em] = 72\end{aligned}$$

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

= $\sqrt{72(72 - 36)(72 - 48)(72 - 60)}$ cm²

= $\sqrt{72 \times 36 \times 24 \times 12}$ cm²

= $\sqrt{746,496}$ cm²

= 864 cm²

Hence, the area of the triangle is 864 cm².

(i) Given:

AB = AC = 4 cm

∠A = 90°

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

= $\frac{1}{2}$ x 4 x 4 cm²

= $\frac{1}{2}$ x 16 cm²

= 8 cm²

(ii) By using the Pythagoras theorem,

AB² + AC² = BC²

⇒ (4)² + (4)² = BC²

⇒ 16 + 16 = BC²

⇒ BC² = 32

⇒ BC = $\sqrt{32}$

⇒ BC = 4 $\sqrt{2}$ cm

Now considering BC as the base of the triangle, altitude AD will be its height.

Area of the triangle will remain the same as before i.e., 8 cm².

Let h be the altitude of triangle

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

∴ 8 = $\frac{1}{2}$ x BC x AD

⇒ 8 = $\frac{1}{2}$ x $4 \sqrt{2}$ x h

⇒ 8 = $2 \sqrt{2}$ x h

⇒ h = $\frac{8}{2 \sqrt{2}}$

⇒ h = $\frac{4}{\sqrt{2}}$

⇒ h = 2.83 cm

Hence, the length of perpendicular from A to BC is 2.83 cm.

Given:

Area = $36\sqrt{3}\text{ sq. cm}$

Let s be the side of equilateral triangle.

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

$$\begin{aligned}⇒ 36\sqrt{3} = \frac{\sqrt{3}}{4} \times s^2\\[1em] ⇒ 36\cancel{\sqrt{3}} = \frac{\cancel{\sqrt{3}}}{4} \times s^2\\[1em] ⇒ 36 = \frac{1}{4} \times s^2\\[1em] ⇒ s^2 = 36 \times 4\\[1em] ⇒ s^2 = 144\\[1em] ⇒ s = \sqrt{144}\\[1em] ⇒ s = 12\end{aligned}$$

Perimeter = 3 x side

= 3 x 12 cm

= 36 cm

Hence, the perimeter is 36 cm.

Given:

Perimeter = 36 cm

Base = 16 cm

Let a be the length of the equal sides of the triangle.

Perimeter = 2 x Equal side + Base

⇒ 36 = 2a + 16

⇒ 2a = 36 - 16

⇒ 2a = 20

⇒ a = $\frac{20}{2}$

⇒ a = 10 cm

Thus, the triangle has two equal sides, each 10 cm long, and a base of 16 cm.

a = 10 cm, b = 10 cm and c = 16 cm.

The semi-perimeter is:

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{10 + 10 + 16}{2}\\[1em] = \frac{36}{2}\\[1em] = 18\end{aligned}$$

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

= $\sqrt{18(18 - 10)(18 - 10)(18 - 16)}$ cm²

= $\sqrt{18 \times 8 \times 8 \times 2}$ cm²

= $\sqrt{2,304}$ cm²

= 48 cm²

Hence, the area is 48 cm².

Given:

Base (b) = 24 cm

Area = 192 cm²

Let s be the length of the equal sides of the triangle.

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

$$\begin{aligned}⇒ 192 = \frac{1}{4} \times 24 \times \sqrt{4s^2 - 24^2}\\[1em] ⇒ 192 = 6 \times \sqrt{4s^2 - 576}\\[1em] ⇒ \sqrt{4s^2 - 576} = \frac{192}{6}\\[1em] ⇒ \sqrt{4s^2 - 576} = 32\\[1em] ⇒ 4s^2 - 576 = 1024\\[1em] ⇒ 4s^2 = 1024 + 576\\[1em] ⇒ 4s^2 = 1600\\[1em] ⇒ s^2 = \frac{1600}{4}\\[1em] ⇒ s^2 = 400\\[1em] ⇒ s = \sqrt{400}\\[1em] ⇒ s = 20\end{aligned}$$

Perimeter = sum of all sides of triangle

= 20 + 20 + 24 cm

= 64 cm

Hence, the perimeter is 64 cm.

For triangle ABC (right-angled triangle),

Using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ 8² + height² = 16²

⇒ 64 + height² = 256

⇒ height² = 256 - 64

⇒ height² = 192

⇒ height = $\sqrt{192}$

⇒ height = 8 $\sqrt{3}$

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

= $\frac{1}{2} \times 8 \times 8 \sqrt{3}$

= $4 \times 8 \sqrt{3}$

= $32 \sqrt{3}$ cm²

For triangle BCD,

Area of an equilateral triangle = $\frac{\sqrt{3}}{4}$ x side²

= $\frac{\sqrt{3}}{4}$ x 8²

= $\frac{\sqrt{3}}{4}$ x 64

= 16 $\sqrt{3}$ cm²

Area of Δ ABD (shaded portion) = Area of Δ ABC - Area of Δ BDC

= 32 $\sqrt{3}$ - 16 $\sqrt{3}$ cm²

= 16 $\sqrt{3}$ cm²

Hence, the area of the shaded portion is 16$\sqrt{3}$ cm² = 27.712 cm².

Given:

AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and ∠DBC = 90°

In Δ BCD,

By using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ BC² + 12² = 13²

⇒ BC² + 144 = 169

⇒ BC² = 169 - 144

⇒ BC² = 25

⇒ BC = $\sqrt{25}$

⇒ BC = 5 cm

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

= $\frac{1}{2}$ x 12 x 5 cm²

= $\frac{1}{2}$ x 60 cm²

= 30 cm²

For Δ ABD,

Let AD = a = 10 cm, BD = b = 12 cm and AB = c = 8 cm.

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{10 + 12 + 8}{2}\\[1em] = \frac{30}{2}\\[1em] = 15\end{aligned}$$

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

= $\sqrt{15(15 - 10)(15 - 12)(15 - 8)}$ cm²

= $\sqrt{15 \times 5 \times 3 \times 7}$ cm²

= $\sqrt{1,575}$ cm²

= 39.7 cm²

Area of quadrilateral ABCD = Area of Δ ABD + Area of Δ BCD

= 39.5 + 30 cm²

= 69.5 cm²

Perimeter of quadrilateral ABCD = Sum of all sides of quadrilateral

= AB + BC + CD + DA

= 8 + 10 + 13 + 5

= 36 cm

Hence, the area of quadrilateral is 69.7 cm² and the perimeter is 36 cm.

Given:

Cost of cultivating the field = ₹ 36.72 per 100 m²

Total cost = ₹ 49,572

Total cost = Area x Cost of cultivating per 100 m²

Area = $\frac{\text{Total cost}}{\text{Cost of cultivating}}$

$$\begin{aligned}= \frac{49,572 \times 100}{36.72}\\[1em] = \frac{49,572 \times 10,000}{3672}\\[1em] = 1,35,000 \text{ m}^2\end{aligned}$$

Let the height of the triangle be h.

Since the base of the field is three times its height, we have:

Base = 3h

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

$$\begin{aligned}⇒ 1,35,000 = \frac{1}{2} \times 3h \times h\\[1em] ⇒ 1,35,000 = \frac{3}{2} \times h^2\\[1em] ⇒ h^2 = \frac{1,35,000 \times 2}{3}\\[1em] ⇒ h^2 = \frac{2,70,000}{3}\\[1em] ⇒ h^2 = 90,000\\[1em] ⇒ h = \sqrt{90,000}\\[1em] ⇒ h = 300\end{aligned}$$

Thus, the height of the triangle is 300 m.

Base = 3 x height = 3 x 300 m = 900 m

Hence, the height is 300 m and the base is 900m.

(i) Given:

The sides of a triangular field are in the ratio 5 : 3 : 4.

Perimeter = 180 m

Let the sides of field be 5a, 3a and 4a.

Perimeter = Sum of all sides of triangular field

⇒ 180 = 5a + 3a + 4a

⇒ 180 = 12a

⇒ a = $\frac{180}{12}$

⇒ a = 15

Thus, sides of field = 5a , 3a and 4a

= 5 x 15, 3 x 15 and 4 x 15

= 75 m, 45 m and 60 m

Let a = 75 m, b = 45 m and c = 60 m.

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{75 + 45 + 60}{2}\\[1em] = \frac{180}{2}\\[1em] = 90\end{aligned}$$

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

= $\sqrt{90(90 - 75)(90 - 45)(90 - 60)}$ m²

= $\sqrt{90 \times 15 \times 45 \times 30}$ m²

= $\sqrt{18,22,500}$ m²

= 1,350 m²

Hence, the area is 1,350 m².

(ii) Area = $\frac{1}{2}$ x base x altitude

Base = 75 m

Let h be the altitude of triangle corresponding to the largest side 75 m,

⇒ 1350 = $\frac{1}{2}$ x 75 x h

⇒ h = $\frac{1350 \times 2}{75}$

⇒ h = $\frac{2,700}{75}$

⇒ h = 36 m

Hence, the altitude of the triangle corresponding to its largest side is 36 m.

(iii) Cost of levelling = ₹ 10 per square metre

Total cost = Area x Cost of levelling

= ₹ 1,350 x 10

= ₹ 13,500

Hence, the total cost of levelling is ₹ 13,500.

Each of equal sides of an isosceles triangle is 4 cm greater than its height.

Let h be the height of the triangle.

Equal sides: AB = AC = h + 4

Base: BC = 24 cm

In Δ ABD and Δ ACD,

AD = AD [∵ Common Side]

∠ ADB = ∠ ADC [∵ Both are 90°]

AB = AC [∵ Δ ABC is isosceles]

∴ Δ ABD ≅ Δ ACD [RHS axiom]

∴ BD = CD [C.P.C.T]

∴ BD = CD = $\frac{\text{BC}}{2}$ = ${\frac{24}{2}}$ = 12 cm

By using the Pythagoras theorem in Δ ABD,

BD² + AD² = AB²

⇒ 12² + h² = (h + 4)²

⇒ 144 + h² = h² + 4² + 2 x h x 4

⇒ 144 + $\cancel{h^2}$= $\cancel{h^2}$+ 16 + 8h

⇒ 144 - 16 = 8h

⇒ 128 = 8h

⇒ h = $\frac{128}{8}$

⇒ h = 16 cm

⇒ a = h + 4 = 16 + 4 = 20 cm

Perimeter of triangle = Sum of all sides

= AB + AC + BC

= (20 + 20 + 24) cm

= 64 cm

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

= $\frac{1}{2}$ x 24 x 16

= 12 x 16

= 192 cm²

Hence, the perimeter is 64 cm and the area is 192 cm².

Given:

Perimeter = 60 cm

Let a be the sides of equilateral triangle.

Perimeter = Sum of all sides

⇒ 60 = a + a + a

⇒ 60 = 3a

⇒ a = $\frac{60}{3}$

⇒ a = 20 cm

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

$$\begin{aligned}= \frac{\sqrt{3}}{4} \times 20^2\\[1em] = \frac{\sqrt{3}}{4} \times 400\\[1em] = 100 \sqrt{3} \\[1em] = 173.2 cm^2\end{aligned}$$

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

$$\begin{aligned}= \frac{\sqrt{3}}{2} \times 20\\[1em] = 10 \sqrt{3}\\[1em] = 17.32 cm\end{aligned}$$

Hence, the area is 173.2 cm² and the height is 17.32 cm.

Δ ABC is shown in the figure below:

Given:

Area = 150 cm²

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

$$\begin{aligned}⇒ 150 = \frac{1}{2} \times x \times (x + 5)\\[1em] ⇒ 150 \times 2 = x^2 + 5x\\[1em] ⇒ 300 = x^2 + 5x\\[1em] ⇒ x^2 + 5x - 300 = 0\\[1em] ⇒ x^2 + 20x - 15x - 300 = 0\\[1em] ⇒ x(x + 20) - 15(x + 20) = 0\\[1em] ⇒ (x + 20)(x - 15) = 0\\[1em] ⇒ x = -20 \text{ or } 15\end{aligned}$$

Since length cannot be negative, AB = 15 cm.

