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In an isosceles trapezoid $ABCD$ the leg $AB$ and the smaller base $BC$ are 2 cm long, and $BD$ is perpendicular to $AB$.Calculate the area of trapezoid.


Let $\angle BAD=\theta,$
$BD=h$,
$\angle ABD=90^\circ$
$\angle CBD=90^\circ-\theta$
$CD=2$because trapezoid is isosceles

Apply cosine law in triangle BDC,

$\cos(90-\theta)=\frac{h^2+2^2-2^2}{2\times 2\times h}=\frac{h}{4}$

$\sin\theta=\frac{h}{4}..(1)$

In right triangle $ABD,\sin \theta=\frac{h}{\sqrt{h^2+4}}..(2)$
From $(1)$ and $(2)$,$h=2\sqrt3$

Area of $ABCD=\frac{1}{2}\times 2\times h+\frac{1}{2}\times 2\times 2\times \sin2\theta=3\sqrt3$
But the answer given is $2\sqrt2(\sqrt{5}+1)$

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2 Answers 2

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I've got $3\sqrt 3$ using a slightly different method. $\angle BAD=\theta$, $\angle ABD =90^\circ$ means $\angle CBD=\angle BDA=90^\circ-\theta$. Since the trapezoid is isosceles, $\angle CDA=\theta$, and you can get $\angle CDB=2\theta -90^\circ$. Since $BC=AB=CD$ you get $\angle CDB=\angle CDB$ or $$2\theta-90^\circ=90^\circ-\theta$$ so $\theta=60^\circ$. Drawing perpendiculars from $B$ and $C$ to $AD$, you can get $AD=BC+2AB\cos 60^\circ=4$, and the height $h=AB\sin60^\circ=\sqrt 3$. Therefore the area is $$\frac12 (BC+AD)\cdot h=\frac12 6\sqrt3=3\sqrt 3$$

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you are correct. The answer should be $3\sqrt{3}$

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