Kumar walks round a rectangular field,the length of which is twice its width.He then walks round another rectangular field half as wide but having the same perimeter as the first field.If the difference in area between the two fields is 432m ,find the length of the second field.[full working]
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l1 =2*b1
area(A1) = l1*b1 = 2*(b1^2)
perimeter(P1) =2(l1+b1) =6*b1
now,
b2 = b1/2
and, P1 =P2
=> 6*(b1) =2(l2+b1/2)
6*(b1) = 2*(l2) + b1
=> l2 =(5/2)*b1 ----------(i)
therefore,
area (A2) = l2*b2 =5/4(b1^2)
also,
A1 - A2 = 432
2*(b1^2) - 5/4(b1^2) = 432
(2 - 5/4)*( b1^2) = 432
=> b1 = 24
hence, l2 =5/2(24) = 60
thus,the length of second field is 60 m.
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area(A1) = l1*b1 = 2*(b1^2)
perimeter(P1) =2(l1+b1) =6*b1
now,
b2 = b1/2
and, P1 =P2
=> 6*(b1) =2(l2+b1/2)
6*(b1) = 2*(l2) + b1
=> l2 =(5/2)*b1 ----------(i)
therefore,
area (A2) = l2*b2 =5/4(b1^2)
also,
A1 - A2 = 432
2*(b1^2) - 5/4(b1^2) = 432
(2 - 5/4)*( b1^2) = 432
=> b1 = 24
hence, l2 =5/2(24) = 60
thus,the length of second field is 60 m.
0
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