Question

State and Prove the relation between areas of two similar triangles.

Answer 1

Given : $△ABC\sim △PQR$

To prove,

$\frac{Area\left(△ABC\right)}{Area\left(△PQR\right)}=\frac{B{C}^2}{Q{R}^2}=\frac{A{B}^2}{P{Q}^2}=\frac{A{C}^2}{P{R}^2}$

Construction : Draw segment $AD\perp$ side $BC\&PS\perp$ side $QR$

Proof: $△ABC\sim △PQR$ (given)

$\therefore \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$..........(1)

$\angle B=\angle Q\&\angle C=\angle R$...........(2)

Now in $\angle ADB\&△PSQ,$

$\angle ABD=\angle PQS$ (from ii)

$\angle ADB=\angle PSQ$ ($={90}^0$)

$\therefore △ADB\sim △PSQ$ (A-A similarity)

$\therefore \frac{AB}{PQ}=\frac{BD}{QS}=\frac{AD}{PS}$

$\therefore$ From (i) and (iii)

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PS}$......(iv)

Area of $△=\frac{1}{2}\times b\times h=\frac{\frac{1}{2}\times AD\times BC}{\frac{1}{2}\times PS\times QR}=\frac{Area\left(△ABC\right)}{Area\left(△PQR\right)}$

$=\frac{AD\times BC}{PS\times QR}=\frac{AD}{PS}\times \frac{BC}{QR}=\frac{BC}{QR}\times \frac{BC}{QR}[\because \frac{AD}{PS}=\frac{BC}{QR}]$

$=\frac{B{C}^2}{Q{R}^2}$

$\therefore \frac{Aof\left(△ABC\right)}{Aof\left(△PQR\right)}=\frac{B{C}^2}{Q{R}^2}=\frac{A{B}^2}{P{Q}^2}=\frac{A{C}^2}{P{R}^2}$

$[\because \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}(fromi\left)\right]$