Circles, Polygons and Area Mensuration Practice Questions for XAT: 147+ Solved Questions with Step-by-Step Solutions

Solve 147+ Circles, Polygons and Area Mensuration practice questions for XAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Geometry & Mensuration

Chapter Journey

Geometry & Mensuration

Circles & Curvilinear Mensuration

Core properties, tangents, touching circles, and inscribed shapes within circular boundaries.

Polygons, Area & Data Sufficiency

Regular/irregular polygons, area calculations for triangles and quadrilaterals, and data sufficiency frameworks.

By the end of this chapter, you will visualize complex geometric setups and calculate areas with absolute precision.

The Geometry of Curves

The Intuition of Curves

Polygons enclose space using straight line segments. Circles enclose space using a continuous curve.

Square

Circle

The Isoperimetric Principle

For any given perimeter, the circle encloses the maximum possible area. Curvature distributes the boundary evenly, eliminating "wasted" corners.

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

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Consider a cyclic quadrilateral $A B C D$ inscribed in a circle of radius $10$ . Determine if the area of $A B C D$ is uniquely determined.

I. Diagonal $A C$ is a diameter of the circle.

II. The length of side $A B$ is $12$ and the length of side $B C$ is $16$ .

A.
Statement I alone is sufficient
B.
Statement II alone is sufficient
C.
Both statements together are sufficient, but neither alone is sufficient
D.
Both statements together are not sufficient

Correct Answer:

Step-by-Step Solution

Key idea: This is a Data Sufficiency Synthesis combining cyclic properties with triangle rigidity. We need to check if the quadrilateral is fixed.

Step 1: Analyze Statement I.

$A C$ is a diameter. This implies $∠ B = 9 0^{\circ}$ and $∠ D = 9 0^{\circ}$ .

However, vertices $B$ and $D$ can slide along the semicircles defined by $A C$ . The shape is flexible. Area is NOT unique.

$\to$ Insufficient.

Step 2: Analyze Statement II.

$A B = 12, B C = 16$ . In a circle of radius 10 (diameter 20), note that $1 2^{2} + 1 6^{2} = 144 + 256 = 400 = 2 0^{2}$ .

This implies $△ A B C$ is a right triangle with hypotenuse $A C = 20$ .

So $A C$ MUST be a diameter.

This fixes $△ A B C$ completely. Area( $A B C$ ) = $0.5 \times 12 \times 16 = 96$ .

BUT, vertex $D$ is still free to move on the other semicircle. Area( $A D C$ ) varies.

Total Area = Area( $A B C$ ) + Area( $A D C$ ). Since Area( $A D C$ ) varies, Total Area is NOT unique.

$\to$ Insufficient.

Step 3: Combine Statements.

From II, we already deduced $A C$ is diameter. Statement I adds no new constraint.

Wait! Re-evaluate.

Does II imply AC is diameter? Yes, because chord length corresponding to right angle is diameter. $12 - 16 - 20$ is a right triple. Diameter is 20. So AC is diameter.

So combining I and II gives exactly the same information as II alone.

Vertex D is still unconstrained.

HOLD ON. Let me re-read carefully.

Is it possible the question implies a specific configuration? No, DS requires uniqueness.

If D can be anywhere on the arc, area varies. Max area when D is midpoint of arc (isosceles right triangle ADC). Min area approaches 0 as D approaches A or C.

Therefore, even combined, the area is NOT uniquely determined.

CORRECTION: The answer should be "Both statements together are not sufficient".

Let me verify standard exam traps. Often students assume "cyclic quad + 3 sides" or similar fixes it. But here we have 2 adjacent sides forming a right triangle. The other two sides are unknown.

Unless... did I miss a constraint? "Cyclic quadrilateral ABCD". Order matters.

With AB=12, BC=16 fixed, and AC=diameter fixed, points A, B, C are fixed relative to each other.

Point D lies on the major arc AC? No, AC is diameter, so D lies on one of the semi-circles.

Without CD or DA or angle D info, D is free.

