Mensuration, 3D and Optimization Practice Questions for CAT: 47+ Solved Questions with Step-by-Step Solutions

Solve 47+ Mensuration, 3D and Optimization practice questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Mensuration, 3D and Optimization

CAT QA Geometry

Mensuration, 3D and Optimization

A short chapter where formulas become fast decision tools.

chapter PYQs

①

📦 t1 — Three-Dimensional Mensuration

You master cuboids, spheres, surface area, volume, space diagonal, and 3D enclosure logic.

1 own-course PYQ Selected topic Formula + algebra

②

📐 t2 — Area Optimization in Plane Figures

You learn how geometric restrictions force a maximum area.

1 own-course PYQ

③

💰 t3 — Perimeter and Cost Optimization

You convert side lengths and area constraints into minimum-cost decisions.

1 own-course PYQ

End goal: identify whether the question is asking for a 3D measure, a maximum area, or a minimum cost, then apply the shortest formula path.

Topic Hero: Three-Dimensional Mensuration

Geometry → Mensuration → t1

Three-Dimensional Mensuration

One-line hook: in 3D, the hidden diagonal often controls the whole solid.

What you'll learn here

Cuboid surface area, volume, and total edge length.
Sphere radius, surface area, and volume basics.
Space diagonal of a rectangular box.
Why an inscribed box has diagonal equal to sphere diameter.
How CAT combines formulas using algebraic identities.

Mensuration, 3D and Optimization: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability MCQ

A closed rectangular box is inscribed in a sphere, so that all eight vertices of the box lie on the sphere. The three distinct face-diagonal lengths of the box are $13$ cm, $35$ cm and $210$ cm. The diameter of the sphere, in cm, is

A.
$\frac{7}{2}$
B.
$7$
C.
$98$
D.
$14$

Correct Answer:

Step-by-Step Solution

Key idea: this is a box-inscribed-in-a-sphere question. The space diagonal of the box is the diameter of the sphere. The twist is that the question gives the three face diagonals, not the edges.

Step 1: Let the box sides be $a, b, c$ . The three distinct face-diagonal squares are:

$a^{2} + b^{2}, b^{2} + c^{2}, c^{2} + a^{2} .$

Step 2: Square the given face diagonals and add them:

$(13)^{2} + (35)^{2} + (210)^{2} = 13 + 45 + 40 = 98.$

Step 3: Relate this sum to the space diagonal $d$ :

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

So $2 d^{2} = 98$ , giving $d^{2} = 49$ and $d = 7$ .

Step 4: Since the box is inscribed in the sphere, the space diagonal is the sphere's diameter.

Answer: $7$ cm.

Common trap: using $98$ directly as $d^{2}$ forgets that each of $a^{2}, b^{2}, c^{2}$ appears twice in the sum of face-diagonal squares. Another trap is converting the space diagonal to radius instead of diameter.

Question 2 · Quantitative Ability NAT

A closed rectangular box has its length, breadth and height in the ratio $1 : 2 : 3$ . It is inscribed in a sphere of radius $14$ cm. What is the volume of the box, in cubic cm?

Correct Answer:

Step-by-Step Solution

Key idea: this is a box-inscribed-in-sphere question. The trigger is that all vertices of the rectangular box lie on the sphere, so the box's space diagonal becomes the sphere's diameter.

Step 1: Let the dimensions be $x, 2 x, 3 x$ .

Step 2: The sphere has radius $14$ , so its diameter is

$214 .$

Step 3: For a rectangular box, the space diagonal satisfies

$d^{2} = x^{2} + (2 x)^{2} + (3 x)^{2} .$

$d^{2} = x^{2} + 4 x^{2} + 9 x^{2} = 14 x^{2} .$

Step 4: Since the box is inscribed in the sphere,

$d = 214 .$

Therefore

$14 x^{2} = (214)^{2} = 4 \cdot 14 = 56.$

$x^{2} = 4 ⟹ x = 2.$

Step 5: The dimensions are $2, 4, 6$ . The volume is

$2 \cdot 4 \cdot 6 = 48.$

Answer: 48.

Question 3 · Quantitative Ability MCQ

A rectangle of maximum possible area is placed inside a semicircle, with its base on the diameter. If the longer side of this rectangle is 6 cm, what is the radius of the semicircle?

A.
3
B.
6
C.
2√3
D.
3√2

Correct Answer:

Step-by-Step Solution

Key idea: this is the maximum-area rectangle inside a semicircle pattern. The trigger is "maximum possible area" together with the rectangle's base on the diameter.

Step 1: Let the half-width of the rectangle be $x$ and its height be $y$ . The full width is $2 x$ .

