# Mathematics Questions and Answers – Application of Heron’s Formula in finding Areas of Quadrilaterals

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This set of Mathematics Assessment Questions for Class 9 focuses on “Application of Heron’s Formula in finding Areas of Quadrilaterals”.

1. An umbrella is made by stitching 8 triangular pieces of cloth of two different colours, each piece measures 60cm, 60cm and 20cm. How much cloth of each colour is required for the umbrella?
a) $$50\sqrt{35} cm^2$$
b) $$25\sqrt{65} cm^2$$
c) $$50\sqrt{45} cm^2$$
d) $$25\sqrt{55} cm^2$$

Explanation: s = $$\frac{a+b+c}{2}=\frac{60+60+20}{2} = \frac{140}{2}$$ = 70
According to heron’s formula, area of the triangle = $$\sqrt{s*(s-a)*(s-b)*(s-c)}$$
= $$\sqrt{70*(70-60)*(70-60)*(70-20)}$$
= $$\sqrt{70*10*10*50}$$
= 100$$\sqrt{35} cm^2$$
This area is the combined area of both colours.
Hence, area of each colour = $$\frac{100\sqrt{35}}{2}$$
= $$50\sqrt{35} cm^2$$.

2. A triangle and a parallelogram has same base and same area as shown in the diagram below. Dimensions of triangle are 28cm, 26cm and 30cm with 28cm being the base. What is the height of the parallelogram?
a) 15cm
b) 10cm
c) 12cm
d) 18cm

Explanation: a = 28, b = 26 and c = 30 cm.
s = $$\frac{a+b+c}{2}=\frac{28+26+30}{2} = \frac{84}{2}$$ = 42
According to heron’s formula, area of the triangle = $$\sqrt{s*(s-a)*(s-b)*(s-c)}$$
= $$\sqrt{42*(42-28)*(42-26)*(42-30)}$$
= $$\sqrt{42*14*16*12}$$
= 336 cm2
It is given that area of the triangle and the parallelogram is same.
Now, we know that area of a parallelogram = base * perpendicular (height of a parallelogram)
Therefore, 28 * height = 336
height = $$\frac{336}{28}$$
height = 12cm.

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