Quadrilaterals, Polygons and Area Ratios Previous Year Questions (PYQs) for CAT: 11+ Solved Questions with Step-by-Step Solutions

Solve 11+ Quadrilaterals, Polygons and Area Ratios previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Quadrilaterals, Polygons and Area Ratios

CAT QA Geometry

Quadrilaterals, Polygons and Area Ratios

A staircase from basic four-sided area to polygon-level pattern recognition.

chapter PYQs

①

📐 t1 — Trapeziums, Parallelograms and Rectangles

You master height-based area, parallel-side logic, trapezium similarity, rectangle slicing, and incircle traps in trapeziums.

6 own-course PYQs High-moderate weight Selected topic

②

◆ t2 — Rhombus and Diagonal Geometry

You learn how perpendicular diagonals, side length, and area interact.

3 own-course PYQs

③

⬡ t3 — Regular Polygons and Hexagons

You move from four-sided figures to symmetry, interior angles, diagonals, and standard polygon areas.

6 own-course PYQs

④

🧩 t4 — Area Ratios and Quadrilateral Constraints

You finish with boundary conditions, side constraints, and ratio-based reasoning.

1 own-course PYQ

End goal: identify the figure type, choose the right height or diagonal relation, and avoid wasting time on unnecessary construction.

Topic Hero: Trapeziums, Parallelograms and Rectangles

Geometry → Quadrilaterals → t1

Trapeziums, Parallelograms and Rectangles

One-line hook: when sides are parallel, area starts obeying simple height rules.

What you'll learn here

Area formulas for trapeziums, parallelograms, and rectangles.
How common height creates area ratios.
How extended sides of a trapezium create similar triangles.
How rectangle slicing produces arithmetic or geometric area patterns.
How incircle conditions work in a trapezium.

Quadrilaterals, Polygons and Area Ratios: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability NAT

A trapezium ABCD has side AD parallel to BC, $∠ B A D = 9 0^{\circ}$ , BC = 3 cm and AD = 8 cm. If the perimeter of this trapezium is 36 cm, then its area, in sq. cm, is

Correct Answer:

Step-by-Step Solution

Identify the shape as a right trapezium. Since $A D ∥ B C$ and $∠ B A D = 9 0^{\circ}$ , $A B$ is perpendicular to both parallel sides, making $A B$ the height ( $h$ ) of the trapezium. Drop a perpendicular from $C$ to $A D$ , meeting it at $E$ . Then $A B C E$ is a rectangle, so $C E = h$ and $A E = B C = 3$ . The remaining segment $D E = A D - A E = 8 - 3 = 5$ . In right $△ C E D$ , the hypotenuse $C D = C E^{2} + D E^{2} = h^{2} + 25$ . The perimeter is given as 36, so $A B + B C + C D + D A = h + 3 + h^{2} + 25 + 8 = 36$ . This simplifies to $h^{2} + 25 = 25 - h$ . Squaring both sides gives $h^{2} + 25 = 625 - 50 h + h^{2}$ , leading to $50 h = 600$ , so $h = 12$ . The area of the trapezium is $\frac{1}{2} \times (B C + A D) \times h = \frac{1}{2} \times (3 + 8) \times 12 = 66$ .

Question 2 · Quantitative Ability NAT

Regular polygons A and B have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then the number of sides of B equals

Correct Answer:

Step-by-Step Solution

This is an AP term-equation question. Recognise it because the problem gives sums of specific, evenly-spaced terms (4th, 7th, 10th) and a sum of the first n terms, and asks for another sum — so everything must be converted into the two AP unknowns: first term $a$ and common difference $d$ .

Step 1: Write the given terms using the nth term formula $T_{n} = a + (n - 1) d$ .

$T_{4} = a + 3 d, T_{7} = a + 6 d, T_{10} = a + 9 d$

Step 2: Use the first condition.

$T_{4} + T_{7} + T_{10} = 3 a + 18 d = 99$

Dividing by 3: $a + 6 d = 33$ ... (i)

Notice $a + 6 d$ is exactly $T_{7}$ , so this tells us $T_{7} = 33$ . This works because 4, 7, 10 are equally spaced (gap of 3 each), so their sum is always 3 times the middle term.

Step 3: Use the second condition.

$S_{14} = \frac{14}{2} (2 a + 13 d) = 7 (2 a + 13 d) = 497$

So $2 a + 13 d = 71$ ... (ii)

Step 4: Solve (i) and (ii) together.

From (i): $a = 33 - 6 d$ . Substitute into (ii):

$2 (33 - 6 d) + 13 d = 71$

$66 - 12 d + 13 d = 71$

$66 + d = 71 \Rightarrow d = 5$

So $a = 33 - 30 = 3$ .

Step 5: Find the required sum.

$S_{5} = \frac{5}{2} (2 a + 4 d) = \frac{5}{2} (6 + 20) = \frac{5}{2} \times 26 = 65$

Trap check: a student might try to guess $a$ and $d$ instead of forming two clean linear equations — this is unreliable. Always convert "sum of a few terms" and "sum of first n terms" into linear equations in $a$ and $d$ , then solve simultaneously.

