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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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