Simple geometry

Geometry Level 3

Let $PQ$ and $RS$ be tangents at the extremities of diameter $PR$ of a circle of radius $r$ . If $PS$ and $RQ$ intersect at some point on the circumference of the circle, then what is the diameter of the circle?

\dfrac{\sqrt{ PQ^2+RS^2 }}{2}

\dfrac{PQ+RS}{2}

\sqrt{PQ \cdot RS}

\dfrac{2PQ+RS}{PQ+RS}

1 solution

Vignesh Suresh
Oct 21, 2015

From figure:

$\tan \theta =\frac{RS}{PR} \ldots (1)$ $\tan(90-\theta)=\frac{PQ}{PR}$ $\cot\theta=\frac{PQ}{PR}$ $\frac{1}{\tan\theta}=\frac{PQ}{PR} \ldots (2)$ From 1 and 2 we get $PR^{2}=PQ\times RS$ Therefore $PR=\sqrt{PQ\times RS}$

Thanks for the solution.

Sandeep Bhardwaj - 5 years, 2 months ago

Simple geometry

1 solution

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