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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.

3 years, 9 months ago

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The 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$$^2$$.

- 3 years, 9 months ago

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

- 3 years, 9 months ago

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

- 3 years, 9 months ago

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

Staff - 3 years, 9 months ago

104 cm2 as i can see!

- 3 years, 9 months ago

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 1/24(PQ + QR + RS + SP)=1/2452=104 sq cm

- 3 years, 9 months ago

I got 104

- 3 years, 9 months ago

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

- 3 years, 9 months ago

That assumption isn't necessary.

Staff - 3 years, 9 months ago