Math Destroyer Test #5 Q # 33

[Bio Dental Mom]

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Given: Three circles of radius one that are tangent to each other and whose centers are joined to form an equilateral triangle. Find the area of the little portion enclosed by the 3 circles.

Answer: (root)3 - pi/2

I'm sorry I don't know how to upload the picture. Can anyone help me understand this, its tough... and the soln in the back doesn't help too much either.


Thanks!

 

[wantVCUdental]

Given: Three circles of radius one that are tangent to each other and whose centers are joined to form an equilateral triangle. Find the area of the little portion enclosed by the 3 circles.

Answer: (root)3 - pi/2

I'm sorry I don't know how to upload the picture. Can anyone help me understand this, its tough... and the soln in the back doesn't help too much either.


Thanks!

The area of triangle is root 3, b/c each side is 2 and it's equilateral

The area of the circles, there are 3 of them, each is 60degrees/360 of the circle. area of 1 circile=pi
60/360=1/6, so 1/6pi is the area of 1 piece of the circle that overlap with the triangle. You have 3 pieces of circles in that triangle, so 3/6 pi =1/2 pi

The small area = area of triangle - area occupied by the circles
= root 3 -pi/2

Hope that helps

 

[Bio Dental Mom]

Thanks, that does! I just wasn't sure where they were getting the root*3 from. I totally forgot about the equilateral triangle formula = S^2*root*3/4

 

[ammarm]

i was actually going through this question right now.. i think there is an error in the solution in the DAT destroyer book, for this question..

pls correct me if i am wrong:
the area of a whole circle in radians is 2pi and not 1pi
there the are of the arc was found to be 60 degrees

in terms of radians it will be pi/3 and NOT pi/6
so if you multiply pi/3 * 3= pi (the area subtended by the three arc all together)
the area of the whole triangle was root3

therefore the area in the little portion will be:
root 3- pi and NOT root 3- pi/2

your suggestions are kindly appreciated... i am not sure if i am right so please correct me i am wrong

 

[bharat008]

Given: Three circles of radius one that are tangent to each other and whose centers are joined to form an equilateral triangle. Find the area of the little portion enclosed by the 3 circles.

Answer: (root)3 - pi/2

I'm sorry I don't know how to upload the picture. Can anyone help me understand this, its tough... and the soln in the back doesn't help too much either.


Thanks!

Haven't you already been accepted to Baylor and Houston?/

 

[rockclock]

i was actually going through this question right now.. i think there is an error in the solution in the DAT destroyer book, for this question..

pls correct me if i am wrong:
the area of a whole circle in radians is 2pi and not 1pi
there the are of the arc was found to be 60 degrees

in terms of radians it will be pi/3 and NOT pi/6
so if you multiply pi/3 * 3= pi (the area subtended by the three arc all together)
the area of the whole triangle was root3

therefore the area in the little portion will be:
root 3- pi and NOT root 3- pi/2

your suggestions are kindly appreciated... i am not sure if i am right so please correct me i am wrong

circumference is 2pi...the area is pi. the solution given above is correct.

 

[Moka11]

Gug up this thread but I cant seem to understand the answer. Can someone help?

What rules/laws should I be applying to this question?

 

[lowkeydr]

You know each side of the triangle because circles are tangent to each other and each corner of the triangles are in the center of the circle and they give you the radius.
so each side would equal to 2r, or 2(1) = 2.

I personally dont have equilateral triangle area formula memorized so I just cut the triangle in half, then I can derive the height of the triangle by using 30 60 90 rule (height should be 1x(sqroot3) ).. So 1/2bh = 1/2 (2) (sqrt3) = sqrt3 is the area of the triangle of just the triangle. But they want the area of the inside, so you have to subtract by the area of the subparts of the circles.

you know its 60 degrees of the circle cut off per part or 1/6 of the area of the circle, ( remember 360 for whole circle) so.. 1/6 x pi x 1^2 = 1/6pi, but theres 3 parts so 1/6pi x 3 = 1/2pi
Answer A: sqrt3 - 1/2pi

There really isnt much laws or rules you need to know except 30 60 90 rule or the equilateral triangle area formula (whichever you prefer).

 

[Moka11]

Awesome! Thanks!