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Post Posted: Sep 13, 2009, 2:54 pm • # 1 


Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD; AB = 3CD; and the area of the quadrilateral is 4. if a circle can be drawn touching all the four sides of the quadrilateral, find its radius.
 
 
Post Posted: Sep 14, 2009, 2:06 am • # 2 


This is my solution:

Let CD = x and radius of circle is R. Easy we get AB = 3x, BN = 2x, AD = 2R, AM = DQ = R

We have BC = BP + PC = BM + CQ = (AB + CD) - (AM + DQ) = 4x - 2R

Aply Pitago theorem for \triangle BCN we get BC^2 = NB^2 + CN^2

\Longleftrightarrow (4x - 2R)^2 = 4R^2 + 4x^2 \Longleftrightarrow (2x - R)^2 = R^2 + x^2 \Longleftrightarrow 3x^2 = 4Rx \Longleftrightarrow x = \frac {4}{3}R

But S_{ABCD} = 4 \Longleftrightarrow \frac {(AB + CD).AD}{2} = 4 \Longleftrightarrow \frac {4x.2R}{2} = 4 \Longleftrightarrow Rx ....

But x = \frac {4}{3}R \Longrightarrow R^2 = \frac {3}{4} \Longleftrightarrow R = \frac {\sqrt {3}}{2}


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Post Posted: Sep 14, 2009, 11:35 am • # 3 


Join all the vertex of the quadrilateral with centre of the in-circle then area of the quadrilateral is
\frac{1}{2}R(AB+BC+CD+DA)=4
Also \frac {(AB + CD).AD}{2} = 4
Now solve for R (easy)
 
 
Post Posted: Sep 15, 2009, 5:58 am • # 4 


thanhnam2902 wrote:
This is my solution:

Let CD = x and radius of circle is R. Easy we get AB = 3x, BN = 2x, AD = 2R, AM = DQ = R

We have BC = BP + PC = BM + CQ = (AB + CD) - (AM + DQ) = 4x - 2R

Hidden Text



good proof and easy to anybody to learn
 
 
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