Question

Class 9 Maths MCQ Questions of Heron’s Formula with Answers?

Answer 1

Class 9 Maths MCQ Questions of Heron’s formula with Answers were prepared depends on the newest exam pattern. We have provided MCQ Questions for class 9 Maths with Answers to assist students understand the concept alright. Students also can ask NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula for better exam preparation and to attain more marks.

Practicing the MCQ Questions with answers will boost your confidence thereby helping you score well in the exam and clear all the fundamentals concepts.

Practice MCQ Questions for Class 9 Maths

1. The area of an equilateral triangle with side 2√3 cm is

(a) 5.196 cm²
(b) 0.866 cm²
(c) 3.496 cm²                       
(d) 1.732 cm²

2. 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

3. Area of the triangle is equal to:

(a) Base x Height
(b) 2(Base x Height)
(c) 1/2(Base x Height)
(d) 1/2(Base + Height)

4. 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

5. 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

6. 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

7. 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

8. When the sum of squares of two sides of a triangle is equal to the square of the length of the third side, then it is called a:

(a) Scalene triangle
(b) Right triangle
(c) Isosceles triangle
(d) Equilateral triangle

9. Length of perpendicular drawn on smallest side of scalene triangle is

(a) Smallest
(b) Largest
(c) No relation
(d) None

10. If the sides of a triangle are doubled, then its area

(a) remains the same
(b) becomes three times
(c) becomes doubled
(d) becomes four times

11. The base of an isosceles right triangle is 30 cm. Its area is

(a) 225 cm²
(b) 225√3 cm²
(c) 225√3 cm²
(d) 450 cm²

12. 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

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

(a) √32 cm
(b) √16 cm
(c) √48 cm
(d) √24 cm

14. 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

15. 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

16. The area of an equilateral triangle of side 6 cm is​

(a) 18 cm²
(b) 9√3 cm²
(c) 56√3 cm²
(d) 58√3 cm²

17. The area of triangle with given two sides 18cm and 10cm respectively and perimeter equal to 42 cm is:

(a) 20√11 cm²
(b) 19√11 cm²
(c) 22√11 cm²
(d) 21√11 cm²

18. 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

19. The area of a right angled triangle if the radius of its circumcircle is 3 cm and altitude drawn to the hypotenuse is 2 cm.

(a) 6 cm²
(b) 3 cm²
(c) 4 cm²
(d) 8 cm²

20. 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

Answer 2

1. Answer: (a) 5.196 cm²

Explanation: Given, side of an equilateral triangle is 2√3 cm.
Area of an equilateral triangle = √3/4 (Side)²
= √3/4 (2√3)² = (√3/4) x 4 x 3
= 3√3 = 3 x 1.732 = 5.196 cm²
Hence, the area of an equilateral triangle is 5.196 cm².

2. Answer: (d) 6 cm

Explanation: Given, area of an equilateral triangle = 9√3 cm²
∴ Area of an equilateral triangle = √3/4(Side)²
=> √3/4 (Side)2 = 9√3
=> (Side)² = 36
∴ Side = 6 cm  [taking positive square root because side is always positive]
Hence, the length of an equilateral triangle is 6 cm.

3. Answer: (c) 1/2(Base x Height)

4. Answer: (b) 24 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.

5. Answer: (c) 1344 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.

6. Answer: (c)  24√5 cm

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

s = (35+54+61)/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”

(1/2)×a×h = 420√5

(1/2)×35×h = 420√5

h= (840√5)/35 = 24√5 cm.

7. Answer: (d) 100√3 m²

Explanation: Given: Perimeter of an equilateral triangle = 60m

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 cm².

8. Answer: (b) Right triangle

9. Answer: (c) No relation

10. Answer: (d) becomes four times

11. Answer: (d) 450 cm²

Explanation: Let ABC be the right triangle in which ∠B = 90°
Now,
base = BC
perpendicular = AB
hypotenuse = AC
BC = 30 cm (given)
△ABC is an isosceles right angled triangle. We know that hypotenuse is the longest side of the right triangle.
So, AB = BC = 30 cm
Area of △ABC = 1/2× base × height
= 1/2× BC × AB
= 1/2× 30 × 30
= 450 cm². 

12. Answer: (a) √15 cm²

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

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

By using Heron’s formula, we get:

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

13. Answer: (a) √32 cm

Explanation: Given that area of isosceles triangle = 8 cm².

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

Hence,

(1/2)×h×h = 8

(1/2)h² =8

h²=16

h= 4 cm

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

Hypotenuse²= 4²+4²

Hypotenuse² = 3²

Hypotenuse = √32 cm

14. Answer: (b) Rs 2.16

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

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

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

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

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

15. Answer: (a) 24 cm²

Explanation: Given: Base = 8 cm and Hypotenuse = 10 cm

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

Therefore, area = (1/2)×b×h = (1/2)×8×6 = 24cm².

16. Answer: (b) 9√3 cm²

Explanation: Given, side of an equilateral triangle is 6 cm.
Area of an equilateral triangle = √3/4 (Side)²
= √3/4 (6)² = (√3/4) x 36
= 9√3 cm²
Hence, the area of an equilateral triangle is 9√3 cm².

17. Answer: (d) 21√11 cm²

Explanation: Perimeter = 42

a+b+c=42

18+10+c=42

c=42-28=14 cm

Semiperimeter, s = 42/2 = 21cm

Using Heron’s formula:

\(A=\sqrt{s(s−a)(s−b)(s−c)}\)
By putting the of s, a, b and c, to get the answer equal to 21√11 cm².

18.  Answer: (a) 1320 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:

\(A=\sqrt{s(s−a)(s−b)(s−c)}\)
By putting the values of s, a, b and c, to get the answer equal to 1320 sq.m.

19. Answer: (a) 6 cm²

Explanation: In the case of a right triangle, the hypotenuse is a diameter of the circumcircle, and its center is exactly at the midpoint of the hypotenuse.
As radius of its circumcircle is 3 cm,
so AC( hyhypotenuse)=3*2=6cm.
BD=2cm
Area of triangle ABC=1/2* hyhypotenuse AC* BD
A= 1/2 x 6 x 2= 6
Area =6cm².

20. Answer: (b) 900√3 cm²

Explanation: Given, Perimeter = 180 cm

3a = 180 (Equilateral triangle)

a = 60 cm

Semi-perimeter = 180/2 = 90cm

Now as per Heron’s formula,

\(A=\sqrt{s(s−a)(s−b)(s−c)}\)
Hence, if we put the values here, we get:

A = 900√3

Click here to practice: – Heron’s Formula MCQ Question for Class 9 Maths