IYMC PAST PROBLEM

Geometry Level 4

$ABCD$ is a cyclic quadrilateral such that its diagonal $BD$ bisects the $\angle B$ . Given that $BD=10cm$ and $\angle B=60°$ . Find the area of $ABCD$ (in $cm^2$ ).

The answer is 43.301.

2 solutions

A Former Brilliant Member
Oct 23, 2017

By ratio and proportion:

$\dfrac{AD}{1}=\dfrac{10}{2}$ $\implies$ $AD=5$

$\dfrac{AB}{\sqrt{3}}=\dfrac{10}{2}$ $\implies$ $AB=5\sqrt{3}$

The area of quadrilateral $ABCD$ is $(AB)(AD)=(5\sqrt{3})(5) \approx 43.30127019$

Niranjan Khanderia
Aug 17, 2015

Since BD is the diameter, A and C are right angles. So ABD and CBD are right angled triangles with common hypotenuse and one angle each of 30. So they are congruent. $\text{These two 30-60-90 triangeles has each area of } \dfrac 1 2 *10^2*\dfrac{\sqrt3} 4=21.6506 \\ Area ~~ABCD=43.301$

It is not given that BD is diameter?

Indraneel Mukhopadhyaya - 5 years, 8 months ago

$\text {Since ABCD is cyclic, and angles ABD=CBD=} 30^o.$ $\implies \text {AD=CD.}$ $\Delta s \text{ ABD, CBD have BD common, AD=CD and angles ABD=CBD=} 30^o.$ $\therefore~ \Delta s \text{ are} \equiv \angle s \text{ at A and C are supplementary but equal.}$ $\implies \text{they are } 90^o. \implies \text{ BD is the diameter.}$

Niranjan Khanderia - 5 years, 8 months ago

IYMC PAST PROBLEM

The answer is 43.301.

2 solutions

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