AP Polycet (Polytechnic) 2025 Previous Question Paper with Answers And Model Papers With Complete Analysis

Polycet Mock Test

11) A tangent PQ at a point P of a circle of radius 9 cm meets a line through the center O at a point Q such that OQ = 15 cm. The length of PQ is:

A) 12cm
B) 13cm
C) 24cm
D) 25cm

View Answer

A) 12cm

Explanation: Tangent PQ, OP = 9 cm, OQ = 15 cm (Right triangle OPQ)
Use Pythagoras:
$PQ^2 = OQ^2 - OP^2 = 15^2 - 9^2 = 225 - 81 = 144 \Rightarrow PQ = \sqrt{144} = 12$
Answer: 12 cm

12)Area of a sector of a circle with radius 4 cm and angle 30° is (use π = 3.14):

A)4.08cm²
B)4cm²
C)4.18cm²
D)41.8cm²

View Answer

C)4.18cm²
Explanation: Area of sector = $\frac{θ}{360} \times πr^2$

$= \frac{30}{360} \times 3.14 \times 4^2 = \frac{1}{12} \times 3.14 \times 16 = \frac{50.24}{12} ≈ 4.18$

Answer: 4.18 cm²

13) Area of a sector of a circle with radius 4 cm and angle 30° is (use π = 3.14):

A) 4.08cm²
B) 4cm²
C) 4.18cm²
D) 41.8cm²

View Answer

C) 4.18cm²

Explanation:Area of sector = $\frac{θ}{360} \times πr^2$
$= \frac{30}{360} \times 3.14 \times 4^2 = \frac{1}{12} \times 3.14 \times 16 = \frac{50.24}{12} ≈ 4.18$
Answer: 4.18 cm²

14) Length of an arc of a sector of angle 45° when the radius of the circle is 3 cm, is:

A) $\frac{5π}{4} \, cm$
B) $\frac{3π}{4} \, cm$
C) $\frac{π}{4} \, cm$
D) $\frac{π}{2} \, cm$

View Answer

B) $\frac{3π}{4} \, cm$

Explanation:Arc length = $\frac{θ}{360} × 2πr$
$= \frac{45}{360} × 2π × 3 = \frac{1}{8} × 6π = \frac{6π}{8} = \frac{3π}{4}$
Answer: $\frac{3π}{4} \, cm$

15) Area of minor segment if a chord of a circle of radius 10 cm subtends a right angle at the centre is (use π = 3.14):

A) 28 cm²
B) 28.5 cm²
C) 27 cm²
D) 27.5 cm²

View Answer

B) 28.5 cm²

Explanation:Area of segment = Area of sector − Area of triangle
Sector (90°): $\frac{90}{360} × π × 10^2 = \frac{1}{4} × 3.14 × 100 = 78.5$ Triangle (right-angled isosceles): $\frac{1}{2} × 10 × 10 = 50$
78.5 − 50 = 28.5
Answer: 28.5 cm²

16) A toy is in the form of a cone of radius r and lateral height l mounted on a hemisphere of the same radius, and the total height of the toy is h, then the total surface area of the toy is:

A) π r (2r + l)
B) 2π r + l
C) π r² l
D) π r² h

View Answer

A) π r (2r + l)

Explanation:Surface Area = Cone + Hemisphere (no base)
$= πrl + 2πr^2 = πr(2r + l)$
Answer: πr(2r + l)

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