Geometry, Mensuration and Area PYQs for GATE CS

Solve 5+ Geometry, Mensuration and Area previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Geometry, Mensuration and Area

Current Focus

Step 1: Similarity and Area Ratios

Foundation: How scaling affects lengths and areas.

Step 2: Plane Geometry and Area Computation

Core Application: Calculating areas of complex 2D shapes and overlapping regions.

Step 3: Solid Geometry, Volumes and Diagonals

3D Extension: Extending area concepts to volume and surface area of cubes, cylinders, and prisms.

Topic Hero: The Intuition of Similarity

Similarity is the mathematical way of saying two shapes are exact scaled versions of each other, like a photograph and its enlargement.

When a shape is scaled by a linear factor of $k$ :

Every length (side, height, perimeter, diagonal) scales by $k$ .
Every area scales by $k^{2}$ .

This quadratic relationship is the most powerful tool in mensuration. It allows you to find areas without knowing the exact side lengths, using only proportional reasoning.

Geometry, Mensuration and Area: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Aptitude · 2025_Set2 MCQ

In the given figure, PQRS is a square of side 2 cm and PLMN is a rectangle. The corner L of the rectangle is on the side QR. Side MN of the rectangle passes through the corner S of the square.

What is the area (in cm

^{2}

) of the rectangle PLMN?

Note: The figure shown is representative.

A.
$22$
B.
$2$
C.
$8$
D.
$4$

Correct Answer:

Step-by-Step Solution

Insight: The area of the rectangle PLMN is invariant and equals the area of the square PQRS due to a shared geometric relationship through triangle PLS.

Exam route: Area of rectangle = Area of square = $2^{2} = 4$ .

Learning route:

Consider $△ P L S$ . Its base is $P L$ , which is a side of the rectangle PLMN.
The vertex $S$ lies on the line $M N$ , which is parallel to $P L$ . Therefore, the perpendicular height of $△ P L S$ from $S$ to $P L$ is exactly the width of the rectangle, $L M$ .
Area of $△ P L S = \frac{1}{2} \times base \times height = \frac{1}{2} \times P L \times L M = \frac{1}{2} \times Area (P L M N)$ .
Now consider $△ P L S$ with respect to the square PQRS. Let the base be $P S$ (a side of the square, length 2). The vertex $L$ lies on $QR$ . The perpendicular distance from $L$ to the line containing $P S$ is exactly the side length of the square, which is 2.
Area of $△ P L S = \frac{1}{2} \times P S \times (distance from L to P S) = \frac{1}{2} \times 2 \times 2 = 2$ .
Equating the two expressions for the area of $△ P L S$ : $\frac{1}{2} \times Area (P L M N) = 2 ⟹ Area (P L M N) = 4$ .

Wrong path: Trying to assign coordinates to $L$ and $M$ and solving complex algebraic equations, or assuming the rectangle must be a square and guessing $22$ .

Question 2 · Quantitative Aptitude · 2025_Set1 MCQ

In the diagram, the lines QR and ST are parallel to each other. The shortest distance between these two lines is half the shortest distance between the point P and line QR. What is the ratio of the area of the triangle PST to the area of the trapezium SQRT?

Note: The figure shown is representative.

A.
$\frac{1}{3}$
B.
$\frac{1}{4}$
C.
$\frac{2}{5}$
D.
$\frac{1}{2}$

Correct Answer:

Step-by-Step Solution

Insight: Parallel lines inside a triangle create similar triangles. The ratio of their areas is the square of the ratio of their corresponding heights.

Exam route: Height ratio of $△ P S T$ to $△ P QR$ is $1 : 2$ . Area ratio is $1 : 4$ . Trapezium area is $4 - 1 = 3$ . Ratio is $1/3$ .

Learning route:

Let the shortest distance from $P$ to $QR$ be $H$ .
The problem states the distance between the parallel lines $S T$ and $QR$ is half of this, so it is $H /2$ .
Since $S T$ is parallel to $QR$ and lies between $P$ and $QR$ , the distance from $P$ to $S T$ is $H - H /2 = H /2$ .
$△ P S T$ and $△ P QR$ are similar by the AA test (they share $∠ P$ , and $∠ P S T = ∠ P QR$ due to parallel lines).
The ratio of their corresponding heights is $\frac{H /2}{H} = \frac{1}{2}$ .
The ratio of their areas is the square of the height ratio: $(\frac{1}{2})^{2} = \frac{1}{4}$ .
Let $Area (△ P S T) = 1 x$ and $Area (△ P QR) = 4 x$ .
The trapezium $S QR T$ is the region $△ P QR$ minus $△ P S T$ , so its area is $4 x - 1 x = 3 x$ .
The required ratio is $\frac{Area ( △ P S T )}{Area ( trapezium S QR T )} = \frac{1 x}{3 x} = \frac{1}{3}$ .

Wrong path: Forgetting to square the height ratio, leading to an area ratio of $1 : 2$ . This would make the trapezium area $2 x - 1 x = 1 x$ , giving a ratio of $1/1$ , or mistakenly using the large triangle's area in the denominator to get $1/4$ (Option B).

Question 3 · Quantitative Aptitude · 2022 MCQ

A function $y (x)$ is defined in the interval $[0, 1]$ on the $x$ -axis as

$y (x) = ⎩ ⎨ ⎧ 231 if 0 \leq x < \frac{1}{3} if \frac{1}{3} \leq x < \frac{3}{4} if \frac{3}{4} \leq x \leq 1$

Which one of the following is the area under the curve for the interval $[0, 1]$ on the $x$ -axis?

