PQRST is a pentagon in which all the interior angles are une

Question

PQRST is a pentagon in which all the interior angles are unequal. A circle of radius ‘r’ is inscribed in each of the vertices. Find the area of portion of circles falling inside the pentagon.

A.πr²

B.1.5πr²

C.3πr²

D.2.5πr²

E.6πr²

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chetan2u · Answer

The sum of angles of n sided polygon = (n-2)*180

The sum of angles of pentagon = (5-2)*180=540.

When we draw a circle from any of the vertex, the portion of circle falling inside the polygon will be equal to the angle made by the vertex.  
If it say x degrees, the portion of the circle inside pentagon will be  \frac{x}{360}*\pi r^2

The radius has to be less than the half of the shortest side, so that we do not have any overlaps. The question should have been worded accordingly. Sketch on the left. 
If the above is not the case, the case will be as shown in the right side.

When we add all these portions, it will be sum of all the angles or 540, so our answer =  \frac{540}{360}*\pi r^2=1.5\pi *r^2

PayelRB · Answer

Can someone please help with a diagram?

KarishmaB · Answer

Screenshot 2021-06-23 at 12.13.22.png This is what it means. Yes, the wording is confusing. The question "Find the area of portion of circles falling inside the pentagon." made me think that perhaps this is what they meant. And then it is a simple matter of assuming all angles same (they need to be distinct but that has little relevance when we talk about not-necessarily integers because the angles could be 108.000000001 degrees, 107.99999999999 degrees, 108.00000000002 degrees etc) The area of the 5 circles lying inside the pentagon will be \frac{108}{360} * \pi*r^2 * 5 = 1.5*\pi*r^2 Answer (B)

GMATWhizTeam · Answer

Solution:

We know that area of any sector of circle  = \frac{central angle of that sector }{ 360} 	imes \pi r^2

There are 5 such circle sectors inside pentagon. 
Total area  = \frac{a}{360} 	imes \pi r^2 + \frac{b}{360} 	imes \pi r^2 + \frac{c}{360} 	imes \pi r^2 + \frac{d}{360} 	imes \pi r^2 + \frac{e}{360} 	imes \pi r^2 
 ⇒\frac{ \pi r^2}{360} (a+b+c+d+e)

a+b+c+d+e is sum of interior angles of the pentagon. We know sum of interior angles of pentagon  = (n-2)	imes 180 = 3	imes 180 = 540

⇒\frac{ \pi r^2}{360}	imes 540 
 ⇒1.5 \pi r^2

Hence the right answer is  Option B .

PQRST is a pentagon in which all the interior angles are une

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PQRST is a pentagon in which all the interior angles are une

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