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 :
The areas of two similar triangles $△ABCand△LMN\text{are}64c{m}^2\text{and}81c{m}^2$ respectively. If $MN=6.3cm$, find $BC.$

Answer 1

Step 1: Note the given data and draw the required figure

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

Draw the required figure

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

.

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

The area of a 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 $△PQR,$

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

Finding the ratio of $\mathrm{ar}(△\mathrm{ABC})$ and $\mathrm{ar}(△PQR)$

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

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

$∆\mathrm{ABM}~∆\mathrm{PQN}$

Step 4: Finding the relation between corresponding sides of $△ABC$ and $△PQR$

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

$\frac{AB}{PQ}=\frac{AM}{PN}$………(4)

Now from equations (3) and (4), 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, $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

Now putting this in equation (5), we get

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

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

$$\begin{gathered} ={\left(\frac{BC}{QR}\right)}^2 \\ ={\left(\frac{AC}{PR}\right)}^2 \end{gathered}$$

Step 5: Finding the length of $BC$

Given that, the areas of $△ABC$ and $△LMN$ are $64c{m}^2$ and $81c{m}^2$respectively.

Also, the length of $MN=6.3cm$

According to the above theorem,

$\frac{ar(△ABC)}{ar(△LMN)}={\left(\frac{BC}{MN}\right)}^2$

Substitute the values and we get

$\frac{64}{81}={\left(\frac{BC}{6.3}\right)}^2$

$$\begin{gathered} \Rightarrow \frac{64\times 6.3\times 6.3}{81}=B{C}^2 \\ \Rightarrow BC=\sqrt{\frac{64\times 6.3\times 6.3}{81}} \\ \Rightarrow BC=\frac{8\times 6.3}{9} \\ \Rightarrow BC=\frac{8\times 6.3}{9} \\ \Rightarrow BC=5.6cm \end{gathered}$$

Hence, the length of side BC of $ar(△ABC)$ is $5.6cm.$