What is the area of an equilateral triangle whose one side length is 60?
A. 300√3
B. 400√3
C. 450√3
D. 600√3
E. 900√3
The OA is E.
I know that I can get the height of the triangle using Pythagoras Theorem, then
$$h_{\triangle}=\sqrt{60^2-30^2}=30\sqrt{3}$$
Then, I can get the area of the triangle,
$$A_{\triangle}=\frac{1}{2}b\cdot h$$ $$A_{\triangle}=\frac{1}{2}b\cdot h=\frac{1}{2}\cdot60\cdot30\sqrt{3}=900\sqrt{3}$$
But, is there another strategic approach to this question? Can any experts help please? Thank you!
A. 300√3
B. 400√3
C. 450√3
D. 600√3
E. 900√3
The OA is E.
I know that I can get the height of the triangle using Pythagoras Theorem, then
$$h_{\triangle}=\sqrt{60^2-30^2}=30\sqrt{3}$$
Then, I can get the area of the triangle,
$$A_{\triangle}=\frac{1}{2}b\cdot h$$ $$A_{\triangle}=\frac{1}{2}b\cdot h=\frac{1}{2}\cdot60\cdot30\sqrt{3}=900\sqrt{3}$$
But, is there another strategic approach to this question? Can any experts help please? Thank you!
