Question

Find the ratio of the areas of two similar triangles shown in the figure.

• A. ${\left(\frac{AB}{PQ}\right)}^2$
• B. ${\left(\frac{BC}{QR}\right)}^2$
• C. ${\left(\frac{AC}{PR}\right)}^2$
• D. $Allofthese$

Answer 1

The correct option is D $Allofthese$

We are given two triangles ABC and PQR such that $\Delta ABC\sim \Delta PQR.$
For finding areas of the two triangles, we draw altitudes AM and PN of the triangles.

$Ar\left(\Delta ABC\right)=\frac{1}{2}\times BC\times AM$ and
$Ar\left(\Delta PQR\right)=\frac{1}{2}\times QR\times PN$

So, $\frac{Ar\left(\Delta ABC\right)}{Ar\left(\Delta PQR\right)}=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times QR\times PN}=\frac{BC\times AM}{QR\times PN}$

Now, in $\Delta ABM$ and $\Delta PQN$,
$\angle B=\angle Q$ $(As\Delta ABC\sim \Delta PQR)$
and $\angle M=\angle N$ (Each is of ${90}^{\circ }$)

So, $\Delta ABM\sim \Delta PQN$ (AA similarity)

Therefore, $\frac{AM}{PN}=\frac{AB}{PQ}$

Also, $\Delta ABC\sim \Delta PQR$

So, $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{PR}$

Therefore,$\frac{Ar\left(\Delta ABC\right)}{Ar\left(\Delta PQR\right)}$
= $\frac{AB}{PQ}\times \frac{AM}{PN}$
= $\frac{AB}{PQ}\times \frac{AB}{PQ}$
= ${\left(\frac{AB}{PQ}\right)}^2$

Since the ratio of sides is the same,

$\frac{Ar\left(\Delta ABD\right)}{Ar\left(\Delta BDC\right)}$
= ${\left(\frac{AB}{PQ}\right)}^2$
= ${\left(\frac{BC}{QR}\right)}^2$
= ${\left(\frac{AC}{PR}\right)}^2$

If $\Delta ABC$ and $\Delta PQR$ are similar, then their areas are in the ratio of the square of ratio of corresponding sides.