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The areas of two similar triangles are 100 sq cm and 64 sq cm respectively. If a median of the smaller triangle is 5.6 cm, then the corresponding median of the other is,

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4 cm

4.5 cm

5 cm

5.5 cm

answer is B.

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Detailed Solution

Given that the areas of two similar triangles are 100 sq cm and 64 sq cm respectively.
Consider the figure,
Question Image

Draw perpendiculars from A and D on BC and EF at M and N respectively.
The triangles

Δ

ABC and ΔDEF are similar.
So,

∠

= ∠

E and ∠C

= ∠

F
In the triangles

Δ

ABM and ΔDEN,
It is known that

∠

= ∠

E.
As the angle, M and angle N are the right angles.
So, both angles are equal that is

∠

= ∠

N.
So, the triangles

Δ

ABM

a n d Δ

DEN are similar by the AA pattern.
Then,

AM DN = AB DE = BM EN .

Using the property of similar triangle,

\Rightarrow \frac{ar (ΔABC)}{ar (ΔDEF)} = \frac{{AB}^{2}}{{DE}^{2}}

\Rightarrow \frac{100}{64} = \frac{5.6 \times 5.6}{{DN}^{2}}

\Rightarrow DN = \sqrt{\frac{64 (31.36)}{100}}

\Rightarrow DN = 4.48 cm ≃ 4.5 cm

The area of DN is 4.5cm.
Therefore, option 2 is correct.

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