Question

if a triangle ABC similar traingle DEF, ar(ABC)/ar(DEF)=9/25,BC=21cm,then EF is equal to ???​

Answer 1

Step-by-step explanation:

Theorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides. This proves that the ratio of areas of two similar triangles is proportional to the squares of the corresponding sides of both the triangles.

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Answer 2

Answer:

The value of EF is 35 cm.

Step-by-step explanation:

Given:-

ΔABC is similar to ΔDEF such that ar(ABC)/ar(DEF) = 9/25. Also, BC = 21 cm.

To find:-

The value of EF.

Recall the property of similar triangles,

The ratio of the areas of the triangles is equal to the ratio of squares of their corresponding sides.

Mathematically,

$\frac{ar(ABC)}{ar(D EF)} = \left(\frac{BC}{EF} \right)^2$ . . . . . (1)

Now,

Substitute the values of ar(ABC)/ar(DEF) and BC in the equation (1) as follows:

⇒ $\frac{9}{25} = \left(\frac{21}{EF} \right)^2$

Simplify the equation as follows:

$\frac{9}{25} = \frac{21^2}{EF^2}$

Cross multiply the equation as follows:

$9\times EF^2=25\times21^2$

$EF^2=\frac{25\times21\times21}{9}$

$EF^2=$ 25 × 7 × 7

$EF^2=35^2$

EF = 35 cm

Therefore, the value EF is 35 cm.

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