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Given:

Height of pole = PQ = $\dfrac{5\sqrt{3}}{2}$ m.

BQ = 0.7m

CQ = 2.4m

We need to find length AP.

Formula used:

Distance from the corner point = $\sqrt{(x^2 + y^2 + h^2)}$

Where, x = 0.7 m, y = 2.4 m, and h = height of the lamp = $\dfrac{5\sqrt{3}}{2}$ m.

Calculation:

We can see AQ² = BQ² + AB² and AB = CQ (Shown in fig.)

AP² = AQ² + PQ²

AP² = BQ2 + AB2 + PQ2

AP = $\sqrt{BQ^2 + AB^2 + PQ^2}$

The distance of the top of the lamp from the corner point (AP) is:

⇒ Distance = $\sqrt{(0.7^2 + 2.4^2 + \left(\dfrac{5\sqrt{3}}{2}\right)^2)}$

⇒ Distance = $\sqrt{(0.49 + 5.76 + \dfrac{75}{4})}$

⇒ Distance = $\sqrt{(0.49 + 5.76 + 18.75)}$

⇒ Distance = $\sqrt{25}$

⇒ Distance = 5 m.

∴ The distance of the top of the lamp from the corner point is 5 meters.

Given:

C is the center of a circle with radius = 20 cm.

AB is a chord of length = 32 cm.

CE = 13 cm, E is a point on AB.

We need to find AE × EB.

Formula used:

In a circle, for any chord AB and point E on AB:

AE × EB = (Radius² - Distance from center to chord²)

Calculation:

Radius = 20 cm, so Radius² = 20² = 400

Distance from center to chord (CE) = 13 cm, so CE² = 13² = 169

⇒ AE × EB = Radius² - CE²

⇒ AE × EB = 400 - 169

⇒ AE × EB = 231

∴ The correct answer is option (1).

Given:

The inside of the bowl is part of a sphere.

Depth of water (d) = d

Radius of the water surface = 2d

Formula used:

Using the geometry of spheres, the radius of the sphere (R) is given by:

R = $\frac{(r^2 + d^2)}{2d}$

Where r = radius of the water surface = 2d

Calculation:

R = $\frac{(r^2 + d^2)}{2d}$

⇒ R = $\frac{((2d)^2 + d^2)}{2d}$

⇒ R = $\frac{(4d^2 + d^2)}{2d}$

⇒ R = $\frac{5d^2}{2d}$

⇒ R = 2.5d

∴ The correct answer is option (1).

Alternate Method

Given:

Depth of water = d

Radius of water surface = 2d

Let Radius of sphere = R

Formula used:

By geometry of sphere cross-section:

$(R - d)^2 + (2d)^2 = R^2$

Calculations:

⇒ (R - d)² + (2d)² = R²

⇒ (R² - 2Rd + d²) + 4d² = R²

⇒ R² - 2Rd + 5d² = R²

⇒ -2Rd + 5d² = 0

⇒ 2Rd = 5d²

⇒ R = (5d / 2) = 2.5d

∴ Radius of sphere = 2.5d

Given:

AB = 2 cm, BC = 4 cm, AC = 3 cm

Formula used:

Angle bisector theorem: If a bisector of an angle divides the opposite side of a triangle, the ratio of the divided parts equals the ratio of the other two sides.

BD/DC = AB/AC

Calculations:

In Δ ABC,

AD is the bisector of ∠A

So, BD/DC = AB/AC = 2/3

Since, BC = 4 cm

So, BD = 2/5 × 4 = 8/5 cm

Now, In Δ ABD

BE is the bisector of ∠B

So, BD/AB = ED/AE

⇒ (8/5) / 2 = ED/AE

⇒ 4/5 = ED/AE

⇒ 5/4 = AE / ED

So, AE : ED = 5 : 4

Hence, the correct answer is option (1).

Given:

AB + AC = 10 cm

BD : DC = 3 : 1

Formula used:

Angle bisector theorem: If a bisector of an angle divides the opposite side of a triangle, the ratio of the divided parts equals the ratio of the other two sides.

BD/DC = AB/AC 

Calculation:

Since AD is the angle bisector of ∠A,

So, AB/AC = BD/DC 

⇒ AB/AC = 3/1

Since AB + AC = 10 cm

So, AC = 1/(1 + 3) × 10 = 10/4 = 2.5 cm

Thus, the length of AC is 2.5 cm.

