Question

Ratio of the areas of two similar triangles is not equal to the ratio of the squares of corresponding sides of the triangles.

• A. True
• B. False

Answer 1

The correct option is B

False

Let $\Delta ABC$ and $\Delta PQR$ are two similar triangles.
To Prove: $\frac{\text{Area of}\Delta ABC}{\text{Area of}\Delta PQR}=\frac{A{B}^2}{P{Q}^2}=\frac{B{C}^2}{Q{R}^2}=\frac{A{C}^2}{P{R}^2}$
Construction: Let us draw a perpendicular AD from A to BC and a perpendicular PM from P to QR.
$\because \Delta ABC\sim \Delta PQR$
$\therefore \frac{AB}{PQ}=\frac{BC}{QR}$ ..... (1)
Again, $\Delta ABD\sim \Delta PQM[\because \angle ADB=\angle PMQ\text{and}\angle B=\angle Q]$
$\therefore \frac{AB}{PQ}=\frac{AD}{PM}$ ...... (2)
Now, from (1) and (2) we get, $\frac{BC}{QR}=\frac{AD}{PM}$ ...... (3)
Then, $\frac{\text{Area of triangle ABC}}{\text{Area of triangle PQR}}=\frac{\frac{1}{2}\times BC\times AD}{\frac{1}{2}\times QR\times PM}$
$=\frac{BC\times AD}{QR\times PM}=\frac{BC}{QR}\times \frac{AD}{PM}$
$=\frac{BC}{QR}\times \frac{BC}{QR}$ [From (3)]
$=\frac{B{C}^2}{Q{R}^2}$ ...... (4)
Again, since $\Delta ABC$ and $\Delta PQR$ are similar to each other.
$\therefore \frac{BC}{QR}=\frac{AB}{PQ}=\frac{AC}{PR}$
or, $\frac{B{C}^2}{Q{R}^2}=\frac{A{B}^2}{P{Q}^2}=\frac{A{C}^2}{P{R}^2}$
Hence, from (4) we get,
$\frac{\text{Area of}\Delta ABC}{\text{Area of}\Delta PQR}=\frac{A{B}^2}{P{Q}^2}=\frac{B{C}^2}{Q{R}^2}=\frac{A{C}^2}{P{R}^2}$