00:00

Unit V: Mensuration — Test 2

Reasoning, proof and modelling · 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)
Loading Question...
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

Brahmagupta’s formula directly gives the area from four side lengths for a:

  1. General spatial four-edge figure
  2. Cyclic quadrilateral
  3. General quadrilateral without any further condition
  4. Circle
Model answer and reasoning

(b) Cyclic quadrilateral

  1. The formula √((s − a)(s − b)(s − c)(s − d)) applies to a cyclic quadrilateral.
  2. The cyclicity condition is essential.

Question 2 · 1 mark

Increasing a sphere’s radius by 10% increases its volume by:

  1. 10%
  2. 21%
  3. 30%
  4. 33.1%
Model answer and reasoning

(d) 33.1%

  1. Volume is proportional to r³.
  2. The new factor is 1.1³ = 1.331; the increase is 0.331 = 33.1%.

Question 3 · 1 mark

A conical cup is filled to half its vertical depth, measured from its apex. The water occupies what fraction of its capacity?

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

(c) 1/8

  1. The water makes a similar cone with every linear dimension half the original.
  2. Its volume factor is (1/2)³ = 1/8.

Question 4 · 4 marks

Describe how a rectangle of sides a and b can be replaced by an equal-area square using a geometric-mean construction. Explain the Pythagorean reasoning related to the classical Sulbasutra “squaring” problem.

Model answer and reasoning

The square side is √(ab), so its area is ab.

  1. On a line mark X, Y, Z with XY = a and YZ = b. Draw a semicircle with diameter XZ.
  2. At Y draw a perpendicular meeting the semicircle at P. Let O be the midpoint of XZ.
  3. OP = (a + b)/2 and OY = |a − b|/2. Pythagoras gives YP² = OP² − OY² = ab.
  4. Construct a square on side YP. Its area equals the rectangle’s. This is an equivalent geometric justification of squaring, not a claim that every historical construction uses this exact semicircle diagram.

Question 5 · 4 marks

Find the area of a cyclic quadrilateral with side lengths 3, 4, 5 and 6 cm using Brahmagupta. Explain algebraically how Heron’s formula is its limiting special case when one side tends to zero.

Model answer and reasoning

Area 6√10 cm²; the limiting formula is √(s(s − a)(s − b)(s − c)).

  1. For the quadrilateral s = (3 + 4 + 5 + 6)/2 = 9.
  2. Area = √((9 − 3)(9 − 4)(9 − 5)(9 − 6)) = √360 = 6√10 cm².
  3. If d tends to 0, the semiperimeter tends to (a + b + c)/2 and the factor s − d tends to s.
  4. The remaining three factors give Heron’s formula. A zero-length side itself is a degenerate boundary case, not a genuine quadrilateral.

Question 6 · 4 marks

A circle has radius 10 cm. Find the area of the minor segment cut off by a chord subtending 90° at the centre, and the corresponding arc length.

Model answer and reasoning

Minor segment area 25π − 50 cm²; arc length 5π cm.

  1. The 90° sector area is π × 10²/4 = 25π cm².
  2. The two radii form a right triangle of area (1/2) × 10 × 10 = 50 cm².
  3. Subtract the triangle from the sector to get the minor segment: 25π − 50 cm².
  4. The arc is one quarter of the circumference: (1/4) × 2π × 10 = 5π cm.

Question 7 · 4 marks

A cylinder and cone have radius 3 cm and height 4 cm. Find each volume, cylinder total surface area and cone curved surface area. Compare their volumes.

Model answer and reasoning

Cylinder V = 36π cm³, TSA = 42π cm²; cone V = 12π cm³, CSA = 15π cm²; ratio 3:1.

  1. Cylinder volume = π × 3² × 4 = 36π.
  2. Cylinder TSA = 2π × 3 × (4 + 3) = 42π.
  3. Cone slant height = √(3² + 4²) = 5; CSA = π × 3 × 5 = 15π.
  4. Cone volume = (1/3) × π × 3² × 4 = 12π; the volumes are in ratio 3:1.

Question 8 · 3 marks

A solid sphere of radius 6 cm is melted without loss and recast as spheres of radius 2 cm. Find the number and compare the combined surface area before and after recasting.

Model answer and reasoning

27 spheres; combined surface area becomes three times the original.

  1. Conserve volume: number = 6³/2³ = 27.
  2. Original area is 4π × 6² = 144π cm².
  3. New combined area is 27 × 4π × 2² = 432π cm². Volume is conserved, not surface area.

Question 9 · 4 marks

A regular square pyramid has base side 8 cm and face slant height 5 cm. Find its vertical height, volume and lateral surface area. What happens to volume and surface area when every length doubles?

Model answer and reasoning

Height 3 cm; volume 64 cm³; lateral area 80 cm²; doubled size gives volume ×8 and area ×4.

  1. The face slant height, vertical height and half-base side form a right triangle: h² + 4² = 5², giving h = 3.
  2. Volume = (1/3) × 8² × 3 = 64 cm³.
  3. Four triangular faces give lateral area 4 × (1/2) × 8 × 5 = 80 cm².
  4. Uniform scale factor k multiplies area by k² and volume by k³; for k = 2 the factors are 4 and 8.

Question 10 · 4 marks

A rectangular garden 12 m × 8 m has a path 1 m wide outside it on all sides. Find the outer perimeter and path area. Describe an algorithm that avoids counting the corner areas twice.

Model answer and reasoning

Outer perimeter 48 m; path area 44 m².

  1. Add twice the width to each dimension: outer dimensions 14 m × 10 m.
  2. Outer perimeter = 2(14 + 10) = 48 m.
  3. Subtract areas: path area = 14 × 10 − 12 × 8 = 140 − 96 = 44 m².
  4. The algorithm forms the complete outer rectangle and subtracts the inner one once; this automatically includes each corner exactly once.