Question

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

Answer 1

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Answer 2

Answer:

Hence proved.

To Prove:

$(\Delta \mathrm{ABC}) /(\Delta \mathrm{PQR})=(\mathrm{AB} / \mathrm{PQ})^{2}=(\mathrm{BC} / \mathrm{QR})^{2}=(\mathrm{CA} / \mathrm{RP})^{2}$

Step-by-step explanation:

Step 1:

Given:

$\Delta \mathbf{A B C} \sim \Delta \mathbf{P Q R}$

Construction: Draw $\mathrm{AM} \perp \mathrm{BC}, \mathrm{PN} \perp \mathrm{QR}$

Step 2:

$(\Delta \mathrm{ABC}) /(\Delta \mathrm{PQR})=(1 / 2 \times \mathrm{BC} \times \mathrm{AM}) /(1 / 2 \times \mathrm{QR} \times \mathrm{PN})$

$=\mathrm{BC} / \mathrm{QR} \times \mathrm{AM} / \mathrm{PN}$..........[i]

Step 3:

In $\Delta ABM \quad and \quad \Delta PQN$,

$\angle \mathrm{B}=\angle \mathrm{Q}(\Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}) \angle \mathrm{M}=\angle \mathrm{N}\left(\text { each } 90^{\circ}\right)$

Step 4:

So, $\Delta \mathrm{ABM} \sim \Delta \mathrm{PQN}$ (AA similarity criterion)

Therefore, $AM/PN = AB/PQ$ ............................ [ii]

But,

$\mathrm{AB} / \mathrm{PQ}=\mathrm{BC} / \mathrm{QR}=\mathrm{CA} / \mathrm{RP}(\Delta \mathrm{ABC} \sim \Delta \mathrm{PQR})$ .......[iii]

Step 5:

Hence, from (i)

$(\Delta \mathrm{ABC}) /(\Delta \mathrm{PQR})=\mathrm{BC} / \mathrm{QR} \times \mathrm{AM} / \mathrm{PN}$

$=\mathrm{AB} / \mathrm{PQ} \times \mathrm{AB} / \mathrm{PQ}$

[From ii and iii]

$=(\mathrm{AB} / \mathrm{PQ})^{2}$

Step 6:

Using (iii)

$(\Delta \mathrm{ABC}) / \mathrm{r}(\Delta \mathrm{PQR})=(\mathrm{AB} / \mathrm{PQ})^{2}=(\mathrm{BC} / \mathrm{QR})^{2}=(\mathrm{CA} / \mathrm{RP})^{2}$

Hence it is proved