CBSE Class 10 2025 Mathematics Set-2 (430-6-2) Question Paper : Download Solution PDF With Answer Key

Solution steps would go here.
In \(\triangle OCA\), \(\angle OCA = 90^\circ\) (radius to tangent is perpendicular).
By Pythagoras theorem: \(OA^2 = OC^2 + AC^2\) \( (3.5)^2 = (2.1)^2 + AC^2 \) \( 12.25 = 4.41 + AC^2 \) \( AC^2 = 12.25 - 4.41 = 7.84 \) \( AC = \sqrt{7.84} = 2.8 \) cm.
Since C is the point of tangency, it bisects the chord AB.
So, \( AB = 2 \times AC = 2 \times 2.8 = 5.6 \) cm. \[ \boxed{5.6 cm} \] Quick Tip: The radius to the point of tangency is perpendicular to the tangent. The perpendicular from the center to a chord bisects the chord. Use Pythagoras theorem in the right-angled triangle formed.

Reason (R): The statement "Sum of all angles of a triangle is \(180^\circ\)" is a fundamental property of triangles and is true.

Assertion (A):

Given \(\triangle ABC \sim \triangle PQR\). This implies that corresponding angles are equal:
\(\angle A = \angle P\), \(\angle B = \angle Q\), \(\angle C = \angle R\).

In \(\triangle ABC\), we are given \(\angle A = 65^\circ\) and \(\angle C = 60^\circ\).

Using the property that the sum of angles in a triangle is \(180^\circ\) (which is Reason R):
\(\angle B = 180^\circ - (\angle A + \angle C)\)
\(\angle B = 180^\circ - (65^\circ + 60^\circ)\)
\(\angle B = 180^\circ - 125^\circ\)
\(\angle B = 55^\circ\).

Since \(\triangle ABC \sim \triangle PQR\), we have \(\angle Q = \angle B\).

Therefore, \(\angle Q = 55^\circ\).

The assertion states "Hence \(\angle Q = 55^\circ\)", which is true.

Explanation:

Reason (R) states that the sum of angles in a triangle is \(180^\circ\).

This fact was used to find \(\angle B\) in \(\triangle ABC\). Once \(\angle B\) was found, the property of similar triangles (\(\angle B = \angle Q\)) allowed us to determine \(\angle Q\). Thus, Reason (R) is essential for deriving the conclusion in Assertion (A).

So, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
\[ \boxed{Option (A)} \] Quick Tip:For similar triangles, corresponding angles are equal. The sum of angles in any triangle is always \(180^\circ\). To evaluate an Assertion-Reason question, first check the truthfulness of the Reason, then the Assertion. Finally, check if the Reason correctly explains the Assertion.

We need to evaluate the expression \(\frac{\cos 45^\circ}{\tan 30^\circ + \sin 60^\circ}\).

We know the standard trigonometric values:
\(\cos 45^\circ = \frac{1}{\sqrt{2}}\)
\(\tan 30^\circ = \frac{1}{\sqrt{3}}\)
\(\sin 60^\circ = \frac{\sqrt{3}}{2}\)


Substitute these values into the expression:
\[ \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2}} \]
First, simplify the denominator:
\[ \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2} = \frac{1 \cdot 2 + \sqrt{3} \cdot \sqrt{3}}{2\sqrt{3}} = \frac{2 + 3}{2\sqrt{3}} = \frac{5}{2\sqrt{3}} \]
Now substitute this back into the main expression:
\[ \frac{\frac{1}{\sqrt{2}}}{\frac{5}{2\sqrt{3}}} = \frac{1}{\sqrt{2}} \times \frac{2\sqrt{3}}{5} = \frac{2\sqrt{3}}{5\sqrt{2}} \]
To rationalize the denominator, multiply the numerator and denominator by \(\sqrt{2}\):
\[ \frac{2\sqrt{3}}{5\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{3 \times 2}}{5 \times 2} = \frac{2\sqrt{6}}{10} = \frac{\sqrt{6}}{5} \] \[ \boxed{\frac{\sqrt{6}}{5}} \]

Let the step lengths of the three friends be \(s_1 = 48\) cm, \(s_2 = 52\) cm, and \(s_3 = 56\) cm.

We need to find the minimum distance that each friend can cover in a whole number of steps. This distance will be the Least Common Multiple (LCM) of their step lengths.


