1) The inner circumference of a 21m wide circular race track is 660m. Then
Answers with explanation:
(A) What is radius of inner edge circle of race track?
(B) What is radius of outer edge circle of race track?
(C) What is perimeter of outer edge circle of race track?
(D) What is area of inner edge circle of race track?
(E) What is area of outer edge circle of race track?
(F) What is area of circular race track?
(G) How much long rope will be required to tie all the poles around both sides of the race tack?
Options for question (A) : (a) 84 m. (b) 105 m. (c) 100 m. (d) 66 m.
Options for question (B) : (a) 84 m. (b) 105 m. (c) 126 m. (d) 66 m.
Options for question (C) : (a) 729 m. (b) 512 m. (c) 1000 m. (d) 880 m.
Options for question (D) : (a) 84125 sq.m. (b) 10025 sq.m. (c) 78250 sq.m. (d) 34650 sq.m.
Options for question (E) : (a) 49896 sq.m. (b) 10025 sq.m. (c) 78250 sq.m. (d) 34650 sq.m.
Options for question (F) : (a) 84125 sq.m. (b) 15246 sq.m. (c) 78250 sq.m. (d) 34650 sq.m.
Options for question (G) : (a) 1415 m. (b) 1052 m. (c) 1389 m. (d) 1766 m.
Answers with explanation:
Equations used to solve this problem are:
Equation of perimeter of a circle = 2πr
Equation of area of circle = πr2
(a2– b2) = (a+b)(a-b) in the form (R2 – r2) = (R+r)(R-r)
Assumptions:
Let the radius of inner circle= r
Radius of outer circle = R
Given data:
Perimeter of inner edge circle of race track = 660m
Width of the race track = 21m
Let us find out radius of inner circle that is answer of question (A).
2πr = 660m.
We know that π = 22/7
So 2 X 22/7 X r = 660
Therefore, r = (660 X 7)/(2 X 22) = 15 X 7 = 105 m
Hence r = 105 m.
Thus the radius of inner edge circle of race track is 105m.
Now radius of outer circle will be easy to find. It will be the answer of (B).
As per given data and assumptions R = r + 21 = 105 + 21 = 126 m.
Hence R = 126 m.
Thus the radius of outer edge circle of race track is 126m.
Let us find out perimeter of outer circle means answer of question (C)
Perimeter of a circle = 2πR = 2 X 22/7 X 126 = 729m.
Hence the perimeter of the outer edge circle of the race track is 729m.
The answer of questions (D) and (E) will be easy to find as we know the radii of both the circles.
The area of inner circle = πr2 = 22/7 X 1052 = 34650 sq. m.
Thus the area of inner edge circle of the race track is 34650 sq. m.
The area of outer circle = πR2 = 22/7 X 1262 = 49896 sq. m.
Thus the area of outer edge circle of race track is 49896 sq. m.
As we know the areas of both circles it is very easy to find out the area of the race track that is the answer of question (F).
Area of race track = area of outer edge circle – area of inner edge circle
= 49896 – 34650 = 15246 sq. m.
There is another way to find out the area of race track by using the 3rdequation as follows:
Area of race track = area of outer edge circle – area of inner edge circle
Area of race track = πR2 - πr2 = π(R2 – r2) = π(R+r)(R-r)
Area of race track = 22/7 (126 + 105)(126 – 105) = 22/7 X 231 X 21 = 22 X 231 X 3
Area of race track = 15246 sq. m.
Thus the area of the race track is 15246 sq. m.
Now it’s turn of last question (G).
The required length of rope will be equal to the perimeters of both the circles.
So Length of the rope = perimeter of outer circle + perimeter of inner circle.
Length of the rope = 729 + 660 = 1389m.
Thus the length of the rope required to tie all the poles around both sides of the race track is 1389m.
Thus you can solve the numericals of this type by using the same way I solved this.
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