NCERT Solutions Class 9 Mathematics Solutions for Heron’s Formula - Exercise 12.2 in Chapter 12 - Heron’s Formula
A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m
and AD = 8 m. How much area does it occupy?
First, construct a quadrilateral ABCD and join BD.
We know that
∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m
The diagram is:
Now, apply Pythagoras theorem in ΔBCD
\begin{aligned} &B D^{2}=B C^{2}+C D^{2}\\ &\Rightarrow B D^{2}=12^{2}+5^{2}\\ &\Rightarrow B D^{2}=169\\ &\Rightarrow B D=13 \mathrm{m}\\ &\text { Now, the area of } \Delta B C D=(1 / 2 \times 12 \times 5)=30 \mathrm{m}^{2} \end{aligned}
The semi perimeter of ΔABD
(s) = (perimeter/2)
= (8 + 9 + 13)/2 m
= 30/2 m = 15 m
Using Heron's formula,
Area of ΔABD
\begin{array}{l} \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{15(15-13)(15-9)(15-8)} m^{2} \\ =\sqrt{15 \times 2 \times 6 \times 7 m^{2}} \\ =6 v 35 \mathrm{m}^{2}=35.5 \mathrm{m}^{2} \text { (approximately) } \end{array}
=30 \mathrm{m}^{2}+35.5 \mathrm{m}^{2}=65.5 \mathrm{m}^{2}
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