# Class 9 Maths Chapter 12 Heron’s Formula MCQs

Class 9 Maths Chapter 12 Heron’s Formula MCQs are available online here, with solved answers. The MCQs are prepared as per the latest exam pattern for Class 9. The objective questions are prepared chapter-wise, as per the CBSE syllabus (2022-2023) and NCERT curriculum. These questions are provided with detailed explanations. Get Chapter-wise Class 9 Maths MCQs at BYJU’S.

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## MCQs on Class 9 Maths Chapter 12 Heron’s Formula

Choose the correct answer and solve the MCQs on Heron’s formula.

1) Area of a triangle is equal to:

a. Base x Height

b. 2(Base x Height)

c. ½(Base x Height)

d. ½ (Base + Height)

2) If the perimeter of an equilateral triangle is 180 cm. Then its area will be:

a. 900 cm2

b. 900√3 cm2

c. 300√3 cm2

d. 600√3 cm2

Explanation: Given, Perimeter = 180 cm

3a = 180 (Equilateral triangle)

a = 60 cm

Semi-perimeter = 180/2 = 90 cm

Now as per Heron’s formula,

$$\begin{array}{l}A = \sqrt{s(s-a)(s-b)(s-c)}\end{array}$$

In the case of an equilateral triangle, a = b = c = 60 cm

Substituting these values in the Heron’s formula, we get the area of the triangle as:

A = √[90(90 – 60)(90 – 60)(90 – 60)]

= √(90× 30 × 30 × 30)

A = 900√3 cm2

3) The sides of a triangle are 122 m, 22 m and 120 m respectively. The area of the triangle is:

a. 1320 sq.m

b. 1300 sq.m

c. 1400 sq.m

d. 1420 sq.m

Explanation: Given,

a = 122 m

b = 22 m

c = 120 m

Semi-perimeter, s = (122 + 22 + 120)/2 = 132 m

Using heron’s formula:

$$\begin{array}{l}A = \sqrt{s(s-a)(s-b)(s-c)}\end{array}$$

= √[132(132 – 122)(132 – 22)(132 – 120)]

= √(132 × 10 × 110 × 12)

= 1320 sq.m

4) The area of a triangle with given two sides 18 cm and 10 cm, respectively and a perimeter equal to 42 cm is:

a. 20√11 cm2

b. 19√11 cm2

c. 22√11 cm2

d. 21√11 cm2

Explanation: Perimeter = 42

a + b + c = 42

18 + 10 + c = 42

c = 42 – 28 = 14 cm

Semi perimeter, s = 42/2 = 21 cm

Using Heron’s formula:

$$\begin{array}{l}A = \sqrt{s(s-a)(s-b)(s-c)}\end{array}$$

= √[21(21 – 18)(21 – 10)(21 – 14)]

= √(21 × 3 × 11 × 7)

= 21√11 cm2

5) The sides of a triangle are in the ratio 12: 17: 25 and its perimeter is 540 cm. The area is:

a. 1000 sq.cm

b. 5000 sq.cm

c. 9000 sq.cm

d. 8000 sq.cm

Explanation: The ratio of the sides is 12: 17: 25

Perimeter = 540 cm

Let the sides of the triangle be 12x, 17x and 25x.

Hence,

12x + 17x + 25x = 540 cm

54x = 540 cm

x = 10

Therefore,

a = 12x = 12 × 10 = 120

b = 17x = 17 × 10 = 170

c = 25x = 25 × 10 = 250

Semi-perimeter, s = 540/2 = 270 cm

Using Heron’s formula:

$$\begin{array}{l}A = \sqrt{s(s-a)(s-b)(s-c)}\end{array}$$

= √[270(270 – 120)(270 – 170)(270 – 250)]

= √(270 × 150 × 100 × 20)

= 9000 sq.cm

6) The equal sides of the isosceles triangle are 12 cm, and the perimeter is 30 cm. The area of this triangle is:

a. 9√15 sq.cm

b. 6√15 sq.cm

c. 3√15 sq.cm

d. √15 sq.cm

Explanation: Given,

Perimeter = 30 cm

Semiperimeter, s = 30/2 = 15 cm

a = b = 12 cm

c = ?

a + b + c = 30

12 + 12 + c = 30

c = 30 – 24 = 6 cm

Using Heron’s formula:

$$\begin{array}{l}A = \sqrt{s(s-a)(s-b)(s-c)}\end{array}$$

= √[15(15 – 12)(15 – 12)(15 – 6)]

= √(15 × 3 × 3 × 9)

= 9√15 sq.cm

7) A quadrilateral whose sides are 3 cm, 4 cm, 4 cm, 5 cm and one of the diagonal is equal to 5 cm as per the below figure. The area of the quadrilateral is:

a. 19.17 sq.cm

b. 15.17 sq.cm

c. 20.17 sq.cm

d. 22.17 sq.cm.

