3D Geometry: Mensuration Practice Problems: Level 01

Sol : Option 4
Let r be the radius, We have
πr² = 4/3 πr³
⇒ r = 3/4

Sol : Option 5
Let L be the length and B is the breadth of the garden. We have L = 3B.
Total area of the garden = 180 x 15 = 2700 sq m. ⇒ LB = 2700
⇒ 3B² = 2700 ⇒ B² = 900 ⇒ B = 30m
Hence the length of the garden = 30 × 3 = 90 m

Sol : Option 2
Perimeter of the equilateral triangle is 96√3 cm.
Each of the side of the equilateral triangle is 96√3 / 3 = 32√3 cm.
The height of the equilateral triangle will be = (√3 / 2) x 32√3 = 48cm

Sol : Option 2
Ratio of radii of 2 spheres is 4: 7.
∴ratio of their volume =4³ : 7³ = 64: 343

Sol : Option 3
Volume of the sphere = 268 (4/21) = 5632 / 21 m³. Now 4/3 π ³ = 5632 / 21 ⇒ r³ = (5632 / 21) x (3/4) x (7/22) = 64 ⇒ r = 4m Hence diameter = 8m

Sol : Option 1
Area of curved surface = πrl
Now r = √(13² - 5²) = √169 - 25 = √144 = 12m
Required Area= (22/7) x 13 x 12 = 490.28m²

Sol : Option 1
Let the side of cube is 'a' and radii of sphere is 'r'.
Now Volume of cube= a³
Volume of sphere= 4/3π x r³
a³ / (4/3)π r³ = 6 / π
a / r = 2 / 1
Hence the answer is 2: 1.

Sol : Option 1
Let ‘r’ be the radius of the circle. We have 2πr = 4πr² ⇒ r = 1/2

Sol : Option 5
Semi perimeter of triangle s = (10 + 6 + 6) / 2 = 11.
Area of triangle = √11 x (11-10) x (11-6) x (11-6) = √11 x 5 x 5 = 5√11cm²

Sol : Option 5
Let 'r' be the radius of the sphere. We have
4πr² = 616
4 × 22/7 × r² = 616
r = 7