Quadrilaterals, Polygons and Area Ratios Practice Questions for CAT: 155+ Solved Questions with Step-by-Step Solutions

Solve 155+ Quadrilaterals, Polygons and Area Ratios practice 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 rhombus has integer side length and integer diagonals. Its area is numerically equal to 24 times its side length. What is the perimeter of this rhombus?

Correct Answer:

100

Step-by-Step Solution

Key idea: This combines rhombus metric relations with Diophantine constraints. The condition "Area = 24 × Side" links the product of diagonals to their Pythagorean combination.

Step 1: Set up equations.

Let side be $s$ . Area $A = 24 s$ .

Also $A = \frac{1}{2} d_{1} d_{2} \Rightarrow d_{1} d_{2} = 48 s$ .

Rhombus diagonal identity: $d_{1}^{2} + d_{2}^{2} = 4 s^{2}$ .

Step 2: Form equation for sum of diagonals.

$(d_{1} + d_{2})^{2} = d_{1}^{2} + d_{2}^{2} + 2 d_{1} d_{2} = 4 s^{2} + 2 (48 s) = 4 s^{2} + 96 s$ .

For $d_{1} + d_{2}$ to be integer, $4 s^{2} + 96 s$ must be a perfect square.

Let $4 (s^{2} + 24 s) = k^{2} \Rightarrow s^{2} + 24 s = m^{2}$ .

Step 3: Solve Diophantine equation.

Complete the square: $(s + 12)^{2} - 144 = m^{2}$ .

$(s + 12)^{2} - m^{2} = 144 \Rightarrow (s + 12 - m) (s + 12 + m) = 144$ .

Let factors be $u, v$ with $uv = 144, v > u$ . Since sum $(s + 12 - m) + (s + 12 + m) = 2 (s + 12)$ is even, $u, v$ must have same parity. Both must be even.

Even factor pairs of 144: $(2, 72), (4, 36), (6, 24), (8, 18), (12, 12)$ .

Step 4: Test candidates for valid geometry.

Pair (2,72): $2 (s + 12) = 74 \Rightarrow s = 25$ .
Pair (4,36): $2 (s + 12) = 40 \Rightarrow s = 8$ .
Pair (6,24): $2 (s + 12) = 30 \Rightarrow s = 3$ .
Pair (8,18): $2 (s + 12) = 26 \Rightarrow s = 1$ .

Check existence: For real diagonals, we need $d_{1}^{2} + d_{2}^{2} \geq 2 d_{1} d_{2} \Rightarrow 4 s^{2} \geq 96 s \Rightarrow s \geq 24$ .

Only $s = 25$ satisfies this. ( $s = 8, 3, 1$ yield imaginary diagonals).

Step 5: Calculate perimeter.

Valid side $s = 25$ . Perimeter = $4 s = 100$ .

(Verification: If $s = 25$ , $d_{1} d_{2} = 1200, d_{1}^{2} + d_{2}^{2} = 2500$ . Diagonals are 30 and 40. Integers. Valid.)

Answer: 100

Question 2 · Quantitative Ability MCQ

A trapezium $A B C D$ has $A B ∥ C D$ . A circle is inscribed in the trapezium, tangent to all four sides. If $A B = 18$ cm, $C D = 8$ cm, and the area of the trapezium is $156$ sq. cm, what is the length, in cm, of each non-parallel side?

A.
13
B.
15
C.
17
D.
20

Correct Answer:

Step-by-Step Solution

Key idea: This uses the Tangential Trapezium Property combined with the Right Triangle Construction for non-parallel sides. Crucially, it requires verifying that the implied isosceles configuration is geometrically consistent.

Step 1: Use Tangential Quadrilateral Property (Pitot Theorem).

For any quadrilateral with an inscribed circle, sums of opposite sides are equal.

$A B + C D = A D + B C$ .

Given $A B = 18, C D = 8 \Rightarrow A D + B C = 26$ .

Step 2: Determine Height.

Area = $\frac{1}{2} (A B + C D) \times h = 156$ .

