These AP 9th Class Maths Important Questions 10th Lesson Heron’s Formula will help students prepare well for the exams.

## AP Board Class 9 Maths 10th Lesson Heron’s Formula Important Questions

### 9th Class Maths Heron’s Formula 2 Marks Important Questions

Question 1.

Find the area of a triangle whose sides are 6 cm, 8 cm and 10 cm respectively.

Solution:

∵ 6^{2} + 8^{2} = 10^{2}

Triangle is right angled.

∴ Area = \(\frac{\text { Base } \times \text { Height }}{2}\) = \(\frac{6 \times 8}{2}\) = 24 cm^{2}

Aliter (Using Heron’s formula)

Here, a = 6 cm, b = 8 cm, c = 10 cm

∴ s = \(\frac{a+b+c}{2}\) = \(\frac{6+8+10}{2}\) = 12 cm

∴ Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{12(12-6)(12-8)(12-10)}\)

= \(\sqrt{12 \times 6 \times 4 \times 2}\) = 24 cm^{2}

Question 2.

Find tlie area of an equilateral triangle whose perjmeter is 18 cm, using Heron’s formula.

Solution:

Perimeter = 18 cm

∴ Side = \(\frac{18}{3}\) cm = 6 cm

Here, a = 6 cm, b = 6 cm, c = 6 cm

∴ s = \(\frac{a+b+c}{2}\) = \(\frac{6+6+6}{2}\) = 9 cm

∴ Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{9(9-6)(9-6)(9-6)}\)

= 9√3 cm^{2}

= 9 × 1.73 cm^{2} = 15.57 cm^{2}

Question 3.

Find the area of a triangle with sides 35 cm, 54 cm and 61 cm.

Solution:

Here, a = 35 cm, b = 54 cm, c = 61 cm

∴ s = \(\frac{a+b+c}{2}\) = \(\frac{35+54+61}{2}\) = \(\frac{150}{2}\)

= 75 cm

∴ Area of the triangle

= 2 × 2 × 3 × 5 × 7 × √5 = 420√5,cm^{2}

Question 4.

The base of an isosceles triangle measures 24 cm and its area is 60 cm^{2}. Find its perimeter.

Solution:

Here, a = 24 cm

Area = \(\frac{a}{4} \sqrt{4 b^2-a^2}\)

⇒ 60 = \(\frac{24}{4} \sqrt{4 \mathrm{~b}^2-(24)^2}\)

⇒ 60 = 6\(6 \sqrt{4 b^2-576}\)

⇒ 10 = \(\sqrt{4 b^2-576}\)

⇒ 100 = 4b^{2} – 576(Squaring both sides)

⇒ 4b^{2} = 676 ⇒ b^{2} = \(\frac{676}{4}\) = 169

⇒ b = \(\sqrt{169}\) = 13 cm

∴ Perimeter = a + b + b = a + 2b

= 24 + 13 + 13

= 24 + 26 = 50 cm

Question 5.

The unequal side of an isosceles triangle is 4 cm and its perimeter is 20 cm. Find its area.

Solution:

a = 4 cm ………… (1)

Perimeter = 20 cm

⇒ a + b + b = 20

⇒ 4 + 2b = 20 (From (1))

⇒ 2b = 16

⇒ b = 8 cm

∴ Area = \(\frac{a}{4} \sqrt{4 b^2-a^2}\)

= \(\frac{4}{4} \sqrt{4(8)^2-(4)^2}\)

= \(\sqrt{256-16}\) = \(\sqrt{240}\)

= \(\sqrt{2 \times 2 \times 2 \times 2 \times 3 \times 5}\)

2 × 2\(\sqrt{15}\) = 4\(\sqrt{15}\) cm^{2}

Question 6.

Each side of an equilateral triangle is 2x cm. If x√3 = 48, then find its area.

Solution:

x√3 = 48 ⇒ x = 16√3 ⇒ 2x = 32√3

∴ a = 32√3 cm

∴ Area = \(\frac{\sqrt{3}}{4}\)a^{2} = \(\frac{\sqrt{3}}{4}\)(32√3)^{2}

= 768√3 cm^{2}

Question 7.

Write the Heron’s formula and explain the terms in it.

Solution:

Heron’s formula:

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

Where a, b, c are sides of the triangle and s = \(\frac{a+b+c}{2}\)

s = semi-perimeter i.e., half the perimeter of the triangle.

### 9th Class Maths Heron’s Formula 4 Marks Important Questions

Question 1.

The sides of a triangle are x, x + 1 and 2x – 1. Its area is x\(\sqrt{10}\). Find the value of x.