Final answer: 65

Question 3 · Quantitative Ability MCQ

ABCD is a trapezium in which AB is parallel to DC, AD is perpendicular to AB, and AB = 3DC. If a circle inscribed in the trapezium touching all the sides has a radius of 3 cm, then the area, in sq. cm, of the trapezium is

A.
48
B.
$303$
C.
$362$
D.
54

Correct Answer:

Step-by-Step Solution

This is an incircle-in-a-right-trapezium question, recognisable because the right angle at A (AD perpendicular to AB) combines with a circle that touches all four sides.

Step 1: Since AB is parallel to DC and AD is perpendicular to AB, the leg AD is exactly the distance between the parallel sides. The inscribed circle touches both parallel sides, so that distance equals the diameter: AD = 2r = 6 cm.

Step 2 (tangential quadrilateral condition): a quadrilateral has an inscribed circle exactly when opposite side sums are equal, so AB + DC = AD + BC.

Step 3: Let DC = x, so AB = 3x. Then 3x + x = 6 + BC, which gives BC = 4x - 6.

Step 4: Drop a perpendicular from C to AB, meeting it at E. Then CE = AD = 6 and EB = AB - DC = 3x - x = 2x. Pythagoras in triangle CEB gives BC^2 = CE^2 + EB^2, so (4x - 6)^2 = 6^2 + (2x)^2.

Step 5: Expand: 16x^2 - 48x + 36 = 36 + 4x^2, so 12x^2 = 48x, hence x = 4 (x is not 0). Therefore DC = 4 and AB = 12.

Step 6: Area = (1/2)(AB + DC) x AD = (1/2)(12 + 4)(6) = 48.

The area is 48 sq. cm, which is option A.

Common wrong path: using the radius 3 as the height (halving the true height), or skipping the opposite-side-sum condition and assuming BC equals AD; both break the slant-side equation.

Question 4 · Quantitative Ability MCQ

If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is

A.
$37 + 13$
B.
$13 + 12$
C.
$\frac{37 + 13}{2}$
D.
$\frac{13 + 12}{2}$

Correct Answer:

Step-by-Step Solution

This is a rhombus recover-the-diagonals question, recognisable because the area gives the product of the diagonals while the side gives the sum of their squares.

Step 1: Area of a rhombus is half the product of the diagonals: (1/2) d1 d2 = 12, so d1 d2 = 24.

Step 2: The diagonals bisect each other at 90 degrees, so the half-diagonals and one side form a right triangle: (d1/2)^2 + (d2/2)^2 = 5^2. Multiplying by 4 gives d1^2 + d2^2 = 100.

Step 3: Build the sum and difference. (d1 + d2)^2 = 100 + 2(24) = 148, so d1 + d2 = 2 sqrt(37). (d1 - d2)^2 = 100 - 2(24) = 52, so d1 - d2 = 2 sqrt(13), taking d1 as the longer diagonal.

Step 4: Add the two equations: 2 d1 = 2 sqrt(37) + 2 sqrt(13), so d1 = sqrt(37) + sqrt(13).

The longer diagonal is sqrt(37) + sqrt(13) cm, which is option A.

Common trap: writing d1^2 + d2^2 = 25 by forgetting the halves in the right triangle, or using d1 d2 = 12 by forgetting the half in the area formula; either error collapses the whole chain.

Question 5 · Quantitative Ability MCQ

If the area of a regular hexagon is equal to the area of an equilateral triangle of side 12 cm, then the length, in cm, of each side of the hexagon is

A.
$46$
B.
$66$
C.
$6$
D.
$26$

Correct Answer:

Step-by-Step Solution

This is an area-equation question comparing a regular hexagon to an equilateral triangle. The trigger is that two different shapes are said to have equal area, and we must find a missing side length — the method is to write both areas using standard formulas and equate them.

Step 1: Recall the area formula for an equilateral triangle of side $a$ :

$Area = \frac{3}{4} a^{2}$

With $a = 12$ :

$Area = \frac{3}{4} (12)^{2} = \frac{3}{4} \times 144 = 363$

Step 2: Recall the area formula for a regular hexagon of side $s$ (it splits into 6 equilateral triangles of side $s$ ):

$Area = \frac{3 3}{2} s^{2}$

Step 3: Set the two areas equal, since the question says they are equal:

$\frac{3 3}{2} s^{2} = 363$

Step 4: Solve for $s^{2}$ . Divide both sides by $3$ first:

$\frac{3}{2} s^{2} = 36$

$s^{2} = 36 \times \frac{2}{3} = 24$

Step 5: Take the square root:

$s = 24 = 4 \times 6 = 26$

Answer: each side of the hexagon is $26$ cm, option D.

Common trap: mixing up the hexagon area formula with the triangle area formula, or forgetting to cancel $3$ from both sides before solving for $s^{2}$ , which leads to a wrong constant under the square root.

Quadrilaterals, Polygons and Area Ratios Previous Year Questions (PYQs) for CAT: 11+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Quadrilaterals, Polygons and Area Ratios

Quadrilaterals, Polygons and Area Ratios

📐 t1 — Trapeziums, Parallelograms and Rectangles

◆ t2 — Rhombus and Diagonal Geometry

⬡ t3 — Regular Polygons and Hexagons

🧩 t4 — Area Ratios and Quadrilateral Constraints

Topic Hero: Trapeziums, Parallelograms and Rectangles

Trapeziums, Parallelograms and Rectangles

What you'll learn here

Quadrilaterals, Polygons and Area Ratios: 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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