Two horses are tied to opposite vertices of a 6m by 6√2m rectangular field with 6m ropes. To find the common grazing area: 1) The horses can graze a combined area formed by two intersecting circles of radius 6m centered at the vertices. 2) The area of each circular segment is calculated as the sector area minus the triangular area. 3) The common grazing area is twice the circular segment area, which equals 12π - 18√3 square meters.

1. Mensuration

2. Mensuration
Two horses are tied to opposite vertices of a rectangular field of
length and breadth 6 and 6√2 meters. If they are tied with rope of
length 6 meters, what is the common area that both horses can
graze?
(a) 6𝜋 - 9 3 (b) 12𝜋 - 18 3
(c) 𝜋 - 2 3 (d) 16𝜋 -8 3

3. Mensuration
Solution :
The shaded region gives the common area that both horses can
graze.
Two horses are tied to opposite vertices of a rectangular field of
length and breadth 6 and 6√2 meters. If they are tied with rope of
length 6 meters, what is the common area that both horses can
graze?
A B
CD
6 6
6 2
6 2

4. Mensuration
In order to get a better idea about this region, join BD
B and D are centers of two intersecting circles of radius 6 cms each.
Length of BD = 6 cms each
Length of BD = 62 + (6 2)2 = 6 3
A B
CD

5. Mensuration
BE = 6 m BD = 6 3 BD =
6 3
2
BD is the r bisector of EF (common chord)
EGB is a right 
EG2 + BG2 = BE2
=> EG = 3m => EF = 6m
 BEF is equilateral
E
G
J DB
H
F

6. Mensuration
Area of segment EJF = area (sector EJFB) – area (BEF)
=
1
6
x  x 62 -
3
4
x 62
= 6 - 9 3
Common area that they can graze
= 2 x (area of segment EJF)
= 2 x (6 - 9 3)
= 12 - 18 3
Answer Choice (B)
B
F E
J

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