Question

State whether the following statement is true or false.

The ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

• A. True
• B. False

Answer 1

The correct option is A True

$△ABC\sim △PQR$ [ Given ]

Draw, $AM\perp BC$ and $PN\perp QR$

$\Rightarrow$ $ar\left(ABC\right)=\frac{1}{2}\times BC\times AM$ ---- ( 1 )

$\Rightarrow$ $ar\left(PQR\right)=\frac{1}{2}\times QR\times PN$ ----- ( 2 )

Dividing ( 1 ) and ( 2 )

$\Rightarrow$ $\frac{ar\left(ABC\right)}{ar\left(PQR\right)}=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times QR\times PN}$

$\Rightarrow$ $\frac{ar\left(ABC\right)}{ar\left(PQR\right)}=\frac{BC\times AM}{QR\times PN}$ ---- ( 3 )

In $△ABM$ and $△PQN$

$\Rightarrow$ $\angle B=\angle Q$ [ As $△ABC\sim △PQR$ and angles of smilar triangles are equal ]

$\Rightarrow$ $\angle M=\angle N$ [ Both ${90}^o$ ]

$\therefore$ $△ABM\sim △PQN$ [ By $AA$ similarity ]

$\therefore$ $\frac{AB}{PQ}=\frac{AM}{PN}$ [ Corresponding sides of similar triangles are proportional ] ----- ( 4 )

Substituting ( 4 ) in ( 3 ),

$\Rightarrow$ $\frac{ar\left(ABC\right)}{ar\left(PQR\right)}=\frac{BC}{QR}\times \frac{AB}{PQ}$ ---- ( 5 )

Now,

$△ABC\sim △PQR$ [ Given ]

$\Rightarrow$ $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

Putting in ( 5 )

$\Rightarrow$ $\frac{ar\left(ABC\right)}{ar\left(PQR\right)}=\frac{AB}{PQ}\times \frac{AB}{PQ}={\left(\frac{AB}{PQ}\right)}^2$

Now, again using $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

$\Rightarrow$ $\frac{ar\left(ABC\right)}{ar\left(PQR\right)}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{AC}{PR}\right)}^2$