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Given:

∠A = 30º

AB = 7 cm

AC = 12 cm

Formula Used:

Area of triangle = (1/2) × AB × AC × sin(∠A)

Calculation:

⇒ Area = (1/2) × 7 × 12 × sin(30º)

⇒ Area = (1/2) × 7 × 12 × 1/2

⇒ Area = (7 × 12) / 4

⇒ Area = 21 cm²

∴ The correct answer is 21 cm².

Given:

ABC is a triangle right-angled at B.

AB = p.

BC = √3p.

BD is perpendicular to AC.

Formula Used:

AC = √(AB² + BC²).

Area of triangle = (1/2) × Base × Height.

Area of triangle = (1/2) × AC × BD.

Equating both expressions for area to find BD.

Calculation:

⇒ AC = √(p² + (√3p)²)

⇒ AC = √(p² + 3p²)

⇒ AC = √4p²

⇒ AC = 2p

Area of triangle ABC = (1/2) × AB × BC

⇒ Area = (1/2) × p × √3p

⇒ Area = (√3p²) / 2

Area of triangle ABC = (1/2) × AC × BD

⇒ (√3p²) / 2 = (1/2) × (2p) × BD

⇒ (√3p²) / 2 = p × BD

⇒ BD = (√3p²) / (2p)

⇒ BD = √3p / 2

∴ The correct answer is √3p / 2.

Given:

Difference between an interior angle and an e×terior angle of a regular polygon = 120º

Formula Used:

Interior angle of a regular polygon = (n - 2) × 180° / n

E×terior angle of a regular polygon = 360° / n

Interior angle - E×terior angle = 120°

Calculation:

⇒ ((n - 2) × 180° / n) - (360° / n) = 120°

⇒ (180n - 360 - 360) / n = 120

⇒ (180n - 720) / n = 120

⇒ 180n - 720 = 120n

⇒ 180n - 120n = 720

⇒ 60n = 720

⇒ n = 720 / 60

⇒ n = 12

∴ The correct answer is 12.

Given:

An angle is exactly one-fourth of its complementary angle.

Formula Used:

Let the angle be x.

Complementary angles add up to 90º: x + 4x = 90º.

Calculation:

⇒ x + 4x = 90º

⇒ 5x = 90º

⇒ x = 90º / 5

⇒ x = 18º

∴ The correct answer is 18º.

Given:

Side 1 = 11 cm

Side 2 = 60 cm

Side 3 = 61 cm

Formula Used:

Area of triangle = √[s(s - a)(s - b)(s - c)]

Where s = (a + b + c) / 2

Area of triangle formed by mid-points = Original Area / 4

Calculation:

⇒ s = (11 + 60 + 61) / 2

⇒ s = 66 cm

⇒ Area = √[66(66 - 11)(66 - 60)(66 - 61)]

⇒ Area = √[66 × 55 × 6 × 5]

⇒ Area = √[108900]

⇒ Area = 330 cm²

⇒ Area of triangle formed by mid-points = 330 / 4

⇒ Area = 82.5 cm²

∴ The correct answer is 82.5 cm².

Given:

Angle of the swing = 30º

Arc length = 55 cm

π = 22/7

Formula Used:

Length of arc (l) = (θ/360) × 2πr

Here, r = length of the pendulum

Calculation:

⇒ 55 = (30/360) × 2 × (22/7) × r

⇒ 55 = (1/12) × (44/7) × r

⇒ 55 = (44/84) × r

⇒ 55 = (11/21) × r

⇒ r = 55 × (21/11)

⇒ r = 105 cm

∴ The correct answer is 105 cm.

Given:

Angle at vertex = 60º

Curved surface area = 100 m²

Formula Used:

Curved Surface Area of cone = π × r × l

Volume of cone = (1/3) × π × r² × h

l = √(r² + h²)

tan(Angle/2) = r/h

Calculation:

⇒ tan(60º/2) = tan(30º) = 1/√3

⇒ r/h = 1/√3

⇒ r = h/√3

Curved surface area = π × r × l

⇒ 100 = π × (h/√3) × √((h/√3)² + h²)

⇒ 100 = π × (h/√3) × √(h²/3 + h²)

⇒ 100 = π × (h/√3) × √(4h²/3)

⇒ 100 = π × (h/√3) × (2h/√3)

⇒ 100 = (2πh²)/(3)

⇒ h² = (100 × 3)/(2π)

⇒ h² = 150/π

⇒ h = √(150/π)

r = h/√3 = √(150/π)/√3 = √(150/(3π)) = √(50/π)

Volume = (1/3) × π × r² × h

⇒ Volume = (1/3) × π × (50/π) × √(150/π)

⇒ Volume = (1/3) × 50 × √(150/π)

⇒ Volume = (50/3) × √(150/π)

⇒ Volume = (50 × √150)/(3√π)

⇒ Volume = (250√6)/(3√π) m³

∴ The correct answer is (250√2)/(√(3π)) m³.

Given:

Volume of a cone = Volume of a hemisphere

Base radius of cone = Radius of hemisphere = r

Formula Used:

Volume of a cone = (1/3)πr²h

Volume of a hemisphere = (2/3)πr³

Calculation:

⇒ (1/3)πr²h = (2/3)πr³

⇒ r²h = 2r³

⇒ h = 2r

∴ The correct answer is 2 : 1.

Given:

Area of equilateral triangle = 36√3 cm²

Formula Used:

Area of equilateral triangle = (√3 / 4) × side²

Perimeter of triangle = 3 × side

Circumference of semicircle = π × radius + diameter

Diameter = 2 × radius

Calculation:

⇒ (√3 / 4) × side² = 36√3

⇒ side² = (36√3 × 4) / √3

⇒ side² = 144

⇒ side = √144

⇒ side = 12 cm

Perimeter of triangle = 3 × side

⇒ Perimeter = 3 × 12

⇒ Perimeter = 36 cm

For semicircle, the wire length = Circumference

⇒ 36 = π × radius + 2 × radius

⇒ 36 = radius(π + 2)

⇒ radius = 36 / (π + 2)

∴ The correct answer is option (4): None of the above

Given:

Radius of the circle = 1 (unit radius)

Side of the square = 1 (unit side length)

Formula Used:

Area of a square = side²

Area of a circle = π × radius²

Calculation:

For the largest square inscribed in a circle:

⇒ Diagonal of the square = Diameter of the circle

⇒ Diagonal = 2 × radius = 2 × 1 = 2

⇒ Side of the square = Diagonal / √2 = 2 / √2 = √2

⇒ Area of the square (x) = (Side)² = (√2)² = 2

For the largest circle inscribed in a square:

⇒ Diameter of the circle = Side of the square = 1

⇒ Radius of the circle = Diameter / 2 = 1 / 2

⇒ Area of the circle (y) = π × (Radius)² = π × (1/2)² = π / 4

Relation between x and y:

⇒ x = 2, y = π / 4

Now, 

x/y = 2/( π / 4)

⇒ x/y = 8/π

⇒ πx = 8y

∴ The correct answer is πx = 8y.