PQR is a triangle such that |PQ| = |QR| = 8cm and QPR = 60º. Find the area of \(\angle\) PQR
-
16\(\sqrt{3}cm^2\)
- 16\(\sqrt{2}cm^2\)
- 8\(\sqrt{3}cm^2\)
- 16\(\sqrt{2}cm^2\)
The correct answer is: A
Explanation
The area of a triangle given two sides and the included angle = \(\frac{1}{2}\) x a x b x Sin\(\theta\)
a = b = 8cm
\(\theta\) = 60º
Area = \(\frac{1}{2}\) x 8 x 8 x Sin 60º = \(\frac{1}{2}\) x 64 x \(\frac{\sqrt{3}}{2}\) ( Sin 60º = \(\frac{\sqrt{3}}{2}\) )
= \(\frac{64\sqrt{3}}{4}\) = 16\(\sqrt{3}cm^2\)
There is an explanation video available .