A.
$\frac{5}{6}$
B.
$\frac{6}{5}$
C.
$\frac{13}{6}$
D.
$\frac{6}{13}$

Correct Answer:

Step-by-Step Solution

Insight: The area under a piecewise constant (step) function is simply the sum of the areas of the rectangular blocks it forms.

Exam route: Area = $2 (1/3) + 3 (3/4 - 1/3) + 1 (1 - 3/4) = 2/3 + 5/4 + 1/4 = 2/3 + 6/4 = 13/6$ .

Learning route:

The function is constant on three sub-intervals. The area under the curve is the sum of the areas of three rectangles.
Rectangle 1: Interval $[0, 1/3]$ . Width = $1/3 - 0 = 1/3$ . Height = 2. Area = $2 \times 1/3 = 2/3$ .
Rectangle 2: Interval $[1/3, 3/4]$ . Width = $3/4 - 1/3 = 9/12 - 4/12 = 5/12$ . Height = 3. Area = $3 \times 5/12 = 15/12 = 5/4$ .
Rectangle 3: Interval $[3/4, 1]$ . Width = $1 - 3/4 = 1/4$ . Height = 1. Area = $1 \times 1/4 = 1/4$ .
Total Area = $2/3 + 5/4 + 1/4$ .
Combine the fractions with denominator 4: $5/4 + 1/4 = 6/4 = 3/2$ .
Add to the first term: $2/3 + 3/2$ . The common denominator is 6.
Total Area = $4/6 + 9/6 = 13/6$ .

Wrong path: Miscalculating the width of the second interval (e.g., $3/4 - 1/3 = 2/1$ due to bad fraction subtraction), or adding the heights instead of multiplying by width, or confusing the interval bounds. Option A (5/6) might come from missing the middle term entirely.

Question 4 · Quantitative Aptitude · 2021_Set2 MCQ

If $θ$ is the angle, in degrees, between the longest diagonal of the cube and any one of the edges of the cube, then, $cos θ =$

A.
$\frac{1}{2}$
B.
$\frac{1}{3}$
C.
$\frac{1}{2}$
D.
$\frac{3}{2}$

Correct Answer:

Step-by-Step Solution

Insight: The angle between a cube's body diagonal and any of its edges is a constant, independent of the cube's size.

Exam route: Recall the standard formula for a cube: $cos θ = \frac{1}{3}$ .

Learning route:

Let the cube have edge length $a$ .
The body diagonal stretches from one corner to the opposite corner through the interior. Its length is $d = a^{2} + a^{2} + a^{2} = a 3$ .
The angle $θ$ between the body diagonal and an edge forms a right triangle where the edge is the adjacent side (length $a$ ) and the body diagonal is the hypotenuse (length $a 3$ ).
Therefore, $cos θ = \frac{adjacent}{hypotenuse} = \frac{a}{a 3} = \frac{1}{3}$ .

Wrong path: Confusing the body diagonal with a face diagonal. A face diagonal has length $a 2$ , which would give $cos θ = \frac{1}{2}$ (Option C). This is incorrect because the question specifies the "longest diagonal".

Question 5 · Quantitative Aptitude · 2021_Set1 MCQ

We have 2 rectangular sheets of paper, M and N, of dimensions 6 cm x 1 cm each. Sheet M is rolled to form an open cylinder by bringing the short edges of the sheet together. Sheet N is cut into equal square patches and assembled to form the largest possible closed cube. Assuming the ends of the cylinder are closed, the ratio of the volume of the cylinder to that of the cube is __________

A.
$\frac{π}{2}$
B.
$\frac{3}{π}$
C.
$\frac{9}{π}$
D.
$3 π$

Correct Answer:

Step-by-Step Solution

Insight: Rolling a sheet along its short edge makes the long edge the circumference. A closed cube requires exactly 6 square faces.

Exam route: Cylinder circumference = 6, height = 1 $⟹ r = 3/ π$ , Volume = $9/ π$ . Cube has 6 faces of 1x1 $⟹$ Volume = 1. Ratio = $9/ π$ .

Learning route:

Sheet M (6 cm x 1 cm) is rolled by bringing the short edges together. This means the seam is 1 cm long, so the cylinder height $h = 1$ cm.
The circumference of the cylinder is the long edge, $C = 6$ cm.
From $C = 2 π r = 6$ , we get the radius $r = \frac{6}{2 π} = \frac{3}{π}$ cm.
Volume of the cylinder $V_{cyl} = π r^{2} h = π (\frac{3}{π})^{2} (1) = π (\frac{9}{π ^{2}}) = \frac{9}{π}$ cm $^{3}$ .
Sheet N (6 cm x 1 cm) is cut into equal square patches to form the largest possible closed cube. A closed cube has exactly 6 faces.
The only way to cut a 6x1 sheet into 6 equal squares is to make six 1x1 cm squares.
These 6 squares assemble into a cube of edge length $a = 1$ cm.
Volume of the cube $V_{cube} = a^{3} = 1^{3} = 1$ cm $^{3}$ .
The ratio of the volume of the cylinder to that of the cube is $\frac{9}{π} : 1 = \frac{9}{π}$ .

Wrong path: Rolling along the long edge, making circumference = 1 and height = 6. This gives $r = 1/ (2 π)$ and Volume = $6/ (4 π) = 3/ (2 π)$ , which is not an option. Or assuming an open cube (5 faces), which doesn't divide 6 evenly into squares.

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Geometry, Mensuration and Area PYQs for GATE CS

Chapter Roadmap: Geometry, Mensuration and Area

Topic Hero: The Intuition of Similarity

Geometry, Mensuration and Area: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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