WAIT. Let me check the provided solution key logic in my head.

Usually, these questions have "Together Sufficient" as answer. Did I construct a trick question where answer is E?

Or did I miss something obvious?

Maybe Statement II meant "AB=12, CD=16"? No, prompt says AB and BC.

Okay, if the answer is truly E, that's a valid Level 4 trap (assuming sufficiency where none exists).

BUT, let's look at the options. Standard DS options.

Let's reconsider Statement II.

If AB=12, BC=16, and R=10.

Chord AB subtends angle at center. Chord BC subtends angle.

Are there two possible configurations for B?

In a circle of R=10, chord 12 is unique (up to rotation). Chord 16 is unique.

Since $1 2^{2} + 1 6^{2} = 2 0^{2}$ , the angle between them MUST be 90 degrees for them to fit in a circle of dia 20?

Actually, if angle B was not 90, AC would not be 20.

Could AC be something else?

By Cosine Rule in $△ A B C$ : $A C^{2} = 1 2^{2} + 1 6^{2} - 2 (12) (16) cos B$ .

Also $A C = 2 R sin B = 20 sin B$ .

$400 sin^{2} B = 400 - 384 cos B$ .

$400 (1 - cos^{2} B) = 400 - 384 cos B$ .

$400 - 400 cos^{2} B = 400 - 384 cos B$ .

$400 cos^{2} B - 384 cos B = 0$ .

$cos B (400 cos B - 384) = 0$ .

Solutions: $cos B = 0 \Rightarrow B = 9 0^{\circ}$ OR $cos B = 384/400 = 0.96$ .

Ah! There are TWO possible triangles ABC.

Case 1: $B = 9 0^{\circ}, A C = 20$ .

Case 2: $cos B = 0.96, A C = 20 sin B = 20 (0.28) = 5.6$ .

So Statement II allows TWO different shapes for $△ A B C$ .

Even if we combine with I ( $A C$ is diameter), we force Case 1.

So Combined: We are in Case 1 ( $B = 90, A C = 20$ ).

But D is STILL FREE.

So Answer is indeed Not Sufficient.

HOWEVER, for a teaching question bank, "Not Sufficient" due to a free vertex is often considered a "trick" rather than a synthesis test unless explicitly testing flexibility.

Let's MODIFY the question slightly to make it Sufficient when combined, to teach the synthesis of fixing the shape.

MODIFICATION: Change Stmt II to "Side $A B = 12$ and Side $C D = 16$ ".

If I: AC is diameter.

If II: AB=12, CD=16.

Combined: AC is diameter $⟹ B = 90, D = 90$ .

In Rt $△ A B C$ , $A B = 12, A C = 20 ⟹ B C = 16$ .

In Rt $△ A D C$ , $C D = 16, A C = 20 ⟹ A D = 12$ .

Now ALL 4 sides are fixed ( $12, 16, 16, 12$ ).

Shape is a Rectangle. Area is uniquely $12 \times 16 = 192$ .

This makes Answer C. This is better pedagogy for L4 synthesis.

REVISED Q3 STATEMENT:

I. Diagonal AC is a diameter.

II. Side AB = 12 and Side CD = 16.

Answer: C.

Step 1: Stmt I alone. AC diameter fixes angles B and D to 90. Sides unknown. Insuff.

Step 2: Stmt II alone. Two chords. Relative position unknown. Insuff.

Step 3: Combined. AC diameter + AB=12 fixes BC=16 (Pythagoras). AC diameter + CD=16 fixes AD=12. Quad is fully determined (Rectangle 12x16). Area unique. Suff.

Answer: Both statements together are sufficient, but neither alone is sufficient.

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Is the area of a planar quadrilateral $A B C D$ uniquely determined?

I. The side lengths are $A B = 3, B C = 4, C D = 12, D A = 13$ .

II. $∠ A B C = 9 0^{\circ}$ .

A.
Statement I alone is sufficient
B.
Statement II alone is sufficient
C.
Both statements together are sufficient, but neither alone is sufficient
D.
Both statements together are not sufficient

Correct Answer:

Step-by-Step Solution

Key idea: A polygon's area is uniquely determined only if its shape is rigid. We must check if the given constraints fix the polygon or allow multiple valid configurations (like convex vs. non-convex).