Thus, AC = x + 5 = 15 + 5 cm = 20 cm

By using the Pythagoras theorem,

AB² + AC² = BC²

⇒ 15² + 20² = BC²

⇒ 225 + 400 = BC²

⇒ 625 = BC²

⇒ BC = $\sqrt{625}$

⇒ BC = 25

Hence, the sides of the triangle are 15 cm, 20 cm and 25 cm.

Given:

Area = 54 cm²

Let the two sides of the right-angled triangle be x cm and (x - 3) cm.

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

⇒ 54 = $\frac{1}{2}$ x BC x AB

⇒ 54 = $\frac{1}{2} \times$ (x - 3) $\times$ x

⇒ 54 $\times$ 2 = (x - 3) $\times$ x

⇒ 108 = x² - 3x

⇒ x² - 3x - 108 = 0

⇒ x² - 12x + 9x - 108 = 0

⇒ x(x - 12) + 9(x - 12) = 0

⇒ (x - 12)(x + 9) = 0

⇒ x = 12 or -9

Since length cannot be negative, AB is 12 cm.

BC = (x - 3) cm = 12 - 3 cm = 9 cm

By using the Pythagoras theorem,

AB² + BC² = AC²

⇒ 12² + 9² = AC²

⇒ 144 + 81 = AC²

⇒ 225 = AC²

⇒ AC = $\sqrt{225}$

⇒ AC = 15

Perimeter of right angled triangle = AB + BC + AC

= 12 + 9 + 15 cm

= 36 cm

Hence, the perimeter of the triangle is 36 cm.

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

For Δ ABD,

Area = $\frac{1}{2}$ x BD x AO

For Δ CBD,

Area = $\frac{1}{2}$ x BD x CO

Area of quadrilateral ABCD = Area of Δ ABD + Area of Δ CBD

= $\frac{1}{2}$ x BD x AO + $\frac{1}{2}$ x BD x CO

= $\frac{1}{2}$ x BD x (AO + CO)

= $\frac{1}{2}$ x BD x AC

Hence, option 2 is the correct option.

Given:

Adjacent sides of rectangle = x and y

Area of rectangle = base x height

= $x \times y$

Hence, option 1 is the correct option.

Area of shaded portion = Area of rectangle ABCD - Area of square DEFG

Area of square = Side²

Area of rectangle = Base x Height

Area of shaded portion = 20 x 16 - 8² cm²

= 320 - 64 cm²

= 256 cm²

Hence, option 3 is the correct option.

Given:

Perimeter = 72 cm

Let s be the side of square.

Perimeter = 4 x side

⇒ 4 x s = 72

⇒ s = $\frac{72}{4}$

⇒ s = 18 cm

Area = Side²

= 18² cm²

= 324 cm²

Hence, option 3 is the correct option.

Given:

Area = 360 cm².

One diagonal = 20 cm

Let d be the other diagonal.

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

⇒ $\frac{1}{2}$ x 20 x d = 360

⇒ 10 x d = 360

⇒ d = $\frac{360}{10}$

⇒ d = 36 cm

Hence, option 4 is the correct option.

The quadrilateral is shown in the figure below:

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

For Δ ABD,

Area = $\frac{1}{2}$ x BD x AM

= $\frac{1}{2}$ x 30 x 19 cm²

= 15 x 19 cm²

= 285 cm²

For Δ CBD,

Area = $\frac{1}{2}$ x BD x CN

= $\frac{1}{2}$ x 30 x 11 cm²

= 15 x 11 cm²

= 165 cm²

Area of quadrilateral ABCD = Area of Δ ABD + Area of Δ CBD

= 285 + 165 cm²

= 450 cm²

Hence, the area of quadrilateral is 450 cm².

Since the diagonals of the quadrilateral intersect at right angles, the area of the quadrilateral is given by:

Area = $\frac{1}{2}$ x Product of diagonals

= $\frac{1}{2}$ x 16 x 13 cm²

= 8 x 13 cm²

= 104 cm²

Hence, the area of the quadrilateral is 104 cm².

For Δ ABD,

By using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ AB² + BD² = AD²

⇒ AB² + (24)² = (26)²

⇒ AB² + 576 = 676

⇒ AB² = 676 - 576

⇒ AB² = 100

⇒ AB = $\sqrt{100}$

⇒ AB = 10 cm

Area of Δ ABD = $\frac{1}{2}$ x AB x BD

= $\frac{1}{2}$ x 10 x 24 cm²

= 5 x 24 cm²

= 120 cm²

Area of equilateral triangle BCD = $\frac{\sqrt{3}}{4}$ x side²

= $\frac{\sqrt{3}}{4}$ x 24²

= $\frac{\sqrt{3}}{4}$ x 576

= 144 ${\sqrt{3}}$ cm²

= 249.41 cm²

Total area of quadrilateral ABCD = Δ ABD + Δ BCD

= 120 + 249.41 cm²

= 369.41 cm²

Hence, the area of quadrilateral ABCD is 369.41 cm².

Quadrilateral ABCD is shown in the figure below:

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

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

= $\frac{1}{2}$ x 24 x 32 cm²

= 12 x 32 cm²

= 384 cm²

By using the Pythagoras theorem,

AD² + AB² = BD²

⇒ 24² + 32² = BD²

⇒ 576 + 1,024 = BD²

⇒ 1,600 = BD²

⇒ BD = $\sqrt{1,600}$

⇒ BD = 40 cm

In triangle BCD,

Let the sides of the triangle be:

a = 40 cm, b = 52 cm and c = 52 cm.

The semi-perimeter s:

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{40 + 52 + 52}{2}\\[1em] = \frac{144}{2}\\[1em] = 72\end{aligned}$$

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

= $\sqrt{72(72 - 40)(72 - 52)(72 - 52)}$ cm²

= $\sqrt{72 \times 32 \times 20 \times 20}$ cm²

= $\sqrt{921,600}$ cm²

= 960 cm²

Therefore, area of quadrilateral ABCD = Area of Δ DAB + Area of triangle BCD

= 384 + 960 cm²

= 1344 cm²

Hence, the area of quadrilateral ABCD is 1344 cm².

Given:

Perimeter = $\frac{3}{5}$ km.

Length of the field = Twice its width.

Let a be the width of the field.

So, the length = 2a

Perimeter of a rectangle = 2(length + width)

$$\begin{aligned}⇒ \frac{3}{5} = 2(2a + a)\\[1em] ⇒ \frac{3}{5} = 2 \times 3a\\[1em] ⇒ \frac{3}{5} = 6a\\[1em] ⇒ a = \frac{3}{5 \times 6}\\[1em] ⇒ a = \frac{3}{30}\\[1em] ⇒ a = \frac{1}{10}\\[1em]\end{aligned}$$

Thus, width = $\frac{1}{10}$ km

= $\frac{1}{10} \times 1,000$ m

= 100 m

Length = 2a = 2 x 100 m

= 200 m

Area = length x width

= 200 x 100 m²

= 20,000 m²

Hence, the area of the rectangle is 20,000 m².

Given:

The length of the rectangular field is 85 m.

The breadth of the rectangular field is 60 m.

The width of the path to be covered with grass is 5 m.

The length of the inner rectangular field = 85 m - 5 m - 5 m = 75 m

The breadth of the inner rectangular field = 60 m - 5 m - 5 m = 50 m

The area of a rectangle = length x breadth

⇒ Area of the inner field = 75 x 50 m² = 3,750 m²

Hence, the area to be covered with grass is 3,750 m².

Given:

Length of the rectangle = 6 cm

Breadth of the rectangle = 4 cm

Area of the rectangle = length x breadth

= 6 x 4 cm²

= 24 cm²

Let h be the height of the triangle.

Base of the triangle = 6 cm

It is given that area of triangle is 3 times the area of the rectangle.

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

⇒ 3 x 24 = $\frac{1}{2}$ x 6 x h

⇒ 3 x 24 = 3 x h

⇒ $\cancel{3}$ x 24 = $\cancel{3}$ x h

⇒ h = 24 cm

Hence, the height of the triangle is 24 cm.

Given:

Dimensions of grass plot (ABCD) = 25 x 13 m

Width of footpath = 2 m

Area of footpath = Area of ABQP + Area of QNOR + Area of RCDS + Area of SHMP

= 2Area of ABQP + 2Area of QNOR

= 2(Area of ABQP + Area of QNOR)

Area of ABQP = l x b = AB x AP

= 25 x 2 m²

= 50 m²

Area of QNOR = l x b = DN x QN

= 17 x 2 m²

= 34 m²

Area of footpath = 2(50 + 34) m²

= 2 x 84 m²

= 168 m²

= 1680000 sq. cm

Total area = Number of tiles x Area covered by 1 tiles

⇒ 1680000 = Number of tiles x 400

⇒ Number of tiles = $\frac{1680000}{400}$

= 4200

Hence, the number of tiles = 4200.

Given:

Length of the garden = 120 m

Cost of fencing = 75 paise per metre = ₹ 0.75 per m

Total cost = ₹ 300

Total cost of fencing = Perimeter x Cost per metre

⇒ ₹ 300 = Perimeter x ₹ 0.75

⇒ Perimeter = $\frac{300}{0.75}$ m

⇒ Perimeter = $\frac{30000}{75}$ m

⇒ Perimeter = 400 m

Let b be the breadth of the garden.

Perimeter = 2(l + b)

⇒ 2(120 + b) = 400

⇒ 120 + b = $\frac{400}{2}$

⇒ 120 + b = 200

⇒ b = 200 - 120

⇒ b = 80 m

Area of the rectangular garden = Length x Breadth

= 120 x 80 m²

= 9600 m²

Hence, the area of the garden is 9600 sq. m.

Given:

Length = x

Width = $\frac{4}{7}$ of the length = $\frac{4}{7} \times x$

Perimeter = $y$

Perimeter of a rectangle = 2(l + b)

$$\begin{aligned}⇒ 2\Big(x + \frac{4}{7}x\Big) = y\\[1em] ⇒ 2\Big(\frac{7 + 4}{7}x\Big) = y\\[1em] ⇒ 2\Big(\frac{11}{7}x\Big) = y\\[1em] ⇒ \frac{22}{7}x = y\\[1em]\end{aligned}$$

The length when y = 4400 cm,

$$\begin{aligned}⇒ \frac{22}{7}x = 4400\\[1em] ⇒ x = \frac{4400 \times 7}{22}\\[1em] ⇒ x = \frac{30800}{22}\\[1em] ⇒ x = 1400 cm = 14 m\end{aligned}$$

Hence, the equation connecting x and y is $y = \frac{22}{7}x$ and the length of the room is 14 m.

(i) Given:

Breadth of the verandah = x

Length of the verandah = x + 3

It is also given that the numerical value of the area is equal to the numerical value of the perimeter.

⇒ l x b = 2(l + b)

⇒ x(x + 3) = 2(x + x + 3)

⇒ x² + 3x = 2(2x + 3)

⇒ x² + 3x = 4x + 6

⇒ x² + 3x - 4x - 6 = 0

⇒ x² - x - 6 = 0

Hence, the equation is x² - x - 6.

(ii) From (i),

⇒ x² - x - 6 = 0

⇒ x² - 3x + 2x - 6 = 0

⇒ x(x - 3) + 2(x - 3) = 0

⇒ (x - 3)(x + 2) = 0

⇒ x = 3 or - 2

Since breadth cannot be negative, x = 3 m.

Length = x + 3 = 3 + 3 m = 6 m

Hence, length = 6 m and breadth = 3 m.

