Question

∆ ABC ~∆PQR if AB:PQ =4:5 find A(∆ABC):A(PQR)​

Answer 1

Please find below solution

Answer 2

Area (∆ ABC) : Area (Δ PQR)​ is 16 : 25.

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Let's understand a few concepts:

To find the ratio of Area (∆ ABC) : Area (Δ PQR)​ we must use the Theorem of Areas of Similar Triangles.

What are similar triangles?

Two triangles are said to be similar if their corresponding angles are equal and their corresponding sides are proportional to each other.

What is the Theorem of Areas of Similar Triangles?

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

For example: if ΔABC and ΔPQR are two similar triangles then we can say that,

$\boxed{\bold{\frac{Area(\triangle \:ABC)}{Area(\triangle \:PQR)} = \bigg(\frac{AB}{PQ} \bigg)^2 = \bigg(\frac{BC}{QR} \bigg)^2 = \bigg(\frac{AC}{PR} \bigg)^2 }}$

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Let's solve the given problem:

Here we have

Area (Δ ABC) ~ Area (Δ PQR)

AB : PQ = 4 : 5

$\implies \frac{AB}{PQ} = \frac{4}{5}$ . . . (1)

By using the above theorem of the areas of similar triangles, we get

$\frac{Area(\triangle \:ABC)}{Area(\triangle \:PQR)} = \bigg(\frac{AB}{PQ} \bigg)^2$

• on substituting the given values from (1)

$\implies \frac{Area(\triangle \:ABC)}{Area(\triangle \:PQR)} = \bigg(\frac{4}{5} \bigg)^2$

$\implies \frac{Area(\triangle \:ABC)}{Area(\triangle \:PQR)} = \frac{(4)^2}{(5)^2}$

• 4² = 16 and 5² = 25

$\implies \frac{Area(\triangle \:ABC)}{Area(\triangle \:PQR)} = \frac{16}{25}$

$\implies \boxed{\boxed{\bold{Area(\triangle \:ABC):Area(\triangle \:PQR)} = 16:25}}}}$

Thus, Area (∆ ABC) : Area (Δ PQR)​ is 16 : 25.

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