CMAT 2026 Speed Challenge: Geometry & Mensuration | Practice Set 2
CMAT 2026: Geometry & Mensuration
Geometry & Mensuration Quick Prep
Geometric Mean (Altitude):
In a right triangle, \(Leg^2 = \text{Adjacent Segment} \times \text{Hypotenuse}\).
Cylinder Volume:
\(V = \pi r^2 h\). If \(h = r\), then \(V = \pi r^3\).
Area Comparisons:
Triangle = \(\frac{1}{2}bh\) | Rect/Parallelogram = \(bh\).
🚀 SPEED GOAL: Complete these 3 questions in 4 minutes! Focus on the “10-90-20” rule—identify the sitters first.
Target: 90 Secs
QT & DI | Question ID: 100005
In triangles ABC and ADC in the above figure, angle BAC = angle ADC = 90 degrees. Also, DC is 2 units long and AC is 4 units long. What is the length of BC?
8 units
10 units
6√3 units
12√3 units
Correct Option: 8 units
Rationale: The problem describes a right-angled triangle ABC (where angle BAC is 90 degrees) with an altitude AD drawn to the hypotenuse BC (creating a right angle at D, so angle ADC is 90 degrees).
In this configuration, we can use the geometric mean theorem (Euclid’s theorem for right triangles), which states that the square of the length of a leg of the right triangle is equal to the product of the length of the adjacent segment of the hypotenuse and the length of the entire hypotenuse.
Specifically for leg AC, the formula is:
\(AC^2 = DC \times BC\)
Substituting the given values:
\(4^2 = 2 \times BC\)
\(16 = 2 \times BC\)
Dividing by 2, we get \(BC = 8\) units.
Why other options wrong: Options 2, 3, and 4 are incorrect values derived from misapplying geometric theorems or Pythagorean calculations (e.g., calculating the altitude AD instead of the full hypotenuse BC). Difficulty: Moderate
Target: 45 Secs
QT & DI | Question ID: 100004
Consider the following figures (I, II and III) and the statements (A, B, C and D) regarding them.
A. The areas of three figures are different
B. The areas of three figures are equal
C. The perimeters of three figures are equal
D. Perimeters of figures (I) and (II) are equal
Choose the correct answer from the options given below:
A and C only
B and D only
B only
B and C only
Correct Option: 3 (B only)
Rationale: First, we calculate the area of each figure:
1. Area of Figure I (Triangle) = \(\frac{1}{2} \times 8 \times 6 = 24\) sq cm.
2. Area of Figure II (Rectangle) = \(6 \times 4 = 24\) sq cm.
3. Area of Figure III (Parallelogram) = \(6 \times 4 = 24\) sq cm.
Since all areas are 24 sq cm, Statement B is correct and Statement A is false.
Perimeters:
– Figure II (Rectangle) Perimeter = \(2(6 + 4) = 20\) cm.
– Figure I (Triangle): If right-angled, sides are 6, 8, 10; Perimeter = 24 cm. Since 24 ≠20, perimeters are not equal. Statement D and C are false.
Why other options wrong: Options 1, 2, and 4 depend on Statement C or D being true, which they are not. Difficulty: Easy
Target: 60 Secs
QT & DI | Question ID: 100006
If height and radius of right circular cylinder are equal and its volume is 176/7 cm\(^3\) then the diameter of the cylinder, is:
4 cm
8 cm
2 cm
5 cm
Correct Option: 4 cm
Rationale: Let radius = \(r\) and height = \(h\). Given \(h = r\).
Volume \(V = \pi r^2 h = \pi r^3\).
\(\frac{176}{7} = \frac{22}{7} \times r^3\)
\(176 = 22 \times r^3\)
\(r^3 = 8 \implies r = 2\) cm. Diameter = \(2r = 2 \times 2 = 4\) cm.
Why other options wrong: Option 3 is the radius (not diameter). Option 2 is \(r^3\). Difficulty: Easy