Given:

Length of the rectangular field = 80 m

Width of the rectangular field = 45 m

Area of shaded path = Area of path ABCD + Area of cross path EFGH - Area of path IJKL

Area of path ABCD = l x b = AB x BC

= 15 x 45 m²

= 675 m²

Area of path EFGH = l x b = EF x FG

= 8 x 80 m²

= 640 m²

Area of path IJKL = l x b = IJ x KL

= 15 x 8 m²

= 120 m²

Area of shaded path = 675 + 640 - 120 m²

= 1,195 m²

Hence, the area of shaded portion is 1,195 m².

Given:

Dimensions of the hall = 45 m x 32 m

Area = l x b

= 45 x 32 m²

= 1,440 m²

Cost per metre of carpet = ₹40

Width of the carpet = 1.20 m

Cost per square metre = $\frac{40}{1.20}$ = $\frac{400}{12}$ = ₹ 33.33

Total cost = Cost per square metre x area of hall

= 33.33 x 1440

= ₹ 48,000

Width of the carpet in metres: 80 cm = 0.8 m

Cost per metre of carpet = ₹25

Cost per square metre = $\frac{25}{0.8}$ = $\frac{250}{8}$ = ₹ 31.25

Total cost = Cost per square metre x area of hall

= 31.25 x 1440

= ₹ 45,000

Hence, the cost of covering the hall with a 1.20 m wide carpet at ₹ 40 per metre is ₹ 48,000 and the cost of carpeting the hall with an 80 cm wide carpet at ₹ 25 per metre is ₹ 45,000.

Given:

Diagonal of the square = 15 m

Let a be the length of side of the square.

Using the Pythagoras theorem for a square,

⇒ Side² + Side² = Diagonal²

⇒ 2 x Side² = Diagonal²

⇒ 2 x a² = 15²

⇒ 2a² = 225

⇒ a² = $\frac{225}{2}$

⇒ a² = 112.5

⇒ a = $\sqrt{112.5}$

⇒ a = 10.61 m

Area of square = side²

= 10.61² m²

= 112.5 m²

Perimeter of square = 4 x side

= 4 x 10.60 m

= 42.43 m

Hence, the area of the square plot is 112.5 m² and the perimeter is 42.43 m.

Given:

Length of the lawn = 30 m

Breadth of the lawn = 12 m

Width of flowerbed = 2 m

New length of lawn = 30 - 2 - 2 m = 30 - 4 m = 26 m

New breadth of lawn = 12 - 2 m = 10 m

Hence, the length of lawn = 26 m and the breadth of lawn = 10 m.

(ii) Area of flowerbed = Area of AMPX + Area of XDCY + Area of NBYO

Area of AMPX = l x b = AM x MP

= 10 x 2 m²

= 20 m²

Area of XDCY = l x b = XD x DC

= 30 x 2 m²

= 60 m²

Area of NBYO = l x b = NB x BY

= 10 x 2 m²

= 20 m²

Area of flowerbed = 20 + 60 + 20 m²

= 100 m²

Hence, the area of flowerbed is 100 m².

Given:

Dimensions of the floor = 15 m x 8 m

Dimensions of the tile = 50 cm x 25 cm = 0.5 m x 0.25 m

Area of floor = l x b

= 15 x 8 m²

= 120 m²

Area of tile = l_tile x b_tile

= 0.5 x 0.25 m²

= 0.125 m²

Let n be the number of tiles.

Area of floor = Area of tile x Number of tiles

⇒ 120 = 0.125 x n

⇒ n = $\frac{120}{0.125}$

⇒ n = $\frac{120000}{125}$

⇒ n = 960

Now, a carpet is laid on the floor, leaving a space of 1 m between its edges and the edges of the floor.

Length of the carpet = 15 - 1 - 1 m = 13 m

Breadth of the carpet = 8 - 1 - 1 m = 6 m

Area of uncovered floor = Area of floor - Area of carpet

= 15 x 8 - 13 x 6 m²

= 120 - 78 m²

= 42 m²

Fraction of floor left uncovered = $\frac{42}{120}$ = $\frac{7}{20}$

Hence, the number of tiles needed is 960 and the fraction of the floor left uncovered is $\frac{7}{20}$.

Step 1 Finding the area of the parallelogram.

Length of one side of the parallelogram (longer side) = 24 cm

Length of the adjacent side (shorter side) = 18 cm

The distance between the longer sides = 12 cm

Area of parallelogram = base x height

= AB x DE

= 24 x 12 cm²

= 288 cm²

Step 2 Finding the sides of the parallelogram.

Let a be the distance between the shorter sides.

Area = base x height

= BC x AF

= 18 x a

Since area of the parallelogram will be same, if we consider AB as the base or BC as the base.

⇒ AB x DE = BC x AF

⇒ 288 = 18 x a

⇒ a = $\frac{288}{18}$

⇒ a = 16 cm

Hence, the distance between the shorter sides of the parallelogram is 16 cm.

By joining diagonal BD, the parallelogram is divided into two triangles. Let the sides of Δ ABD be:

a = 28 cm, b = 26 cm and c = 30 cm.

The semi-perimeter s:

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

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

= $\sqrt{42(42 - 28)(42 - 26)(42 - 30)}$ cm²

= $\sqrt{42 \times 14 \times 16 \times 12}$ cm²

= $\sqrt{112,896}$ cm²

= 336 cm²

Area of parallelogram ABCD = 2 x area of Δ ABD

= 2 x 336 cm²

= 672 cm²

Let h be the distance between the shorter sides.

Area of the parallelogram = base x height

⇒ 26 x h = 672

⇒ h = $\frac{672}{26}$

⇒ h = 25.84 cm

Hence, the area of the parallelogram is 672 cm² and the distance between the shorter sides is 25.84 cm.

(i) Given:

Area of rhombus = 216 sq. cm

One diagonal = 24 cm

Let d be the other diagonal of rhombus.

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

⇒ 216 = $\frac{1}{2}$ x 24 x d

⇒ 216 = 12 x d

⇒ d = $\frac{216}{12}$

⇒ d = 18

Hence, the length of other diagonal is 18 cm.

(ii) The rhombus is shown in the figure below:

Diagonal AC = 24 cm

Then, OA = OC = $\frac{24}{2}$ = 12 cm

Diagonal BD = 18 cm

Then, OB = OD = $\frac{18}{2}$ = 9 cm

Since the diagonals of a rhombus bisect at 90°.

Applying pythagoras theorem in Δ AOB, we get:

AB² = OA² + OB²

⇒ AB² = (12)² + (9)²

⇒ AB² = 144 + 81

⇒ AB² = 225

⇒ AB = $\sqrt{225}$

⇒ AB = 15 cm

Hence, the length of each side of the rhombus is 15 cm.

(iii) Perimeter of the rhombus = 4 x side

= 4 x 15 cm

= 60 cm

Hence, the perimeter of the rhombus 60 cm.

(i) Given:

Perimeter of the rhombus = 52 cm

One diagonal BD = 24 cm

Let a be the length of a side of the rhombus.

Perimeter of a rhombus = 4 x Side

⇒ 4 x a = 52

⇒ a = $\frac{52}{4}$

⇒ a = 13 cm

BD = 24 cm

Then, OB = OD = $\frac{24}{2}$ = 12 cm

Since the diagonals of a rhombus bisect at 90°.

Applying pythagoras theorem in triangle AOB, we get:

AB² = OA² + OB²

⇒ (13)² = OA² + (12)²

⇒ 169 = OA² + 144

⇒ OA² = 169 - 144

⇒ OA² = 25

⇒ OA = $\sqrt{25}$

⇒ OA = 5 cm

AC = 2 x OA = 2 x 5 cm = 10 cm

Hence, the length of the other diagonal is 10 cm.

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

= $\frac{1}{2}$ x 24 x 10 cm²

= 12 x 10 cm²

= 120 cm²

Hence, the area of the rhombus is 120 cm².

Given:

Perimeter of the rhombus = 46 cm

Height of the rhombus = 8 cm

Let a be the length of a side of the rhombus.

Perimeter = 4 x Side

⇒ 46 = 4 x a

⇒ a = $\frac{46}{4}$

⇒ a = 11.5 cm

Area = base x height

= 11.5 x 8 cm²

= 92 cm²

Hence, the area of the rhombus is 92 cm².

Given:

AB = 1.8 m, CD = 0.6 m, DE = 0.8 m, EF = 0.3 m and AF = 1.2 m

Area of ABCDEF = Area of rectangle AGEF + Area of rectangle GHCD + Area of triangle HBC.

Area of rectangle AGEF = l x b = AF x EF

= 1.2 x 0.3 m²

= 0.36 m²

Area of rectangle GHCD = l x b = GH x HC

= 0.6 x 2 m² (HC = AF + ED = 1.2 + 0.8 = 2 cm)

= 1.2 m²

Area of triangle HBC = $\frac{1}{2}$ x b x h = $\frac{1}{2}$ x HB x HC

= $\frac{1}{2}$ x 0.9 x 2 m² (HB = AB - AH = 1.8 - 0.9 = 0.9)

= 0.9 x 1 m²

= 0.9 m²

Now, area of ABCDEF = 0.36 m² + 1.2 m² + 0.9 m²

= 2.46 m²

Hence, the area of cross-section is 2.46 sq. m.

Area of figure ABCEF = Area of trapezium ABCF + Area of triangle CEF

Area of triangle CEF = $\frac{1}{2}$ x b x h = $\frac{1}{2}$ x CF x DE

= $\frac{1}{2}$ x 25 x 12 m²

= 25 x 6 m²

= 150 m²

In triangle BGC,

Base² + Height² = Hypotenuse²

⇒ CG² + BG² = CB²

Let h be the length of BG.

⇒ 10² + h² = 26² (∵CG = CF - GF = CF - BA = 25 - 15 = 10)

⇒ 100 + h² = 676

⇒ h² = 676 - 100

⇒ h² = 576

⇒ h = $\sqrt{576}$

⇒ h = 24 cm

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

= $\frac{1}{2}$ (AB + CF) x BG

= $\frac{1}{2}$ (15 + 25) x 24 m²

= $\frac{1}{2}$ x 40 x 24 m²

= 20 x 24 m²

= 480 m²

So, total area of figure ABCEF = 150 + 480 m²

= 630 m²

Hence, the area is 630 m².

Area of figure ABCDE = Area of Δ ABF + Area of Δ AED + Area of Δ DHC + Area of trapezium BFHC

Area of Δ ABF = $\frac{1}{2}$ x b x h = $\frac{1}{2}$ x AF x BF

= $\frac{1}{2}$ x 50 x 50 m²

= 25 x 50 m²

= 1250 m²

Area of Δ AED = $\frac{1}{2}$ x b x h = $\frac{1}{2}$ x AD x GE

= $\frac{1}{2}$ x (AF + FG + GH + HD) x GE

= $\frac{1}{2}$ x (50 + 40 + 15 + 25) x 60 m²

= 130 x 30 m²

= 3900 m²

Area of Δ DHC = $\frac{1}{2}$ x b x h = $\frac{1}{2}$ x HD x CH

= $\frac{1}{2}$ x 25 x 25 m²

= 12.5 x 25 m²

= 312.5 m²

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

= $\frac{1}{2}$ (BF + CH) x FH

= $\frac{1}{2}$ (BF + CH) x (FH + GH)

= $\frac{1}{2}$ (50 + 25) x (40 + 15) m²

= $\frac{1}{2}$ x 75 x 55 m²

= 37.5 x 55 m²

= 2062.5 m²

Thus, area of figure ABCDE = 1250 + 3900 + 312.5 + 2062.5 m²

= 7525 m²

Hence, the area is 7525 sq. m.

Let AB = BC = a cm and triangle ABC is a right triangle,

Using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ BC² + AB² = AC²

⇒ a² + a² = ($14\sqrt{2}$)²

⇒ 2a² = 392

⇒ a² = $\frac{392}{2}$

⇒ a² = 196

⇒ a = $\sqrt{196}$

⇒ a = 14 cm

Thus, AB = BC = 14 cm.

