Area and Perimeter
- A person observed that he required 30 s time to cross a circular ground along its diameter than to cover it once along the boundary. If his speed was 30m/min, than the radius of the circular ground is (take π = 22/7)
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Let the radius of circular field = r m.
Speed of person in m/s = 30/60 = 1/2m/s
According to the question,
[(2πr) /(1/2)] - [(2r)/(1/2)] = 30Correct Option: B
Let the radius of circular field = r m.
Speed of person in m/s = 30/60 = 1/2m/s
According to the question,
[(2πr) /(1/2)] - [(2r)/(1/2)] = 30
⇒ 4πr - 4r = 30
⇒ [4 x (22/7) - 4]r =30
⇒ (125 - 4)r = 30
⇒ (8.5)r = 30
⇒ r = 30/8.5 = 3.5 m
- The area of a rectangular field is 15 time the sum of its length and breadth. If the length of that field is 40 m, what is the breadth of that field ?
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Length ot rectangle = 40 m
Let breadth of = k
Then, according of the question,
(40 + k)15 = 40kCorrect Option: A
Length ot rectangle = 40 m
Let breadth of = k
Then, according of the question,
(40 + k)15 = 40k
⇒ 600 + 15k = 40k
⇒ 25k = 600
∴ k = 24 m
- The Different between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is
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Let the length of rectangle = L m
∴ Breadth of rectangle = B m
Using conditions from the question,
L - B = 23 ....(i)
2(L + B) = 206
L + B = 103 ....(ii)
Then , area of rectangle = L x BCorrect Option: A
Let the length of rectangle = L m
∴ Breadth of rectangle = B m
Using conditions from the question,
L - B = 23 ....(i)
2(L + B) = 206
L + B = 103 ....(ii)
On adding Eqs. (i) and (ii), we get
2L = 126
⇒ L = 63 m
⇒ B = 103 - 63 = 40 m
Then , area of rectangle = L x B
= 63 x 40
= 2520 m2 .
- The area of rectangle whose length is 5 more than twice its width is 75 sq units. What is the perimeter of the rectangle ?
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Let the width of the rectangle = k units
∴ Length = (2k + 5) units.
According to the question,
Area = k(2k + 5)Correct Option: A
Let the width of the rectangle = k units
∴ Length = (2k + 5) units.
According to the question,
Area = k(2k + 5)
⇒ 75 = 2k2 + 5k
⇒ 2k2 + 5k - 75 = 0
⇒ 2k2 + 15k - 10k - 75 = 0
⇒ k(2k + 15) - 5(2k + 15) = 0
⇒ (2k + 15) (k - 5) = 0
⇒ k = 5 and -15/2
Width cannot be negative.
∴ Width = 5 units
∴ Length = 2x + 5 = 2 x 5 + 5 = 15 unit
∴ perimeter of the rectangle = 2(15 + 5)
= 40 units
- A rectangle has 30 cm as its length and 720 sq cm as its area. Its area is increased to 11/4 times its original area by increasing only its length. Its new perimeter is
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Original breadth of rectangle = 720/30 = 24 cm
Now , area of rectangle = (5/4) x 720 = 900 cm2
∴ New length of rectangle = 900/24 = 37.5 cm
∴ New perimeter of rectangle = 2(l+ b)Correct Option: A
Original breadth of rectangle = 720/30 = 24 cm
Now , area of rectangle = (5/4) x 720 = 900 cm2
∴ New length of rectangle = 900/24 = 37.5 cm
∴ New perimeter of rectangle = 2(l+ b)
= 2(37.5 + 24 )
= 2 x 61.5
= 123 cm