$\frac{1}{2} (26) h = 156 \Rightarrow 13 h = 156 \Rightarrow h = 12$ .

Step 3: Analyze Non-Parallel Sides.

The question asks for "the length of EACH non-parallel side", implying $A D = B C$ .

If $A D = B C$ , then $2 A D = 26 \Rightarrow A D = 13$ .

Step 4: Verify Consistency (Crucial Step).

In an isosceles trapezium, the projection of the slant side onto the base is $\frac{A B - C D}{2} = \frac{18 - 8}{2} = 5$ .

By Pythagoras, required height for leg 13 and projection 5 is $1 3^{2} - 5^{2} = 169 - 25 = 144 = 12$ .

This matches the calculated height $h = 12$ exactly.

Thus, the isosceles configuration is valid and the side length is indeed 13.

Answer: A

Question 3 · Quantitative Ability MCQ

In trapezium $A B C D$ , sides $A B$ and $C D$ are parallel with $A B = 12$ cm and $C D = 18$ cm. The non-parallel sides $A D$ and $B C$ are extended to meet at point $P$ . If the area of $△ P A B$ is $72$ sq. cm, what is the area of the trapezium $A B C D$ , in sq. cm?

A.
90
B.
162
C.
234
D.
306

Correct Answer:

Step-by-Step Solution

Key idea: This is a Trapezium Extension Similarity question. When non-parallel sides of a trapezium meet, they form two similar triangles sharing a common vertex. The ratio of their areas equals the square of the ratio of their parallel bases.

Step 1: Identify the similar triangles.

Since $A B ∥ C D$ , $△ P A B \sim △ P D C$ . The correspondence is $A B \leftrightarrow C D$ .

Step 2: Determine the similarity ratio $k$ .

$k = \frac{C D}{A B} = \frac{18}{12} = \frac{3}{2}$

Step 3: Relate the areas using $k^{2}$ .

$\frac{Area ( △ P D C )}{Area ( △ P A B )} = k^{2} = (\frac{3}{2})^{2} = \frac{9}{4}$

Given $Area (△ P A B) = 72$ :

$Area (△ P D C) = 72 \times \frac{9}{4} = 18 \times 9 = 162 sq. cm$

Step 4: Calculate trapezium area by subtraction.

The trapezium $A B C D$ is the region between the two triangles.

$Area (A B C D) = Area (△ P D C) - Area (△ P A B)$

$Area (A B C D) = 162 - 72 = 90 sq. cm$

Wait — re-evaluating calculation in Step 3/4.

Area(PDC) = 162. Area(PAB) = 72.

Trapezium = 162 - 72 = 90.

Let me re-read the options. Option A is 90. Option B is 162.

My manual calc says 90. Let me double check the "added layer".

Ah, usually questions ask for the larger triangle or the trapezium.

If Area(PAB)=72, ratio=1.5, Area(PDC)=162. Trap is answering 162 (area of large triangle) instead of trapezium.

Correct answer for Trapezium is indeed 90.

Self-Correction during drafting: I must ensure the answer key matches the calculation.

Calculation: $72 \times (2.25 - 1) = 72 \times 1.25 = 90$ .

Therefore, correct option is A. I will adjust the answer key below to A.

Answer: The area of trapezium $A B C D$ is 90 sq. cm.

Question 4 · Quantitative Ability MCQ

In parallelogram $A B C D$ , point $E$ is the midpoint of side $C D$ . Line segment $A E$ intersects diagonal $B D$ at point $F$ . What fraction of the area of parallelogram $A B C D$ is the area of $△ D E F$ ?

A.
1/12
B.
1/8
C.
1/6
D.
1/4

Correct Answer:

Step-by-Step Solution

Key idea: This is a midpoint-intersection area ratio question. The trigger is a line connecting a vertex to a midpoint of the opposite side intersecting a diagonal. This creates similar triangles and fixed area ratios independent of the parallelogram's specific dimensions.

Step 1: Identify similar triangles.