The diameter of the semicircle is BC = 14 cm, so the radius r is:

r = $\frac{d}{2}$ = $\frac{14}{2}$ = 7 cm

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

$$\begin{aligned}= \frac{1}{2} \times \frac{22}{7} \times 7^2\\[1em] = \frac{1 \times 22}{2 \times 7}\times 49\\[1em] = \frac{22}{14}\times 49\\[1em] = \frac{1,078}{14}\\[1em] = 77 \text{cm}^2\end{aligned}$$

Thus, the area of the shaded portion (the semicircle) is 77 cm².

Hence, option 1 is the correct option.

Given:

Diameter of the original circle = 14 cm

Radius of the original circle = r = $\frac{d}{2}$ = $\frac{14}{2}$ = 7 cm

Perimeter of the original circle = 2πr

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

When diameter is doubled, new diameter = 2 x 14 cm = 28 cm

New radius, r = $\frac{d}{2}$ = $\frac{28}{2}$ = 14 cm

New perimeter of circle:

$$\begin{aligned}= 2 \times \frac{22}{7} \times 14\\[1em] = \frac{44}{7} \times 14\\[1em] = \frac{616}{7}\\[1em] = 88 \text{ cm}\\[1em]\end{aligned}$$

Thus, the new perimeter is doubled compared to the original perimeter.

Hence, option 2 is the correct option.

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

Area of shaded portion = Area of square - $\frac{1}{4}$ Area of quarter circle

$$\begin{aligned}= Side^2 - \frac{1}{4} πr^2\\[1em] = 14^2 - \frac{1}{4} \times \frac{22}{7} \times 14^2\\[1em] = 196 - \frac{22}{28}\times 14 \times 14\\[1em] = 196 - \frac{22}{2} \times 14\\[1em] = 196 - \frac{308}{2} \\[1em] = 196 - 154\\[1em] = 42\text{cm}^2\end{aligned}$$

The area of the shaded portion is 42 cm².

Hence, option 3 is the correct option.

Let r be the radius of circle.

Area of circle = πr²

When the radius is doubled, new radius = 2r

Area of new circle = π x (2r)²

= π x 4r²

= 4πr²

The new area is four times the area of the original circle.

Hence, option 3 is the correct option.

Area of shaded portion = Area of bigger circle - Area of smaller circle

$$\begin{aligned}= π \times R^2 - π \times r^2\\[1em] = \frac{22}{7} \times 28^2 - \frac{22}{7} \times 14^2\\[1em] = \frac{22}{7} \times (28^2 - 14^2)\\[1em] = \frac{22}{7} \times (784 - 196)\\[1em] = \frac{22}{7} \times 588\\[1em] = \frac{12,936}{7}\\[1em] = 1,848 \text{ cm}^2\end{aligned}$$

The area of shaded portion is 1848 cm².

Hence, option 1 is the correct option.

Given:

Diameter = d = 28 cm

Radius = r = $\frac{d}{2}$ = $\frac{28}{2}$ = 14 cm

(i) Circumference of a circle = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 14\\[1em] = \frac{44}{7} \times 14 \\[1em] = \frac{616}{7}\\[1em] = 88 \text{ cm}\end{aligned}$$

Hence, the circumference of a circle is 88 cm.

(ii) Area of a circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 14^2\\[1em] = \frac{22}{7} \times 196\\[1em] = \frac{4,312}{7}\\[1em] = 616 \text{ cm}^2\end{aligned}$$

Hence, the area of a circle is 616 cm².

(i) Let r be the radius of the circle.

The circumference of a circle = 308 m

As we know, the circumference of a circle = 2πr

$$\begin{aligned}⇒ 2 \times \frac{22}{7} \times r = 308\\[1em] ⇒ \frac{44}{7} \times r = 308\\[1em] ⇒ r = \frac{308 \times 7}{44}\\[1em] ⇒ r = \frac{2,156}{44}\\[1em] ⇒ r = 49 \text{ m}^2\end{aligned}$$

Hence, the radius of a circle is 49 m.

(ii) Area of a circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 49^2\\[1em] = \frac{22}{7} \times 2,401\\[1em] = \frac{52,822}{7}\\[1em] = 7,546 \text{ m}^2\end{aligned}$$

Hence, the area of a circle is 7,546 m².

Given:

Circumference of a circle + Diameter of a circle = 116 cm

$$\begin{aligned}⇒ 2πr + 2r = 116\\[1em] ⇒ 2r(π + 1) = 116\\[1em] ⇒ 2r\Big(\frac{22}{7} + 1\Big) = 116\\[1em] ⇒ 2r\Big(\frac{22 + 7}{7}\Big) = 116\\[1em] ⇒ 2r\Big(\frac{29}{7}\Big) = 116\\[1em] ⇒ r\Big(\frac{58}{7}\Big) = 116\\[1em] ⇒ r = \frac{116 \times 7}{58}\\[1em] ⇒ r = \frac{812}{58}\\[1em] ⇒ r = 14 \text{ cm}\end{aligned}$$

Hence, the radius of a circle is 14 cm.

Circumference of a circle = 2πr

For the first circle,

r₁ = 25 cm

Circumference₁ = 2 x π x 25 = 50π cm

For the second circle,

r₂ = 18 cm

Circumference₂ = 2 x π x 18 = 36π cm

Total circumference = Circumference₁ + Circumference₂

= 50π + 36π

= 86π cm

Let the radius of the new circle be R.

Circumference of the new circle = 2πR

86π = 2πR

86 = 2R

R = $\frac{86}{2}$

R = 43 cm

Hence, the radius of the new circle is 43 cm.

Circumference of a circle = 2πr

For the first circle,

r₁ = 48 cm

Circumference₁ = 2 x π x 48 = 96π cm

For the second circle,

r₂ = 13 cm

Circumference₂ = 2 x π x 13 = 26π cm

Total circumference = Circumference₁ - Circumference₂

= 96π - 26π

= 70π cm

Let the radius of the new circle be R.

Circumference of the new circle = 2πR

70π = 2πR

70 = 2R

R = $\frac{70}{2}$

R = 35 cm

And, area of the new circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 35^2\\[1em] = \frac{22}{7} \times 1,225\\[1em] = \frac{26,950}{7}\\[1em] = 3,850 \text{ cm}^2\end{aligned}$$

Hence, the area of new circle is 3,850 cm².

For the first circle,

d₁ = 32 cm

r₁ = $\frac{d_1}{2}$ = $\frac{32}{2}$ = 16 cm

Area₁ = π x 16² = 256π cm²

For the second circle,

d₂ = 24 cm

r₂ = $\frac{d_2}{2}$ = $\frac{24}{2}$ = 12 cm

Area₂ = π x 12² = 144π cm²

Total area = Area₁ + Area₂

= 256π + 144π

= 400π cm

Let the radius of the new circle be R.

Area of the new circle = πR²

400π = πR²

400 = R²

R = $\sqrt{400}$

R = 20 cm

Hence, the radius of the new circle is 20 cm.

Given:

Radius of original circle = r = 5 m

Area of new circle = 49 times area of original circle.

Let R be the radius of new circle.

$$\begin{aligned}⇒ πR^2 = 49 \times πr^2\\[1em] ⇒ \frac{22}{7} \times R^2 = 49 \times \frac{22}{7} \times 5^2\\[1em] ⇒ \cancel{\frac{22}{7}} \times R^2 = 49 \times \cancel{\frac{22}{7}} \times 5^2\\[1em] ⇒ R^2 = 49 \times 25\\[1em] ⇒ R^2 = 1,225\\[1em] ⇒ R = \sqrt{1,225}\\[1em] ⇒ R = 35 \text{ m}^2\end{aligned}$$

Circumference of new circle = 2πR

$$\begin{aligned}= 2 \times \frac{22}{7} \times 35\\[1em] = \frac{44}{7} \times 35\\[1em] = \frac{1,540}{7}\\[1em] = 220 \text{ m}^2\end{aligned}$$

Hence, the circumference of new circle is 220 m.

Given:

The dimensions of rectangular piece of card-board are:

Length = 55 cm

Width = 42 cm

The largest circle that can be cut from the rectangle will have a diameter equal to the shorter side of the rectangle.

Diameter = Width = 42 cm

∵ Radius = r = $\frac{d}{2}$ = $\frac{42}{2}$ = 21 cm

Area of the circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 21^2\\[1em] = \frac{22}{7} \times 441\\[1em] = \frac{9,704}{7}\\[1em] = 1,386 \text{ cm}^2\end{aligned}$$

Area of the rectangular piece of cardboard = 55 x 42 cm²

= 2,310 cm²

Therefore, area of remaining cardboard = Area of rectangle - Area of circle

= (2,310 - 1,386) cm²

= 924 cm²

So, the ratio between the area of the circle cut and the area of the remaining card-board = 1,386 : 924

= 231 : 154

= 3 : 2

Hence, the ratio between the area of the circle cut and the area of the remaining cardboard is 3 : 2.

Given:

Side of square ABCD = 28 cm

Side of square = 2 x diameter of circle

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

Radius of circle = $\frac{d}{2}$ = $\frac{14}{2}$ = 7 cm

Area of the remaining card-board = Area of square - 4 x Area of 1 circle

= side² - 4 x πr²

$$\begin{aligned}= 28^2 - 4 \times \frac{22}{7} \times 7^2\\[1em] = 28^2 - 4 \times \frac{22}{7} \times 7 \times 7\\[1em] = 28^2 - 4 \times \frac{22}{\cancel{7}} \times \cancel{7} \times 7\\[1em] = 28^2 - 4 \times 22 \times 7\\[1em] = 784 - 616\\[1em] = 168 \text{ cm}^2\end{aligned}$$

Hence, the area of remaining cardboard is 168 cm².

Let the two radii of two circles be 3a and 8a.

The difference between their areas = 2695 π cm²

⇒ π x (8a)² - π x (3a)² = 2695 π

⇒ π x 64a² - π x 9a² = 2695 π

⇒ π x (64a² - 9a²) = 2695 π

⇒ $\cancel{π}$ x (64a² - 9a²) = 2695 $\cancel{π}$

⇒ 64a² - 9a² = 2695

⇒ 55a² = 2695

⇒ a² = $\frac{2695}{55}$

⇒ a² = 49

⇒ a = $\sqrt{49}$

⇒ a = 7 cm

The radii are 3a and 8a = 3 x 7 cm and 8 x 7 cm = 21 cm and 56 cm

Area of smaller circle = π x (21)²

$$\begin{aligned}= \frac{22}{7} \times 441\\[1em] = \frac{9,702}{7} \\[1em] = 1,386 \text{ cm}^2\end{aligned}$$

Hence, the area of smaller circle is 1,386 cm².

Let the diameters of the three circles be 3a, 5a and 6a.

Radius of three circles = $\frac{3a}{2}$, $\frac{5a}{2}$ and $\frac{6a}{2}$

Circumference of a circle = 2πr

For the first circle,

r₁ = $\frac{3a}{2}$ cm

Circumference₁ = 2 x π x $\frac{3a}{2}$ = 3aπ cm

For the second circle,

r₂ = $\frac{5a}{2}$ cm

Circumference₂ = 2 x π x $\frac{5a}{2}$ = 5aπ cm

For the third circle,

r₂ = $\frac{6a}{2}$ cm

Circumference₂ = 2 x π x $\frac{6a}{2}$ = 6aπ cm

Total circumference = Circumference₁ + Circumference₂ + Circumference₃

⇒ 3aπ + 5aπ + 6aπ = 308

⇒ 14aπ = 308

⇒ 14 x a x $\frac{22}{7}$ = 308

⇒ 2 x a x 22 = 308

⇒ 44a = 308

⇒ a = $\frac{308}{44}$

⇒ a = 7 cm

Radius of three circles = $\frac{3}{2}$ x 7 cm, $\frac{5}{2}$ x 7 cm and $\frac{6}{2}$ x 7 cm

= 10.5 cm, 17.5 cm and 21 cm

Difference between the area of the largest and the smallest circles = π(21)² - π(10.5)²

= 441π - 110.25π cm²

= 330.75π cm²

= 330.75 x $\frac{22}{7}$ cm²

= 47.25 x 22 cm²

= 1039.5 cm²

Hence, the difference in the area = 1039.5 cm².

