Question

I just need 30 please

Answer 1

In the smaller triangle, let $y$ be the length of the smallest leg. Then the smallest leg in the larger triangle is $14-y$.

Within the scope of the smaller triangle, we have $y$ such that

$x^2+y^2=13^2\implies y=\sqrt{13^2-x^2}$

Then within the larger the triangle, we would have

$x^2+(14-y)^2=15^2\iff x^2+\left(14-\sqrt{13^2-x^2}\right)^2=15^2$

Now we can solve for $x$:

$x^2+14^2-28\sqrt{13^2-x^2}+(13^2-x^2)=15^2$

$\implies x=12$