Question

seg SP ⊥ side YK and
seg YT ⊥ seg SK
If SP = 6, YK = 13, YT = 5 and TK = 12
then find
$A(∆SYK):A(∆YTK)$

Answer 1

We have:
SP = 6, YK = 13, YT = 5 and TK = 12
We know that the ratio of the areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.
Thus, we get:

$$\begin{gathered} A\left(△SYK\right):A\left(△YTK\right) \\ =\frac{A\left(△SYK\right)}{A\left(△YTK\right)} \\ =\frac{YK\times SP}{TK\times YT} \\ =\frac{13\times 6}{12\times 5} \\ =\frac{13}{10} \\ =13:10 \end{gathered}$$