The correct answer is option (1).

Given:

AB + BC = 7.1 cm

BC + CA = 12.1 cm

CA + AB = 7.2 cm

Calculation:

Let the sides of the triangle be:

AB = a

BC = b

CA = c

We are given the following system of equations:

a + b = 7.1

b + c = 12.1

c + a = 7.2

(a + b) + (b + c) + (c + a) = 7.1 + 12.1 + 7.2

2a + 2b + 2c = 26.4

a + b + c = 13.2

c = 13.2 - 7.1 = 6.1

a = 13.2 - 12.1 = 1.1

b = 13.2 - 7.2 = 6.0

Heron’s formula for the area of a triangle is given by:

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

Where s is the semi-perimeter of the triangle, given by:

s = $\frac{a + b + c}{2} = \frac{1.1 + 6.0 + 6.1}{2} = \frac{13.2}{2} = 6.6$ cm

$\text{Area} = \sqrt{6.6(6.6 - 1.1)(6.6 - 6.0)(6.6 - 6.1)}$

$\text{Area} = \sqrt{6.6 \times 5.5 \times 0.6 \times 0.5}$

$\text{Area} = \sqrt{6.6 \times 5.5 \times 0.3}$

$\text{Area} = \sqrt{10.89}$

$\text{Area} = 3.3 , \text{cm}^2$

Therefore, the area of the triangle is: 3.3 cm²

Given:

Adjacent sides of a parallelogram: 10 cm and 8 cm

Angle between them: 150^∘

Formula used:

Area of parallelogram = a × b × sin(θ)

Where:

a = one side, b = adjacent side, θ = angle between sides

Calculation:

Area = 10 × 8 × sin(150^∘)

⇒ sin(150^∘) = sin(30^∘) = 1/2

⇒ Area = 10 × 8 × (1/2)

⇒ Area = 40 cm²

∴ The correct answer is option (2).

Given:

Triangle ABC

Bisectors of angles A and C intersect and form an angle = 130°

Find angle B

Formula used:

Angle between bisectors of A and C inside triangle = 90° + B/2

Calculations:

Given, angle between bisectors = 130°

Using formula: 90 + B/2 = 130

⇒ B/2 = 130 - 90

⇒ B/2 = 40

⇒ B = 80°

∴ Angle B = 80°.

Given:

$\log_{10}2000 + \log_{10}400 + 4\log_{10}25 + 5\log_{10}20$

Formula used:

$\log a + \log b = \log(ab)$

$n\log a = \log(a^n)$

Calculation:

$\log_{10}2000 + \log_{10}400 + 4\log_{10}25 + 5\log_{10}20$

= $\log_{10}(2000 \times 400 \times 25^4 \times 20^5)$

⇒ $2000 = 2^4 \times 5^3$,   $400 = 2^4 \times 5^2$,   $25^4 = (5^2)^4 = 5^8$,   $20^5 = (2^2 \times 5)^5 = 2^{10} \times 5^5$

⇒ Product = $(2^4 \times 5^3) \times (2^4 \times 5^2) \times (5^8) \times (2^{10} \times 5^5)$

⇒ $= 2^{4+4+10} \times 5^{3+2+8+5}$

⇒ $= 2^{18} \times 5^{18}$

⇒ $= (2 \times 5)^{18} = 10^{18}$

Therefore,

$\log_{10}(10^{18}) = 18$

∴ The value is 18.

Given:

$\frac{\log_{10}(100001 - 4^x)}{5 - x} = 1$

Formula used:

log a = b ⇒ a = 10^b

Calculation:

⇒ log₁₀(100001 - 4^x) = 5 - x

⇒ 100001 - 4^x = 10^5-x

⇒ 100001 - 4^x = 10^5-x

Check all the options:

x = 0

⇒ 100001 - 40 = 105-0

⇒ 100001 - 1 = 100000

x = 0 satisfies the condition.

x = 1

⇒ 100001 - 41 = 105 - 1

⇒ 100001 - 4 = 10000

⇒ 99997 ≠ 10000

x = 1 does not satisfy the condition.

x = 10

⇒ 100001 - 410 = 105 - 10

⇒ 100001 - 410 ≠ 10- 5

x = 10 does not satisfy the condition.

∴ The correct answer is option 1.