# Area Of Similar Triangles

Before learning the area of a similar triangle, let us first recall the similarity conditions. The two triangles are similar Â to each other if,

• Corresponding angles of the triangles are equal
• Corresponding sides of Â the triangles are in Â proportion

If there are two triangles say Î”ABC and Î”PQR, then they are similar if,

• âˆ A=âˆ P, âˆ B=âˆ Q and âˆ C=âˆ R
• $$\begin{array}{l} \frac {AB}{PQ} Â =\frac {BC}{QR} Â = \frac {AC}{PR} \end{array}$$

If we have two similar triangles, then not only their angles and sides share a relationship but also the ratio of their perimeter, altitudes, angle bisectors, areas and other aspects are in ratio.

In the upcoming discussion, the relation between the area of two similar triangles is discussed.

## Area of Similar Triangles Theorem

Theorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

To prove this theorem, consider two similar triangles Î”ABC and Î”PQR;

According to the stated theorem,

$$\begin{array}{l}\large \frac {area ~of~ Î”ABC}{area~ of~ Î”PQR} = \large \left(\frac{AB}{PQ}\right)^2 Â =\large \left(\frac{BC}{QR}\right)^2 = Â \large \left(\frac{CA}{RP}\right)^2 \end{array}$$

As, Area of triangle = (1/2) Ã— Base Ã— Height

To find the area of Î”ABC and Î”PQR, draw the altitudes AD and PE from the vertex A and P of Î”ABC andÎ”PQR, respectively, as shown in the figure given below:

Now,

$$\begin{array}{l} \text{Area ofÂ triangle ABC} =\frac{1}{2} \times BC \times AD\end{array}$$

area of Î”PQR = (1/2)Â Ã— QR Ã— PE

The ratio of the areas of both the triangles can now be given as:

$$\begin{array}{l}\large \frac{area~ of~ Î”ABC}{area~ of~ Î”PQR} Â = Â \large \frac{ Â \frac 12 Ã— BC Ã— AD}{ \frac 12 Ã— QR Ã— PE}\end{array}$$
$$\begin{array}{l} \large \frac{area~ of~ Î”ABC}{area ~of~ Î”PQR} Â = \large \frac{BC~Ã—~AD}{QR~Ã—~PE}..(1) \end{array}$$

Now in âˆ†ABD and âˆ†PQE, it can be seen that:

âˆ ABC = âˆ PQR (Since Î”ABC ~ Î”PQR)

âˆ ADB = âˆ PEQ (Since both the angles are 90Â°)

From AA criterion of similarity âˆ†ADB ~ âˆ†PEQ

$$\begin{array}{l} \frac {AD}{PE}= \frac {AB}{PQ}…(2)\end{array}$$

Since it is known that Î”ABC~ Î”PQR,

$$\begin{array}{l} \frac {AB}{PQ}= Â \frac {BC}{QR} = \frac {AC}{PR}…(3)\end{array}$$

Substituting this value in equation (1), we get

$$\begin{array}{l} \large \frac{area~ of~ Î”ABC}{area ~of~ Î”PQR} Â =\large \frac {AB}{PQ}Ã—\frac {AD}{PE}\end{array}$$

Using equation (2), we can write

$$\begin{array}{l} \large \frac{area~ of~ Î”ABC}{area ~of~ Î”PQR} =\large \frac {AB}{PQ} Ã— Â \frac {AB}{PQ}\end{array}$$
$$\begin{array}{l} \large \frac{area~ of~ Î”ABC}{area ~of~ Î”PQR}=\large \left( \frac {AB}{PQ}\right)^2\end{array}$$

Also from equation (3),

$$\begin{array}{l} \large \frac {area ~of~ Î”ABC}{area~ of~ Î”PQR} = \large \left(\frac{AB}{PQ}\right)^2 =\large \left(\frac{BC}{QR}\right)^2 =Â \large \left(\frac{CA}{RP}\right)^2 \end{array}$$

This proves that the ratio of the area of two similar triangles is proportional to the squares of the corresponding sides of both the triangles.

#### To Know how to Find the Area Of Similar Triangles, Watch The Below Video:

To have a better insight consider the following example.

### Example

Example 1: In Î”ABC andÎ”APQ, the length of the sides are given as AP = 5 cm , PB = 10cm and BC = 20 cm. Find the ratio of the areas of Î”ABC and Î”APQ.

Solution: Â In Î”ABC and Î”APQ , âˆ PAQ is common and âˆ APQ = âˆ BAC (corresponding angles)

â‡’ Î”ABC ~ Î”APQ Â (AA criterion for similar triangles)

Since both the triangles are similar, using the theorem for areas of similar triangles we have,

$$\begin{array}{l} \large \frac {area ~of~ Î”ABC}{area~ of~ Î”APQ} = Â \large \left( \frac {AB}{AP} \right)^2 Â = \large \left( \frac {15}{5} \right)^2Â = 9 \end{array}$$