This set of Class 9 Maths Chapter 12 Multiple Choice Questions & Answers (MCQs) focuses on “Heron’s Formula”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.
1. Which of the following formula is used for finding the area of triangle?
a) base * height
b) \(\frac{1}{2}\) * base * height
c) 2 * base * height
d) base2 * height
View Answer
Explanation: As shown in the diagram below, area of the triangle is given by \(\frac{1}{2}\) * base * height where, base = BC and height is AD.
2. In heron’s formula \(\sqrt{s*(s-a)*(s-b)*(s-c)}\), what is the value of s if a, b and c are sides of the triangle?
a) \(\frac{a+b+c}{4}\)
b) a+b+c
c) \(\frac{a+b+c}{2}\)
d) 2a+2b+2c
View Answer
Explanation: In heron’s formula \(\sqrt{s*(s-a)*(s-b)*(s-c)}\), s is the half perimeter of the triangle.
Perimeter of the triangle having sides a, b and c is a + b + c.
Hence, s = half perimeter of the triangle = \(\frac{a+b+c}{2}\).
3. What is the area of the triangle having sides equal to 10 cm, 16 cm and 24 cm?
a) \(25\sqrt{15} cm^2\)
b) \(35\sqrt{17} cm^2\)
c) \(15\sqrt{13} cm^2\)
d) \(15\sqrt{15} cm^2\)
View Answer
Explanation: a = 10, b = 16 and c = 24
s = \(\frac{a+b+c}{2}=\frac{10+16+24}{2} = \frac{50}{2}\) = 25
According to heron’s formula, area of the triangle = \(\sqrt{s*(s-a)*(s-b)*(s-c)}\)
= \(\sqrt{25*(25-10)*(25-16)*(25-24)}\)
= \(\sqrt{25*15*9*1}\)
= \(15\sqrt{15} cm^2\)
4. The sides of a triangle are in the proportion of 2 : 3 : 5 and its perimeter is 200 cm. The area of this triangle is __________ cm2.
a) 375 \(\sqrt{23}\)
b) 375 \(\sqrt{21}\)
c) 345 \(\sqrt{23}\)
d) 345 \(\sqrt{21}\)
View Answer
Explanation: Let the side of a triangles be a = 2x, b = 3x and c = 5x.
Perimeter of the triangle is 200 cm.
Therefore a + b + c = 2x + 3x + 5x = 200
10x = 200
Hence, x = 20
a = 2x = 2(20) = 40 cm
b = 3x = 3(30) = 90 cm
c = 5x = 5(20) = 100 cm
Now, s = \(\frac{a+b+c}{2}=\frac{40+90+100}{2} = \frac{230}{2}\) = 115
According to heron’s formula, area of the triangle = \(\sqrt{s*(s-a)*(s-b)*(s-c)}\) = \(\sqrt{115*(115-40)*(115-90)*(115-100)}\)
= \(\sqrt{115*75*25*15}\)
= \(375 \sqrt{23}\)
5. A triangular garden has sides 90m, 140m and 80m. A fence is to be put all around the garden. What will be the total cost of fencing at the rate of Rs 15 per metre? A 5m wide space is to be left on the one side for gate opening.
a) Rs 4525
b) RS 4975
c) Rs 4575
d) Rs 4230
View Answer
Explanation: 90 + 140 + 80 = 310m
Considering 5m space for gate opening, length of fencing = 310 – 5 = 305m.
Cost of 1m fencing = Rs 15,
Then the cost for 305m fencing = 305 * 15
= Rs 4575.
6. A triangular board having sides 45m, 30m and 35m is used for advertising. One company uses this board for its advertisement for 4 months. How much rent will the company has to pay if the rent is Rs 3500 per m2?
a) Rs 611800
b) Rs 1835400
c) Rs 611900
d) Rs 1835500
View Answer
Explanation: a = 45, b = 30 and c = 35
We know that s = \(\frac{a+b+c}{2} = \frac{45+30+35}{2} = \frac{110}{2}\) = 55
According to heron’s formula, area of the triangle = \(\sqrt{s*(s-a)*(s-b)*(s-c)}\)
= \(\sqrt{55*(55-45)*(55-30)*(55-35)}\)
= \(\sqrt{55*10*25*20}\)
= 524.40 m2
Now the rent for 1 m2 = Rs 3500,
then rent for 992.15 m2 = 524.40 * 3500 = Rs 18,35,400
The rent which we got above is for one year.
The company used the board for 4 months.
Therefore rent for 4 months = Rs \(\frac{18,35,400*4}{12}\)
= Rs 6,11,800.
More MCQs on Class 9 Maths Chapter 12:
To practice all chapters and topics of class 9 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
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