Since $A B ∥ C D$ , we have $A B ∥ D E$ . Thus, $△ A B F \sim △ E D F$ .

The ratio of similarity is determined by the parallel segments $A B$ and $D E$ .

Since $E$ is the midpoint of $C D$ and $A B = C D$ , we have $D E = \frac{1}{2} A B$ .

Therefore, the similarity ratio $k = \frac{D E}{A B} = \frac{1}{2}$ .

Step 2: Relate heights and bases.

Let $H$ be the height of the parallelogram (distance between $A B$ and $C D$ ).

The intersection $F$ divides the vertical distance in the ratio $1 : 2$ (same as similarity ratio).

So, the height of $△ D E F$ with respect to base $D E$ is $h_{D E F} = \frac{1}{1 + 2} H = \frac{1}{3} H$ .

Step 3: Calculate area of $△ D E F$ .

Base $D E = \frac{1}{2} C D = \frac{1}{2} A B$ .

Height $h_{D E F} = \frac{1}{3} H$ .

$Area (D E F) = \frac{1}{2} \times base \times height = \frac{1}{2} \times (\frac{1}{2} A B) \times (\frac{1}{3} H) = \frac{1}{12} (A B \times H)$

Step 4: Compare to parallelogram area.

$Area (A B C D) = A B \times H$ .

Ratio = $\frac{\frac{1}{12} ( A B \times H )}{A B \times H} = \frac{1}{12}$ .

Answer: 1/12.

Question 5 · Quantitative Ability NAT

In parallelogram $A B C D$ , point $E$ lies on side $B C$ such that $B E : E C = 1 : 2$ . Point $F$ lies on side $C D$ such that $C F : F D = 1 : 3$ . If the area of $△ A E F$ is $39$ sq. cm, what is the area, in sq. cm, of parallelogram $A B C D$ ?

Correct Answer:

104

Step-by-Step Solution

Key idea: This is an area subtraction problem inside a parallelogram. Instead of finding $△ A E F$ directly, we find the areas of the three surrounding corner triangles ( $△ A B E$ , $△ E C F$ , $△ F D A$ ) as fractions of the total parallelogram area, then subtract from 1.

Step 1: Let Area( $A B C D$ ) = $S$ .

Step 2: Express corner triangle areas in terms of $S$ .

$△ A B E$ : Base $B E = \frac{1}{3} B C$ . Height relative to base $B C$ is same as parallelogram height. Area( $△ A B C$ ) = $S /2$ . Area( $△ A B E$ ) = $\frac{B E}{B C} \times Area (△ A B C) = \frac{1}{3} \times \frac{S}{2} = \frac{S}{6}$ .
$△ F D A$ : Base $D F = \frac{3}{4} C D$ . Area( $△ A D C$ ) = $S /2$ . Area( $△ F D A$ ) = $\frac{D F}{C D} \times Area (△ A D C) = \frac{3}{4} \times \frac{S}{2} = \frac{3 S}{8}$ .
$△ E C F$ : Use the corner product rule. Area( $△ B C D$ ) = $S /2$ . $C E = \frac{2}{3} B C$ , $C F = \frac{1}{4} C D$ . Area( $△ E C F$ ) = $\frac{C E}{B C} \times \frac{C F}{C D} \times Area (△ B C D) = \frac{2}{3} \times \frac{1}{4} \times \frac{S}{2} = \frac{S}{12}$ .

Step 3: Sum of excluded areas = $\frac{S}{6} + \frac{3 S}{8} + \frac{S}{12}$ .

Common denominator is 24: $\frac{4 S}{24} + \frac{9 S}{24} + \frac{2 S}{24} = \frac{15 S}{24} = \frac{5 S}{8}$ .

Step 4: Area( $△ A E F$ ) = $S - \frac{5 S}{8} = \frac{3 S}{8}$ .

Given $\frac{3 S}{8} = 39 \Rightarrow S = \frac{39 \times 8}{3} = 13 \times 8 = 104$ .

Answer: 104

Quadrilaterals, Polygons and Area Ratios Practice Questions for CAT: 155+ 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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