Search Previous Year Questions

Given:

Top diameter of the frustum (d₁) = 2k

Bottom diameter of the frustum (d₂) = 2.5k

Height of the frustum (h) = k

Formula used:

Radius of top (r₁) = d₁ / 2

Radius of bottom (r₂) = d₂ / 2

Slant height of frustum (l) = $\sqrt{h^2 + (r_2 - r_1)^2}$

Whole Surface Area of Frustum = Area of top base + Area of bottom base + Lateral Surface Area

Area of top base = $\pi r_1^2$

Area of bottom base = $\pi r_2^2$

Lateral Surface Area = $\pi(r_1 + r_2)l$

Calculations:

Calculate radii:

r₁ = 2k / 2 = k

r₂ = 2.5k / 2 = 1.25k

Calculate the difference in radii:

r₂ - r₁ = 1.25k - k = 0.25k

Calculate the slant height (l):

l = $\sqrt{k^2 + (0.25k)^2}$

⇒ l = $\sqrt{k^2 + 0.0625k^2}$

⇒ l = $\sqrt{1.0625k^2}$

⇒ l = k$\sqrt{1.0625}$

Now, calculate the surface areas:

Area of top base = $\pi r_1^2$ = $\pi (k)^2$ = $\pi k^2$

Area of bottom base = $\pi r_2^2$ = $\pi (1.25k)^2$ = $\pi (1.5625k^2)$ = 1.5625$\pi k^2$

Lateral Surface Area = $\pi(r_1 + r_2)l$

⇒ Lateral Surface Area = $\pi(k + 1.25k)k\sqrt{1.0625}$

⇒ Lateral Surface Area = $\pi(2.25k)k\sqrt{1.0625}$

⇒ Lateral Surface Area = 2.25$\pi k^2\sqrt{1.0625}$

Whole Surface Area = $\pi k^2$ + 1.5625$\pi k^2$ + 2.25$\pi k^2\sqrt{1.0625}$

⇒ Whole Surface Area = $\pi k^2 (1 + 1.5625 + 2.25\sqrt{1.0625})$

⇒ Whole Surface Area = $\pi k^2 (2.5625 + 2.25 \times 1.030776...)$

⇒ Whole Surface Area = $\pi k^2 (2.5625 + 2.319246...)$

⇒ Whole Surface Area = $\pi k^2 (4.881746...)$

Let's use the fraction form for sqrt(1.0625) for precision:

1.0625 = 10625 / 10000 = 425 / 400 = 85 / 80 = 17/16 (Error in calculation here, 10625/10000 = 17/16 was for 1.0625, but 0.25k is (1/4)k)

0.25k = (1/4)k

(0.25k)² = (1/16)k²

l = $\sqrt{k^2 + \frac{1}{16}k^2}$ = $\sqrt{\frac{16k^2 + k^2}{16}}$ = $\sqrt{\frac{17k^2}{16}}$ = $\frac{k\sqrt{17}}{4}$

Lateral Surface Area = $\pi(k + 1.25k)\frac{k\sqrt{17}}{4}$

⇒ Lateral Surface Area = $\pi(2.25k)\frac{k\sqrt{17}}{4}$

⇒ Lateral Surface Area = $\pi(\frac{9}{4}k)\frac{k\sqrt{17}}{4}$

⇒ Lateral Surface Area = $\frac{9\pi k^2\sqrt{17}}{16}$

Whole Surface Area = $\pi k^2$ + $\pi (\frac{5}{4}k)^2$ + $\frac{9\pi k^2\sqrt{17}}{16}$

⇒ Whole Surface Area = $\pi k^2$ + $\frac{25\pi k^2}{16}$ + $\frac{9\pi k^2\sqrt{17}}{16}$

⇒ Whole Surface Area = $\pi k^2 \left(1 + \frac{25}{16} + \frac{9\sqrt{17}}{16}\right)$

⇒ Whole Surface Area = $\pi k^2 \left(\frac{16}{16} + \frac{25}{16} + \frac{9\sqrt{17}}{16}\right)$

⇒ Whole Surface Area = $\frac{\pi k^2}{16} (16 + 25 + 9\sqrt{17})$

⇒ Whole Surface Area = $\frac{\pi k^2}{16} (41 + 9\sqrt{17})$

Using $\sqrt{17} \approx 4.123$

⇒ Whole Surface Area = $\frac{\pi k^2}{16} (41 + 9 \times 4.123)$

⇒ Whole Surface Area = $\frac{\pi k^2}{16} (41 + 37.107)$

⇒ Whole Surface Area = $\frac{\pi k^2}{16} (78.107)$

⇒ Whole Surface Area ≈ 4.8816 $\pi k^2$

The exact expression is $\frac{\pi k^2}{16} (41 + 9\sqrt{17})$

∴ The whole surface area of the frustum is $\frac{\pi k^2}{16} (41 + 9\sqrt{17})$.

