Area & Perimeter of Plane Figures — Exercise 19(A)

Length of a side and corresponding altitude of a triangle is doubled. The area of triangle will become :

1. double

2. half

3. four times

4. one-fourth

Answer

The area of a triangle is given by:

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

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

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

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

= 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.

If each side of an equilateral triangle is halved, its area will be :

1. halved

2. four times

3. unaltered

4. one-fourth

Answer

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

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

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

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

= $\dfrac{1}{4}$ 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.

The area of a triangle is 37.5 cm². If its base is 12.5 cm; the corresponding altitude is :

1. 3 cm

2. 25 cm

3. 6 cm

4. 12 cm

Answer

Given:

Area = 37.5 cm²

Base = 12.5 cm

Let h be the altitude of triangle.

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

⇒ 37.5 = $\dfrac{1}{2}$ x 12.5 x h

⇒ h = $\dfrac{37.5 \times 2}{12.5}$

⇒ h = $\dfrac{75}{12.5}$

⇒ h = 6 cm

Hence, option 3 is the correct option.

ABC is a triangle with AB = AC = 12 cm and ∠A = 90°, the area of the triangle ABC is :

1. 144 cm²

2. 36 cm²

3. 72 cm²

4. 108 cm²

Answer

Given:

AB = AC = 12 cm

∠A = 90°

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

= $\dfrac{1}{2}$ x 12 x 12 cm²

= $\dfrac{1}{2}$ x 144 cm²

= 72 cm²

Hence, option 3 is the correct option.

The sides of a triangle are 9 cm, 12 cm and 15 cm; the area of the triangle is :

1. 54 cm²

2. 96 cm²

3. 108 cm²

4. 135 cm²

Answer

Let the sides of the triangle be:

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

The semi-perimeter s:

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

∵ 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.

Find the area of a triangle whose sides are 18 cm, 24 cm and 30 cm.

Also, find the length of altitude corresponding to the largest side of the triangle.

Answer

Let the sides of the triangle be:

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

The semi-perimeter s:

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

∵ 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 = $\dfrac{1}{2}$ x base x altitude

Let h be the altitude:

⇒ 216 = $\dfrac{1}{2}$ x 30 x h

⇒ 216 = 15 x h

⇒ h = $\dfrac{216}{15}$

⇒ 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.

The lengths of the sides of a triangle are in the ratio 3 : 4 : 5. Find the area of the triangle if its perimeter is 144 cm.

Answer

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 = $\dfrac{144}{12}$

⇒ 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:

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

∵ 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.

ABC is a triangle in which AB = AC = 4 cm and ∠A = 90°. Calculate :

(i) the area of Δ ABC,

(ii) the length of perpendicular from A to BC.

Answer

(i) Given:

AB = AC = 4 cm

∠A = 90°

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

= $\dfrac{1}{2}$ x 4 x 4 cm²

= $\dfrac{1}{2}$ 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 = $\dfrac{1}{2}$ x base x height

∴ 8 = $\dfrac{1}{2}$ x BC x AD

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

⇒ 8 = $2 \sqrt{2}$ x h

⇒ h = $\dfrac{8}{2 \sqrt{2}}$

⇒ h = $\dfrac{4}{\sqrt{2}}$

⇒ h = 2.83 cm

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

The area of an equilateral triangle is $36\sqrt{3}\text{ sq. cm}$. Find its perimeter.

Answer

Given:

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

Let s be the side of equilateral triangle.

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

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

Perimeter = 3 x side

= 3 x 12 cm

= 36 cm

Hence, the perimeter is 36 cm.

Find the area of an isosceles triangle with perimeter 36 cm and base 16 cm.

Answer

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 = $\dfrac{20}{2}$

⇒ 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:

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

∵ 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.

The base of an isosceles triangle is 24 cm and its area is 192 sq. cm. Find its perimeter.

Answer

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 = $\dfrac{1}{4} \times b \times \sqrt{4s^2 - b^2}$

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

Perimeter = sum of all sides of triangle

= 20 + 20 + 24 cm

= 64 cm

Hence, the perimeter is 64 cm.

The given figure shows a right-angled triangle ABC and an equilateral triangle BCD. Find the area of the shaded portion.

Answer

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 = $\dfrac{1}{2}$ x base x height

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

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

= $32 \sqrt{3}$ cm²

For triangle BCD,

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

= $\dfrac{\sqrt{3}}{4}$ x 8²

= $\dfrac{\sqrt{3}}{4}$ 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.

Find the area and the perimeter of quadrilateral ABCD, given below; if, AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and ∠DBC = 90°.

Answer

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 = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x 12 x 5 cm²

= $\dfrac{1}{2}$ x 60 cm²

= 30 cm²

For Δ ABD,

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

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

∵ 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.

The base of a triangular field is three times its height. If the cost of cultivating the field at ₹ 36.72 per 100 m² is ₹ 49,572; find its base and height.

Answer

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 = $\dfrac{\text{Total cost}}{\text{Cost of cultivating}}$

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

Let the height of the triangle be h.

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

Base = 3h

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

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

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.

The sides of a triangular field are in the ratio 5 : 3 : 4 and its perimeter is 180 m. Find :

(i) its area.

(ii) altitude of the triangle corresponding to its largest side.

(iii) the cost of levelling the field at the rate of ₹ 10 per square metre.

Answer

(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 = $\dfrac{180}{12}$

⇒ 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.

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

∵ 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 = $\dfrac{1}{2}$ x base x altitude

Base = 75 m

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

⇒ 1350 = $\dfrac{1}{2}$ x 75 x h

⇒ h = $\dfrac{1350 \times 2}{75}$

⇒ h = $\dfrac{2,700}{75}$

⇒ 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. If the base of the triangle is 24 cm; calculate the perimeter and the area of the triangle.

Answer

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 = $\dfrac{\text{BC}}{2}$ = ${\dfrac{24}{2}}$ = 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 = $\dfrac{128}{8}$

⇒ 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 = $\dfrac{1}{2}$ x base x height

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

= 12 x 16

= 192 cm²

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

Calculate the area and the height of an equilateral triangle whose perimeter is 60 cm.

Answer

Given:

Perimeter = 60 cm

Let a be the sides of equilateral triangle.

Perimeter = Sum of all sides

⇒ 60 = a + a + a

⇒ 60 = 3a

⇒ a = $\dfrac{60}{3}$

⇒ a = 20 cm

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

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

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

$= \dfrac{\sqrt{3}}{2} \times 20\\[1em] = 10 \sqrt{3}\\[1em] = 17.32 cm$

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

In triangle ABC; angle A = 90°, side AB = x cm, AC = (x + 5) cm and area = 150 cm². Find the sides of the triangle.

Answer

Δ ABC is shown in the figure below:

Given:

Area = 150 cm²

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

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

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.

If the difference between the sides of a right angled triangle is 3 cm and its area is 54 cm²; find its perimeter.

Answer

Given:

Area = 54 cm²

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

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

⇒ 54 = $\dfrac{1}{2}$ x BC x AB

⇒ 54 = $\dfrac{1}{2} \times$ (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.