Question

The areas of two similar triangles ABC and DEF are 36 cm² and 81 cm² respectively. fEF = 6.75 cm, find BC. (a) 6cm (b) 9cm (c) 5.5cm (d) 4.5cm please tell me the answer and show the step also​

Answer 1

Step-by-step explanation:

jab triangle Siimilar hote Hain hain to unki sides k square ka ratio unke triangles ke area ke ratio ke equal hota hai

answer=5 (d), option

Answer 2

Given:

The areas of two similar triangles ABC and DEF are 36 cm² and 81 cm² respectively. fEF = 6.75 cm, find BC. (a) 6cm (b) 9cm (c) 5.5cm (d) 4.5cm

To find:

The length of BC

Solution:

The area of Δ ABC = 36 cm²

The area of Δ DEF = 81 cm²

The length of EF = 6.75 cm

We know that,

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

Here we have  

Δ ABC similar to Δ DEF

So, based on the above theorem, we get

$\frac{Area \:(\triangle \:ABC)}{Area \:(\triangle \:D EF)} = \bigg(\frac{BC}{EF} \bigg)^2$

On substituting the given values, we get

$\implies \frac{36\:cm^2}{81\:cm^2} = \bigg(\frac{BC}{6.75\:cm} \bigg)^2$

on taking square roots on both sides, we get

$\implies \sqrt{\frac{36}{81} } = \sqrt{\bigg(\frac{BC}{6.75\:cm} \bigg)^2}$

$\implies \frac{6}{9} = \frac{BC}{6.75\:cm}$

$\implies \frac{2}{3} = \frac{BC}{6.75\:cm}$  

$\implies BC = \frac{2}{3} \times 6.75\:cm$  

$\implies \bold{BC = 4.5\:cm}$ ← option (d)

Thus, the length of BC is → 4.5 cm.

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