In a rectangular park, there is an elliptical flower bed in the middle and the border. The side leng

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20 Nov 2015 Number of bricks

In a rectangular park, there is an elliptical flower bed in the middle and the border. The side lengths of the park are 30m and 20m. Major axis & minor axis are of length 14m and 10m respectively. Around the flower bed in the middle there is a walking path 1.4m wide and in the middle of each side there is entrance of 2m width to the park upto the walking path. At the border the width of flower bed is 1m. Rest of the park is grass area. In the entire walking path is covered by bricks of dimension 10*20 cm^2, find the approximate number of bricks required.

OPtion

1) 3978
2) 4858
3) 5928
4) 4688
5) 3656
6) 4988
7) 3898
8) 3624
9) 4868
10)4818

Solution

By using figure solve it
The length of walking path from entrance of smaller side to the walking path around the flower bed is
= 30-14-1.4-1.4 / 2 = 6.6 m
and the length of other straight walking path is
= 20-10-1.4-1.4 / 2 = 3.6 m
Area of straight waling path (approx)
= 2(6.6*2) + 2(3.6*2) = 40.8 m^2
Area of elliptical (waling path)
= 22/7 (7+1.4)(5+1.4)-(22/7) *5*7
= 168.96 - 110 = 58.96 m^2
Total walking area = 58.96 + 40.8 = 99.76 m^2
Area covered by each brick = 200*10^-4 m^2 = 0.02 m^2
Number of bricks required = 99.76/0.02 = 4988 (approx)
Option 6)

Correct Option: 6 Best Solution (2)

Jaikumar Bhatia   9 years AGO

Inner elliptical of pathway
a1=14/2=7m
b1=10/2=5m
A1=Pi*a1*b1=Pi*7*5=110m^2.............................(i)
Outer elliptical of pathway
a2=[(14+1.4+1.4)/2]=8.4m
a2=[(10+1.4+1.4)/2]=6.4m
A2=Pi*a2*b2=Pi*8.4*6.4=168.96m^2........................(ii)
Therefore
Area of elliptical Pathway=A(p1)
=A2-A1
=168.96-110.....................................from(i)&(ii)
=58.96m^2=A(p1)................................................(iii)
Sides of Rectangular park
x=30m
y=20m
Lenght of horizontal entrance
L(h)=x-(14+1.4+1.4)=13.2m........................(iv)
Length of vertical entrance
L(v)=y-(10+1.4+1.4)=7.2m...........................(v)
w=width of entrance=2m...........................(vi)
Therefore
Area of entrance pathway=A(p2)
=[L(h)*w]+[L(v)*w]
=[13.2*2]+[7.2*2].............from(iv),(v),(vi)
=26.4+14.4
=40.8=A(p2)...........................................(vii)
Therefore
Total pathway area is
A(p)=A(p1)+A(p2)
=58.96+40.8...................from(iii)&(vii)
=99.76m^2
Area of one brick=A(br)
=10*20=200cm^2=0.02m^2
Therefore
No of bricks=A(p)/A(br)
=99.76/0.02
=4988 nos
Ans is 4988 nos

so option is (6)

Like? Yes (2) | No   

bhaskar   9 years AGO

(pi((9.4*7.4)-(8*6))+(30-18.8+20-14.8)*2)/.02=4988

Like? Yes (1) | No