Question

In the figure $segSP\perp sideYKandsegYT\perp segSK$. If $SP=6,YK=13,YT=5andTK=12$ then
$A\left(△SYK\right):A\left(△YTK\right)=13:x$ .find $x$.

Answer 1

$\frac{A\left(△SYK\right)}{A\left(△YTK\right)}$ = $\frac{\frac{1}{2}SP\times YK}{\frac{1}{2}YT\times TK}$ (Area of triangle = $\frac{1}{2}$ base $\times$ height)
$\frac{A\left(△SYK\right)}{A\left(△YTK\right)}=\frac{\frac{1}{2}(6\times 13)}{\frac{1}{2}(5\times 12)}$
$\frac{A\left(△SYK\right)}{A\left(△YTK\right)}=\frac{13}{10}$