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$ABCD$ is a square. $F$ is the mid-point of $AB$. $BE$ is one third of $BC$. If the area of $∆\ FBE\ =\ 108\ cm^2$, find the length of $AC$.
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Academic Mathematics NCERT Class 10

Given:

$ABCD$ is a square.

$F$ is the mid-point of $AB$.

$BE$ is one-third of $BC$.

Area of $∆\ FBE\ =\ 108\ cm^2$,

To do:

We have to find the length of $AC$.

Solution:

Let the length of the sides of the square be $x$.

This implies,

$AB = BC = CD = DA = x\ cm$

$AF = FB = \frac{x}{2}\ cm$ ($F$ is the mid-point of $AB$)

$BE = \frac{x}{3}\ cm$ ($BE$ is one-third of $BC$)

Area of $∆FBE = \frac{1}{2} \times BE \times FB$

$108 = \frac{1}{2} \times (\frac{x}{3}) \times (\frac{x}{2})$

$\frac{x^2}{12} = 108$

$x^2= 108\times12$

$x^2=1296$

$x = \sqrt{1296}$

$x = 36\ cm$

In $∆ABC$,

By Pythagoras theorem,

$AC^2 = AB^2 + BC^2$

$AC^2 = x^2 + x^2$

$AC^2=2x^2$

$AC^2 = 2(36)^2$

$AC = \sqrt{2(36)^2}$

$AC= 36 \times \sqrt{2}$

$AC = 36\times1.414 = 50.904\ cm$

Therefore, the length of $AC$ is $50.904\ cm$.

Updated on: 10-Oct-2022

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