Given:

Top diameter of the frustum (d₁) = 2k

Bottom diameter of the frustum (d₂) = 2.5k

Height of the frustum (h) = k

Formula used:

Radius of top (r₁) = d₁ / 2

Radius of bottom (r₂) = d₂ / 2

Volume of frustum (V) = $\frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1 r_2)$

Calculations:

Calculate radii:

r₁ = 2k / 2 = k

r₂ = 2.5k / 2 = 1.25k

Substitute the values into the volume formula:

V = $\frac{1}{3}\pi (k) (k^2 + (1.25k)^2 + k \times 1.25k)$

⇒ V = $\frac{1}{3}\pi k (k^2 + 1.5625k^2 + 1.25k^2)$

⇒ V = $\frac{1}{3}\pi k (1k^2 + 1.5625k^2 + 1.25k^2)$

⇒ V = $\frac{1}{3}\pi k (k^2(1 + 1.5625 + 1.25))$

⇒ V = $\frac{1}{3}\pi k (3.8125k^2)$

⇒ V = $\frac{3.8125}{3}\pi k^3$

To express 3.8125 as a fraction:

3.8125 = 38125 / 10000 = 1525 / 400 = 305 / 80 = 61 / 16

So, V = $\frac{1}{3}\pi k \left(\frac{61}{16}k^2\right)$

⇒ V = $\frac{61}{48}\pi k^3$

∴ The volume of the frustum is $\frac{61}{48}\pi k^3$.

Given:

p + b + h = 24

p + b - h = 4

Formula used:

The radius (r) of the inscribed circle in a right-angled triangle is given by the formula:

r = $\frac{p + b - h}{2}$

Where: p and b are the perpendicular sides of the triangle

Area of triangle = radius (r) of the inscribed circle × semi - perimeter.

c is the hypotenuse of the triangle

Calculation:

p + b + h = 24 cm

p + b - h = 4 cm

The radius (r) of the inscribed circle in a right-angled triangle is given by the formula:

r = $\frac{p + b - h}{2}$

r = 4/2 = 2 cm

Area of triangle = 2 × 24/2 = 24 cm²

∴ The correct anser is option 2.

Given:

p + b + h = 24

p + b - h = 4

Formula used:

The radius (r) of the inscribed circle in a right-angled triangle is given by the formula:

r = $\frac{a + b - c}{2}$

Where: a and b are the perpendicular sides of the triangle

c is the hypotenuse of the triangle

Calculation:

p + b + h = 24 cm

p + b - h = 4 cm

The radius (r) of the inscribed circle in a right-angled triangle is given by the formula:

r = $\frac{p + b - h}{2}$

r = 4/2 = 2 cm

Therefore, the radius of the inscribed circle is: 2 cm.

Given:

Radius of circle M = 8 cm

Radius of circle N = 16 cm

Distance between centers = 8 + 16 = 24 cm

Formula used:

Length of direct common tangent = $\sqrt{d^2 - (r_1 - r_2)^2}$

Calculation:

⇒ d = 24, r₁ = 16, r₂ = 8

⇒ PQ = √(24² - (16 - 8)²)

⇒ PQ = √(576 - 64)

⇒ PQ = √512

⇒ PQ = 16√2 cm

∴ The length of common tangent PQ = 16√2 cm.

Given:

Circle M radius = 8 cm (centre U).

Circle N radius = 16 cm (centre V).

They touch externally ⇒ UV = 8 + 16 = 24 cm.

PQ = length of common external tangent = 16√2 cm.

Formula used:

If two radii to tangent points are perpendicular to the tangent, those two radii are parallel.

Area of a trapezoid with parallel sides a and b separated by distance d = (a + b)/2 × d.

Calculations:

⇒ UP = radius of M = 8 cm (UP ⟂ PQ).

⇒ VQ = radius of N = 16 cm (VQ ⟂ PQ).

⇒ UP ∥ VQ and the distance between these parallel segments = length PQ = 16√2 cm.

⇒ Area of quadrilateral P – Q – V – U = (UP + VQ)/2 × PQ

⇒ Area = (8 + 16)/2 × 16√2

⇒ Area = 24/2 × 16√2 = 12 × 16√2

⇒ Area = 192√2 cm²

∴ Area of quadrilateral P Q V U = 192√2 cm².

