Find the radius

Geometry Level 3

Let A B C D ABCD be a quadrilateral in which A B AB is parallel to C D CD and perpendicular to A D AD , B A = 3 C D BA = 3CD and the area of the quadrilateral is 4 square units. If a circle can be drawn touching all sides if the quadrilateral, then its radius is:

[3^(1/2)]/2 4 2[3^(1/2)] 2

Swastik Sahoo by Swastik Sahoo , 21, India

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2 solutions

Marta Reece
Jun 14, 2017

From △ O C E \triangle OCE we can get tan ⁡ θ = R 3 a − R \tan\theta=\dfrac{R}{3a-R}

From △ B F C \triangle BFC likewise tan ⁡ 2 θ = 2 R 2 a = R a \tan2\theta=\dfrac{2R}{2a}=\dfrac Ra

The formula for a tangent of a double angle is tan ⁡ 2 θ = 2 tan ⁡ θ 1 − tan ⁡ 2 θ \tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}

Substituting into it R a = 2 × R 3 a − R 1 − ( R 3 a − R ) 2 \dfrac Ra=\dfrac{2\times\dfrac{R}{3a-R}}{1-\left(\dfrac{R}{3a-R}\right)^2}

This simplifies to a = 4 R 3 a=\dfrac{4R}{3}

Are of A B C D ABCD is [ A B C D ] = 2 R × a + 3 a 2 = 4 a R = 4 [ABCD]=2R\times \dfrac{a+3a}{2}=4aR=4

From that we get another relationship between a a and R R , namely a = 1 R a=\dfrac 1R

Combining the two, we get R = 3 4 R=\boxed{\dfrac{\sqrt3}{4}}

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Ahmad Saad
Nov 30, 2016

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