Explanation: Using Pythagoras theorem, in ΔABC,

AC2 = AB2 + BC2

⇒ 52 = 32 + 42

⇒ 25 = 25

Hence, ABC is a right triangle.

Area of ΔABC = ½ x 3 x 4 = 6 sq.cm

Semiperimeter of ΔACD = s = (5+5+4)/2 = 14/2 = 7 cm

Area of ΔACD can be determined by using Heron’s formula.

$$\begin{array}{l}A = \sqrt{s(s-a)(s-b)(s-c)}\end{array}$$

= √[7(7 – 5)(7 – 5)(7 – 4)]

= √(7 × 2 × 2 × 3)

= 9.17 sq.cm

Therefore, the area of quad.ABCD = Area of ΔABC + Area of ΔACD = (6 + 9.17) sq.cm = 15.17 sq.cm

8) The area of an equilateral triangle having side length equal to √3/4 cm (using Heron’s formula) is:

a. 2/27 sq.cm

b. 2/15 sq.cm

c. 3√3/64 sq.cm

d. 3/14 sq.cm

Explanation: Here, a = b = c = √3/4

Semiperimeter = (a + b + c)/2 = 3a/2 = 3√3/8 cm

Using Heron’s formula,

$$\begin{array}{l}A = \sqrt{s(s-a)(s-b)(s-c)}\end{array}$$

= √[(3√3/8) (3√3/8 – √3/4)(3√3/8 – √3/4)(3√3/8 – √3/4)]

= 3√3/64 sq.cm

9) The sides of a parallelogram are 100 m each and the length of the longest diagonal is 160 m. The area of a parallelogram is:

a. 9600 sq.m

b. 9000 sq.m

c. 9200 sq.m

d. 8800 sq.m

Explanation: The diagonal divides the parallelogram into two equivalent triangles. Hence, its area will be equal to the sum of the area of the two triangles.

Thus, the sides of one triangle will be 100 m, 160 m, and 100 m.

So, a = 100 m, b = 160 m, c = 100 m

Semiperimeter (s) = (a + b + c)/2 = (100 + 160 + 100)/2 = 360/2 = 180 m

Using Heron’s formula,

$$\begin{array}{l}A = \sqrt{s(s-a)(s-b)(s-c)}\end{array}$$

= √[180(180 – 100)(180 – 160)(180 – 100)]

= √(180 × 80 × 20 × 80)

= 4800 sq.m

Thus, the area of the parallelogram = 2 × 4800 sq.m = 9600 sq.m

10) The sides of a triangle are in the ratio of 3: 5: 7 and its perimeter is 300 cm. Its area will be:

a. 1000√3 sq.cm

b. 1500√3 sq.cm

c. 1700√3 sq.cm

d. 1900√3 sq.cm

Explanation:

The ratio of the sides is 3: 5: 7

Perimeter = 300 cm

Let the sides of the triangle be 3x, 5x and 7x.

Hence,

3x + 5x + 7x = 300 cm

15x = 300 cm

x = 20

Therefore,

a = 3x = 3 × 20 = 60

b = 5x = 5 × 20 = 100

c = 7x = 7 × 20 = 140

Semi-perimeter, s = 300/2 = 150 cm

Using Heron’s formula:

$$\begin{array}{l}A = \sqrt{s(s-a)(s-b)(s-c)}\end{array}$$

= √[150(150 – 60)(150 – 100)(150 – 140)]

= √(150 × 90 × 50 × 10)

= 1500√3 sq.cm

11) The base of a right triangle is 8 cm and the hypotenuse is 10 cm. Its area will be

(a) 24 cm2

(b) 40 cm2

(c) 48 cm2

(d) 80 cm2

Explanation:

Given: Base = 8 cm and Hypotenuse = 10 cm

Hence, height = √[(102 – 82) = √36 = 6 cm

Therefore, area = (½)×b×h = (½)×8×6 = 24 cm2.

