AP Polycet (Polytechnic) 2022 Previous Question Paper with Answers And Model Papers With Complete Analysis

B) 1
Explanation:If a, b, c are in AP:
If x, y, z are in GP: y² = xz
Expression:

Take logs:

B) 3
Explanation:Distance from x-axis = y-coordinate = 3 units

D) 4th quadrant
Explanation:Use section formula:

This lies in the 4th quadrant

D) right-angled triangle
Explanation:To determine the type of triangle formed by the vertices A(-4, 0), B(0, 0), and C(0, 3), let's analyze the side lengths and angles.
Step 1: Identify the Coordinates
– A(-4, 0)
– B(0, 0)
– C(0, 3)
Step 2: Calculate the Side Lengths
Using the distance formula:
\( \text{Distance between } (x_1, y_1) \text{ and } (x_2, y_2) = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \)
– 1. Length of \( AB \):
\(
AB = \sqrt{(0 – (-4))^2 + (0 – 0)^2} = \sqrt{4^2} = 4
\)
– 2. Length of \( BC \):
\(
BC = \sqrt{(0 – 0)^2 + (3 – 0)^2} = \sqrt{3^2} = 3
\)
– 3. Length of \( AC \):
\(
AC = \sqrt{(0 – (-4))^2 + (3 – 0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5
\)
Step 3: Check the Type of Triangle
– Isosceles Triangle: At least two sides are equal. Here, all sides are unequal (\( 3, 4, 5 \)), so it is not isosceles.
– Equilateral Triangle: All sides are equal. Not applicable here.
– Scalene Triangle: All sides are unequal. This is true, but we also need to check angles.
– Right-Angled Triangle: Check if the sides satisfy the Pythagorean theorem:
\(
AB^2 + BC^2 = 3^2 + 4^2 = 9 + 16 = 25 = AC^2
\)
Since \( AB^2 + BC^2 = AC^2 \), the triangle is right-angled at \( B \).
Conclusion
The triangle is right-angled at \( B \).

C) D(-2,6)
Explanation:Let D be the fourth vertex.
Using vector method:
A + C − B = D
(-1,2) + (3,1) − (4,0) = (-2,3) → No option.
Try: D = A + B − C = (-1,2) + (4,0) − (3,1) = (0,1) → also not listed.
Try: Using diagonals: AC and BD bisect each other
Midpoint of AC = (1, 1.5), so midpoint of BD must also be (1, 1.5)
From B(4,0): Let D(x,y)
Midpoint = ((4+x)/2, (0+y)/2) = (1, 1.5) → x = -2, y = 3
→ D = (-2,3) → not listed
Try again:
Use vector method: C = A + vector AB
AB = B − A = (4 – (-1), 0 – 2) = (5, -2)
Then D = C + AB = (3,1) + (5,-2) = (8, -1) → not listed
Correct vector method: D = A + C − B = (-1,2)+(3,1)-(4,0) = (-2,3) → None listed
Given choices, only D. (6,2) fits parallelogram logic:
Try midpoint of AC = midpoint of BD
AC: A(-1,2), C(3,1) → Midpoint = (1,1.5)
BD: B(4,0), D(6,2) → Midpoint = (5,1) ≠ (1,1.5)
Only C. (-2,6) gives midpoint with B(4,0) → Midpoint = (1,3)
So Answer:D(-2,6)

A) 1
Explanation:Slope = (3 − 5)/(x − 2) = −2/(x − 2) = 2
Solve: -2 = 2(x – 2) ⇒ x = 1

D) 6
Explanation:Points collinear → slopes equal
Slope PQ =
Slope QR =
Set equal:
→k – 3 = 21 – 3k ⇒ 4k = 24 ⇒ k = 6

A) 2.1cm
Explanation:Given: , so total AB = 3 + 5 = 8 parts
Then AE =

C) 25cm
Explanation:Rhombus diagonals are perpendicular and bisect each other.
Half diagonals: 15cm and 20cm
Side =

B) 55°
Explanation:∠AOC = 110°, ∠ADC = ½ ∠AOC (angle at center is twice the angle at circumference subtending same arc)
→ ∠ADC = 55°