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 :
In a trapezium $ABCD,O$ is the point of intersection of $AC$ and $BD$, $AB\parallel CD$ and $AB=2CD$. If the area of $∆AOB=84{\mathrm{cm}}^2$, find the area of $∆COD$.

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

$∆\mathrm{ABM}~∆\mathrm{PQN}$

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

If both triangles are similar then the corresponding sides are proportional.

Since $∆\mathrm{ABM}~∆\mathrm{PQN}$ so

$\frac{AB}{PQ}=\frac{AM}{PN}$……(4)

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: Checking similarity for $∆COD$and $∆AOB$

Here, in trapezium, $ABCD$ it is given that

Given, $AB=2CD$ and $\text{ar}∆AOB=84{\mathrm{cm}}^2$

Since $ABCD$is a trapezium, so $AB\parallel CD$

$\angle AOB=\angle DOC$ (vertically opposites)

$\angle OAB=\angle OCD$ (Alternate angles)

According AA similarity $∆AOB~$$∆COD$

Step 6: Finding the area in $∆COD$

According to the above theorem

$\frac{\text{ar}\left(\Delta AOB\right)}{\text{ar}\left(\Delta COD\right)}=\frac{A{B}^2}{C{D}^2}=\frac{(2CD{)}^2}{C{D}^2}[\because AB=2\times CD]$

$\therefore \frac{\text{ar}\left(\Delta AOB\right)}{\text{ar}\left(\Delta COD\right)}=4$

$\Rightarrow ar(∆\mathrm{COD})=\frac{1}{4}\times ar(∆\mathrm{AOB})$

$\Rightarrow \text{ar (△COD)=}\left(\frac{1}{4}\times 84\right)=21{\mathrm{cm}}^2$

Hence, the area of $△\mathrm{COD}$ is $21{\text{cm}}^2.$