Question : Find the area of triangle whose sides are 10 cm, 12 cm, and 18 cm.
Option 1: $22 \sqrt{2} \mathrm{~cm}^2$
Option 2: $30 \sqrt{2} \mathrm{~cm}^2$
Option 3: $28 \sqrt{2} \mathrm{~cm}^2$
Option 4: $40 \sqrt{2} \mathrm{~cm}^2$
Correct Answer: $40 \sqrt{2} \mathrm{~cm}^2$
Solution :
The area of a triangle by Heron's formula = $\sqrt{s(s - a)(s - b)(s - c)}$
where $a$, $b$, and $c$ are the sides of the triangle, and $s$ is the semi-perimeter of the triangle.
Now, $s = \frac{a + b + c}{2}= \frac{10 + 12 + 18}{2} = 20$
So, the area $=\sqrt{20(20 - 10)(20 - 12)(20 - 18)} = \sqrt{20 \times 10 \times 8 \times 2} = 40 \sqrt{2} \text{ cm}^2$
Hence, the correct answer is $40 \sqrt{2} \text{ cm}^2$.
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