CAT 2026 Slot 2 Quant: Geo Mensu Questions with Solutions

Answer: 3

1. The triangle is right-angled because one side is horizontal and another is vertical.
2. The two perpendicular sides are:
3. 15 − 3 = 12
13 − 4 = 9
4. Using the 9-12-15 Pythagorean triplet, the hypotenuse is 15.
5. For a right-angled triangle, the inradius is:
6. (radius) = (sum of the two perpendicular sides − hypotenuse) ÷ 2
7. So,
8. r = (12 + 9 − 15) ÷ 2
r = 6 ÷ 2
r = 3
9. Answer: 3 (TITA).

Why this method works:
The vertices with common y-coordinate give the horizontal leg, and those with common x-coordinate give the vertical leg. Their meeting point is the right angle. Translation of the triangle does not change its side lengths or inradius.

Independent check and common trap:
An independent area check gives area=12×9/2=54 and semiperimeter=(12+9+15)/2=18. Since area=inradius×semiperimeter, r=54/18=3. This verifies the right-triangle shortcut.

G Strategy shortcut

Spot: Geometry – Coordinate Geometry.
Apply: Read the legs 9 and 12; hypotenuse 15. Inradius=(9+12−15)/2=3.
Check: An independent area check gives area=12×9/2=54 and semiperimeter=(12+9+15)/2=18. Since area=inradius×semiperimeter, r=54/18=3. This verifies the right-triangle shortcut.

Answer: 18

1. Since AB is parallel to CD, triangles EAB and EDC are similar.
2. Their corresponding bases are 3 cm and 2 cm. So their perimeters are also in the ratio 3 : 2.
3. The two slant sides of the trapezium together measure:
4. 10 − 3 − 2 = 5 cm.
5. Let the perimeter of the larger triangle EAB be P. The smaller triangle's perimeter is 2P/3.
6. The difference between these two perimeters is 5 + (3 − 2) = 6 cm.
7. So,
8. P − 2P/3 = 6
P/3 = 6
P = 18 cm.
9. Answer: 18 (TITA).

Why this method works:
Parallel bases create similar triangles with vertex E. Because AB is the longer base, triangle EAB is larger and ECD is smaller. The trapezium’s slant sides are the differences between the corresponding triangle sides.

Independent check and common trap:
AD+BC=5 and the larger triangle’s two slant sides are three times that difference, totaling 15. Add AB=3 to obtain perimeter 18. This is a second route to the same result without solving for either slant side separately.

G Strategy shortcut

Spot: Geometry – Trapezium.
Apply: Scale ratio 3: 2. Slant-side difference=10−3−2=5; base difference 1. Triangle perimeter difference 6 is one part →larger perimeter 3×6=18.
Check: AD+BC=5 and the larger triangle’s two slant sides are three times that difference, totaling 15. Add AB=3 to obtain perimeter 18. This is a second route to the same result without solving for either slant side separately.

Answer: C — (7 + 4√3) : 1

1. The centres of the three equal circles form an equilateral triangle. If each circle has radius r, the side of this triangle is 2r.
2. The distance from the centre of the equilateral triangle to any vertex is 2r/√3.
3. For the larger circle X, its radius is:
4. R = r + 2r/√3
5. For the smaller circle Y, its radius is:
6. S = 2r/√3 − r
7. Therefore,
8. R/S = (1 + 2/√3)/(2/√3 − 1)
9. Multiplying the numerator and denominator by (2/√3 + 1), we get:
10. R/S = (1 + 2/√3)² / (4/3 − 1)
11. Simplifying gives:
12. R/S = 7 + 4√3
13. Answer: (7 + 4√3) : 1.

Why this method works:
The common centre of the outer and inner circles is the centre of the equilateral triangle of the original centres. Its distance d to each original centre is 2r/√3. Internal tangency gives R−r=d; external tangency gives s+r=d.

Independent check and common trap:
Therefore R/s=(2+√3)/(2−√3). Multiply top and bottom by 2+√3: denominator=4−3=1 and numerator=4+4√3+3. The complete ratio is (7+4√3):1, so parentheses apply to the entire first term.

G Strategy shortcut

Spot: Geometry – Circles.
Apply: Centres form an equilateral triangle of side 2r. Centre-to-vertex distance d=2r/√3. Outer/inner radii ratio=(d+r)/(d−r)=7+4√3.
Check: Therefore R/s=(2+√3)/(2−√3). Multiply top and bottom by 2+√3: denominator=4−3=1 and numerator=4+4√3+3. The complete ratio is (7+4√3):1, so parentheses apply to the entire first term.