12) The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm2 is

(a) Rs 2.00

(b) Rs 2.16

(c) Rs 2.48

(d) Rs 3.00

Explanation:

Given: a = 6 cm, b = 8 cm, c = 10 cm.

s = (6 + 8 + 10)/2 = 12 cm

Hence, by using Heron’s formula, we can write:

A = √[12(12 – 6)(12 – 8)(12 – 10)] = √[(12)(6)(4)(2)] = √576 = 24 cm2

Therefore, the cost of painting at a rate of 9 paise per cm2 = 24 × 9 paise = Rs. 2.16

13) An isosceles right triangle has an area of 8 cm2. The length of its hypotenuse is

(a) √32 cm

(b) √16 cm

(c) √48 cm

(d) √24 cm

Explanation:

Given that area of the isosceles triangle = 8 cm2.

As the given triangle is isosceles triangle, let base = height = h

Hence,

(½)×h×h = 8

(½)h2 = 8

h= 16

h = 4 cm

Since it is isosceles right triangle, Hypotenuse2 = Base2+Height2

Hypotenuse= 4+ 42

Hypotenuse2 = 32

Hypotenuse = √32 cm

14) The area of an isosceles triangle having a base 2 cm and the length of one of the equal sides 4 cm, is

(a) √15 cm2

(b) √(15/2) cm2

(c) 2√15 cm2

(d) 4√15 cm2

Explanation:

Given that a = 2 cm, b= c = 4 cm

s = (2 + 4 + 4)/2 = 10/2 = 5 cm

By using Heron’s formula, we get:

A =√[5(5 – 2)(5 – 4)(5 – 4)] = √[(5)(3)(1)(1)] = √15 cm2.

15) The perimeter of an equilateral triangle is 60 m. The area is

(a) 10√3 m2

(b) 15√3 m2

(c) 20√3 m2

(d) 100√3 m2

Explanation:

Given: Perimeter of an equilateral triangle = 60 m

3a = 60 m (As the perimeter of an equilateral triangle is 3a units)

a = 20 cm.

We know that area of equilateral triangle = (√3/4)a2 square units

A = (√3/4)202

A = (√3/4)(400) = 100√3 m2.

16) The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude

(a) 16√5 cm

(b) 10√5 cm

(c)  24√5 cm

(d) 28 cm

Explanation:

Given: a = 35 cm, b = 54cm, c = 61cm

s = (35 + 54 + 61)/2 = 150/2 = 75 cm.

Hence, by using Heron’s formula, A = √[75(75 – 35)(75 – 54)(75 – 61)] = √(882000) = 420√5 cm2

The area of triangle with longest altitude “h” is given as”

(½)×a×h = 420√5 {a is less than b and c}

(½)×35×h = 420√5

h = (840√5)/35 = 24√5 cm.

17) The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is

(a) 1322 cm2

(b) 1311 cm2

(c) 1344 cm2

(d) 1392 cm2

Explanation:

Since, all the sides of a triangle are given, we can find the area of a triangle using Heron’s formula.

Let a = 56 cm, b= 60 cm, c = 52 cm

s = (56+60+52)/2 = 84 cm.

Area of triangle using Heron’s formula, A = √[s(s-a)(s-b)(s-c)] square units

A = √[84(84-56)(84-60)(84-52)] = √(1806336) =1344 cm2.

18) If the area of an equilateral triangle is 16√3 cm2, then the perimeter of the triangle is

(a) 48 cm

(b) 24 cm

(c) 12 cm

(d) 36 cm

Explanation:

Given: Area of equilateral triangle = 16√3 cm2

(√3/4)a2 = 16√3

a2 = [(16√3)(4)]/3

a2 = 64

a = 8cm
Therefore, perimeter = 3(8) = 24 cm.

19) The area of an equilateral triangle with sides 2√3 cm is

(a) 5.196 cm2

(b) 0.866 cm2

(c) 3.496 cm2

(d) 1.732 cm2

Explanation:

Given: Side = 2√3 cm

We know that, area of equilateral triangle = (√3/4)a2 square units

A = (√3/4)(2√3)2 = (√3/4)(12) = 3√3 = 3(1.732) = 5.196 cm.

20) The length of each side of an equilateral triangle having an area of 9√3 cm2 is

(a) 8 cm

(b) 36 cm

(c) 4 cm

(d) 6 cm

Explanation:

Given: Area of equilateral triangle = 9√3 cm2

Hence, (√3/4)a2 = 9√3

a2 = [(9√3)(4)]/√3

a2 = 36

a = 6 cm.

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