Two large circles (each of radius 0.5 cm) and two smaller circles are

Question

https://gmatclub.com/forum/download/file.php?id=62335 Two large circles (each of radius 0.5 cm) and two smaller circles are inscribed into a larger circle of radius 1cm. Find area of shaded part of the figure. A) \frac{5\pi}{18} B) 2\pi C) 5\pi D) 4\pi E) 2\pi/3 Screenshot_7.png

chetan2u · Answer

Of course we can find the radius by joining the Center of each circle.

But the choices give us answer immediately. A 20 seconds analysis.

The area of the entire figure =  \pi r^2=\pi * 1=\pi 
B, C and D can be eliminated immediately.

Clearly, the unshaded part is less than the unshaded part, thus the area of shaded part has to be less than  \frac{\pi }{2}

Only A left.

CrackverbalGMAT · Answer

Solution:

Area of the shaded part here = Area of the large circle containing all smaller circles - Area of all smaller circles

Radius of the large circle = Diameter of the second large circle which is 2*0.5=1 cm

Thus area of this large circle = pi * (1)^2 = pi

And we know that the area of shaded part of the figure cannot be greater than this value of (pi) and hence we must eliminate B,C and D.

It is clear from the figure that the unshaded part occupies more than half of the big circle and hence the shaded part should occupy less than half of the big circle in terms of area. Since the bigger circle has an area of (pi) cm^2, we cannot have E as an answer choice as it is 2/3rd of pi or close to 66.67% of the large circles area which is way above the 50% mark.

That leaves us with A.  (option a)

Devmitra Sen
GMAT SME

andreagonzalez2k · Answer

I don't see how the radius of the smaller circle is calculated. Could you explain it? Thanks a lot.

chetan2u · Answer

Let us join the centre A and C to get hypotenuse AC. 
Draw a perpendicular on the diameter to meet at B

So ABC is a right angled triangle, where AC=0.5+r, BC=0.5 and AB=larger radius-r=1-r

Pythagorus theorem tells us  (1-r)^2+0.5^2=(0.5+r)^2........1+r^2-2r+0.5^2=0.5^2+r+r^2......r=\frac{1}{3}

Area of all the 4 circles = 2*\pi *\frac{1}{2}^2+2*\pi *\frac{1}{3}^2=\pi (\frac{1}{2}+\frac{2}{9})=\frac{13\pi }{18}

Thus area of shaded region =  \pi *1^2-\frac{13\pi }{18}=\frac{5\pi }{18}

A

andreagonzalez2k

PyjamaScientist · Answer

Hi, I have attached an image below highlighting the method to solve for the radius of smaller circle. I am not a pro at this, so pls do with my bad hand writing.

bumpbot · Answer

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Two large circles (each of radius 0.5 cm) and two smaller circles are

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