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 to the following :

$ABC$ is an isosceles triangle right-angled at $B$. Similar triangles $ACD$ and $ABE$ are constructed on sides $AC$ and $AB$ . Find the ratio between the areas of $△ABE$ and $△ACD$.

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: Note the given data and draw a diagram that represents the given problem

Given that, $ABC$ is an isosceles triangle right-angled at $B$. Similar triangles $ACD$ and $ABE$ are constructed on sides $AC$ and $AB$ .

Let $AB=BC=xcm$

Pythagoras theorem: Square of the opposite side to right angle equals the sum of squares of the other two sides.

According to Pythagoras theorem,

$$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ \Rightarrow A{C}^2=x^2+x^2 \\ \Rightarrow A{C}^2=2x^2 \\ \Rightarrow AC=\sqrt{2}x \end{gathered}$$

The diagram for the given problem is

Step 5: Find the ratio between the areas of $△ABE$ and $△ACD$.

AAA similarity: If in two triangles, the corresponding angles are equal, then the triangles are similar.

We know that each angle of an equilateral triangle is $60°$.

So the angles of $△ABE$ and $△ACD$ are equal to each other. According to AAA similarity $△ABE~$ $△ACD$.

Relationship between areas and sides of similar triangles: The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides”

$$\begin{gathered} \frac{\text{ar}(△ABE)}{\text{ar}(△ACD)}={\left(\frac{AB}{AC}\right)}^2 \\ \Rightarrow \frac{\text{ar}(△ABE)}{\text{ar}(△ACD)}={\left(\frac{x}{\sqrt{2}x}\right)}^2 \\ \Rightarrow \frac{\text{ar}(△ABE)}{\text{ar}(△ACD)}=\frac{1}{2} \\ \Rightarrow \text{ar}(△ABE):\text{ar}(△ACD)=1:2 \end{gathered}$$

Hence, the ratio between the areas of $△ABE$ and $△ACD$ is $1:2$.