- Question:

There is grassy square plot. One horse is tied at the mid point of one side of the field and another horse is tied at the adjescent side of the first one. The rope is just long enough for the horses to reach the centre of the square. Find the ratio of the area of the commonly grazed part to the area of the square field.

a) (pi-2)/8

b) (pi-2)/6

three other options were also there. - Can anyone help how to find the remainder of 34! when divided by 71On Thu, Dec 3, 2009 at 2:05 AM, chayank <chayankag@...> wrote:
Question:

There is grassy square plot. One horse is tied at the mid point of one side of the field and another horse is tied at the adjescent side of the first one. The rope is just long enough for the horses to reach the centre of the square. Find the ratio of the area of the commonly grazed part to the area of the square field.

a) (pi-2)/8

b) (pi-2)/6

three other options were also there.

--

SRIRAM|

SURAT | MOBILE :- 9909902378

- hi.......this is a simple question......consider a square with s>0,s<=a,y>0,y<=athe 2 semicirles will hv cetre at (a/2,0) and (0,a/2) with a/2 as radiusfind their pt of intersection and use integration to calculate its area........it will be in terms of a^2and the area of the square is a^2................the aswer comes out to be pi-2/8
**From:**chayank <chayankag@...>**To:**ascent4cat@yahoogroups.com**Sent:**Thu, 3 December, 2009 3:35:56 PM**Subject:**[2IIM CAT Prep] CAT 09 29 Nov 3:30pm Batch question

Question:

There is grassy square plot. One horse is tied at the mid point of one side of the field and another horse is tied at the adjescent side of the first one. The rope is just long enough for the horses to reach the centre of the square. Find the ratio of the area of the commonly grazed part to the area of the square field.

a) (pi-2)/8

b) (pi-2)/6

three other options were also there.

The INTERNET now has a personality. YOURS! See your Yahoo! Homepage. there is no need to apply integration this problem can be solved by simple geometry. Draw a square and draw two semi circles taking a/2 as a radius along the adjacent sides. u will observe thatthe commonly grazed area can be found by area of semicircle - area of right angle triangle taking a/2 as a radius. hence the answer will be (pi-2)/8 --- On

**Mon, 14/12/09, Pankaj Jain**wrote:*<pankajain1990@...>*

From: Pankaj Jain <pankajain1990@...>

Subject: Re: [2IIM CAT Prep] CAT 09 29 Nov 3:30pm Batch question

To: ascent4cat@yahoogroups.com

Date: Monday, 14 December, 2009, 11:59 PMhi.......this is a simple question.... ..consider a square with s>0,s<=a,y>0,y<=athe 2 semicirles will hv cetre at (a/2,0) and (0,a/2) with a/2 as radiusfind their pt of intersection and use integration to calculate its area........ it will be in terms of a^2and the area of the square is a^2......... .......the aswer comes out to be pi-2/8

**From:**chayank <chayankag@yahoo. com>**To:**ascent4cat@yahoogro ups.com**Sent:**Thu, 3 December, 2009 3:35:56 PM**Subject:**[2IIM CAT Prep] CAT 09 29 Nov 3:30pm Batch question

There is grassy square plot. One horse is tied at the mid point of one side of the field and another horse is tied at the adjescent side of the first one. The rope is just long enough for the horses to reach the centre of the square. Find the ratio of the area of the commonly grazed part to the area of the square field.

a) (pi-2)/8

b) (pi-2)/6

three other options were also there.

The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.

The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.