A geometry problem by Aareyan Manzoor

By the pythagorean theorem, we have

$x=\sqrt{20^2-16^2}=12$

The perimeter of the triangle is $20+20+24=64$ . Computing for the other side length of the rectangle, we have

$64=2L+2(5)$ $\implies$ $2L=54$ $\implies$ $\color{#D61F06}\boxed{L=27}$

the triangle half will form 2 congruent Right-angle triangle

let the height be adjacent. let the 20 m line be hypotenuse so the below line will be 2 opposite. o = √(h^2-a^2 ) o= √ 20 20 - 16 16

o=12 whole line =24 perimetre = 24 + 20 +20 p= 64

rectangle 1st line= 5 2nd =5 other line = 64- 10)/2 = 54/2 =27 (ans)