Given:

Perimeter of triangle ABC = 105 cm

Ratio of altitudes AD : BE : CF = 3 : 5 : 6

Formula used:

The area of a triangle (Area) can be expressed as: Area = $(\dfrac{1}{2}) × base × height.$

For a triangle with sides a, b, c and corresponding altitudes ha, hb, hc, we have:

2 × Area = a × ha = b × hb = c × hc

This implies that the sides of a triangle are inversely proportional to their corresponding altitudes. That is, $a : b : c = (\dfrac{1}{h_a}) : (\dfrac{1}{h_b}) : (\dfrac{1}{h_c})$

Calculations:

Let the sides of the triangle be a, b, c, where:

a = BC (side opposite to vertex A)

b = CA (side opposite to vertex B)

c = AB (side opposite to vertex C)

Let the altitudes be ha, hb, hc, where:

ha = AD

hb = BE

hc = CF

Given the ratio of altitudes: ha : hb : hc = 3 : 5 : 6.

Using the property that sides are inversely proportional to altitudes:

$a : b : c = (\dfrac{1}{3}) : (\dfrac{1}{5}) : (\dfrac{1}{6})$

⇒ $a : b : c = (\dfrac{1}{3}) × 30 : (\dfrac{1}{5}) × 30 : (\dfrac{1}{6}) × 30$

⇒ a : b : c = 10 : 6 : 5

This means the ratio of the sides BC : CA : AB is 10 : 6 : 5.

Therefore, AB : BC : CA = c : a : b = 5 : 10 : 6.

∴ The correct answer is option 2.

Given:

Perimeter of triangle ABC = 105 cm

Ratio of sides AB : BC : CA = 5 : 10 : 6

Formula used:

Heron's Formula for the area of a triangle:

Area = $(\sqrt{s(s-a)(s-b)(s-c)})$

Where a, b, c are the lengths of the sides of the triangle, and

s is the semi-perimeter s = $(\dfrac{\text{a + b + c}}{2})$

Calculations:

From the previous problem, we established

The ratio of the sides AB : BC : CA = 5 : 10 : 6.

Let the sides be AB = 5x, BC = 10x, and CA = 6x.

The perimeter is the sum of the sides:

Perimeter = AB + BC + CA = 5x + 10x + 6x = 21x

Given Perimeter = 105 cm.

⇒ 21x = 105

⇒ x = $(\dfrac{105}{21})$

⇒ x = 5

a (BC) = 10x = 10 × 5 = 50 cm

b (CA) = 6x = 6 × 5 = 30 cm

c (AB) = 5x = 5 × 5 = 25 cm

The semi-perimeter (s): s = $(\dfrac{\text{Perimeter}}{2})$ = $(\dfrac{105}{2})$ = 52.5 cm

Area = $(\sqrt{s(s-a)(s-b)(s-c)})$

Area = $(\sqrt{52.5 \times (52.5 - 50) \times (52.5 - 30) \times (52.5 - 25)})$

Area = $(\sqrt{52.5 \times 2.5 \times 22.5 \times 27.5})$

Area = $(\sqrt{99738.28125})$

Area ≈ 284.975 cm²

Area ≈ 285 cm²

∴ The correct answer is option 4.

Given:

Inner radius of the hollow sphere (R) = 20 cm

Height of the pot (h) = 30 cm

Calculations:

Let 'r' be the inner radius of the circular opening of the pot.

Draw a vertical line from the center of the sphere to the center of the circular opening. This line segment will be perpendicular to the plane of the circular opening.

Let the center of the sphere be O. Let the center of the circular opening be C'.

The radius of the sphere (R) goes from O to any point on the surface of the sphere, including the edge of the circular opening.

Consider a right-angled triangle formed by:

The radius of the sphere (R) as the hypotenuse, from the center of the sphere to any point on the edge of the circular opening.

d = |R - h| = |20 - 30| = |-10| = 10 cm.

Now, substitute the values into the Pythagorean theorem:

R² = r² + d²

20² = r² + 10²

400 = r² + 100

r² = 400 - 100

r² = 300

r = $\sqrt{300}$

r = $\sqrt{100 \times 3}$

r = 10$\sqrt{3}$ cm

∴ The inner radius of the circular opening of the pot is 10$\sqrt{3}$ cm.

Given:

Inner radius of the hollow sphere (R) = 20 cm

Height of the pot (h) = 30 cm

From the previous calculation, the radius of the circular opening (r) = 10$\sqrt{3}$ cm

From the previous calculation, the perpendicular distance from the center of the sphere to the plane of the circular opening (d) = 10 cm

Calculations:

The perpendicular distance from the center of the sphere to the plane of the circular opening (d) as the adjacent side to the angle θ (this is the vertical line segment from the center to the center of the opening).

The inner radius of the circular opening (r) as the opposite side to the angle θ.

We can use trigonometric ratios:

cos(θ) = Adjacent / Hypotenuse = d / R

cos(θ) = d / R = 10 cm / 20 cm = 1/2

The angle whose cosine is 1/2 is 60°.

⇒ θ = 60° = π/3

∴ The angle made by the line joining the center of the sphere and any point on the rim of the circular opening with a vertical line passing through the center is 60°.