CAT 2026 Slot 3 Quant: Geo Mensu Questions with Solutions

Answer: A — (6, 2)

1. Complete squares: (x+2)²+(y−2)²=16. Centre O=(−2,2), radius=4.

2. From an external point P, equal tangents and equal radii give two congruent right triangles. OP bisects the 60° angle.

3. In either triangle, sin 30°=4/OP. Hence OP=8.

4. The locus is (x+2)²+(y−2)²=64. Substitute x=6: 64+(y−2)²=64, so y=2.

5. The single intersection is (6,2), option A.

Why this method works:
Let T be a point of tangency. The radius OT is perpendicular to PT, making OPT right-angled. The segment from the external point to the centre bisects the angle between equal tangents, so ∠OPT=30°.

Independent check and common trap:
In that right triangle sin 30°=OT/OP=4/OP, hence OP=8. The locus is centred at(−2,2) and its rightmost x-coordinate is−2+8=6. The vertical tangent touches at the same y-coordinate as the centre, giving(6,2).

G Strategy shortcut

Spot: Coordinate Geometry and Circles.
Apply: Complete the square, then OP=r/sin(θ/2)=4/sin 30°=8. The rightmost point of the locus is (−2+8,2)=(6,2).
Check: In that right triangle sin 30°=OT/OP=4/OP, hence OP=8. The locus is centred at(−2,2) and its rightmost x-coordinate is−2+8=6. The vertical tangent touches at the same y-coordinate as the centre, giving(6,2).

Answer: C — 9:8

1. In triangles ABE and CDE, the angles at E are vertically opposite, hence have equal sines.

2. By the sine rule, AE/CE=(AB/CD)×[sin∠ABE/sin∠CDE].

3. Because E lies on BD, these angles are ∠ABD and ∠CDB. In the same circumcircle their sines are AD/(2R) and CB/(2R).

4. Therefore AE/CE=(AB×AD)/(CD×CB)=(3/2)(3/4)=9/8.

5. Answer 9:8, option C. The intersecting-chords product alone does not determine this ratio.

Why this method works:
Use the same circumradius R when converting chord lengths to sines. In triangles ABE and CDE the angles at E are vertical, so the sine-rule expressions have a common factor that cancels. This is why the product of adjacent-side ratios appears.

Independent check and common trap:
AB/CD=3/2 and AD/CB=3/4; the second ratio is the reciprocal of the supplied BC/AD. Their product is 9/8, giving AE: CE=9:8. Using 4/3 instead would invert the relevant side pairing.

G Strategy shortcut

Spot: Circles and Cyclic Quadrilaterals.
Apply: For E=AC∩BD in a cyclic quadrilateral, AE/CE=(AB×AD)/(CB×CD). Substitute the two ratios: (3/2)×(3/4)=9/8.
Check: AB/CD=3/2 and AD/CB=3/4; the second ratio is the reciprocal of the supplied BC/AD. Their product is 9/8, giving AE: CE=9:8. Using 4/3 instead would invert the relevant side pairing.

Answer: B — 4/3 cm

1. Area ABC=½×6×8=24 cm². Area ABP=24/1.5=16 cm².

2. All three triangles use bases on BC and height AB=6. Thus area=3×base and BP=16/3.

3. Their areas in the stated order are in AP, so their bases BP, BQ, BC are in AP: 2BQ=BP+BC.

4. BQ=(16/3+8)/2=20/3. Therefore PQ=BQ−BP=4/3 cm, option B.

Why this method works:
Every triangle has vertex A and a base on BC, so all share the perpendicular height AB. Their areas therefore scale directly with BP, BQ, BC. An arithmetic progression of areas gives an arithmetic progression of those bases in the same order.

Independent check and common trap:
The three areas are 16,20,24 square centimetres; the bases are 16/3,20/3,24/3 centimetres. Their consecutive differences equal 4/3. In particular P and Q lie inside BC and PQ=4/3, consistent with the stated geometry.

G Strategy shortcut

Spot: Geometry and Arithmetic Progression.
Apply: Same height means BP: BQ: BC is an AP. BP=(2/3)BC=16/3; Q is halfway from P to C, so PQ=(8−16/3)/2=4/3.
Check: The three areas are 16,20,24 square centimetres; the bases are 16/3,20/3,24/3 centimetres. Their consecutive differences equal 4/3. In particular P and Q lie inside BC and PQ=4/3, consistent with the stated geometry.