First, find the prime factorisation of each step length:
\(48 = 2 \times 24 = 2 \times 2 \times 12 = 2 \times 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3^1\)
\(52 = 2 \times 26 = 2 \times 2 \times 13 = 2^2 \times 13^1\)
\(56 = 2 \times 28 = 2 \times 2 \times 14 = 2 \times 2 \times 2 \times 7 = 2^3 \times 7^1\)


The LCM is the product of the highest powers of all prime factors that appear in any of the numbers:

Highest power of 2: \(2^4\)

Highest power of 3: \(3^1\)

Highest power of 7: \(7^1\)

Highest power of 13: \(13^1\)


LCM(48, 52, 56) = \(2^4 \times 3^1 \times 7^1 \times 13^1\)

LCM = \(16 \times 3 \times 7 \times 13\)

LCM = \(48 \times 7 \times 13\)

LCM = \(336 \times 13\)

336

\(\times\) 13

-----
1008 (\(336 \times 3\))

3360 (\(336 \times 10\))

-----
4368

So, the minimum distance they can all cover in complete steps is 4368 cm.
The question asks: "What is the minimum distance each should walk so that each can cover the same distance in complete steps ten times?"
This phrasing can be interpreted as: what is the minimum common distance (D) that they can all cover in complete steps? The "ten times" part implies that this action of covering distance D is repeated ten times. The question asks for D itself.

If the question meant that the number of steps taken by each person to cover the common distance must be a multiple of 10, the problem would be different (LCM of 10*48, 10*52, 10*56). However, the phrasing "cover the same distance ... ten times" suggests the "same distance" is the LCM, and this is done 10 times. The "minimum distance" refers to this "same distance".

Therefore, the minimum distance is 4368 cm.

Number of steps for each friend to cover this distance:

Friend 1: \(4368 / 48 = 91\) steps.

Friend 2: \(4368 / 52 = 84\) steps.

Friend 3: \(4368 / 56 = 78\) steps.

Each covers 4368 cm in complete steps. They can do this ten times. The minimum such distance is 4368 cm.
\[ \boxed{4368 cm} \] Quick Tip: \textbf{Quick Tip:} The minimum distance that multiple entities can cover in complete units of their respective measures is the LCM of those measures. Prime factorise each number, then find the LCM by taking the highest power of each prime factor present in any of the numbers. Carefully interpret what "ten times" refers to. In this context, it likely refers to repeating the action of covering the LCM distance.

Let \(h\) be the height of each pole (AB and CD).

Let P be the point on the road between the poles.

The width of the road BD = 85 m.

Let the distance of the point P from pole AB be \(x\) m. So, BP = \(x\).

Then the distance of the point P from pole CD is \((85-x)\) m. So, DP = \(85-x\).

The angles of elevation from P to the top of poles A and C are \(60^\circ\) and \(30^\circ\) respectively.

Let \(\angle APB = 60^\circ\) and \(\angle CPD = 30^\circ\).

(Diagram: Two vertical poles AB, CD of height h. Road BD of width 85.
Point P on BD. BP=x, DP=85-x. Lines AP, CP forming angles of elevation.)

In right-angled \(\triangle ABP\):
\(\tan 60^\circ = \frac{AB}{BP} = \frac{h}{x}\)
\(\sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}\) --- (1)


In right-angled \(\triangle CDP\):
\(\tan 30^\circ = \frac{CD}{DP} = \frac{h}{85-x}\)
\(\frac{1}{\sqrt{3}} = \frac{h}{85-x} \Rightarrow h = \frac{85-x}{\sqrt{3}}\) --- (2)


Equating the expressions for \(h\) from (1) and (2):
\(x\sqrt{3} = \frac{85-x}{\sqrt{3}}\)
\(x\sqrt{3} \times \sqrt{3} = 85-x\)
\(3x = 85-x\)
\(3x + x = 85\)
\(4x = 85\)
\(x = \frac{85}{4} = 21.25\) m.


So, the distance of the point from one pole is \(x = 21.25\) m.
The distance of the point from the other pole is \(85-x = 85 - 21.25 = 63.75\) m.


Now, find the height \(h\) using equation (1):
\(h = x\sqrt{3} = 21.25 \times \sqrt{3}\).

Given \(\sqrt{3} = 1.73\).
\(h = 21.25 \times 1.73\).

21.25

\(\times\) 1.73

--------
6375 (21.25 \(\times\) 0.03)

148750 (21.25 \(\times\) 0.70)

2125000 (21.25 \(\times\) 1.00)

--------
36.7625

So, \(h = 36.7625\) m.

The height of the poles is 36.7625 m.

The distances of the point from the poles are 21.25 m and 63.75 m.
\[ \boxed{Height of poles = 36.7625 m; Distances from point = 21.25 m and 63.75 m} \] Quick Tip: Draw a clear diagram representing the situation. Use trigonometric ratios (tan, sin, cos) in right-angled triangles. \(\tan \theta = \frac{opposite}{adjacent}\). Set up equations based on the given information and solve them simultaneously. Remember standard values: \(\tan 60^\circ = \sqrt{3}\), \(\tan 30^\circ = 1/\sqrt{3}\).