Step 2: The top corner of the rectangle lies on the semicircle, so

$x^{2} + y^{2} = r^{2} .$

Step 3: The rectangle area is

$A = (2 x) y = 2 x y .$

For a fixed $x^{2} + y^{2}$ , the product $x y$ is maximised when $x = y$ . Therefore, at maximum area,

$x = y .$

Step 4: The longer side of this rectangle is its full width, $2 x$ . We are given

$2 x = 6,$

$x = 3.$

Since $x = y$ , we also have $y = 3$ .

Step 5: Use the circle constraint:

$r^{2} = x^{2} + y^{2} = 3^{2} + 3^{2} = 18.$

Thus

$r = 18 = 32 .$

Answer: $32$ .

Question 4 · Quantitative Ability MCQ

A rectangle is drawn inside a semicircle of radius $5$ cm, with its base on the diameter. If the height of the rectangle must be at least $4$ cm, the maximum possible area of the rectangle, in sq cm, is

A.
$24$
B.
$25$
C.
$20$
D.
$30$

Correct Answer:

Step-by-Step Solution

Key idea: this is the largest-rectangle-inside-a-semicircle pattern, but with an extra constraint. The constraint must be checked before using the usual unconstrained maximum.

Step 1: Let the half-width of the rectangle be $x$ and its height be $y$ . The full width is $2 x$ .

Step 2: Since the top corner lies on the semicircle of radius $5$ ,

$x^{2} + y^{2} = 25.$

The area is

$A = 2 x y = 2 y 25 - y^{2} .$

Step 3: Without the extra condition, maximize $A^{2}$ :

$A^{2} = 4 y^{2} (25 - y^{2}) .$

Let $t = y^{2}$ . Then $A^{2} = 4 t (25 - t)$ , a downward parabola with maximum at

$t = \frac{25}{2}, y = \frac{5}{2} \approx 3.54.$

Step 4: But the question requires $y \geq 4$ . The unconstrained maximizing height $3.54$ is not allowed.

Step 5: For $t \geq 16$ , the parabola $4 t (25 - t)$ is decreasing, so the maximum under $y \geq 4$ occurs at the smallest allowed height, $y = 4$ .

Step 6: If $y = 4$ , then

$x^{2} = 25 - 16 = 9 ⟹ x = 3.$

Therefore,

$A = 2 x y = 2 (3) (4) = 24.$

Answer: $24$ sq cm.

Common trap: answering $25$ , the unconstrained maximum area for radius $5$ . That maximum occurs at height $5/ 2$ , which violates the condition $y \geq 4$ .

Question 5 · Quantitative Ability NAT

The lowest possible cost of fencing a rectangular plot of area $A$ sq m is ₹ $48000$ . The cost is ₹ $300$ per m for one side and ₹ $100$ per m for each of the other three sides. Find $A$ .

Correct Answer:

7200

Step-by-Step Solution

Key idea: this is the unequal-side fencing pattern in reverse. You are given the minimum cost and must recover the fixed area.

Step 1: Define variables. Let the expensive side have length $x$ , and the other dimension be $y$ .

Step 2: Write the total cost. One side of length $x$ costs ₹ $300$ per m. The opposite side of length $x$ costs ₹ $100$ per m. The two sides of length $y$ each cost ₹ $100$ per m. Hence

$C = 300 x + 100 x + 100 y + 100 y = 400 x + 200 y .$

Step 3: Use the fixed-area condition:

$x y = A ⟹ y = \frac{A}{x} .$

Then

$C (x) = 400 x + 200 \cdot \frac{A}{x} = 400 x + \frac{200 A}{x} .$

Step 4: Minimum of $p x + \frac{q}{x}$ is $2 pq$ . Here $p = 400$ and $q = 200 A$ , so

$C_{m i n} = 2 400 \cdot 200 A = 2 80000 A .$

Step 5: Use the given minimum cost:

$48000 = 2 80000 A .$

Divide by 2:

$24000 = 80000 A .$

Square:

$576000000 = 80000 A .$

Therefore

$A = \frac{576000000}{80000} = 7200.$

Answer: 7200

Mensuration, 3D and Optimization Practice Questions for CAT: 47+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Mensuration, 3D and Optimization

Mensuration, 3D and Optimization

📦 t1 — Three-Dimensional Mensuration

📐 t2 — Area Optimization in Plane Figures

💰 t3 — Perimeter and Cost Optimization

Topic Hero: Three-Dimensional Mensuration

Three-Dimensional Mensuration

What you'll learn here

Mensuration, 3D and Optimization: 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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