In triangle XYZ above, the length of XZ is 6/7 of the length of altitude h. What is the area of ∆ XYZ in terms of h?
$$A.\ \frac{h^{2}}{3}$$
$$B.\ \frac{3h^{2}}{7}$$
$$C.\ \frac{3h}{7}$$
$$D.\ \frac{6h^{2}}{7}$$
$$E.\ \frac{12h^{2}}{7}$$
The OA is B.
We know that the area of a triangle is given by
$$A=\frac{1}{2}\cdot b\cdot h$$
In this case
$$b=XZ=\frac{6}{7}\cdot h$$
Then the area of the triangle XYZ in terms of h will be
$$A=\frac{1}{2}\cdot\frac{6}{7}\cdot h\cdot h\ =\frac{3h^2}{7}$$
Is there another way to solve this PS question? Can someone help? Thanks!
