Semi circles

Are you saying the area of the small semi circle is pi/4? Shouldn't it be pi/8?

Agree w madeinafrica, answer is E.

@mendelay: small circle's radius is 1/2, so area = pi * r^2 = pi / 4.

So the semi circle should then be pi/8, correct?

Absolutly !!

and the answer is Pi/8-(Pi/6-root(3)/4)

Then the answer should be D.

Nice explanation by "madeinafrica" ...thanks.

madeinafrica,

Do you concur with D?

Doing my calculations again gave me this. and it shoud be C i think...

\(\pi\)/ 8 - (\(\pi\)/ 6-\(\sqrt{3}\)/4) = \(\sqrt{3}\)/4 - \(\pi\)/ 24

See also in your post, the sign for Pi/24 is negative... so it is C (and not D)

Wow! A nice question. Though I would love to see such questions on the test, trust me it took nearly 5 minutes to come up with a logic and solve it. 5 minutes is too high.

My answer: D. Here is how I approached the problem - may not be an efficient one, but I stuck to basics. If someone has some really cool shortcut to solve it even more quickly, please educate me (and us).

Triangle (Equilateral): From the chord measuring 1 cm draw a chord which goes to the centre of the bigger circle - do this at both the extremes of the chord. You are actually drawing the radii of the bigger circle and this would be of length 1 cm. The area of this triangle may now be computed as \sqrt{3}/4.

Area of the Sector: Remember we are dealing with equilateral triangle - the two radii would intersect at point O (centre) on the bigger circle and the angle measure at that point would be 60 degrees. There the area of the sector = (1/2)*(1^2)*Pi*(60/360) = Pi/12

Area of the unshaded portion between the bigger semi-circle and smaller semi-circle:Area of the sector - Area of the triangle. (Pi/12) - \sqrt{3}/4

Area of the smaller semi-circle:(1/2)*((1/2)^2)*Pi = Pi/8

Area of the lume: (Area of the smaller semi-circle) - (Area of the unshaded portion between the bigger semi-circle and smaller semi-circle) = (Pi/8) - (Pi/12) + \sqrt{3}/4 (please note the change in sign before the square root)
= solving this equation would lead to (Pi + 6 \sqrt{3})/24 => (Pi/24) + \sqrt{3}/4.

And so I rest the case with D.

amazing problem. Thanks for posting.

madeinafrica,

Yes, C. My bad! Thanks for pointing that out. That's +1 kudos for you.

hello guys
wow! so hard quetion !!!
Bunuel, may be you explain the solution of this question!
thanks.

1/2 * pi (1/2)^2 - Total lune

= pi/8

Area of triangle on joining two radii = root(3)/4 * (1)^2

Are of sector = 1/6 * pi * (1)^2

So area of lune = pi/8 - (pi/6 - root(3)/4 )

= root(3)/4 - pi/24

Answer - C

answer is [root(3)/4-pi/24]
so alternative c.....

Please see the attached figure:



We need to find the area of the grey shade.

Area of semicircle = \(\frac{\pi*r^2}{2}\)

Area of bigger semicircle(radius=1) = \(\frac{\pi*(1)^2}{2}=\frac{\pi}{2}\)

Area of smaller semicircle(radius=1/2) = Area of Grey+ Area of Turquoise= \(\frac{\pi*(\frac{1}{2})^2}{2}=\frac{\pi}{8}\)

Area of lune = Area of grey shade = Area of smaller semicircle - Area of turquoise shade

We already know the area of the smaller semicircle. If we find the area of the turquoise shade, we will have our area of the lune.

Area of turquoise shade = Area of sector AOB(Turquoise+Orange) - Area of triangle AOB(Orange)

Area of sector AOB(Turquoise+Orange) = \(\frac{\theta}{360}*\pi*r^2=\frac{\theta}{360}*\pi*(1)^2=\frac{\theta}{360}*\pi\)

\(\theta=\angle AOB=60^{\circ}\) as \(\triangle AOB\) is an equilateral triangle as AO=1(radius), BO=1(radius), AB=1(given)

Area of sector AOB(Turquoise+Orange) = \(\frac{\theta}{360}*\pi=\frac{60}{360}*\pi=\frac{\pi}{6}\).

Area of equilateral triangle AOB(Orange) = \(\frac{\sqrt{3}}{4}*r^2=\frac{\sqrt{3}}{4}*1^2=\frac{\sqrt{3}}{4}\)

Area of turquoise shade = \(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\)

Area of lune = Area of smaller semicircle - Area of turquoise shade = \(\frac{\pi}{8}-(\frac{\pi}{6}-\frac{\sqrt{3}}{4})=\frac{\pi}{8}-\frac{\pi}{6}+\frac{\sqrt{3}}{4}= -\frac{\pi}{24}+\frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4} - \frac{\pi}{24}\)

Ans: "C"

thank you fluke, it's a prefect solution!
+1 kudos