The height projected to the base of the isosceles triangle is equal to H and is twice as large as its projection on the side. Find the area of the triangle.
My attempt at answering : What I understand from the question is that the triangle here is an equilateral triangle because the projection of the height to the base is twice as large as its projection to the side. Then I used the formula $ A =\sqrt(3/4)*a^2$ and got $(H^2\sqrt3)/3$.
The answer given is $H^2\sqrt3$.
Can somebody maybe point out where I'm wrong here. Thanks.
Best Answer
$|AD|=H,|AE|=H/2$. We have $AE/AD=\cos BAD$, so $\angle BAD=60^o$.
$\angle ADB=90^o$. So $BD=\sqrt3\ AD$. Hence the area of the triangle (1/2 base x height) is $\sqrt3\ H^2$.