Geometry, Mensuration, Trigonometry, Statistics & Probability — Practice

Question 1

In a circle with centre O, AB is a chord. C is a point on the major arc. If ∠AOB = 140°, find ∠ACB.

Accepted Answer

By the Angle at Centre Theorem: ∠AOB = 2 × ∠ACB (angle at centre = twice the angle at circumference, subtended by the same arc AB). Therefore ∠ACB = ∠AOB / 2 = 140° / 2 = 70°.

Question 2

A toy is made by mounting a cone of radius 5 cm and height 12 cm on a hemisphere of the same radius. Find the total surface area and volume of the toy.

Accepted Answer

Cone: r = 5, h = 12. Slant height l = √(5² + 12²) = √(25 + 144) = √169 = 13 cm. Hemisphere: r = 5 cm. TOTAL SURFACE AREA: Only the CURVED surface of the cone + CURVED surface of hemisphere (the flat base of cone and flat face of hemisphere are joined — hidden). TSA = πrl + 2πr² = π(5)(13) + 2π(25) = 65π + 50π = 115π = 115 × 22/7 ≈ 361.4 cm². VOLUME = V_cone + V_hemisphere = (1/3)πr²h + (2/3)πr³ = (1/3)π(25)(12) + (2/3)π(125) = 100π + (250/3)π = (300π + 250π)/3 = 550π/3 ≈ 576.2 cm³.

Question 3

From the top of a cliff 100 m high, the angles of depression of the top and bottom of a tower are 30° and 60° respectively. Find the height of the tower and its distance from the base of the cliff.

Accepted Answer

Let tower height = h, distance from cliff to tower = d. From top of cliff (height 100m): angle of depression to BOTTOM of tower = 60°. So tan60° = 100/d → √3 = 100/d → d = 100/√3 metres. Angle of depression to TOP of tower = 30°. Vertical distance from top of cliff to top of tower = 100 − h. tan30° = (100 − h)/d → 1/√3 = (100 − h) / (100/√3) → 1/√3 × 100/√3 = 100 − h → 100/3 = 100 − h → h = 100 − 100/3 = 200/3 ≈ 66.67 m. Distance d = 100/√3 = 100√3/3 ≈ 57.74 m. Tower height = 200/3 m ≈ 66.7 m.

Practice — Geometry, Mensuration, Trigonometry, Statistics & Probability

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