CAT 2026 Slot 1 Quant: Geometry & Mensuration Questions with Solutions

3 questions: Q1, Q4, Q20. Solve each one here: tap the answer, press Check, then open the solution.

Question 1Geometry – Polygons

MCQ · +3 / −1

A and B are two regular polygons with $a$ and $b$ sides respectively. The interior angle of polygon B is $\tfrac{4}{3}$ times the interior angle of polygon A. If $b = 3a$, what is each interior angle of a polygon with $a + 3b$ sides?

Answer and solution

Correct answer: D

Solution

Interior angle of A $= \dfrac{180(a-2)}{a}$, of B $= \dfrac{180(b-2)}{b}$.
With $b = 3a$: $\dfrac{180(3a-2)}{3a} = \dfrac{4}{3}\cdot\dfrac{180(a-2)}{a}$
$3a - 2 = 4a - 8 \Rightarrow a = 6,\; b = 18$
Sides $= a + 3b = 6 + 54 = 60$
Interior angle $= \dfrac{180 \times 58}{60} = 174^\circ$

Question 4Geometry – Triangles

MCQ · +3 / −1

In triangle ABC, $AB = 12$ cm, $AC = 15$ cm and $BC = 18$ cm. Find the length of the median AD (in cm).

Answer and solution

Correct answer: A

Solution

D is the midpoint of BC, so $BD = 9$.
Apollonius' theorem: $AB^2 + AC^2 = 2(AD^2 + BD^2)$
$144 + 225 = 2(AD^2 + 81) \Rightarrow 369 = 2AD^2 + 162$
$AD^2 = 103.5 \Rightarrow AD = \sqrt{103.5}$ cm

Question 20Geometry – Quadrilaterals

MCQ · +3 / −1

Three sides of a quadrilateral are 3 cm, 2 cm and 6 cm, and the fourth side is an integer. How many lengths are possible for the fourth side?

Answer and solution

Correct answer: C

Solution

Any side must be less than the sum of the other three. Let the fourth side be $x$.
$x < 3 + 2 + 6 = 11$
$6 < 3 + 2 + x \Rightarrow x > 1$
$x \in \{2, 3, \ldots, 10\}$ → $9$ lengths

An expected paper by Cetking, not the actual CAT paper. CAT is conducted by the IIMs.