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Geometry Mensuration CAT 2025 Slot 2 Actual Questions

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Visual Lens G Strategy | Hexagon Area | Moderate

Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD, respectively.
Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is _______

Answer & Explanation

Correct answer is B) 5 : 24

Let area of the hexagon is 6E, where E represents the area of any one of the six congruent central triangles, such as triangle OBC.

The area of trapezium PBCQ is obtained by decomposing it into simpler regions. It consists of the full area of triangle OBC along with suitable fractional parts of the triangles adjacent to the base BC.

This decomposition gives: Area(PBCQ) = 5E/4.

Therefore, the required ratio of areas is: (5E/4) ÷ 6E = 5/24.

Visual Lens G Strategy | Triangle

In a triangle ABC, points D and E are on the sides BC and AC, respectively. BE and AD intersect at point T such that AD : AT = 4 : 3, and BE : BT = 5 : 4. Point F lies on AC such that DF is parallel to BE. Then, BD : CD is

A) 9 : 4

B) 7 : 4

C) 11 : 4

D) 15 : 4

Answer & Explanation

Correct answer: C) 11:4

Given, AD : AT = 4 : 3 ⇒ AT : TD = 3 : 1 and BE : BT = 5 : 4 ⇒ BT : TE = 4 : 1

By the mass points method, masses are assigned inversely to the segments.

From AT : TD = 3 : 1 ⇒ Mass at A : Mass at D = 1 : 3.

From BT : TE = 4 : 1, ⇒ Mass at B : Mass at E = 1 : 4.

Choosing consistent masses,

A = 5/4, D = 15/4 and B = 1, E = 4, so mass at T = 5.

Since D lies on BC, Mass at D = Mass at B + Mass at C

15/4 = 1 + Mass at C ⇒ Mass at C = 11/4.

Therefore, BD : DC = Mass at C : Mass at B ⇒ (11/4) : 1 ⇒ 11 : 4.

Concept of Angles of Circles | Moderate

Two tangents drawn from a point P touch a circle with center O at points Q and R. Points A and B lie on PQ and PR, respectively, such that AB is also a tangent drawn to the same circle. If ∠AOB = 50°, then ∠APB is _____

Answer & Explanation

Correct Answer 80

From P, PQ and PR are tangents to the circle with centre O, so OQ ⟂ PQ and OR ⟂ PR.

AB is also a tangent, touching the circle at S, hence OS ⟂ AB.

From an external point, the line joining the centre bisects the angle between tangents.

Thus OA bisects ∠QOS and OB bisects ∠ROS.

Let AOQ = AOS = x and BOR = BOS = y.

Given ∠AOB = x + y = 50°.

Then ∠QOS = 2x, ∠ROS = 2y and hence ∠QOR = 2(x + y) = 100°.

In quadrilateral ORPQ, angles at Q and R are 90°, so

∠QOR + ∠APB = 180°.

Therefore, ∠APB = 180° − 100° = 80°.

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