Solution:

a = x, b = x + 1, c = 2x – 1

According to the question,

Area = x\(\sqrt{10}\)

⇒ \(x \sqrt{2(x-1)}\) = x\(\sqrt{10}\)

⇒ \(\sqrt{2(\mathrm{x}-1)}\) = \(\sqrt{10}\)

⇒ 2(x – 1) = 10

⇒ x – 1 = 5

⇒ x = 6

Question 2.

If the area of right angled triangle is 216 cm^{2} and base is 24 cm, find the perimeter of the triangle.

Solution:

Area = \(\frac{1}{2}\) × Base × Height

⇒ 216 = \(\frac{1}{2}\) × 24 × Height

⇒ 216 = 12 × Height

⇒ Height = \(\frac{216}{12}\) = 18 cm

∴ Hypotenuse

= \(\sqrt{(\text { Base })^2+(\text { Height })^2}\)

= \(\sqrt{(24)^2+(18)^2}\) = \(\sqrt{576 + 324}\)

= \(\sqrt{900}\) = 30 cm

∴ Perimeter

= Base + Height + Hypotenuse

= 24 + 18 + 30 = 72 cm

Question 3.

The perimeter of a triangular field is 450 m and its sides are in the ratio 13 : 12 : 5. Find the area of the field.

Solution:

2s = a + b + c = 450 m

a : b : c = 13 : 12 : 5

Sum of ratios = 13 + 12 + 5 = 30

∴ a = \(\frac{13}{30}\) × 450 = 195 m

b = \(\frac{12}{30}\) × 450 = 180 m

c = \(\frac{5}{30}\) × 450 = 75 m

∵ 75^{2} + 180^{2} = 195^{2}

∴ Triangle is right angled.

(By converse of Pythagoras Theorem)

∴ Area = \(\frac{\text { Base } \times \text { Perpendicular }}{2}\)

= \(\frac{75 \times 180}{2}\) = 75 × 90 = 6750 cm^{2}

Question 4.

Perimeter of a triangle is 144 m and the ratio of sides is 3 : 4 : 5. Find the area of the triangle.

Solution:

2s = a + b + c = 144 m ⇒ s = 72 m

a : b : c = 3 : 4 : 5

Sum of ratios = 3 + 4 + 5 = 12

∴ a = \(\frac{3}{12}\) × 144 = 36 m

b = \(\frac{4}{12}\) × 144 – 48 m

c = \(\frac{5}{12}\) × 144 = 60 m

∵ 36^{2} + 48^{2} = 1296 + 2304

= 3600 = 60^{2}

∴ Triangle is right angled. (By converse of Pythagoras Theorem)

∴ Area = \(\frac{1}{2}\) × Base × Height

= \(\frac{1}{2}\) × 36 × 48 = 864 m^{2}

Question 5.

The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. Find the area of the triangle.

Solution:

Let the smaller side be x cm. Then, two sides are (x + 4) cm and (2x – 6) cm

∴ Perimeter = x + (x + 4) + (2x – 6) = (4x – 2) cm

According to the question,

4x – 2 = 50

⇒ 4x = 52

⇒ x = 13 cm

∴ Sides cure 13 cm, 17 cm and 20 cm.

Let a = 13 cm, b = 17 cm, c = 20 cm

Question 6.

ΔABC is isosceles with AB = AC, AD ⊥ BC. AB = 10 cm and BC = 16 cm.

Based on the above information, answer the following questions :

(i) Length of AC is

A) 16 cm

B) 8 cm

C) 20 cm

D) 10 cm

Answer:

D) 10 cm

ii) Semi-perimeter of ΔABC is

A) 36 cm

B) 24 cm

C) 18 cm

D) 72 cm

Answer:

C) 18 cm

iii) Area of ΔABC is

A) 180 cm^{2}

B) 48 cm^{2}

C) 120 cm^{2}

D) 80 cm^{2}

Answer:

B) 48 cm^{2}

iv) Length of AD is

A) 6 cm

B) 8 cm

C) 12 cm

D) 9 cm

Answer:

A) 6 cm

v) If each side is doubled area will become

A) 4 times

B) 8 times

C) 2 times

D) 16 times

Answer:

A) 4 times

### 9th Class Maths Heron’s Formula 8 Marks Important Questions

Question 1.

A park in the shape of a quadrilateral ABCD has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy ?

Solution:

Join BD.

Area of right angled triangle BCD.

