Question

Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Apply the above theorem on the following:

$ABC$ is a triangle and $PQ$ is a straight line meeting $AB$ in $P$ and $AC$ in $Q$. If $\mathrm{AP}=1\mathrm{cm},\mathrm{PB}=3\mathrm{cm},\mathrm{AQ}=1.5\mathrm{cm},\mathrm{QC}=4.5\mathrm{cm}$. Prove that the area of $∆APQ$ is one-sixteenth of the area of $∆ABC$.

Answer 1

Step 1: Note the given data and draw a diagram

Let $ABC$ and $PQR$ be two triangle similar triangle.

Since the triangles are similar, so

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$ …..(i)

Let $AD$ and $PS$ be altitudes of $△ABC$ and $△PQR$ respectively.

Step 2: Check similarity for $△ADB$ and $△PSQ$

AA similarity: If two angles of a triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

$\angle ADB=\angle PSQ=90°$

$\angle ABD=\angle PQS\left[\because △ABC~△PQR\right]$

According to AA similarity $△ADB~△PSQ$

So,$\frac{AD}{PS}=\frac{AB}{PQ}$ ……(ii)

From (i) and (ii) we get

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{AD}{PS}$……(iii)

Step 3: Finding the ratio of areas of $△ABC$ and $△PQR$

The area of a triangle is $=\frac{1}{2}\times Base\times height$

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

$\frac{ar\left(△ABC\right)}{ar\left(△PQR\right)}=\frac{BC\times BC}{QR\times QR}$ $\left[\because \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{AD}{PS}\right]$

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

Hence proved

Step 4: Check the similarity of $△APQ$ and $△ABC$

Given that $\mathrm{AP}=1\mathrm{cm},\mathrm{PB}=3\mathrm{cm},\mathrm{AQ}=1.5\mathrm{cm},\mathrm{QC}=4.5\mathrm{cm}$

The length of $AB=AP+BP=1cm+3cm=4cm$

The length of $AC=AQ+QC=1.5+4.5=6cm$

The ratio of the length of $AB$ and $AP$ is $\frac{AB}{AP}=\frac{4}{1}=4$

The ratio of the length of $AC$ and $AQ$ is $\frac{AC}{AQ}=\frac{6}{1.5}=4$

SAS similarity: If one angle of a triangle is equal to one angle of the other and the sides forming these angles are proportional the triangle similar.

In $△APQ\&△ABC$

$\angle A=\angle A$ (common angle)

$\frac{AB}{AP}=\frac{AC}{AQ}=4$

According to SAS similarity

$\therefore △APQ~△ABC$

Step 5: Finding the ratio of $area(△\mathrm{APQ})$ and $area(△\mathrm{ABC})$

So, by the given theorem

$\frac{area(△\mathrm{APQ})}{area(△\mathrm{ABC})}=\frac{{\mathrm{AP}}^2}{{\mathrm{AB}}^2}$

$\Rightarrow \frac{area(△\mathrm{APQ})}{area(△\mathrm{ABC})}=\frac{1^2}{4^2}$ $\left[\because \frac{AB}{AP}=4\Rightarrow \frac{AP}{AB}=\frac{1}{4}\right]$

$\Rightarrow \frac{area(△\mathrm{APQ})}{area(△\mathrm{ABC})}=\frac{1}{16}$

$\therefore area(△\mathrm{APQ})=\frac{1}{16}\times area(△\mathrm{ABC})$

Hence proved.