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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) The 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)
Answer: c
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
Answer: b
Explanation: Given, Perimeter = 180 cm
3a = 180 (Equilateral triangle)
a = 60 cm
Semi-perimeter = 180/2 = 90 cm
Now as per Heron’s formula,
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
Answer: a
Explanation: Given,
a = 122 m
b = 22 m
c = 120 m
Semi-perimeter, s = (122 + 22 + 120)/2 = 132 m
Using Heron’s formula:
= √[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
Answer: d
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:
= √[21(21 – 18)(21 – 10)(21 – 14)]
= √(21 × 3 × 11 × 7)
= 21√11 cm2
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There are 20 questions in RD Sharma Class 9 Solutions for Chapter 12 MCQs.