Cyclic Quadrilateral

Geometry Level 3

In the figure shown above, $OB=80~cm$ , $OA=120~cm$ and $OD=150~cm$ . Which of the following is the area of the cyclic quadrilateral in $cm^2$ to the nearest integer?

$\text{note}:$ $cm$ means $centimeters$

7222 2722 2572 2272

1 solution

A Former Brilliant Member
May 15, 2017

Relevant wiki: Cyclic Quadrilaterals

$OB(OA)=OC(OD)$ $\implies$ $80(120)=OC(150)$ $\implies$ $OC=64$

$A_{ABCD}=A_{AOD}-A_{BOC}=\dfrac{1}{2}(150)(120)(\sin 25)-\dfrac{1}{2}(64)(80)(\sin 25)\approx 2722~cm^2$

A great problem!!.

Sathvik Acharya - 4 years ago

Cyclic Quadrilateral

1 solution

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