In $ \triangle \mathrm{ABC}, \angle \mathrm{A}=90^{\circ} $ and $ \mathrm{AM} $ is an altitude. If $ \mathrm{BM}=6 $ and $ \mathrm{CM}=2 $, find the perimeter of $ \triangle \mathrm{ABC} $.

Given:

In \( \triangle \mathrm{ABC}, \angle \mathrm{A}=90^{\circ} \) and \( \mathrm{AM} \) is an altitude. If \( \mathrm{BM}=6 \) and \( \mathrm{CM}=2 \).

To do:

We have to find the perimeter of \( \triangle \mathrm{ABC} \).

Solution:

In $\triangle ABC$,

By Pythagoras theorem,

$BC^2 =AB^2+AC^2$

$(2+6)^2=AB^2+AC^2$

$(8)^2=AB^2+AC^2$

$64=AB^2+AC^2$.........(i)

Similarly,

In $\triangle ABM$,

By Pythagoras theorem,

$AB^2 =AM^2+BM^2$

$AB^2=AM^2+6^2$......(ii) 

In $\triangle AMC$,

By Pythagoras theorem,

$AC^2 =MC^2+AM^2$

$AC^2=(2)^2+AM^2$......(iii)

From (i), (ii) and (iii),

$64=36+AM^2+4+AM^2$

$64-40=2AM^2$

$24=2AM^2$

$AM^2=12$

$AM=\sqrt{12}=2\sqrt{3}$

Therefore,

$AB^2=12+36$

$=48$

$\Rightarrow AB=\sqrt{48}$

$=4\sqrt{3}$

$AC^2=4+12$

$=16$

$\Rightarrow AC=\sqrt{16}$

$=4$

Therefore, 

Perimeter of triangle ABC $=AB+BC+AC$

$=4\sqrt3+8+4$

$=12+4\sqrt{3}$