Step 1: Statement I alone gives four side lengths. A quadrilateral with given sides is flexible (it can hinge), so its area is not determined.

Step 2: Statement II alone gives one angle but no side lengths, which is obviously insufficient.

Step 3: Combining both, we know the sides are $3, 4, 12, 13$ and $∠ A B C = 9 0^{\circ}$ .

Step 4: Draw diagonal $A C$ . In $△ A B C$ , since $∠ B = 9 0^{\circ}$ , $A C = 3^{2} + 4^{2} = 5$ .

Step 5: Now consider $△ A C D$ . Its sides are $A C = 5, C D = 12, D A = 13$ . Since $5^{2} + 1 2^{2} = 1 3^{2}$ , $△ A C D$ is a right triangle with $∠ A C D = 9 0^{\circ}$ .

Step 6: The quadrilateral is composed of two rigid right triangles: $△ A B C$ (area $= 6$ ) and $△ A C D$ (area $= 30$ ).

Step 7: However, these two triangles can be joined along $A C$ in two valid planar ways:

Convex configuration: The triangles are on opposite sides of $A C$ . Total area $= 6 + 30 = 36$ .
Non-convex (dart) configuration: The smaller triangle is folded inside the larger one. Total area $= 30 - 6 = 24$ .

Step 8: Since both configurations satisfy all given conditions, the area is not uniquely determined.

Answer: Both statements together are not sufficient

Question 3 · Quantitative Aptitude and Data Interpretation (QA & DI) NAT

Two circles with centers $O_{1}$ and $O_{2}$ touch externally at point $P$ . A direct common tangent touches the circles at $A$ and $B$ respectively. If the length of the tangent segment $A B$ is $12$ cm and the radius of the smaller circle is $4$ cm, find the radius of the larger circle (in cm).

Correct Answer:

Step-by-Step Solution

Key idea: This is a Touching Circles & Tangent Construction problem. The key is constructing a right triangle by translating the tangent segment to the line of centers.

Step 1: Visualize the Construction.

Let $r_{1} = 4$ and $r_{2}$ be the unknown radius.

Draw radii $O_{1} A$ and $O_{2} B$ perpendicular to tangent $A B$ .

Draw a line through $O_{1}$ parallel to $A B$ , meeting $O_{2} B$ at $C$ .

This forms a rectangle $A B C O_{1}$ and a right triangle $△ O_{1} C O_{2}$ .

Step 2: Identify Triangle Dimensions.

In $△ O_{1} C O_{2}$ :

Hypotenuse $O_{1} O_{2} = r_{1} + r_{2} = 4 + r_{2}$ (since circles touch externally).
Leg $O_{1} C = A B = 12$ (opposite sides of rectangle).
Leg $O_{2} C = ∣ r_{2} - r_{1} ∣ = r_{2} - 4$ (assuming $r_{2} > r_{1}$ ).

Step 3: Apply Pythagoras.

$(O_{1} O_{2})^{2} = (O_{1} C)^{2} + (O_{2} C)^{2}$

$(r_{2} + 4)^{2} = 1 2^{2} + (r_{2} - 4)^{2}$

Step 4: Solve for $r_{2}$ .

Expand both sides:

$r_{2}^{2} + 8 r_{2} + 16 = 144 + r_{2}^{2} - 8 r_{2} + 16$

Subtract $r_{2}^{2} + 16$ from both sides:

$8 r_{2} = 144 - 8 r_{2}$

$16 r_{2} = 144$

$r_{2} = 9 cm$

Alternative Shortcut: For external touching circles, $A B = 2 r_{1} r_{2}$ .

$12 = 2 4 \cdot r_{2} \Rightarrow 6 = 4 r_{2} \Rightarrow 36 = 4 r_{2} \Rightarrow r_{2} = 9$ .

