Question

Prove that the ratio of areas of two similar triangles is equal to the ratio of square of the ratio of their corresponding sides.

Answer 1

Given: $\Delta ABC\sim \Delta PQR$

To Prove : $\frac{A\left(\Delta ABC\right)}{A\left(\Delta PQR\right)}=\frac{A{B}^2}{P{Q}^2}=\frac{B{C}^2}{Q{R}^2}=\frac{A{C}^2}{P{R}^2}$

Construction: Draw seg AM $\perp$ side BC, and seg PN $\perp$ side QR

Proof: $\Delta ABC\sim \Delta PQR$ ....... (Given)

$\therefore \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$ ....... (c.s.s.t) ...... (1)

and $\angle ABC$$\simeq \angle PQR$ ....... (c.s.s.t)

i.e., $\angle ABM\simeq \angle PQN$ ...... (2)

In $\Delta ABM$ and $\Delta PQN$

$\angle ABM\simeq =\angle PQN$ [From (2)]

$\angle AMB\simeq \angle PNQ$ ...... (Both are right angles)

$\therefore \Delta AMB\sim \Delta PNQ$ ..... (AA test for similarity)

$\therefore \frac{AM}{PN}=\frac{MB}{NQ}=\frac{AB}{PQ}$ ...... (c.s.s.t) ..... (3)

$\therefore \frac{AM}{PN}=\frac{AB}{PQ}=\frac{BC}{QR}$ [From (1) and (3)] ..... (4)

Now, $\frac{A\left(\Delta ABC\right)}{A\left(\Delta PQR\right)}=\frac{BC\times AM}{QR\times PN}$

$=\frac{BC}{QR}\times \frac{AM}{PN}$

$=\frac{BC}{QR}\times \frac{BC}{QR}$ [From (4)]

$\therefore \frac{A\left(\Delta ABC\right)}{A\left(\Delta PQR\right)}=\frac{B{C}^2}{Q{R}^2}$ ........ (5)

Similarly, we can prove that

$\frac{A\left(\Delta ABC\right)}{A\left(\Delta PQR\right)}=\frac{A{B}^2}{P{Q}^2}$ and

$\frac{A\left(\Delta ABC\right)}{A\left(\Delta PQR\right)}=\frac{A{C}^2}{P{R}^2}$ ...... (6)

$\frac{A\left(\Delta ABC\right)}{A\left(\Delta PQR\right)}=\frac{B{C}^2}{Q{R}^2}=\frac{A{B}^2}{P{Q}^2}=\frac{A{C}^2}{P{R}^2}$ From (5) and (6) $\left[henceproved\right]$