# Mathematics Questions and Answers – Heron’s Formula & Area of a Triangle by Heron’s Formula

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Heron’s Formula & Area of a Triangle by Heron’s Formula”.

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

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

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) $$15\sqrt{15} cm^2$$

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 = $$\sqrt{s*(s-a)*(s-b)*(s-c)}$$
= $$\sqrt{25*(25-10)*(25-16)*(25-24)}$$
= $$\sqrt{25*15*9*1}$$
= $$15\sqrt{15} cm^2$$
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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 $$\sqrt{21}$$

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

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

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.

Sanfoundry Global Education & Learning Series – Mathematics – Class 9.