Question

A door of width $6$ meter has an arch above it having a height of $2m$. Find the radius of the arch.

Answer 1

Given: Width of door $=AE=6m$

Let $C$ be the centre of the semicircle that makes the arch.

Then $AC=CD=$ Radius of the semicircle

Let $r$ be the radius of the arch.

Then $BC=CD-BD$

$=$ radius $-$ height of the arch

$=r-h$

In right-angled triangle $ABC,$

$A{C}^2=A{B}^2+B{C}^2$ (By Pythagoras theorem)

$\Rightarrow r^2=3^2+(r-2{)}^2$

$\Rightarrow r^2=9+r^2-4r+4$

$\Rightarrow 4r=13$

$\Rightarrow r=\frac{13}{4}=3.25m$