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
circle
This topic has expert replies
- PussInBoots
- Master | Next Rank: 500 Posts
- Posts: 157
- Joined: Tue Oct 07, 2008 5:47 am
- Thanked: 3 times
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.
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 ?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.
-
- Master | Next Rank: 500 Posts
- Posts: 197
- Joined: Sun May 18, 2008 2:47 am
- Thanked: 12 times
Problem 1)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
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!
-
- Master | Next Rank: 500 Posts
- Posts: 392
- Joined: Thu Jan 15, 2009 12:52 pm
- Location: New Jersey
- Thanked: 76 times
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: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
(32sqrt(3))^(2)*sqrt(3)/4=32^2*3*sqrt(3)/4=8*32*3sqrt(3)=768sqrt(3).