Question

The ratio of the areas of similar triangles is equal to the ratio of the squares on the corresponding sides. Prove.

Using the above theorem, prove that the area of the equilateral triangle describe on the side of a square is half of the area of the equilateral triangle described on its diagonal.

Answer 1

Step 1: Find a relation between the ratio of sides

Let $ABC$ and $PQR$ be two triangles such that $△ABC~△PQR$

Construction: Draw $AD\perp BC,PS\perp QR$

If two triangles are similar, then the ratio of their corresponding sides is also equal

Here $△ABC~△PQR$

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

Step 2: Apply AA similarity in $△ABD$ and $△PQS$

If two angles in one triangle are congruent to two angles in another triangle, the two triangles are similar.

$$\begin{gathered} \angle B=\angle Q\left(△ABC~△PQR\right) \\ \angle ADB=\angle PSQ=90°\left(\mathrm{By}\mathrm{construction}\right) \\ △ABD~△PQS\left(\mathrm{AA}\mathrm{similarity}\right) \end{gathered}$$

If two triangles are similar, then the ratio of the corresponding sides is equal.

$\therefore \frac{AB}{PQ}=\frac{AD}{PS}.....................................\left(2\right)$

Step 3: Find the ratio of areas of $△ABD$ and $△PQS$

Area of a triangle$=\frac{1}{2}\times base\times height$

$$\begin{gathered} \therefore ar(△ABC)=\frac{1}{2}\times BC\times AD......................\left(3\right) \\ \therefore ar(△PQR)=\frac{1}{2}\times QR\times PS......................\left(4\right) \end{gathered}$$

$$\begin{gathered} \frac{ar(△ABC)}{ar(△PQR)}=\frac{\frac{1}{2}\times BC\times AD}{\frac{1}{2}\times QR\times PS} \\ \Rightarrow \frac{ar(△ABC)}{ar(△PQR)}=\frac{BC\times AD}{QR\times PS} \\ \Rightarrow \frac{ar(△ABC)}{ar(△PQR)}=\frac{BC\times AB}{QR\times PQ}\left(fromequation2\right) \\ \Rightarrow \frac{ar(△ABC)}{ar(△PQR)}=\frac{AB\times AB}{PQ\times PQ}\left(fromequation1\right) \\ \Rightarrow \frac{ar(△ABC)}{ar(△PQR)}=\frac{A{B}^2}{P{Q}^2}=\frac{A{C}^2}{P{R}^2}=\frac{B{C}^2}{Q{R}^2} \end{gathered}$$

Hence, the ratio of the areas of similar triangles is equal to the ratio of the squares on the corresponding sides.

Step 4: Draw a diagram for the given problem

Let $ABCD$ be a square.

$AC$ and $BD$ are diagonals of the square.

Draw two equilateral triangles which described on the diagonal of the square and on side of the square

Let $a$ be the length of the side of the square $ABCD$.

Then, the length of the diagonal$=a\sqrt{2}$units.

Step 5: Check whether the $ar\left(△AFD\right)$ is half of $ar\left(△ACE\right)$

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

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

According to AA similarity $△AFD~△ACE$

Apply the given theorem,

$\frac{ar(△ABE)}{ar(△ADF)}=\frac{A{C}^2}{A{D}^2}$

Putting the value of $AC$ and $AD$

$$\begin{gathered} \Rightarrow \frac{ar(△ABE)}{ar(△ADF)}=\frac{{\left(a\sqrt{2}\right)}^2}{a^2} \\ \Rightarrow \frac{ar(△ABE)}{ar(△ADF)}=\frac{2a^2}{a^2} \\ \Rightarrow \frac{ar(△ABE)}{2}=ar\left(△ADF\right) \end{gathered}$$

Therefore, the area of the equilateral triangle described on the side of a square is half of the area of the equilateral triangle described on its diagonal.

Hence proved.