circle

Expert replies

maihuna: Legendary Member; Posts: 1578; Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members; Thanked: 82 times; GMAT Score:720

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

Like this post

Source: Beat The GMAT — Problem Solving | Thu Aug 06, 2009 11:12 am

PussInBoots: Master | Next Rank: 500 Posts; Posts: 157; Joined: Tue Oct 07, 2008 5:47 am; Thanked: 3 times

by PussInBoots » Thu Aug 06, 2009 12:54 pm

1) 4 * 4 - 4 * { 1 * pi / 4 } - 1 * pi = 16 - 2 * pi

Like this post

tom4lax: Master | Next Rank: 500 Posts; Posts: 175; Joined: Mon Feb 09, 2009 3:57 pm; Thanked: 4 times

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.

Like this post

deepak115: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Mon Aug 03, 2009 2:19 pm

by deepak115 » Fri Aug 07, 2009 5:32 am

and since PI ~ 22/7
16-2*22/7 = 68/7

Like this post

adssaini: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Tue Aug 04, 2009 6:18 pm; Thanked: 2 times

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 ?

Like this post

shahdevine: Master | Next Rank: 500 Posts; Posts: 197; Joined: Sun May 18, 2008 2:47 am; Thanked: 12 times

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!

Like this post

truplayer256: Master | Next Rank: 500 Posts; Posts: 392; Joined: Thu Jan 15, 2009 12:52 pm; Location: New Jersey; Thanked: 76 times

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).

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

560 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review