Question

prove that the ratio of the areas of two similar triangle is equal to the square of the ratio of their corresponde sides ​

Answer 1

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Answer 2

GIVEN :-

$\sf\triangle ABC \sim \triangle PQR$

TO PROVE :-

$\sf\dfrac {Area \ of \ \triangle ABC }{Area \ of \ \triangle PQR}=\dfrac{AB^2}{PQ^2}= \dfrac{BC^2}{QR^2}=\dfrac{AC^2}{PR^2}$

SOLUTION :-

$\longrightarrow\sf \dfrac{Area \ of \ \triangle ABC}{Area \ of \ \triangle PQR}= \dfrac{\dfrac{1}{2}\times AD \times BC}{\dfrac{1}{2}\times PS \times QR} \\\\\longrightarrow\sf \dfrac{Area \ of \ \triangle ABC}{Area \ of \ \triangle PQR}= \dfrac{AD}{PS} \times \dfrac{BC}{QR} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ........................(1)$

We know ,

$\sf\triangle ABC \sim \triangle PQR$

$\longrightarrow\sf \dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AC}{PR} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ........................(2)$

Consider ΔABD and ΔPQS,

∠B = ∠Q   ( ∵ ΔABC ~ ΔPQR )

∠ADB = ∠PSQ ( 90° each )

∴ ΔABD ~  ΔPQS ( AA similarity )

$\sf\therefore \dfrac{AB}{PQ}=\dfrac{BD}{QS}=\dfrac{AD}{PS}$

From (2) , we have

$\sf\dfrac{AB}{PQ}= \dfrac{BC}{QR}= \dfrac{AC}{PR}=\dfrac{AD}{PS}$

Substitute in equation (1),

$\longrightarrow\sf \dfrac{Area \ of \ \triangle ABC}{Area \ of \ \triangle PQR}= \dfrac{AB}{PQ}\times \dfrac{AB}{PQ}= \dfrac{BC}{QR}\times \dfrac{BC}{QR}=\dfrac{AC}{PR}\times \dfrac{AC}{PR} \\\\\\\longrightarrow\bf \dfrac{Area \ of \ \triangle ABC}{Area \ of \ \triangle PQR}= \dfrac{AB^2}{PQ^2}=\dfrac{BC^2}{QR^2}=\dfrac{AC^2}{PR^2}$

Hence proved .