Question

∆ABC ~∆PQR, A(∆ABC) =16,A(∆PQR) =25 then find the value of ratio AB_PQ​

Answer 1

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Given:

∆ABC ∽ ∆PQR

Area of ∆ABC = 16

Ares of ∆PQR = 25

To Find:

The value of the ratio of AB:PQ

Solution:

Since we are given that ∆ABC and ∆PQR are similar to each other, so we can state the following theorem.

The ratio of the two similar triangle is equal to the square of the ratio of their corresponding sides.

here,

AB and PQ are the corresponding sides of the similar triangle.

∴$\frac{area(angle \: abc)}{area \: (angle \: pqr)} = \frac{ab}{pq}^{2}$

subtitling the given values of area of

∆ABC & ∆PQR

$= \frac{16}{25} = ( \frac{ab}{pq} {}^{2} )$

taking square root on both sides

$= > \: \sqrt{ \frac{16}{25} } = \sqrt{ \frac{ab}{pq} } {}^{2} \\ = > \: \frac{4}{5} = \frac{ab}{pq}$

=> AB:PQ = 4:5

Thus,the value of ratio of AB:PQ is 4:5

Answer 2

Answer:

∆ABC ~∆PQR, A(∆ABC) =16,A(∆PQR) =25 then find the value of ratio AB_PQ