Given:

r₁ = 20 cm

r₂ = 15 cm

Area of ring shaped region = π(r₁² - r₂²)

= π(20² - 15²) cm²

= π(400 - 225) cm²

= π x 175 cm²

= $\frac{22}{7}$ x 175 cm²

= 22 x 25 cm²

= 550 cm²

Hence, the area of ring shaped region is 550 cm².

Let r be the radius of the circular park.

Circumference = 2πr

⇒ 2πr = 55

⇒ 2 x $\frac{22}{7}$ x r = 55

⇒ $\frac{44}{7}$ x r = 55

⇒ r = $\frac{7 \times 55}{44}$

⇒ r = $\frac{7 \times 5}{4}$

⇒ r = $\frac{35}{4}$

⇒ r = 8.75 m

Radius of outer park = radius of park + width of the path

= 8.75 + 3.5 m

= 12.25 m

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

= π (12.25)² - π (8.75)² m²

= 150.0625π - 76.5625π m²

= 73.5 π m²

= 73.5 x $\frac{22}{7}$ m²

= 10.5 x 22 m²

= 231 m²

Hence, the area of the path is 231 m².

Let r be the radius of the circular garden A.

Circumference of garden A = 2πr

⇒ 2πr = 1.760 km

⇒ 2πr = 1760 m

⇒ 2 x $\frac{22}{7}$ x r = 1760 m

⇒ $\frac{44}{7}$ x r = 1760 m

⇒ r = $\frac{7 \times 1760}{44}$ m

⇒ r = 7 x 40 m

⇒ r = 280 m

Area of garden A = πr²

= π x (280)²

= 78400π

Let R be the radius of garden B.

It is given that the area of garden B is 25 times the area of garden A.

⇒ πR² = 25 x πr²

⇒ πR² = 25 x 78400π

⇒ $\cancel{π}$ R² = 25 x 78400 $\cancel{π}$

⇒ R² = 25 x 78400

⇒ R² = 1960000

⇒ R = $\sqrt{1960000}$

⇒ R = 1400 m

Circumference of garden B = 2πR

= 2 x $\frac{22}{7}$ x 1400 m

= 2 x 22 x 200 m

= 8800 m = 8.8 km

Hence, the circumference of garden B is 8.8 km.

Diameter of the wheel = 84 cm

Radius of the wheel = $\frac{d}{2}$ = $\frac{84}{2}$ = 42 cm

Circumference of the wheel = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 42\\[1em] = 2 \times 22 \times 6\\[1em] = 264 \text{ cm}\end{aligned}$$

Let the wheel make n revolutions.

Total distance traveled by the wheel = n x Circumference of the wheel

⇒ 3.168 km = n x 264 cm

⇒ 316,800 cm = n x 264 cm

⇒ n = $\frac{316,800}{264}$

⇒ n = 1200

Hence, the wheel makes 1,200 revolutions.

Diameter of wheel = 80 cm

Radius of the wheel = $\frac{d}{2}$ = $\frac{80}{2}$ = 40 cm

Circumference of the wheel = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 40\\[1em] = \frac{1,760}{7} \text{ cm}\end{aligned}$$

Let the wheel make n number of revolutions.

Speed of the car = 66 km per hour

Time = 10 min

Total distance = Speed x Time

= $\frac{66 \times 100,000}{60}$ x 10 cm

= 11 x 100,000 cm

= 1,100,000

Total distance = Number of revolutions x Circumference of the wheel

⇒ 1,100,000 cm = n x $\frac{1,760}{7}$ cm

⇒ n = $\frac{1,100,000 x 7}{1,760}$

⇒ n = $\frac{7,700,000 }{1,760}$

⇒ n = 4,375

Hence, the wheel makes 4,375 complete revolutions.

Diameter of wheel = 42 cm

Radius of wheel = $\frac{d}{2}$ = $\frac{42}{2}$ = 21 cm

Circumference of wheel = 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 21\\[1em] = 2 \times 22 \times 3\\[1em] = 132 \text{ cm}\end{aligned}$$

Total distance covered by one wheel in 1 minute = Number of revolutions x Circumference of wheel

= 1,200 x 132 cm

= 158,400 cm

= 1.584 km

Speed of the train = $\frac{\text{Distance}}{\text{Time}}$

= $\frac{1.584}{\frac{1}{60}}$

= $\frac{1.584 \times 60}{1}$

= 95.04 km/hr

Hence, the speed of the train is 95.04 km/hr.

Length of the minute hand (radius of the circle) = 8 cm

Time interval between 9.05 a.m. and 8.30 a.m. = 35 minutes

Area swept by the minute hand in 1 hr = πr²

$$\begin{aligned}= \frac{22}{7} \times 8^2\\[1em] = \frac{22}{7} \times 64\\[1em] = \frac{1,408}{7} \text{ cm}^2\end{aligned}$$

Area of the circle in 60 minutes = $\frac{1,408}{7}$ cm²

Area of the circle in 1 minute = $\frac{1,408}{7 \times 60}$ cm²

= $\frac{352}{105}$ cm²

Area of the circle in 35 minutes = $\frac{352 \times 35}{105}$ cm²

= $\frac{352}{3}$ cm²

= $117\frac{1}{3}$ cm²

Hence, the area swept by the minute hand between 8:30 a.m. and 9:05 a.m. is $117\frac{1}{3}$ cm².

Circumference of the circle = 2πr

For outer circle,

$$\begin{aligned}⇒ 2 \times \frac{22}{7} \times R = 396\\[1em] ⇒ \frac{44}{7} \times R = 396\\[1em] ⇒ R = \frac{396 \times 7}{44}\\[1em] ⇒ R = \frac{9 \times 7}{1}\\[1em] ⇒ R = 63 \text{ cm}\end{aligned}$$

For inner circle,

$$\begin{aligned}⇒ 2 \times \frac{22}{7} \times r = 374\\[1em] ⇒ \frac{44}{7} \times r = 374\\[1em] ⇒ r = \frac{374 \times 7}{44}\\[1em] ⇒ r = \frac{17 \times 7}{2}\\[1em] ⇒ r = 59.5 \text{ cm}\end{aligned}$$

Area of shaded portion = π(R² - r²)

= π(63² - 59.5²) cm²

= π(3,969 - 3,540.25) cm²

= π x 428.75 cm²

= $\frac{22}{7}$ x 428.75 cm²

= 22 x 61.25 cm²

= 1,347.5 cm²

Hence, the area of the shaded portion is 1,347.5 cm².

Given:

Base of the triangle = 8 cm

Hypotenuse of the triangle = 10 cm

Let h be the height of the triangle.

In a right-angled triangle, using the Pythagoras theorem,

⇒ Base² + Height² = Hypotenuse²

⇒ (8)² + h² = (10)²

⇒ 64 + h² = 100

⇒ h² = 100 - 64

⇒ h² = 36

⇒ h = ​$\sqrt{36}$

⇒ h = 6 cm.

By formula,

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

= $\frac{1}{2}$ x 8 x 6 cm²

= 24 cm².

Hence, option 1 is the correct option.

Given,

Area of equilateral triangle = $4\sqrt{3}$ cm²

By formula,

Area of the equilateral triangle = $\frac{\sqrt{3}}{4}$ x side²

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

By formula,

Perimeter of equilateral triangle = 3 x side

= 3 x 4 cm

= 12 cm.

Hence, option 3 is the correct option.

Given,

Perimeter of a square = 80 cm

⇒ 4 x side = 80

⇒ side = $\frac{80}{4}$

⇒ side = 20 cm

By formula,

Area of square = side²

= 20² cm²

= 400 cm².

So, given area is false.

Hence, option 2 is the correct option.

Given, area of a trapezium = 32 cm²

Distance between its parallel sides = 8 cm

By formula,

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

⇒ 32 = $\frac{1}{2}$ x (Sum of parallel sides) x 8

⇒ 32 = Sum of parallel sides x 4

⇒ Sum of parallel sides = $\frac{32}{4}$

⇒ Sum of parallel sides = 8 cm.

Hence, option 3 is the correct option.

Given,

Sides of a triangular board are 8 cm, 6 cm and 10 cm.

⇒ 10² = 100, 8² = 64 and 6² = 36.

⇒ 100 = 64 + 36

⇒ 10² = 8² + 6²

This confirms the triangle with sides 6 cm, 8 cm, and 10 cm is a right triangle, where 10 cm is the hypotenuse.

Thus, sides of the triangle are 8 cm and 6 cm.

So, statement 2 is true.

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

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

Rate of painting = ₹ 10 per square cm

Cost of painting = Area x rate of painting

= $\frac{1}{2}$ x 8 x 6 x ₹ 10

So, statement 1 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Given,

The perimeter of a rhombus = 40 cm.

One diagonal = 12 cm.

As we know, the perimeter of the rhombus = 4 x side

⇒ 4 x side = 40 cm

⇒ side = $\frac{40}{4}$ cm

⇒ side = 10 cm

Let ABCD be a rhombus. From figure,

AB = 10 cm

Let diagonal AC = 12 cm

We know that,

Diagonals of rhombus bisect each other.

Then, OA = OC = $\frac{12}{2}$ = 6 cm

Let OB be a cm.

Since the diagonal of a rhombus bisect at 90°.

Applying pythagoras theorem in triangle AOB, we get:

⇒ AB² = OA² + OB²

⇒ (10)² = (6)² + a²

⇒ a² = (10)² - (6)²

⇒ a² = 100 - 36

⇒ a² = 64

⇒ a = $\sqrt{64}$

⇒ a = 8

So, OB = 8 cm

BD = 2 x OB = 2 x 8 cm = 16 cm.

∴ Statement 1 is true.

As solved above,

If the other diagonal of this rhombus = x cm,

Then $\frac{x}{2}$ will be equal to 10² - 6².

∴ Statement 2 is false.

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

Hence, option 3 is the correct option.

Perimeter of the figure = sum of all the sides

= 7 + 13 + 12 + x

= (32 + x) cm

So, assertion (A) is true.

From figure,

AB = AC - BC = 12 - 7 = 5 cm

Since, ABE is a right angled triangle. Using pythagoras theorem,

⇒ AE² = AB² + BE²

⇒ 13² = 5² + x²

⇒ 169 = 25 + x²

⇒ x² = 169 - 25

⇒ x² = 144

⇒ x = $\sqrt{144}$

⇒ x = 12 cm

Perimeter = (32 + x) = (32 + 12) cm.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Given, BC = 14 cm

From figure,

Diameter (d) = 2r = 14 cm

AB has 4 circles (3 circles and 2 semi-circles).

AB = 4 x diameter of one circle = 4 x 2r = 4 x 14 cm = 56 cm.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

ΔABC is shown in the figure below:

Area of isosceles triangle ABC =

$$\begin{aligned}= \frac{b}{4} \sqrt{4a^2 - b^2}\\[1em] = \frac{36}{4} \sqrt{4 \times 30^2 - 36^2}\\[1em] = 9 \sqrt{4 \times 30^2 - 36^2}\\[1em] = 9 \sqrt{4 \times 900 - 1,296}\\[1em] = 9 \sqrt{3600 - 1,296}\\[1em] = 9 \sqrt{2,304}\\[1em] = 9 \times 48 \\[1em] = 432 \text{ cm}^2\end{aligned}$$

∠ BOC = ∠ COD = 45° (∵ AD divide ∠ BOC in 2 equal halves)

Let OB = OC = x.

