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CAT QuantmediumGeometry & Mensuration

Geometry and Mensuration

Updated 22 October 2026

Five questions on triangles, circles, coordinate geometry, and areas and volumes.

Question 1 · Multiple choice (+3 / −1)

A triangle has sides 13 cm, 14 cm and 15 cm. What is the radius of its circumcircle?

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Answer: B. 65/8 cm

  1. Semi-perimeter s = (13 + 14 + 15)/2 = 21.
  2. Area by Heron's formula = √(21 × 8 × 7 × 6) = √7,056 = 84 cm².
  3. Circumradius R = abc/(4 × area) = (13 × 14 × 15)/(4 × 84) = 2,730/336 = 65/8 cm.
  4. For comparison, the inradius is area/s = 84/21 = 4 cm.
Question 2 · Type in the answer (+3 / 0)

What is the area of the region in the xy-plane defined by |x − 3| + 2|y| ≤ 8? Enter the answer in square units.

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Answer: 64

  1. The boundary |x − 3| + 2|y| = 8 is a rhombus centred at (3, 0).
  2. Setting y = 0: |x − 3| = 8, so the horizontal vertices are (−5, 0) and (11, 0), a diagonal of length 16.
  3. Setting x = 3: 2|y| = 8, so the vertical vertices are (3, 4) and (3, −4), a diagonal of length 8.
  4. Area of a rhombus = ½ × d₁ × d₂ = ½ × 16 × 8 = 64 square units.
  5. Shortcut: |x|/a + |y|/b ≤ 1 has area 2ab; here a = 8 and b = 4, giving 64.
Question 3 · Multiple choice (+3 / −1)

Two circles of radii 9 cm and 4 cm touch each other externally, and both rest on a straight line that is a common tangent. A third circle is drawn in the gap between them so that it touches both circles and the same line. What is the radius of the third circle?

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Answer: B. 1.44 cm

  1. For two circles of radii R and r touching externally and resting on a line, the distance between their points of contact with the line is 2√(Rr).
  2. So for 9 and 4 this distance is 2√36 = 12; for the small circle of radius x it is 2√(9x) = 6√x from the big circle and 2√(4x) = 4√x from the other.
  3. The two smaller distances add up to the larger: 6√x + 4√x = 12, so √x = 1.2.
  4. x = 1.44 cm.
  5. Shortcut: 1/√x = 1/√9 + 1/√4 = 1/3 + 1/2 = 5/6, so √x = 6/5.
Question 4 · Type in the answer (+3 / 0)

A hemispherical bowl of internal radius 9 cm is full of juice. The juice is poured into cylindrical bottles, each of internal radius 1.5 cm and height 4 cm. How many bottles can be filled completely? Enter the number of bottles.

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Answer: 54

  1. Volume of the bowl = (2/3)πr³ = (2/3)π × 729 = 486π cm³.
  2. Volume of one bottle = πr²h = π × 2.25 × 4 = 9π cm³.
  3. Number of bottles = 486π ÷ 9π = 54.
  4. π cancels, so there is no need to substitute a value for it.
Question 5 · Multiple choice (+3 / −1)

In triangle ABC, angle A = 90°, AB = 8 cm and AC = 6 cm. The bisector of angle A meets BC at D. What is the length of AD in cm?

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Answer: C. 24√2/7

  1. Place A at (0, 0), B at (8, 0) and C at (0, 6). BC = 10.
  2. By the angle bisector theorem, BD : DC = AB : AC = 8 : 6, so D divides BC in the ratio 4 : 3.
  3. D = (3 × 8 + 4 × 0)/7, (3 × 0 + 4 × 6)/7 = (24/7, 24/7), which lies on y = x as expected.
  4. AD = √2 × 24/7 = 24√2/7 cm (about 4.85 cm).
  5. Shortcut: AD = 2bc cos(A/2)/(b + c) = 2 × 8 × 6 × (1/√2)/14 = 24√2/7.

Full worked solution

  1. Heron's formula: area = √(s(s − a)(s − b)(s − c)); circumradius R = abc/(4 × area); inradius r = area/s.
  2. |x − h|/a + |y − k|/b ≤ 1 is a rhombus with diagonals 2a and 2b, area 2ab.
  3. Two circles touching externally and a common line: contact points are 2√(Rr) apart; a circle in the gap satisfies 1/√x = 1/√R + 1/√r.
  4. Recasting or pouring: volume is conserved. Hemisphere (2/3)πr³, cylinder πr²h, cone (1/3)πr²h.
  5. Angle bisector theorem: the bisector from A divides BC in the ratio AB : AC; its length is 2bc cos(A/2)/(b + c).