Question 7.

Let 'h' be the height of the triangle ABC, semiperimeter(S) $= \frac{4+4+r+4+4+r}{2} = 8+r$, $a=4+r, b=4+r, c=8$

Area of triangle ABC $=\sqrt{\ s\cdot\left(s-a\right)\left(s-b\right)\left(s-c\right)}=$ $\sqrt{\left(\ 8+r\right)\times\ 4\times\ 4\times\ r}$ = $\ \frac{\ 1}{2}\times \left(4+4\right)\times height$

Height (h) = $\sqrt{\left(8+r\right)r}$

Now, $h+4 = 4 \longrightarrow \sqrt{8+r} + r = 4$ (Considering the height of the triangle)

$\sqrt{\left(8+r\right)r}$=4-r

16r=16

r=1 Alternatively,

$AE^@ + EC^2 = AC^2 \longrightarrow 4^2 + (4-r)^2 = (4+r)^2 \longrightarrow \longrightarrow \longrightarrow r=1$