In Δ BOC, by using the Pythagoras theorem,

OB² + OC² = BC²

⇒ x² + x² = (36)²

⇒ 2x² = 1,296

⇒ x² = $\frac{1,296}{2}$

⇒ x² = 648

⇒ x = $\sqrt{648}$

⇒ x = 18 $\sqrt{2}$

Now the area of triangle BOC = $\frac{1}{2}$ x base x height

$$\begin{aligned}= \frac{1}{2} \times 18 \sqrt{2} \times 18 \sqrt{2}\\[1em] = \frac{1}{2} \times 18 \times 18 \times 2\\[1em] = 18 \times 18\\[1em] = 324 \text{ cm}^2\end{aligned}$$

Area of quadrilateral ABOC = Area of Δ ABC - Area of Δ BOC

= 432 - 324 cm²

= 108 cm²

Hence, the area of quadrilateral ABOC is 108 cm².

Given:

The length of the rectangular field = 30 m

The breadth of the rectangular field = 24 m

Area of path = 360 m²

Let the width of the path be x m.

The length of the smaller rectangular field = 30 m - x m - x m

= 30 - 2x m

The breadth of the smaller rectangular field = 24 m - x m - x m

= 24 - 2x m

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

⇒ Area of the larger rectangular field = 30 x 24 m²

= 720 m²

⇒ Area of the smaller rectangular field = (30 - 2x) x (24 - 2x) m²

= (720 - 108x - 4x²) m²

Area of the path = Area of larger rectangular field - Area of smaller rectangular field

⇒ 360 = 720 - (720 - 108x - 4x²)

⇒ 360 = 720 - 720 + 108x + 4x²

⇒ 4x² + 108x - 360 = 0

⇒ x² + 27x - 90 = 0

⇒ x² + 30x - 3x - 90 = 0

⇒ x(x + 30) - 3(x + 30) = 0

⇒ (x + 30)(x - 3) = 0

⇒ x = - 30 or 3

Since the width of the path cannot be negative,

x = 3 m

Hence, the width of the path is 3 m.

(i) Area of the square = 484 m²

Let a be the length of side of the square.

⇒ a² = 484

⇒ a = $\sqrt{484}$

⇒ a = 22 m

Total length of the wire = Perimeter of the square = 4 x 22m = 88 m

Perimeter of the square = Perimeter of equilateral triangle.

⇒ 3 x side = 88 m

⇒ side = $\frac{88}{3}$ m

⇒ side = 29.3 m

Area of equilateral triangle = $\frac{\sqrt{3}}{4}$ x side²

= $\frac{\sqrt{3}}{4}$ x (29.3)² m²

= $\frac{\sqrt{3}}{4}$ x 858.49 m²

= 372.57 m²

Hence, the area of the equilateral triangle is 372.57 sq. m.

(ii) Given:

Length of the rectangle = 16 m

Let b be the breadth of the rectangle.

Perimeter of the rectangle = Perimeter of the square

⇒ 2(l + b) = 88 m

⇒ 2(16 + b) = 88 m

⇒ 16 + b = $\frac{88}{2}$ m

⇒ 16 + b = 44 m

⇒ b = 44 - 16 m

⇒ b = 28 m

Area of the rectangle = l x b

= 16 x 28 m²

= 448 m²

Hence, the area of the rectangle is 448 sq. m.

Draw CE parallel to DA which meets AB at point E.

Since, AB || DC thus AE || DC and DA || CE.

Since, opposite sides are parallel, thus AECD is a parallelogram.

Opposite sides of a parallelogram are equal, thus CE = AD = 10 cm and AE = DC = 12 cm.

In isosceles triangle EBC,

Draw CF ⊥ EB.

In an isosceles triangle, the perpendicular from the common vertex to the base, bisects it.

Thus, EF = $\frac{EB}{2} = \frac{8}{2}$ = 4 cm.

In right triangle CEF,

By pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

EC² = CF² + EF²

10² = CF² + 4²

100 = CF² + 16

CF² = 100 - 16

CF² = 84

CF = $\sqrt{84}$ = 9.16 cm

By formula,

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

Area of trapezium ABCD = $\frac{1}{2} \times (AB + DC) \times CF$

$$\begin{aligned}= \frac{1}{2} \times (20 + 12) \times 9.16 \\[1em] = \frac{1}{2} \times 32 \times 9.16 \\[1em] = 146.56 \text{ cm}^2.\end{aligned}$$

Hence, area of trapezium ABCD = 146.56 cm².

Draw CO such that CO is perpendicular to AB.

AO = DC = 8 cm

AB = AO + OB

⇒ 14 = 8 + OB

⇒ OB = 14 - 8 = 6 cm

Area of trapezium ABCD = $\frac{1}{2}$ x (sum of parallel sides) x height

In triangle BCO, by using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ (6)² + Height² = 10²

⇒ 36 + Height² = 100

⇒ Height² = 100 - 36

⇒ Height² = 64

⇒ Height = $\sqrt{64}$

⇒ Height = 8 cm

Area of trapezium ABCD = $\frac{1}{2}$ x (8 + 14) x 8

= $\frac{1}{2}$ x 22 x 8 sq. cm

= 11 x 8 sq. cm

= 88 sq. cm

Hence, the area of trapezium ABCD is 88 sq. cm.

Draw CE parallel to DA which meets AB at point E.

AE = DC = 20 cm

AB = AE + EB

⇒ 32 = 20 + EB

⇒ EB = 32 - 20 = 12 cm

And, DA = CE = 10 cm

For the Δ EBC,

Let the sides of the triangle be:

a = 10 cm, b = 12 cm and c = 16 cm.

The semi-perimeter s:

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{10 + 12 + 16}{2}\\[1em] = \frac{38}{2}\\[1em] = 19\end{aligned}$$

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

= $\sqrt{19(19 - 10)(19 - 12)(19 - 16)}$ cm²

= $\sqrt{19 \times 9 \times 7 \times 3}$ cm²

= $\sqrt{3,591}$ cm²

= 59.9 cm²

Let h be the height of Δ EBC,

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

⇒ $\frac{1}{2}$ x 12 x height = 59.9

⇒ 6 x height = 59.9

⇒ height = $\frac{59.9}{6}$

⇒ height = 9.98 cm

Area of trapezium ABCD = $\frac{1}{2}$ x (sum of parallel sides) x height

= $\frac{1}{2}$ x (20 + 32) x 9.98

= $\frac{1}{2}$ x 52 x 9.98 sq. cm

= 26 x 9.98 sq. cm

= 259.65 sq. cm

Hence, the area of trapezium ABCD is 259.65 sq. cm.

Draw DE and CF perpendicular to AB.

ABCD is an isosceles trapezium.

Let EF = FB = x cm

DC = EF = 18 cm

AB = AE + EF + FB

⇒ 30 = x + 18 + x

⇒ 30 = 2x + 18

⇒ 2x = 30 - 18

⇒ 2x = 12

⇒ x = $\frac{12}{2}$

⇒ x = 6 cm

As EDA is a right angled triangle, by using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ (6)² + height² = 12²

⇒ 36 + height² = 144

⇒ height² = 144 - 36

⇒ height² = 108 cm

⇒ height = $\sqrt{108}$ cm

⇒ height = 10.39 cm

Area of trapezium ABCD = $\frac{1}{2}$ x (sum of parallel sides) x height

= $\frac{1}{2}$ x (18 + 30) x 10.39

= $\frac{1}{2}$ x 48 x 10.39 sq. cm

= 24 x 10.39 sq. cm

= 249.41 sq. cm

Hence, the area of trapezium ABCD is 249.41 sq. cm.

Given:

Perimeter of the rectangular board = 70 cm

Length = x cm

Let b be the width of the rectangular board.

Perimeter of a rectangle = 2(l + b)

⇒ 2(x + b) = 70

⇒ x + b = $\frac{70}{2}$

⇒ x + b = 35

⇒ b = 35 - x

Area = l x b

⇒ x $\times$ (35 - x) = 300

⇒ 35x - x² = 300

⇒ 35x - x² - 300 = 0

⇒ x² - 35x + 300 = 0

⇒ x² - 20x - 15x + 300 = 0

⇒ x(x - 20) - 15(x - 20) = 0

⇒ (x - 20)(x - 15) = 0

⇒ x = 20 or 15

Therefore, b = (35 - x)

b = 15 or 20

Hence, the width of rectangular board is (35 - x) and the dimensions of the board are 20 cm and 15 cm.

Given:

Area of the rectangle = 640 m²

Length = x meters

Let b be the width of the rectangle.

Area of the rectangle = l x b

⇒ x × b = 640

⇒ b = $\frac{640}{x}$ m

By formula,

Perimeter of the rectangle = 2(l + b)

⇒ $2\Big(x + \frac{640}{x}\Big)$ = 104

⇒ $\Big(x + \frac{640}{x}\Big) = \frac{104}{2}$

⇒ $\Big(x + \frac{640}{x}\Big)$ = 52

⇒ x² + 640 = 52x

⇒ x² - 52x + 640 = 0

⇒ x² - 32x - 20x + 640 = 0

⇒ x(x - 32) - 20(x - 32) = 0

⇒ (x - 20)(x - 32) = 0

⇒ x - 20 = 0 or x - 32 = 0

⇒ x = 20 m or x = 32 m.

Now,

⇒ b = $\frac{640}{x}$

Case 1 : x = 20 m

⇒ b = $\frac{640}{20}$ = 32 m.

Case 2 : x = 32 m

⇒ b = $\frac{640}{32}$ = 20 m.

Hence, the width of rectangle is $\frac{640}{x}$ and the dimensions of the rectangle are 20 m and 32 m.

Given:

Let s be the side of the square.

Length of the rectangle = 2 x Side of the square

⇒ l = 2s

Width of the rectangle = Side of the square + 6

⇒ w = s + 6

Area of the rectangle = 3 x area of the square

⇒ l x w = 3s²

⇒ 2s x (s + 6) = 3s²

⇒ 2s² + 12s = 3s²

⇒ 2s² + 12s - 3s² = 0

⇒ - s² + 12s = 0

⇒ s² - 12s = 0

⇒ s(s - 12) = 0

⇒ s = 0 or 12

Since the side cannot be zero, s = 12 cm.

The dimensions of the rectangle:

l = 2 x s = 2 x 12 cm = 24 cm

w = s + 6 = 12 + 6 = 18 cm

Hence, the length and width of the rectangle are 24 cm and 18 cm, and the side of the square is 12 cm.

Square ABCD is shown in the figure below:

Given:

$\frac{\text{Area of Δ ABP}}{\text{Area of trapezium APCD}} = \frac{1}{5}$

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

= $\frac{1}{2}$ x 12 x (12 - CP)

Area of trapezium APCD = $\frac{1}{2}$ x (sum of parallel sides) x height

= $\frac{1}{2}$ x (12 + CP) x 12

$$\begin{aligned}⇒ \frac{\frac{1}{2} \times 12 \times (12 - CP)}{\frac{1}{2} \times (12 + CP) \times 12} = \frac{1}{5}\\[1em] ⇒ \frac{\cancel{\frac{1}{2}} \times 12 \times (12 - CP)}{\cancel{\frac{1}{2}} \times (12 + CP) \times 12} = \frac{1}{5}\\[1em] ⇒ \frac{\cancel{12} \times (12 - CP)}{ (12 + CP) \times \cancel{12}} = \frac{1}{5}\\[1em] ⇒ \frac{(12 - CP)}{ (12 + CP)} = \frac{1}{5}\\[1em] ⇒ 5 \times (12 - CP) = 1 \times (12 + CP)\\[1em] ⇒ 60 - 5CP = 12 + CP\\[1em] ⇒ 60 - 12 = 5CP + CP\\[1em] ⇒ 48 = 6CP\\[1em] ⇒ CP = \frac{48}{6}\\[1em] ⇒ CP = 8 \text { cm}\end{aligned}$$

Hence, the length of CP is 8 cm.

