Find the area of a quadrilateral $ABCD$ in which $AD = 24\ cm, \angle BAD = 90^o$ and $BCD$ forms an equilateral triangle whose each side is equal to $26\ cm$. (Take $\sqrt{3}= 1.73$)

Given:

A quadrilateral $ABCD$ in which $AD = 24\ cm, \angle BAD = 90^o$ and $BCD$ forms an equilateral triangle whose each side is equal to $26\ cm$.

To do:

We have to find the area of the quadrilateral.

Solution:

In right $\triangle ABD$,

$BD^2 = AB^2+ AD^2$

$(26)^2 = AB^2 + (24)^2$

$676 = AB^2 + 576$

$AB^2 = 676 - 576$

$= 100$

$= (10)^2$

$\Rightarrow AB = 10\ cm$

Area of right angled triangle $ABD=\frac{1}{2} \text base \times altitude$

$=\frac{1}{2} \times 10 \times 24$

$=120 \mathrm{~cm}^{2}$

Area of equilateral triangle $\mathrm{BCD}=\frac{\sqrt{3}}{4} \times(side)^{2}$

$=\frac{1.73}{4} \times 26 \times 26$

$=292.37 \mathrm{~cm}^{2}$

Total area of the quadrilateral $\mathrm{ABCD}=120+292.37$

$=412.37 \mathrm{~cm}^{2}$.

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Updated on: 10-Oct-2022

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