Perimeter Of A Simple Triangle

Question

Here in $ \triangle ABC$ $ AC=4 , DE= EF =1, \angle ABC=90^{\circ} $. The perimeter of the triangle $ \triangle ABC$ can be written as $ \sqrt {m } + n $ where $m$ and $n$ are non-negative integers.What is the value of $m+n$?

Answer 1

let, $AD=x$ and $FC=y$
and we know $ \angle ABC=90^{\circ}$
we can write this equation(using Pythagoras):$x+1)^2 + (y+1)^2=16$..........$(1)$
again $ \triangle ADE$ and $\triangle EFC$ is similar.
So,using similarity we acn say $ \frac{1}{y}= \frac{x}{1} $ or $xy=1$......$(2)$
Now we have to solve $(1)$ and $(2)$ equation.
from equation $1$,
$x^2+2x+1+y^2+2y+2 = 16 $
or,$x^2+y^2+2(x+y)+2=16$
or,$(x+y)^2-2xy+2(x+y)+2=16$
or,$(x+y)^2-2*1+2(x+y)+2=16 $ [using equation $(2)$ ]
or,$(x+y)^2-2+2(x+y)+2=16 $
or,$(x+y)^2+2(x+y)-16=0 $
or,$a^2+2a-16=0$ [ Let $a=x+y$]
so,$a=\sqrt{17}-1 or,a=-\sqrt{17}-1$[ we can't take this value as it is negative]
so ,$x+y=\sqrt{17}-1$
hence,perimeter of this triangle is $ \triangle ABC $
$ =x+y+DB+BF+AC = \sqrt{17}-1+1+1+4=\sqrt{17} +5$
here $ m=17 $ and $ n=5$
finally,$m+n=17+5=\boxed{22}$