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

Answer:

refer the attachment

Step-by-step explanation:

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Answer 2

Given :

ΔABC∼ΔPQR...(I)

$\dfrac{AB}{PQ}$ = $\dfrac{BC}{QR}$ = $\dfrac{CA}{RP}$  --------(II)

​∠B=∠Q...(III)  

To prove :

Ratio of areas of ΔABC and ΔPQR is equal to the square of the ratio of their corresponding sides.

SOLUTION:

In ΔABC and ΔPQS  we get:

∠B=∠Q [from (3)]      

∠ADB=∠PSQ=90°

 ∴ΔABD∼ΔPQS [ By AA similarity]

⇒  $\dfrac{AB}{PQ}$= $\dfrac{AD}{PS}$ -----(IV)

$\dfrac{\textrm{Area of}\ \Delta \textrm{ ABC}}{\textrm{Area of}\ \Delta\textrm{ PRQ}}$

= $\dfrac{\frac{1}{2} \times BC \times AD }{\frac{1}{2} \times QR \times PS}$

= $\dfrac{ BC \times AD }{QR \times PS}$

From equation (II) and (IV) we get:

$\dfrac{AB}{PQ}$$\TIMES$ $\times$$\dfrac{AB}{PQ}$

=$(\dfrac{AB}{PQ}) ^{2}$

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

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