Given:

The boundary wall is built all around the rectangular plot at a distance of 1 meter, which means the wall forms a larger rectangle around the plot as shown in the figure below:

Dimensions of the outer rectangle (including the boundary wall):

Length of the outer rectangle = (45 + 1 + 1) m = 47 m

Width of the outer rectangle = (30 + 1 + 1) m = 32 m

Perimeter of the outer rectangle = 2(length + breadth)

= 2(47 + 32)

= 2 x 79 = 158 m

Area of the inner surface of the boundary wall is the perimeter of the outer rectangle multiplied by the height of the wall.

Inner surface area of boundary wall = Perimeter of outer rectangle x height of wall

= 158 m x 2.4 m

= 379.2 m²

Hence, the area of the inner surface of the boundary wall is 379.2 m².

(i) Area of the square = 576 cm²

Let a be the length of side of the square.

⇒ a² = 576

⇒ a = $\sqrt{576}$

⇒ a = 24 cm

Total length of the wire = Perimeter of the square = 4 x 24 cm = 96 cm

Perimeter of the square = Perimeter of equilateral triangle.

⇒ 3 x side = 96 cm

⇒ side = $\frac{96}{3}$ cm

⇒ side = 32 cm

Area of equilateral triangle = $\frac{\sqrt{3}}{4}$ x side²

= $\frac{\sqrt{3}}{4}$ x (32)² cm²

= $\frac{\sqrt{3}}{4}$ x 1,024 cm²

= 256 $\sqrt{3}$ cm²

Hence, the area of equilateral triangle is 256 $\sqrt{3}$ cm².

(ii) Given:

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

l - b = 4 ...............(1)

Perimeter of rectangle = Perimeter of square

⇒ 2(l + b) = 96 cm

⇒ 2(l + b) = 96 cm

⇒ l + b = $\frac{96}{2}$ cm

⇒ l + b = 48 cm ...............(2)

Add equation (1) and (2), we get

⇒ (l - b) + (l + b) = 4 + 48

⇒ l - b + l + b = 52

⇒ 2l = 52

⇒ l = $\frac{52}{2}$

⇒ l = 26 cm

So, b = l - 4 = 26 - 4 = 22 cm

Area of rectangle = l x b

= 26 x 22 cm²

= 572 cm²

Hence, the area of rectangle is 572 cm².

Given:

Area of the first parallelogram = y cm²

Height of the first parallelogram = h cm

Area of the first parallelogram = base x height

⇒ y = base x h

⇒ base of the first parallelogram = $\frac{y}{h}$ cm

The base of the second parallelogram = $\Big(\frac{y}{h} + x\Big)$ cm

Area of the second parallelogram = base x height

⇒ $2y = \Big(\frac{y}{h} + x\Big)$ x height

⇒ height = $\frac{2hy}{y + hx}$

Hence, the height of the second parallelogram is $\frac{2hy}{y + hx}$.

Given:

Let the given trapezium be as shown in the figure below:

The distance between parallel sides of a trapezium, (i.e., height) = 15 cm

The length of the line segment joining the mid-points of its non-parallel sides, EF = 26 cm

Since EF is half of the sum of the lengths of the parallel sides.

EF = $\frac{1}{2}$ x (AD + BC)

⇒ $\frac{1}{2}$ x (AD + BC) = 26

⇒ AD + BC = 26 x 2 cm = 52 cm

Area of trapezium = $\frac{1}{2}$ x (Sum of parallel sides) x height

= $\frac{1}{2}$ x (AD + BC) x height

= $\frac{1}{2}$ x 52 x 15

= 26 x 15

= 390 cm²

Hence, the area of the trapezium is 390 cm².

Given:

Diagonal of the rectangular plot = 34 m

Perimeter of the rectangular plot = 92 m

Let l be the length and b be the breadth of the rectangular plot.

Perimeter = 2(l + b)

⇒ 2(l + b) = 92

⇒ l + b = $\frac{92}{2}$

⇒ l + b = 46

Squaring both side,

⇒ (l + b)² = 46²

⇒ l² + b² + 2lb = 2,116 ...............(1)

As all the angles of rectangle are 90°, we can use the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ l² + b² = 34²

⇒ l² + b² = 1,156

Using equation (1), we get

⇒ 2,116 - 2lb = 1,156

⇒ 2,116 - 1,156 = 2lb

⇒ 960 = 2lb

⇒ lb = $\frac{960}{2}$

⇒ lb = 480

Hence, the area of the rectangular plot is 480 m².

Given:

Rate of fencing = ₹ 240 per metre

Total cost = ₹ 52,800

Let r be the radius of the field.

Total cost = Circumference of field x Rate of fencing

$$\begin{aligned}⇒ 52,800 = 2πr \times 240\\[1em] ⇒ 2πr = \frac{52,800}{240}\\[1em] ⇒ 2πr = 220\\[1em] ⇒ 2 \times \frac{22}{7} \times r = 220\\[1em] ⇒ \frac{44}{7} \times r = 220\\[1em] ⇒ r = \frac{220 \times 7}{44}\\[1em] ⇒ r = \frac{5 \times 7}{1}\\[1em] ⇒ r = 35 \text{ m}\end{aligned}$$

Area of the field = πr²

$$\begin{aligned}= \frac{22}{7} \times 35^2\\[1em] = \frac{22}{7} \times 1225\\[1em] = 22 \times 175\\[1em] = 3,850 \text{ m}^2\end{aligned}$$

Rate of ploughing = ₹ 12.50 per m²

Total cost of ploughing = Area of the field x Rate of ploughing

= 3,850 x 12.50

= ₹ 48,125

Hence, the total cost of ploughing is ₹ 48,125.

Let r₁ and r₂ be the radii of the two circles.

Sum of the areas = 58π

⇒ A₁ + A₂ = 58π

⇒ πr₁² + πr₂² = 58π

⇒ π(r₁² + r₂²) = 58π

⇒ $\cancel{π}$ (r₁² + r₂²) = 58 $\cancel{π}$

⇒ r₁² + r₂² = 58 ...............(1)

Also, since the circles touch externally, the sum of their radii equals the distance between their centers.

r₁ + r₂ = 10

⇒ r₁ = 10 - r₂

Substituting the value of r₁ in equation (1),

⇒ (10 - r₂)² + r₂² = 58

⇒ 10² + r₂² - 2 x 10 x r₂ + r₂² = 58

⇒ 100 + 2r₂² - 20r₂ = 58

⇒ 100 + 2r₂² - 20r₂ - 58 = 0

⇒ 2r₂² - 20r₂ - 42 = 0

⇒ r₂² - 10r₂ - 21 = 0

⇒ r₂² - 3r₂ - 7r₂ - 21 = 0

⇒ r₂(r₂ - 3) - 7(r₂ - 3) = 0

⇒ (r₂ - 3)(r₂ - 7) = 0

⇒ r₂ = 3 cm or 7 cm

Using equation (1), we find r₁:

If r₂ = 3, then r₁ = 10 - 3 = 7 cm

If r₂ = 7, then r₁ = 10 - 7 = 3 cm

Hence, the radii of the two circles are 7 cm and 3 cm.

ABCD is a rectangle. So, ∠ABC = 90°.

By using the Pythagoras theorem,

Let h be the diagonal of the rectangle.

Base² + Height² = Diagonal²

⇒ 28² + 21² = h²

⇒ 784 + 441 = h²

⇒ 1,225 = h²

⇒ h = $\sqrt{1,225}$

⇒ h = 35 cm

Diagonal of rectangle = Diameter of circle.

Radius = $\frac{d}{2}$ = $\frac{35}{2}$ = 17.5 cm

Area of shaded portion = Area of circle - Area of rectangle

= πr² - lb cm²

= $\frac{22}{7}$ x 17.5² - 28 x 21 cm²

= $\frac{22}{7}$ x 306.25 - 28 x 21 cm²

= 22 x 43.75 - 588 cm²

= 962.5 - 588 cm²

= 374.5 cm²

Hence, area of the shaded portion = 374.5 cm².

Diagonal of the square = Diameter of the circle.

Radius of circle = 7 cm

Diameter of circle = 2 x 7 cm = 14 cm

Diagonal of square = 14 cm

Let s be the side of square.

By using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ s² + s² = 14²

⇒ 2s² = 196

⇒ s² = $\frac{196}{2}$

⇒ s² = 98 cm

⇒ s = $\sqrt{98}$ cm

Area of square = Side²

= $(\sqrt{98})^2$ cm²

= 98 cm²

Hence, the area of the square is 98 cm².

Given:

Area of the equivalent triangle = $484{\sqrt3} \text{ cm}^2$

Let a be the length of the equivalent triangle.

Area of equivalent triangle = $\frac{\sqrt{3}}{4} \times a^2$

$$\begin{aligned}⇒ \frac{\sqrt{3}}{4} \times a^2 = 484{\sqrt3} \text{ cm}^2\\[1em] ⇒ \frac{\cancel{\sqrt{3}}}{4} \times a^2 = 484 \cancel{\sqrt{3}} \text{ cm}^2\\[1em] ⇒ \frac{1}{4} \times a^2 = 484 \text{ cm}^2\\[1em] ⇒ a^2 = 484 \times 4 \text{ cm}^2\\[1em] ⇒ a^2 = 1,936 \text{ cm}^2\\[1em] ⇒ a = \sqrt{1,936} \text{ cm}^2\\[1em] ⇒ a = 44 \text{ cm}\end{aligned}$$

Perimeter of the triangle = 3 x side

= 3 x 44 cm = 132 cm

Perimeter of triangle = Circumference of circle

$$\begin{aligned}⇒ 2πr = 132\\[1em] ⇒ 2 \times \frac{22}{7} \times r = 132\\[1em] ⇒ \frac{44}{7} \times r = 132\\[1em] ⇒ r = \frac{7 \times 132}{44}\\[1em] ⇒ r = \frac{7 \times 3}{1}\\[1em] ⇒ r = 21\end{aligned}$$

Area of circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 21^2\\[1em] = \frac{22}{7} \times 441\\[1em] = 22 \times 63\\[1em] = 1,386 \text{ cm}^2\end{aligned}$$

Hence, the area of the circle is 1,386 cm².

Given:

Perimeter of the triangle = 450 m

Ratio of the sides = 12 : 5 : 13.

Let the sides of triangle be 12a, 5a and 13a.

Perimeter of the triangle = Sum of sides of the triangle

⇒ 12a + 5a + 13a = 450

⇒ 30a = 450

⇒ a = $\frac{450}{30}$

⇒ a = 15

Thus, sides are 12 x 15, 5 x 15 and 13 x 15 m

= 180 m, 75 m and 195 m

The sides of the triangle are:

a = 180 m, b = 75 m and c = 195 m.

The semi-perimeter s:

$$\begin{aligned}∵ s = \frac{a + b + c}{2}\\[1em] = \frac{180 + 75 + 195}{2}\\[1em] = \frac{450}{2}\\[1em] = 225\end{aligned}$$

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

= $\sqrt{225(225 - 180)(225 - 75)(225 - 195)}$ m²

= $\sqrt{225 \times 45 \times 150 \times 30}$ m²

= $\sqrt{45,562,500}$ m²

= 6,750 m²

Hence, the area of the triangle is 6,750 m².

Let the sides of the triangle be:

a = 26 cm, b = 28 cm and c = 30 cm.

The semi-perimeter s of the triangle is:

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

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

= $\sqrt{42(42 - 26)(42 - 28)(42 - 30)}$ cm²

= $\sqrt{42 \times 16 \times 14 \times 12}$ cm²

= $\sqrt{112,896}$ cm²

= 336 cm²

Area of triangle = Area of parallelogram

(∵ Area of parallelogram = base x height)

Let h be the height of parallelogram.

