Heron’s Formula Class 9 MCQ with Answers – Maths Class 9 MCQ Online Test are covered in this Article. Heron’s Formula Class 9 MCQ Test contains 15 questions. Answers to MCQ on Heron’s Formula Class 9 are available after clicking on the answer. MCQ Questions for Class 9 with Answers have been made for Class 9 students to help check the concept you have learnt from detailed classroom sessions and application of your knowledge.
| Board | CBSE |
| Textbook | Maths (NCERT) |
| Class | Class 9 |
| Chapter | Chapter 12 Heron’s Formula |
| Category | MCQ Questions for Class 9 Maths with Answers |
Heron’s Formula Class 9 MCQ with Answers
1.Heron’s formula for area of a triangle is
(a) √((s-a)(s-b)(s-c))
(b) √((s-a)(s-b)(s+c))
(c) √(s(s-a)(s-b)(s-c))
(d) √(s(s + a)(s + b)(s + c))
Answer
Answer: (c) √(s(s-a)(s-b)(s-c))
2. s in Heron’s formula is
(a)perimeter of the triangle
(b)half of perimeter of triangle
(c)half of perimeter of square
(d)perimeter of rectangle
Answer
Answer: (b)half of perimeter of triangle
Explanation: s = (a+b+c)/2, where a, b and c are sides of the triangle.
3.Area of a triangle is
(a)(1/3)×base× height
(b) (1/3)×base× perimeter
(c) ½×base× height
(d) ½×base× perimeter
Answer
Answer: (c) ½×base× height
4.If perimeter of an equilateral triangle is 360 cm. Then area of the triangle is
(a)200 cm2
(b) 3600√3 cm2
(c) 600√3 cm2
(d) 360√3 cm2
Answer
Answer: (b) 3600√3 cm2
Explanation: All sides of an equilateral triangle are equal.
3a = 360
a = 120
semi-perimeter s = 360/2 = 180 cm
area of equilateral triangle = √(s(s-a)(s-b)(s-c))
= √(180(180-120)(180-120)(180-120))
=√(180(60)(60)(60))
= 3600√3 cm2
5. Find the area of a triangle, two sides of which are 12 cm and 13 cm and the perimeter is 30 cm
(a) 30 cm2
(b) 900 cm2
(c) 300 cm2
(d) 90 cm2
Answer
Answer: (a) 30 cm2
Explanation:
Perimeter = a + b + c
30 = 12 + 13 + c
c = 30 – 25
c = 5 cm
s = 30/2 = 15cm
area of triangle = √(s(s-a)(s-b)(s-c))
= √(15(15-12)(15-13)(15-5))
= √(15(3)(2)(10))
= √900 = 30 cm2
6.The sides of triangular table are 4 cm, 4 cm and 6cm. The cost of painting the table at the rate of 12 rupees per cm2 is
(a) Rs. 53.48
(b) Rs. 60.48
(c) Rs 95.40
(d) Rs. 66.48
Answer
Answer: (c) Rs 95.40
Explanation: s = (4+4+6)/2 = 14/2 = 7 cm
area of triangle = √(s(s-a)(s-b)(s-c))
= √(7(7-4)(7-4)(7-6))
= √ 7x3x3x1
= 3√ 7
= 3x 2.65
= 7.95 (approx)
Cost of painting = 7.95 x 12
= Rs 95.40
Heron’s Formula Class 9 MCQ with Answers
7. The sides of a triangle are in the ratio of 7:3:5 and its perimeter are 450 m. Then s =
(a) 225 cm
(b) 220 cm
(c) 205 cm
(d) 200 cm
Answer
Answer: (a) 225 cm
Explanation: s = semi-perimeter = 450/2 = 225 cm
8. An isosceles triangle has perimeter 40 cm and one side is 6 cm. Then area of the triangle is
(a) 60√70cm2
(b) 6√70cm2
(c) 6√60cm2
(d) 36√70cm2
Answer
Answer: (b) 6√70cm2
Explanation: Let the equal sides be x.