= \(\frac{1}{2}\) × Base × Height = \(\frac{1}{2}\) × 5 × 12 = 30 m^{2}

Right triangle BCD,

BD^{2} + BC^{2} + CD^{2} (By Pythagoras Theorem)

= (12)^{2} + (5)^{2} = 144 + 25 = 169

BD = \(\sqrt{169}\) = 13 m

For Δ ABD,

a = 13 m, b = 8 m, c = 9 m

∴ s = \(\frac{a+b+c}{2}\) = \(\frac{13+8+9}{2}\) = \(\frac{30}{2}\) = 15 m

∴ Area of the ΔABD

= \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{15(15-13)(15-8)(15-9)}\)

= \(\sqrt{15(2)(7)(6)}\)

= \(\sqrt{(3 \times 5)(2)(7)(2 \times 3)}\)

= 3 × 2\(\sqrt{35}\) = 6\(\sqrt{13}\) m^{2}

= 6 × 5.916 = 35.5 m^{2} (approx.)

∴ Area of the quadrilateral ABCD

= Area of Δ BCD + Area of Δ ABD

= 30 m^{2} + 35.5 m^{2} = 65.5 m^{2} (approx.)

Hence, the park occupies the area 65.5 m^{2} (approx.)

Question 2.

Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.

Solution:

For Triangular Area I:

a = 5 cm, b = 5 cm, c = 1 cm

∴ s = \(\frac{a+b+c}{2}\) = \(\frac{5+5+1}{2}\) = \(\frac{11}{2}\) = 5.5 cm

∴ Area I:

= \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{5.5(5.5-5)(5.5-5)(5.5-1)}\)

= \(\sqrt{5.5(.5)(.5)(4.5)}\) = (.5)\(\sqrt{(5.5)(4.5)}\)

= (.5)\(\sqrt{(.5)(11)(.5)(9)}\) = (.5) (.5)(3) \(\sqrt{11}\)

= 0.757\(\sqrt{11}\) = 0.75 (3.3) (approx.)

= 2.5 cm^{2} (approx.)

Area II = 6.5 × 1 = 6.5 cm^{2}.

For Area III:

Area III = (a) + (b) + (c)

Area IV = \(\frac{6 \times 1.5}{2}\) = 4.5 cm^{2}

Area V = \(\frac{6 \times 1.5}{2}\) = 4.5 cm^{2}

∴ Total area of the paper used = Area I + Area II + Area III + Area IV + Area V

= 2.5 cm^{2} + 6.5 cm^{2} + 1.3 cm^{2} + 4.5 cm^{2} + 4.5 cm^{2} = 19.3 cm^{2}

Question 3.

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Solution:

Let the given field be in the shape of a trapezium ABCD in which AB = 25 m, CD = 10 m, BC = 13 m and AD = 14 m From D, draw DE || BC meeting AB at E.

Also, draw DF ⊥ AB.

∴ DE = BC = 13 m

AE = AB – EB = AB – DC

= 25 – 10 = 15 m

= 7 × 3 × 2 × 2 = 84 m^{2}

\(\frac{1}{2}\) × AE × DF = 84

⇒ \(\frac{1}{2}\) × 15 × DF = 84

DF = \(\frac{84 \times 2}{15}\)

⇒ DF = \(\frac{56}{5}\) m = 11.2 m 5

⇒ Height of the trapezium is 11.2 m.

∴ Area of parallelogram EBCD = Base × Height

= EB × DF = 10 × \(\frac{56}{5}\) = 112 m^{2}

∴ Area of the field = Area of ΔAED + Area of parallelogram EBCD

= 84 m^{2} + 112 m^{2}

= 196 m^{2}

Question 4.

If a cloth seller sells triangular shaped clothes of sides 18 units, 24 units, 30 units and 16 units, 26 units, 22 units to the customer at the same price, which is better deal for the customer?

Solution:

For first triangular shaped cloth

a = 18 units, b = 24 units, c = 30 units

∴ s = \(\frac{a+b+c}{2}\) = \(\frac{18+24+30}{2}\) = \(\frac{72}{2}\) = 36 units

∴ Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)

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

= \(\sqrt{36 \times 18 \times 12 \times 6}\)

= \(\sqrt{36 \times 18 \times 2 \times 6 \times 6}\)

= \(\sqrt{36 \times 36 \times 6 \times 6}\)

= 36 × 6 = 216 square units

For second triangular shaped cloth :

a = 16 units, b = 26 units, c = 22 units

∴ s = \(\frac{a+b+c}{2}\) = \(\frac{16+26+22}{2}\)

= \(\frac{64}{2}\) = 32 units

∴ Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{32(32-16)(32-26)(32-22)}\)

= \(\sqrt{32 \times 16 \times 6 \times 10}\)

= \(\sqrt{16 \times 2 \times 16 \times 2 \times 3 \times 2 \times 5}\)

= 16 × 2 × \(\sqrt{15}\)

= 32\(\sqrt{15}\)

= 32 × 3.87

= 123.84 square units (approx.)

∵ 216 > 123.84

∴ First deal is better.