⇒ 336 = 28 x h

⇒ h = $\frac{336}{28}$

⇒ h = 12 cm

Hence, the height of the parallelogram is 12 cm.

Draw a perpendicular line CP on AB, such that CP = 9 m and AP = 23 m.

Let PB = a m

In Δ CPB, by using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ a² + (9)² = (15)²

⇒ a² + 81 = 225

⇒ a² = 225 - 81

⇒ a² = 144

⇒ a² = 144

⇒ a = $\sqrt{144}$

⇒ a = 12 m

So, AB = AP + PB = 23 + 12 = 35 m

Area of trapezium = $\frac{1}{2}$ (sum of parallel sides) x height

= $\frac{1}{2}$ (23 + 35) x 9

= $\frac{1}{2}$ x 58 x 9

= 29 x 9

= 261 m²

Hence, the area of the trapezium is 261 m².

Given:

Sum of the areas of two squares = 400 cm².

The difference of their perimeters = 16 cm

Let a and b be the sides of 2 square.

Sum of Areas,

⇒ a² + b² = 400 ...............(1)

Difference of Perimeters,

⇒ 4a - 4b = 16

⇒ 4(a - b) = 16

⇒ a - b = $\frac{16}{4}$

⇒ a - b = 4

⇒ a = 4 + b

Substituting the value of a in equation (1), we get

⇒ (4 + b)² + b² = 400

⇒ 4² + b² + 2 x 4 x b + b² = 400

⇒ 16 + b² + 8b + b² = 400

⇒ 16 + 2b² + 8b - 400 = 0

⇒ 2b² + 8b - 384 = 0

⇒ b² + 4b - 192 = 0

⇒ b² + 16b - 12b - 192 = 0

⇒ b(b + 16) - 12(b + 16) = 0

⇒ (b + 16)(b - 12) = 0

⇒ b = - 16 or 12

Since the side of a square cannot be negative, b = 12 cm.

So, a = 4 + b = 16 cm

Hence, the sides of the squares are 16 cm and 12 cm.

Diagonal of the square = 24 cm

Let a be the side of square.

By using the Pythagoras theorem,

AB² + BC² = AC²

⇒ a² + a² = (24)²

⇒ 2a² = 576

⇒ a² = $\frac{576}{2}$

⇒ a² = 288

⇒ a = $\sqrt{288}$

⇒ a = 12 $\sqrt{2}$

⇒ a = 12 x 1.41 = 16.92 cm

Area of the square = side²

= ($\sqrt{288}$)² cm²

= 288 cm²

Perimeter of the square = 4 x side

= 4 x 16.92 cm

= 67.68 cm

Hence, the area of the square is 288 cm² and the perimeter is 67.68 cm.

Given:

Area of the square = 121 cm²

Let s be the side of the square.

As we know, the area of the square = side²

⇒ s² = 121

⇒ s = $\sqrt{121}$

⇒ s = 11

Total length of the wire = Perimeter of the square

As we know, the perimeter of the square = 4 x side

= 4 x 11

= 44 cm

Perimeter of the square = Circumference of the circle

Let r be the radius of the circle.

⇒ 2πr = 44

$$\begin{aligned}⇒ 2 \times \frac{22}{7} \times r = 44\\[1em] ⇒ \frac{44}{7} \times r = 44\\[1em] ⇒ r = \frac{7 \times 44}{44}\\[1em] ⇒ r = \frac{308}{44}\\[1em] ⇒ r = 7\end{aligned}$$

Area of the circle = πr²

$$\begin{aligned}= \frac{22}{7} \times 7^2\\[1em] = \frac{22}{7} \times 49\\[1em] = \frac{1,078}{7}\\[1em] = 154\end{aligned}$$

Hence, the area of the circle is 154 cm².

Given:

Perimeter of the semicircular plate = 108 cm

Let r be the radius of the plate.

The perimeter of the semicircular plate includes the curved part (half the circumference of a circle) and the diameter. So,

⇒ πr + 2r = Perimeter

⇒ πr + 2r = 108

⇒ r(π + 2) = 108

⇒ r $\Big(\frac{22}{7} + 2\Big)$ = 108

⇒ r $\Big(\frac{22}{7} + \frac{14}{7}\Big)$ = 108

⇒ r $\Big(\frac{22 + 14}{7} \Big)$ = 108

⇒ r $\Big(\frac{36}{7}\Big)$ = 108

⇒ r = $\Big(\frac{7 \times 108}{36}\Big)$

⇒ r = 7 x 3 = 21 cm

Area of semicircular plate = $\frac{1}{2}$ πr²

$$\begin{aligned}= \frac{1}{2} \times \frac{22}{7} \times 21^2\\[1em] = \frac{22}{14} \times 441\\[1em] = \frac{9,702}{14} \\[1em] = 693 \text{ cm}^2\end{aligned}$$

Hence, the area of the semicircular plate is 693 cm².

Given:

Sum of areas of two circles = 130π cm²

The distance between their centres = 14 cm

Let $r_1$ and $r_2$ be the radii of 2 circles.

Area of the first circle + Area of the second circle = 130π

⇒ A₁ + A₂ = 130π

⇒ πr₁² + πr₂² = 130π

⇒ π(r₁² + r₂²) = 130π

⇒ $\cancel{π}$ (r₁² + r₂²) = 130 $\cancel{π}$

⇒ r₁² + r₂² = 130 ...............(1)

Also, since the circles touch externally, the sum of their radii equals the distance between their centers.

r₁ + r₂ = 14

⇒ r₁ = 14 - r₂

Putting the value of r₁ in equation (1),

⇒ (14 - r₂)² + r₂² = 130

⇒ 14² + r₂² - 2 x 14 x r₂ + r₂² = 130

⇒ 196 + 2r₂² - 28r₂ = 130

⇒ 196 + 2r₂² - 28r₂ - 130 = 0

⇒ 2r₂² - 28r₂ + 66 = 0

⇒ r₂² - 14r₂ + 33 = 0

⇒ r₂² - 11r₂ - 3r₂ + 33 = 0

⇒ r₂(r₂ - 11) - 3(r₂ - 11) = 0

⇒ (r₂ - 11)(r₂ - 3) = 0

⇒ r₂ = 11 cm or 3 cm

Using equation (1), we find r₁:

If r₂ = 11, then r₁ = 14 - 11 = 3cm

If r₂ = 3, then r₁ = 14 - 3 = 11cm

Hence, the radii of the two circles are 11 cm and 3 cm.

(i) Given:

Diameter of the rear wheel = 1.54 m

Diameter of the front wheel = 0.63 m

Radius of the rear wheel = $\frac{d}{2}$ = $\frac{1.54}{2}$ = 0.77 m

Radius of the front wheel = $\frac{d}{2}$ = $\frac{0.63}{2}$ = 0.315 m

Distance travelled by the tractor in 1 revolution of the rear wheel = Circumference of the rear wheel

= 2πr

$$\begin{aligned}= 2 \times \frac{22}{7} \times 0.77\\[1em] = \frac{44}{7} \times 0.77\\[1em] = 44 \times 0.11\\[1em] = 4.84 \text{ m}\end{aligned}$$

The number of revolutions per minute for the rear wheel is = $24\frac{6}{11}$ = $\frac{270}{11}$

So, the distance traveled by the rear wheel in one minute is = 4.84 x $\frac{270}{11}$

= 0.44 x 270

= 118.8 m

Let x be the number of revolutions made by the front wheel.

The total distance travelled by tractor in 1 min = Number of revolutions made by the front wheel in 1 min x Circumference of wheel

$$\begin{aligned}⇒ x \times 2 \times \frac{22}{7} \times 0.315 = 118.8\\[1em] ⇒ x \times 44 \times 0.045 = 118.8\\[1em] ⇒ x \times 1.98 = 118.8\\[1em] ⇒ x = \frac{118.8}{1.98}\\[1em] ⇒ x = 60\end{aligned}$$

Hence, the number of revolutions made by the front wheel is 60.

(ii) Distance travelled by the tractor in 40 minutes = Number of revolutions made by the rear wheel in 40 min x Circumference of the rear wheel

$$\begin{aligned}= \frac{270}{11} \times 40 \times 4.84 \\[1em] = 270 \times 40 \times 0.44 \\[1em] = 4,752 \text{ m}\end{aligned}$$

Hence, the distance travelled by the tractor in 40 minutes is 4,752 m = 4.752 km.

Given:

Sum of area = 74π cm²

The distance between their centres = 12 cm

Let r₁ and r₂ be the radius of 2 circles.

⇒ A₁ + A₂ = 74π

⇒ πr₁² + πr₂² = 74π

⇒ π(r₁² + r₂²) = 74π

⇒ $\cancel{π}$ (r₁² + r₂²) = 74 $\cancel{π}$

⇒ r₁² + r₂² = 74 ...............(1)

And, r₁ + r₂ = 12

⇒ r₁ = 12 - r₂

Putting the value of r₁ in equation (1),

⇒ (12 - r₂)² + r₂² = 74

⇒ 12² + r₂² - 2 x 12 x r₂ + r₂² = 74

⇒ 144 + 2r₂² - 24r₂ = 74

⇒ 144 + 2r₂² - 24r₂ - 74 = 0

⇒ 2r₂² - 24r₂ + 70 = 0

⇒ r₂² - 12r₂ + 35 = 0

⇒ r₂² - 7r₂ - 5r₂ + 35 = 0

⇒ r₂(r₂ - 7) - 5(r₂ - 7) = 0

⇒ (r₂ - 7)(r₂ - 5) = 0

⇒ r₂ = 7 cm or 5 cm

⇒ r₁ = 12 - r₂ = 12 - 7 or 12 - 5 = 5 cm or 7 cm

Hence, the radius of 2 circles = 7 cm and 5 cm.

Let a be the side of the square.

By using the Pythagoras theorem,

AB² + BC² = AC²

⇒ a² + a² = AC²

⇒ AC² = 2a²

⇒ AC = a $\sqrt{2}$

Diagonal of the square = Diameter of the circle

d = a $\sqrt{2}$

Radius, r = $\frac{d}{2}$ = $\frac{a \sqrt{2}}{2}$

Now, the ratio of the area of the circle to the area of the square is:

$$\begin{aligned}= \frac{\text{Area of circle}}{\text{Area of square}}\\[1em] = \frac{πr^2}{\text{side}^2}\\[1em] = \frac{π \times \Big(\frac{a \sqrt{2}}{2}\Big)^2}{a^2}\\[1em] = \frac{π \times a^2 \times 2}{4a^2}\\[1em] = \frac{π \times a^2}{2a^2}\\[1em] = \frac{π}{2}\\[1em] = \frac{22}{14}\\[1em] = \frac{11}{7}\\[1em]\end{aligned}$$

Hence, the ratio of the area of the circle to the area of the square is 11 : 7.

(i) Given,

Radius of circle C₁ = 6 cm.

Radius of circle C₂ = 2 × 6 = 12 cm.

Circumference of circle = 2πr

= 2π × 12

= 24π cm.

Hence, Option (b) is the correct option.

(ii) Given,

Radius of circle C₁ = 6 cm.

Radius of circle C₂ = 2 × 6 = 12 cm.

Radius of circle C₃ = 2.5 × 6 = 15 cm.

Radius of circle C₄ = 3.5 × 6 = 21 cm.

Area of ring = π(R² - r²)

Calculating the area between C₁ and C₂,

Area = π(12² - 6²)

= π(144 - 36)

= 108π cm².

Calculating the area between C₃ and C₄,

Area = π(21² - 15²)

= π(441 - 225)

= 216π cm².

Total area painted = 108π + 216π = 324π cm².

Hence, Option (c) is the correct option.