This is a geometry problem which I seem is tough. See if you can solve it and give me the proper explanation and answer as I am studying for NECM level-II exam. Please give me as fast as possible.

Sides of PQRS touch a circle of 8cm diameter, as shown in the figure. Find the area of PQRS if its perimeter is 52cm.

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## Comments

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TopNewestThe area of quadrilateral \(PQRS\) is \(104\) cm\(^2\).

Let the bases of the four triangles \(\triangle OPQ\), \(\triangle OQR\), \(\triangle ORS\) and \(\triangle OSP\), \(PQ, QR, RS\) and \(SP\), be \(a, b, c\) and \(d\) respectively. Since all the bases are tangential to the circles, this means that the heights \(h\) of the four triangles are equal to the radius of the circle which is \(4\) cm. Therefore the area \(A\) of the quadrilateral \(PQRS\):

\(A = \frac {1}{2} (a+b+c+d) h = \frac {1}{2} \times\) perimeter \(\times\) radius \(= \frac {1}{2} \times 52 \times 4 = \boxed{104} \) cm\(^2\).

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It is 104 because radius is 4 cm. Rest I too did the same.

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Sorry, I got the radius wrong. It should be 4 cm and hence the area should be 104 \(cm^2\).

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FYI - You can edit your comment directly by clicking on the pencil on the right (it's to the left of the time stamp).

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104 cm2 as i can see!

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since PQ + QR + RS + SP =52 & area of the quadrilateral is equal to area of four triangles named ( if O is the center of the circle) [ORQ] + [OQP] +[OSP] + [ORS]. and height of four triangles is equal to the radius of the circle= 4 cm so the area is 1/2

4(PQ + QR + RS + SP)=1/2452=104 sq cmLog in to reply

I got 104

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answer is 104 cm^2 if the diagonals of the quadrilateral meet at center of the circle.

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That assumption isn't necessary.

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