Let \(r_i\) be the inner radius of the bowl, \(r_i = 10\) cm.

Let \(r_o\) be the outer radius of the bowl, \(r_o = 10.5\) cm.


(i) Dimensions of the cuboidal box.

Since the hemispherical bowl "just fits" in the box, the box must accommodate the outer dimensions of the bowl. Assuming the circular base of the hemisphere rests on the base of the cuboid:

Length of the box (\(l\)) = Outer diameter of the bowl = \(2 \times r_o = 2 \times 10.5 = 21\) cm.

Width of the box (\(w\)) = Outer diameter of the bowl = \(2 \times r_o = 2 \times 10.5 = 21\) cm.

Height of the box (\(h\)) = Outer radius of the bowl = \(r_o = 10.5\) cm.

Dimensions are 21 cm \(\times\) 21 cm \(\times\) 10.5 cm.
\[ \boxed{Dimensions: 21 cm (length), 21 cm (width), 10.5 cm (height)} \]

(ii) Total outer surface area of the box.

The box is a cuboid with \(l=21\) cm, \(w=21\) cm, \(h=10.5\) cm.

Total Surface Area (TSA) = \(2(lw + wh + hl)\)

TSA = \(2((21 \times 21) + (21 \times 10.5) + (10.5 \times 21))\)

TSA = \(2(441 + 220.5 + 220.5)\)

TSA = \(2(441 + 441)\)

TSA = \(2(882)\)

TSA = \(1764\) cm\(^2\). \[ \boxed{Total outer surface area of the box = 1764 cm^2} \]

(iii) (a) Difference between the capacity of the bowl and the volume of the box.

Capacity of the bowl = Inner volume of the hemisphere = \(\frac{2}{3} \pi r_i^3\).

Capacity = \(\frac{2}{3} \times 3.14 \times (10)^3 = \frac{2}{3} \times 3.14 \times 1000 = \frac{2 \times 3140}{3} = \frac{6280}{3} \approx 2093.33\) cm\(^3\).


Volume of the box = \(l \times w \times h = 21 \times 21 \times 10.5 = 441 \times 10.5\).
\(441 \times 10.5 = 441 \times (10 + 0.5) = 4410 + 441 \times 0.5 = 4410 + 220.5 = 4630.5\) cm\(^3\).


Difference = Volume of the box - Capacity of the bowl

Difference = \(4630.5 - 2093.333...\)

Difference = \(4630.5 - \frac{6280}{3} = \frac{13891.5 - 6280}{3} = \frac{7611.5}{3} \approx 2537.166...\) cm\(^3\).

As \(2093.33\) is an approximation, using fractions:

Difference = \(4630.5 - \frac{6280}{3} = \frac{9261}{2} - \frac{6280}{3} = \frac{27783 - 12560}{6} = \frac{15223}{6} \approx 2537.166...\) cm\(^3\).
\[ \boxed{Difference \approx 2537.17 cm^3 (Box volume is greater)} \]

OR

(iii) (b) The inner surface of the bowl and the thickness (rim area) is to be painted. Find the area to be painted.

Inner surface area of the bowl (hemisphere) = \(2 \pi r_i^2 = 2 \times 3.14 \times (10)^2 = 2 \times 3.14 \times 100 = 628\) cm\(^2\).

Area of the rim (top annular surface representing thickness) = Area of outer circle - Area of inner circle

Area of rim = \(\pi r_o^2 - \pi r_i^2 = \pi (r_o^2 - r_i^2)\)

Area of rim = \(3.14 \times ((10.5)^2 - (10)^2) = 3.14 \times (110.25 - 100) = 3.14 \times 10.25\).
\(3.14 \times 10.25 = 3.14 \times (10 + 0.25) = 31.4 + 3.14 \times 0.25 = 31.4 + 0.785 = 32.185\) cm\(^2\).

Total area to be painted = Inner surface area + Area of the rim
Total area = \(628 + 32.185 = 660.185\) cm\(^2\).
\[ \boxed{Area to be painted = 660.185 cm^2} \] Quick Tip: "Just fits" implies the smallest containing box matching the object's maximal dimensions in each axis. Volume of hemisphere = \(\frac{2}{3}\pi r^3\). Surface area of hemisphere (curved) = \(2\pi r^2\). TSA of cuboid = \(2(lw+wh+hl)\). Area of an annulus (ring) = \(\pi (R^2 - r^2)\).