CAT — Mensuration

Given:
A chord of length 10√3 m subtends an angle of 120° at the center of a circle.
Find the area of the smaller region cut off by the chord.

Step 1: Find the radius of the circle
For a chord,
Chord length = 2r sin(θ/2)
Here,
10√3 = 2r sin 60°
= 2r × √3/2
= r√3
Therefore,
r = 10 m

Step 2: Find the area of the sector
The smaller region corresponds to the sector of angle 120°.
Area of the sector
= (120/360) × π × 10²
= 100π/3

Step 3: Find the area of the triangle
The triangle formed by the two radii and the chord has
two sides = 10 m
included angle = 120°.
Area
= (1/2) × 10 × 10 × sin 120°
= 50 × √3/2
= 25√3

Step 4: Find the area of the smaller region
Area of the smaller region
= Area of sector − Area of triangle
= 100π/3 − 25√3
= 25(4π/3 − √3)

Final Answer
Answer:
D. 25(4π/3 − √3)

Step 1: Let dimensions be l, b and h.
Sum of all 12 edges: 4(l + b + h) = 144 → l + b + h = 36
Surface area: 2(lb + bh + hl) = 846 → lb + bh + hl = 423

Step 2: Find the sum of squares of the dimensions
(l + b + h)² = l² + b² + h² + 2(lb + bh + hl)
36² = l² + b² + h² + 2 × 423
1296 = l² + b² + h² + 846
l² + b² + h² = 450

Step 3: Find the radius of the sphere
Diagonal² = l² + b² + h² = 450
Diagonal = √450 = 15√2
Radius = (15√2)/2

Step 4: Find the volume of the sphere
Volume = (4/3)πr³ = (4/3)π × ((15√2)/2)³
= (4/3)π × (3375 × 2√2)/8
= (4/3)π × (3375√2)/4
= 1125π√2