Question

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

Answer 1

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

To prove this theorem, consider two similar triangles $\Delta ABC$ and $\Delta PQR$

According to the stated theorem

$\frac{ar△ABC}{ar△PQR}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{CA}{RP}\right)}^2$

Since area of triangle $=\frac{1}{2}\times base\times altitude$

To find the area of $\Delta ABC$ and $\Delta PQR$ draw the altitudes $AD$ and $PE$ from the vertex $A$ and $P$ of $\Delta ABC$ and $\Delta PQR$

Now, area of $\Delta ABC$ $=\frac{1}{2}\times BC\times AD$

area of $\Delta PQR$ $=\frac{1}{2}\times QR\times PE$

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

$\frac{ar△ABC}{ar△PQR}=\frac{\frac{1}{2}\times BC\times AD}{\frac{1}{2}\times QR\times PE}$

$\frac{ar△ABC}{ar△PQR}=\frac{BC\times AD}{QR\times PE}$

Now in $△ABD$ and $△PQE$ it can be seen

$\angle ABC=\angle PQR$ (Since $\Delta ABC\sim \Delta PQR$ )

$\angle ADB=\angle PEQ$ (Since both the angles are $90°$)

From $AA$ criterion of similarity $\Delta ADB\sim \Delta PEQ$

$\frac{AD}{PE}=\frac{AB}{PQ}$

Since it is known that $\Delta ABC\sim \Delta PQR$

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}$

Substituting this value in equation , we get

$\frac{ar△ABC}{ar△PQR}=\frac{AB}{PQ}\times \frac{AD}{PE}$

we can write

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

Similarly we can prove

$\frac{ar△ABC}{ar△PQR}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{CA}{RP}\right)}^2$