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Unit V: Mensuration — Test 1

Concepts and applications · 10 questions · 30 practice marks · 60 minutes.
MCQs are scored automatically; written answers use step-wise self/teacher review.

Section A
Question 1 (1 Mark)
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Full question paper and step-wise model answers

Practice marks are not official examination weightage. Written answers require self- or teacher assessment.

Question 1 · 1 mark

If the radius of a circle doubles, its circumference:

  1. Stays fixed
  2. Doubles
  3. Triples
  4. Quadruples
Model answer and reasoning

(b) Doubles

  1. Circumference C = 2πr.
  2. Replacing r by 2r multiplies C by 2.

Question 2 · 1 mark

For a right circular cone with radius 3 cm and height 4 cm, the slant height is:

  1. 1 cm
  2. 5 cm
  3. 7 cm
  4. 12 cm
Model answer and reasoning

(b) 5 cm

  1. The radius, height and slant height form a right triangle.
  2. l = √(3² + 4²) = 5 cm.

Question 3 · 1 mark

When all dimensions of a cuboid are multiplied by 2, its volume is multiplied by:

  1. 2
  2. 4
  3. 6
  4. 8
Model answer and reasoning

(d) 8

  1. V = lwh.
  2. Each of the three factors doubles, so V becomes 2³V = 8V.

Question 4 · 4 marks

Explain the constant circumference-to-diameter ratio π and distinguish exact π from approximations. Name approximations associated with Archimedes, Aryabhata and Zu Chongzhi. Using 22/7, find the length of a 90° arc of radius 14 cm.

Model answer and reasoning

π is irrational; the arc length is approximately 22 cm.

  1. For any circle C/d = π; π cannot equal any rational fraction exactly.
  2. Historical approximations include Archimedes’ bounds 223/71 < π < 22/7, Aryabhata’s 3.1416, and Zu Chongzhi’s 355/113.
  3. An arc of central angle θ has length (θ/360°) × 2πr.
  4. For θ = 90° and r = 14, using 22/7 gives (1/4) × 2 × (22/7) × 14 = 22 cm. A right-angle arc is one quarter of the circumference.

Question 5 · 4 marks

A triangle has base 12 cm and height 7 cm. Find its area. Prove that a median from the opposite vertex divides it into two equal-area triangles; compare the rectangle and parallelogram formulas.

Model answer and reasoning

Area 42 cm²; each half has area 21 cm².

  1. Triangle area is half the area of a parallelogram with the same base and perpendicular height: A = bh/2.
  2. For the given values, A = 12 × 7/2 = 42 cm².
  3. A median bisects the base. The two smaller triangles have equal bases and the same altitude, so their areas are equal, each 21 cm².
  4. A rectangle has area length × width. A parallelogram can be cut and rearranged into a rectangle, giving base × perpendicular height.

Question 6 · 3 marks

Use Heron’s formula to find the area of a triangle with sides 13 cm, 14 cm and 15 cm.

Model answer and reasoning

84 cm².

  1. The semiperimeter is s = (13 + 14 + 15)/2 = 21 cm.
  2. Area = √(s(s − a)(s − b)(s − c)) = √(21 × 8 × 7 × 6).
  3. The product is 7056 and its square root is 84 cm².

Question 7 · 4 marks

Explain the sector-rearrangement derivation of a circle’s area. Then find the area of a 60° sector of radius 6 cm, leaving π exact.

Model answer and reasoning

Circle area πr²; sector area 6π cm².

  1. Cut the circle into many narrow equal sectors and alternate them to approximate a rectangle/parallelogram.
  2. As the sectors become thinner, the limiting base is half the circumference, πr, and height is r.
  3. Thus the limiting area is πr × r = πr².
  4. A 60° sector is 1/6 of the circle, so its area is (1/6) × π × 6² = 6π cm².

Question 8 · 4 marks

Compare a cube of side 6 cm and a cuboid 9 cm × 6 cm × 4 cm. Find their volumes and total surface areas. State the cylinder formulas for radius r and height h.

Model answer and reasoning

Both volumes 216 cm³; cube TSA 216 cm², cuboid TSA 228 cm². Cylinder CSA 2πrh, TSA 2πr(h + r), V = πr²h.

  1. Cube: V = 6³ = 216 and TSA = 6 × 6² = 216.
  2. Cuboid: V = 9 × 6 × 4 = 216 and TSA = 2(54 + 24 + 36) = 228.
  3. A cube is a cuboid whose length, width and height are equal. Equal volumes need not give equal surface areas.
  4. Unrolling a cylinder gives curved area circumference × height, 2πrh; add two bases for TSA. Volume is base area × height.

Question 9 · 4 marks

A right square pyramid has base side 6 cm, vertical height 4 cm, and triangular-face slant height 5 cm. Find its volume and total surface area. Identify its base and apex and compare its volume rule with a cone.

Model answer and reasoning

Volume 48 cm³; total surface area 96 cm².

  1. The base is the square; the apex is the common vertex of the four triangular faces above it.
  2. Volume = (1/3) × base area × vertical height = (1/3) × 36 × 4 = 48 cm³.
  3. Each triangular face has area (1/2) × 6 × 5 = 15 cm². TSA = 36 + 4 × 15 = 96 cm².
  4. Both pyramid and cone volumes are one third of base area × vertical height; the cone uses circular base area πr².
Right square pyramid with vertical height four, face slant height five and half-base length threeh = 4slant = 5half side = 3apex

Question 10 · 4 marks

For a sphere and a hemisphere each of radius 3 cm, find sphere surface area and volume, hemisphere curved area, total area and volume.

Model answer and reasoning

Sphere: 36π cm² and 36π cm³. Hemisphere: CSA 18π cm², TSA 27π cm², volume 18π cm³.

  1. Sphere SA = 4πr² = 36π and V = (4/3)πr³ = 36π.
  2. Hemisphere CSA is half the sphere area, 2πr² = 18π.
  3. Include the circular base for hemisphere TSA: 3πr² = 27π.
  4. Hemisphere volume is half the sphere volume: (2/3)πr³ = 18π. Distinguish square and cubic units.