Question

Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Using the above result :

To prove that the area of the equilateral triangle described on the side of a right-angled isosceles triangle is half the area of the equilateral triangle described on its hypotenuse.

Answer 1

Step 1: Note the given data and draw the diagram

Let $△\mathrm{ABC}$ and $△\mathrm{PQR}$ be two similar triangles.

Construction:

Draw perpendiculars $AM$ and $PN$ on the sides $BC$ and $QR$ of the $∆\mathrm{ABC}$ and $∆\mathrm{PQR}$.

i.e., $AM\perp BC,PN\perp QR$

.

Step 2: Finding the ratio of the areas of $△ABC$ and $△PQR$

The area of the triangle is $\frac{1}{2}\text{×Base}\times \text{Height}$

For $△ABC$,

$\text{ar(△ABC)=}\frac{1}{2}\text{×Base}\times \text{Height}$

$=\frac{1}{2}\times \mathrm{BC}\times \mathrm{AM}$……..(1)

Similarly, for $△PQR,$

$\text{ar(△ABC)=}\frac{1}{2}\text{×QR}\times \text{PN}$……..(2)

Finding the ratios of the areas of $△ABC$ and $△PQR$

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

$\frac{\text{ar △}\left(\text{ABC}\right)}{\text{ar △}\left(\text{PQR}\right)}=\frac{\text{BC}\times \text{AM}}{\text{QR}\times \text{PN}}$……….(3)

Step 3: Finding the relation between $∆\mathrm{ABM}$ and $∆\mathrm{PQN}$

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

In$∆\mathrm{ABM}$ and $∆\mathrm{PQN}$,

$\angle B=\angle Q$(All corresponding angles are equal in two similar triangles)

$\angle M=\angle N$(Both are ${90}^o$)

Therefore, by $\text{AA}$-Similarity criteria

$∆\mathrm{ABM}~∆\mathrm{PQN}$

If both triangles are similar then the corresponding sides are proportional

$\frac{AB}{PQ}=\frac{AM}{PN}$……(4)

Step 4: Finding the relation between corresponding sides of $△\mathrm{AB}C$ and $△\mathrm{PQ}R$ to their areas

Substitute the value of $\frac{AM}{PN}$ in equation (4) and we get

$\frac{\text{ar}△\left(ABC\right)}{\text{ar}△\left(PQR\right)}=\frac{BC}{QR}\times \frac{AB}{PQ}$……(5)

Since $∆\mathrm{ABC}~∆\mathrm{PQR}$.

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

Substitute the value of $\frac{BC}{QR}$ in equation (5) and we get

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

Similarly, $\frac{\text{ar △}\left(ABC\right)}{\text{ar △}\left(PQR\right)}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{AC}{PR}\right)}^2$.

Hence the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Step 5: Draw a diagram for the given problem

Let $ABC$ be an isosceles triangle and $AB=BC=x$.

$△ABD$ and $△ACE$ are two equilateral triangles described on side of $△ABD$ and on the hypotenuse of $△ACE$ respectively.

Pythagorean theorem: The square of the hypotenuse is equal to the sum of the square of the other two sides.

According to the Pythagorean theorem,

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

The diagram for the given data is

Step 6: Finding the ratio of $△ABD$ and $△ACE$

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

Since $△ABD$ and $△ACE$ both are equilateral triangles, so their angles are equal.

According to $AAA$ similarity,

$△ABD~△ACE$

Similar Triangles property: If two triangles are similar, then their corresponding sides are proportional.

Ssince, $△ABD~△ACE$

$\therefore \frac{AB}{AC}=\frac{DB}{EC}=\frac{AD}{AE}$

According to the theorem,

$\frac{ar\left(\Delta \mathrm{ABD}\right)}{ar\left(\Delta ACE\right)}=\frac{{\mathrm{AB}}^2}{A{C}^2}=\frac{x^2}{(x\sqrt{2}{)}^2}=\frac{{\mathrm{x}}^2}{2x^2}=\frac{1}{2}$

$\Rightarrow \text{area of}△ABD=\frac{1}{2}\times \text{area of}△CAE$

Hence proved.