In the given figure, △ABC∼△PQR. Then, area of △ ABCarea of △

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In The Given Figure, △ABC∼△PQR. Then, area of △...

In the given figure, △ A B C \sim △ P Q R . Then, \frac{a r e a o f △ A B C}{a r e a o f △ P Q R} equals

Options:

A . AB2PQ2

B . BC2QR2

C . AC2PR2

D . All of the above.

Answer: Option D
:
D
We are given two triangles ABC and PQR such that

△ A B C \sim △ P Q R

.
For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles ABC and PQR respectively, as shown below.

In The Given Figure, △ABC∼△PQR. Then, area of △...

Now,

area of △ A B C = \frac{1}{2} \times B C \times A M and area of △ P Q R = \frac{1}{2} \times Q R \times P N

So,

\frac{a r e a o f △ A B C}{a r e a o f △ P Q R} = \frac{\frac{1}{2} \times B C \times A M}{\frac{1}{2} \times Q R \times P N} = \frac{B C \times A M}{Q R \times P N} \dots (1)

Now, in

△ A B M

and

△ P Q N

,

∠ B = ∠ Q

(As

△ A B C \sim △

PQR)
and

∠ A M B = ∠ P N Q = 90^{\circ}

.
So,

△ A B M \sim △ P Q N

( By AA similarity criterion)

∴ \frac{A M}{P N} = \frac{A B}{P Q} \dots (2)

Also,

△ A B C \sim △ P Q R

(Given)
So,

\frac{A B}{P Q} = \frac{A C}{P R} = \frac{B C}{Q R} \dots (3)

From (1) and (3), we get,

\frac{a r e a o f (A B C)}{a r e a o f (P Q R)} = \frac{A B \times A M}{P Q \times P N} = \frac{A B \times A B}{P Q \times P Q} = \frac{A B^{2}}{P Q^{2}}

Now using (3), we get,

\frac{a r e a o f △ A B C}{a r e a o f △ P Q R} = \frac{A B^{2}}{P Q^{2}} = \frac{B C^{2}}{Q R^{2}} = \frac{A C^{2}}{P R^{2}}

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