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circle

by maihuna » Thu Aug 06, 2009 11:12 am
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut at centre. Find the area of the remaining portion of the square.

2
4/7
68/7
75/7
89/7

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle. Find the area of the triangle ABC
Charged up again to beat the beast :)
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by PussInBoots » Thu Aug 06, 2009 12:54 pm
1) 4 * 4 - 4 * { 1 * pi / 4 } - 1 * pi = 16 - 2 * pi
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by tom4lax » Fri Aug 07, 2009 5:11 am
Agreed w/ 1. Answer is C.

For #2. We are left with an equalateral triangle that has radius 32 extending to the center from each point. Picture this as 3 triangles. Break again in 6 seperate triangles, which are now 30-60-90 w/ hypotenuse of 32. Solve from there.

Answer is 768rt3. Each smaller triangles has area 128rt3. Multiply by 6 for each triangle.
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by deepak115 » Fri Aug 07, 2009 5:32 am
and since PI ~ 22/7
16-2*22/7 = 68/7
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by adssaini » Fri Aug 07, 2009 5:59 am
tom4lax wrote:Agreed w/ 1. Answer is C.

For #2. We are left with an equalateral triangle that has radius 32 extending to the center from each point. Picture this as 3 triangles. Break again in 6 seperate triangles, which are now 30-60-90 w/ hypotenuse of 32. Solve from there.

Answer is 768rt3. Each smaller triangles has area 128rt3. Multiply by 6 for each triangle.
Can you draw a picture of how to do the second problem... I am not getting the picture of equliateral trinagle. Whether triangle is inscribed with points ends touching circle ?
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Re: circle

by shahdevine » Fri Aug 07, 2009 6:07 am
maihuna wrote:From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut at centre. Find the area of the remaining portion of the square.

2
4/7
68/7
75/7
89/7

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle. Find the area of the triangle ABC
Problem 1)

This seems straight forward.

Calculate total square area then subtract 4 circle quadrants areas and area of circle in middle.

(4*4)-(22/7)-(22/7)=68/7

Problem 2)

tricky at first.

area of equilateral triangle is side*sqroot(3)/4

inscribed angle = 60 so central angle = 120.

because we have isosceles triangle with 2 lengths of 32 that means opposite angles of those lengths are equal. So 180=120+2x--> x=30

within that iscosceles triangle we can create right triangle with special property of 30-60-90 and corresponding ratios of 1:sqroot(3):2
so 32=2y --> y=16 --> 16*sqroot(3)*2 =32*sqroot(3) = side of equilateral triangle.

plug that value into area of equilateral triangle 32*sqroot(3)*sqroot(3)/4

8x3=24

24 is answer

you got this, man!
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by truplayer256 » Fri Aug 07, 2009 7:25 am
Problem 2)

tricky at first.

area of equilateral triangle is side*sqroot(3)/4

inscribed angle = 60 so central angle = 120.

because we have isosceles triangle with 2 lengths of 32 that means opposite angles of those lengths are equal. So 180=120+2x--> x=30

within that iscosceles triangle we can create right triangle with special property of 30-60-90 and corresponding ratios of 1:sqroot(3):2
so 32=2y --> y=16 --> 16*sqroot(3)*2 =32*sqroot(3) = side of equilateral triangle.

plug that value into area of equilateral triangle 32*sqroot(3)*sqroot(3)/4

8x3=24

24 is answer
Shahdevine, the area of an equilateral triangle is (side)^2*sqrt(3)/4. I think you accidentally forgot to square the 32sqrt(3) when plugging it into the equilateral triangle area formula. The answer should really be:

(32sqrt(3))^(2)*sqrt(3)/4=32^2*3*sqrt(3)/4=8*32*3sqrt(3)=768sqrt(3).
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