Mensuration
- If the radius of a circle is increased by 20% then how much will its area be increased by ?
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Required increase = 20 + 20 + 20 × 20 % 100
= 44%Correct Option: B
Required increase = 20 + 20 + 20 × 20 % 100
= 44%
- A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water (in litres) will fall into the sea in a minute?
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Volume of water flowed in an hour = 2000 × 40 × 3 cubic metre
= 240000 cubic metre
∴ Volume of water flowed in 1 minute= 240000 = 4000 cubic metre 60
= 4000000 litreCorrect Option: B
Volume of water flowed in an hour = 2000 × 40 × 3 cubic metre
= 240000 cubic metre
∴ Volume of water flowed in 1 minute= 240000 = 4000 cubic metre 60
= 4000000 litre
- The height of an equilateral triangle is 4√3cm. The ratio of the area of its circumcircle to that of its in-circle is
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Radius of the circumcircle = 2 × 4√3 cm = 2r cm 3 Radius of the in circle = 1 × 4√3 cm = r cm 10
∴ Required ratio = π(2r)² : πr² = 4 : 1Correct Option: B
Radius of the circumcircle = 2 × 4√3 cm = 2r cm 3 Radius of the in circle = 1 × 4√3 cm = r cm 10
∴ Required ratio = π(2r)² : πr² = 4 : 1
- A circle is inscribed in a square whose length of the diagonal is 12√2 cm. An equilateral triangle is inscribed in that circle. The length of the side of the triangle is
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Side of square = 1 × 12√2 cm = 12 cm √2 ∴ Radius of circle = 12 = 6 cm 2
AB = 2x cm
∴ FH = x cm
∴ From ∆OFH,cos 30° = FH OF ⇒ √3 = x 2 6 ⇒ x = 6 × √3 = 3√3 2
∴ Length of side = 6√3 cmCorrect Option: C
Side of square = 1 × 12√2 cm = 12 cm √2 ∴ Radius of circle = 12 = 6 cm 2
AB = 2x cm
∴ FH = x cm
∴ From ∆OFH,cos 30° = FH OF ⇒ √3 = x 2 6 ⇒ x = 6 × √3 = 3√3 2
∴ Length of side = 6√3 cm
- The radius of the incircle of a triangle whose sides are 9 cm, 12 cm and 15 cm is
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9² + 12² = 15²
∴ The triangle is right angled
AB = 9, BC = 12 cm
OD = OE = OF = x cm
AD = AF = 9 – x
EC = CF = 12 – x
∴ AC = AF + FC
= 9 – x + 12 – x = 15
⇒ 21– 2x = 15
⇒ 2x = 21 – 15 = 6⇒ x = 6 = 3 cm. 2 Correct Option: C
9² + 12² = 15²
∴ The triangle is right angled
AB = 9, BC = 12 cm
OD = OE = OF = x cm
AD = AF = 9 – x
EC = CF = 12 – x
∴ AC = AF + FC
= 9 – x + 12 – x = 15
⇒ 21– 2x = 15
⇒ 2x = 21 – 15 = 6⇒ x = 6 = 3 cm. 2