x + x + 6 = 40
2x = 34
x = 17 cm
s = 40/2 = 20 cm
area of triangle = √(s(s-a)(s-b)(s-c))
= √(20(20-17)(20-17)(20-6))
= √(20(3)(3)(14))
= 6√70cm2
9.The sides of a triangle are in the ratio of 9:10:11 and its perimeter are 900 m. Then side of the triangle are
(a)270cm, 300cm, 350cm
(b) 280cm, 300cm, 330cm
(c) 270cm, 350cm, 330cm
(d) 270cm, 300cm, 330cm
Answer
Answer: (d) 270cm, 300cm, 330cm
Explanation: Let the sides of the triangle be 9x, 10x and 11x.
9x + 10x + 11x = 900
30x = 900
x = 30
Sides of the triangle are 30×9 = 270 cm
30×10 = 300 cm
30 × 11 = 330 cm
10.The sides of a triangle are 50 cm, 100cm and 140cm. Then area of triangle is
(a) 75√51cm2
(b) 7√551cm2
(c) 75√551cm2
(d) 7√55cm2
Answer
Answer: (c) 75√551cm2
Explanation: s = (50+100+140 )/2 = (290 )/2 = 145 cm
area of triangle = √(s(s-a)(s-b)(s-c))
= √(145(145-50)(145-100)(145-140))
= √(145(95)(45)(5))
= 75√551cm2
11.Sides of an equilateral triangle is 8 cm. Then area of equilateral triangle
(a) 16√3cm2
(b)160√3 cm2
(c)6√3 cm2
(d)162 cm2
Answer
Answer: (a) 16√3cm2
Explanation: s = (8+8+8 )/2 = 12 cm
area of equilateral triangle = √(s(s-a)(s-b)(s-c))
= √(12(12-8)(12-8)(12-8))
= √(12(4)(4)(4))
= 16√3cm2
12.The sides of a right -angled triangle are 3cm, 4cm and 5cm. Then area of the triangle is
(a) 4 cm2
(b) 6 cm2
(c) 8 cm2
(d) 16 cm2
Answer
Answer: (b) 6 cm2
Explanation: Area of triangle = 1/2×base× height
= 1/2×3× 4
= 6 cm2
13.The base of a right- angled triangle is 5 cm and hypotenuse is 13cm. Then area of triangle is
(a)130 cm2
(b) 150 cm2
(c) 30 cm2
(d) 60 cm2
Answer
Answer: (c) 30 cm2
Explanation: By Pythagoras theorem
Height2 = Hypotenuse2 – base2
Height = √(169-25) =√144 = 12cm
Area of triangle = 1/2×base× height
= 1/2×5× 12
= 30 cm2
14.In an isosceles triangle two equal sides are 6cm and the perimeter is 20 cm. Then the area of triangle is
(a) 8√5cm2
(b) 8√25cm2
(c) 4√5cm2
(d) 8√15cm2
Answer
Answer: (a) 8√5cm2
Explanation: s = 20/2 = 10cm
Let the third side be x.
6 + 6 + x = 20
x = 8cm
area of triangle = √(s(s-a)(s-b)(s-c))
= √(10(10-8)(10-6)(10-6))
= √(10(2)(4)(4))
= 8√5cm2
15.Perimeter of an equilateral triangle is 60cm. Then product of sides is
(a) 4000 cm
(b) 8000 cm
(c) 5000 cm
(d) 6000 cm
Answer
Answer: (b) 8000 cm
Explanation: Let sides of the equilateral triangle be x.
x + x +x = 60
3x = 60
x = 20cm
Product of sides = 20×20×20 = 8000 cm
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Frequently Asked Questions on Heron’s Formula Class 9 MCQ with Answers
1. Are these MCQ on Heron’s Formula Class 9 are based on 2021-22 CBSE Syllabus?
Yes. There are 15 MCQ’s on this Chapter in this blog.
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