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) base^{2} * 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 __________ cm^{2}.

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 m^{2}?

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 m

^{2}

Now the rent for 1 m

^{2}= Rs 3500,

then rent for 992.15 m

^{2}= 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**.

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