Answer: 9

Question 4 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

A rectangular parallelepiped has a main diagonal of length $13$ units. The sum of the lengths of all its $12$ edges is $72$ units. When the volume of this parallelepiped is maximized, what is its total surface area (in square units)?

A.
145
B.
155
C.
165
D.
Cannot be determined

Correct Answer:

Step-by-Step Solution

Key idea: This is an algebraic identity problem disguised as a 3D geometry optimization problem. The surface area is uniquely determined by the given constraints, making the "maximize volume" condition a distractor.

Step 1: Let the dimensions of the parallelepiped be $a, b, c$ .

The length of the main diagonal is $D = a^{2} + b^{2} + c^{2} = 13$ .

Squaring both sides: $a^{2} + b^{2} + c^{2} = 169$ .

Step 2: The sum of all 12 edges is $4 (a + b + c) = 72$ .

Dividing by 4: $a + b + c = 18$ .

Step 3: We need the total surface area, which is $A = 2 (ab + b c + c a)$ .

We can find this using the algebraic identity:

$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + b c + c a)$ .

Step 4: Substitute the known values into the identity:

$1 8^{2} = 169 + A$ .

$324 = 169 + A$ .

$A = 324 - 169 = 155$ .

Step 5: The surface area is exactly $155$ regardless of the individual values of $a, b, c$ , as long as they satisfy the sum of squares and sum of linear terms. The condition "when the volume is maximized" is irrelevant to the surface area.

Answer: 155

Question 5 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

A cone is inscribed in a sphere of radius $R$ . When the volume of the cone is maximized, what is the ratio of the volume of the cone to the volume of the sphere?

A.
8/27
B.
4/9
C.
16/27
D.
32/81

Correct Answer:

Step-by-Step Solution

Key idea: This is a 3D optimization problem. We need to express the volume of the cone in terms of a single variable, find its maximum using calculus or AM-GM, and then compute the ratio.

Step 1: Let the height of the cone be $H$ and the base radius be $r$ . The sphere has radius $R$ .

By considering the cross-section, the relationship between $r, H,$ and $R$ is:

$r^{2} = R^{2} - (H - R)^{2} = 2 R H - H^{2}$ .

Step 2: The volume of the cone is $V = \frac{1}{3} π r^{2} H = \frac{1}{3} π (2 R H^{2} - H^{3})$ .

Step 3: To maximize $V$ , we take the derivative with respect to $H$ and set it to 0:

$\frac{d V}{d H} = \frac{1}{3} π (4 R H - 3 H^{2}) = 0$ .

Since $H \neq = 0$ , we have $4 R - 3 H = 0 ⟹ H = \frac{4 R}{3}$ .

Step 4: Substitute $H = \frac{4 R}{3}$ back into the expression for $r^{2}$ :

$r^{2} = 2 R (\frac{4 R}{3}) - (\frac{4 R}{3})^{2} = \frac{8 R ^{2}}{3} - \frac{16 R ^{2}}{9} = \frac{8 R ^{2}}{9}$ .

Step 5: Calculate the maximum volume of the cone:

$V_{ma x} = \frac{1}{3} π (\frac{8 R ^{2}}{9}) (\frac{4 R}{3}) = \frac{32}{81} π R^{3}$ .

Step 6: The volume of the sphere is $V_{s p h er e} = \frac{4}{3} π R^{3} = \frac{108}{81} π R^{3}$ .

The ratio is $\frac{V _{ma x}}{V _{s p h er e}} = \frac{32/81}{108/81} = \frac{32}{108} = \frac{8}{27}$ .

Answer: 8/27

Circles, Polygons and Area Mensuration Practice Questions for XAT: 147+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Geometry & Mensuration

Chapter Journey

Circles & Curvilinear Mensuration

Polygons, Area & Data Sufficiency

The Geometry of Curves

The Intuition of Curves

The Isoperimetric Principle

Circles, Polygons and Area Mensuration: 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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