Mensuration
- One acute angle of a right angled triangle is double the other. If the length of its hypotenuse is 10 cm, then its area is
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From ∆ABCsin 30° = AB AC ⇒ 1 = AB 2 AC ⇒ AB = 1 AC = 1 × 10 = 5 cm. 2 2
∴ BC = √AC² - AB² = √10² - 5²
= √100 - 25 = √75
= 5√3 cm
∴ Area of ∆ABC= 1 × AB × BC 2 = 1 × 5 × 5√3 = 25√3 cm² 2 2 Correct Option: A
From ∆ABCsin 30° = AB AC ⇒ 1 = AB 2 AC ⇒ AB = 1 AC = 1 × 10 = 5 cm. 2 2
∴ BC = √AC² - AB² = √10² - 5²
= √100 - 25 = √75
= 5√3 cm
∴ Area of ∆ABC= 1 × AB × BC 2 = 1 × 5 × 5√3 = 25√3 cm² 2 2
- A solid metallic cone of height 10 cm, radius of base 20 cm is melted to make spherical balls each of 4 cm diameter. How many such balls can be made ?
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Let number of balls = n
∴ Volume of n balls = Volume of cone⇒ n × 4 πr³ = 1 πR²h 3 3 ⇒ n × 4 (2)³ = 1 × (20)² × 10 3 3
⇒ n = 125Correct Option: D
Let number of balls = n
∴ Volume of n balls = Volume of cone⇒ n × 4 πr³ = 1 πR²h 3 3 ⇒ n × 4 (2)³ = 1 × (20)² × 10 3 3
⇒ n = 125
- The surface area of a sphere is 616 cm². The volume of the sphere would be :
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Surface area of sphere = 4πr²
⇒ 4πr² = 616⇒ 4 × 22 × r² = 616 7 ⇒ r² = 616 × 7 = 49 4 × 22
⇒ r = √49 = 7 cm∴ Volume of sphere = 4 πr³ 3 = 4 × 22 × 7 × 7 × 7 3 7 = 4312 = 1437 1 cu.cm. 3 3 Correct Option: A
Surface area of sphere = 4πr²
⇒ 4πr² = 616⇒ 4 × 22 × r² = 616 7 ⇒ r² = 616 × 7 = 49 4 × 22
⇒ r = √49 = 7 cm∴ Volume of sphere = 4 πr³ 3 = 4 × 22 × 7 × 7 × 7 3 7 = 4312 = 1437 1 cu.cm. 3 3
- A cistern 6 m long and 4 m wide, contains water up to a depth of 1 m 25 cm. The total area of the wet surface is
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Required total area = Area of four walls + Area of the base
= 2 × 1.25 (6 + 4) + 6 × 4
= 2.5 × 10 + 24 = 49 m².Correct Option: D
Required total area = Area of four walls + Area of the base
= 2 × 1.25 (6 + 4) + 6 × 4
= 2.5 × 10 + 24 = 49 m².
- A cylinder with base radius 8 cm and height 2 cm is melted to form a cone of height 6 cm. The radius of the cone will be
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Volume of cylinder = πr²h
= (π × 8 × 8 × 2) cu. cm.
= 128 p cu. cm.
If the radius of the base of cone be R cm. then1 πR²H = 128π 3
→ R² × 6 = 128 × 3⇒ R² = 128 × 3 = 64 6
⇒ R = √64 = 8 cm.Correct Option: B
Volume of cylinder = πr²h
= (π × 8 × 8 × 2) cu. cm.
= 128 p cu. cm.
If the radius of the base of cone be R cm. then1 πR²H = 128π 3
→ R² × 6 = 128 × 3⇒ R² = 128 × 3 = 64 6
⇒ R = √64 = 8 cm.
