CBSE Class 10 Mathematics Basic Question Paper 2024 Available- Download Solution PDF with Answer Key (Set 2- 430/1/2)

Step 1: Perform the prime factorization of 850 and 500:
850 = 2 × 5² × 17,
500 = 2² × 5³
Step 2: Find the LCM: The LCM is the product of the highest powers of all prime factors appearing in the numbers:
LCM = 2² × 5³ × 17
Step 3: Simplify:
LCM = 17 × 500
Thus, the LCM is 17 × 500.

For a quadratic equation ax² + bx + c = 0, the condition for real and equal roots is:
Δ = b² - 4ac = 0
For the given quadratic equation 4x² - 5x + k = 0, we have:
- a = 4
- b = -5
- c = k
Substitute the values into the discriminant formula:
Δ = (-5)² - 4(4)(k) = 0
25 - 16k = 0
Solving for k:
16k = 25 ⟹ k = 25/16

In statistics, the relationship between the mean, median, and mode is given by:
Mode = 3 × Median - 2 × Mean
Substitute the given values:
Mode = 3 × 23 - 2 × 21 = 69 - 42 = 27
Thus, the mode of the data is 27.

The slant height l of a right circular cone can be calculated using the Pythagoras theorem, as it forms a right triangle with the height and radius:
l = √(r² + h²)
Substitute the given values r = 7 cm and h = 24 cm:
l = √(7² + 24²) = √(49 + 576) = √625 = 25 cm
Thus, the slant height is 25 cm.

Let the quadratic polynomial be f(x) = (α - 1)x² + αx + 1.
Given that one of the zeroes is -3, we can substitute x = -3 into the equation:
f(-3) = (α - 1)(-3)² + α(-3) + 1 = 0
Simplifying:
(α - 1)(9) - 3α + 1 = 0
9α - 9 - 3α + 1 = 0
6α - 8 = 0
Solve for α:
6α = 8 ⟹ α = 8/6 = 4/3

There are 2 red queens in a deck of 52 cards (one from hearts and one from diamonds). The probability of drawing a red queen is:
2/52 = 1/26
Thus, the probability is 1/26.

The value of x that divides the statistical data into two equal parts is called the median. It is the middle value in an ordered data set.

To find the probability of getting exactly one tail, we first look at all possible outcomes when three coins are tossed. The total number of outcomes is 2³ = 8, because each coin has two possible outcomes (heads or tails). The possible outcomes are:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
We need to find the number of outcomes with exactly one tail. The outcomes with one tail are: HHT, HTH, and THH. So, there are 3 favorable outcomes.
Thus, the probability of getting exactly one tail is the ratio of favorable outcomes to total outcomes, which is:
P(exactly one tail) = 3/8

We are given sin θ = 1/3. We need to find sec θ.
Recall the following trigonometric identities:
sec θ = 1/cos θ
sin² θ + cos² θ = 1
From sin θ = 1/3, we can find cos θ using the identity sin² θ + cos² θ = 1:
(1/3)² + cos² θ = 1 ⟹ 1/9 + cos² θ = 1
cos² θ = 1 - 1/9 = 8/9
Thus:
cos θ = √8/3
Now, we can find sec θ:
sec θ = 1/cos θ = 1/(√8/3) = 3/√8 = 3/(2√2)
Thus, the correct answer is:
Answer: (b) 3/(2√2)

The formula for the outer surface area (curved surface area) of a cylinder is given by:
A = 2πr h
where:
- r is the radius of the base of the cylinder,
- h is the height of the cylinder.
Given:
- Radius r = 7 cm
- Height h = 10 cm
Substituting the values into the formula:
A = 2π × 7 × 10 = 140π
Using π = 3.14:
A = 140 × 3.14 = 439.6 sq cm ≈ 594 sq m
Thus, the outer surface area is approximately 594 sq cm.
Therefore, the correct answer is option (b) 594 sq m.

The probability of getting a 6 on a fair die is 1/6.
Thus, the probability of losing the game (i.e., not getting a 6) is the complement of the probability of success:
P(Losing) = 1 - P(Success) = 1 - 1/6 = 5/6
Thus, the probability of losing the game is 5/6.

The distance between two points (x₁, y₁) and (x₂, y₂) is given by the distance formula:
d = √((x₂ - x₁)² + (y₂ - y₁)²)
Substitute the coordinates (x₁, y₁) = (2, -3) and (x₂, y₂) = (-2, 3):
d = √((-2 - 2)² + (3 - (-3))²) = √((-4)² + (6)²) = √(16 + 36) = √52
d = 2√13 units
Thus, the distance is 2√13 units.

We are given the equation:
sin²θ + sinθ + cos²θ = 2
Since cos²θ + sin²θ = 1 (the Pythagorean identity), the equation simplifies to:
1 + sinθ = 2
sinθ = 1
The sine of θ is 1 when θ = 90°.
Thus, the value of θ is 90°.

Given the diameter length is 6 cm, the endpoints of the diameter are (-4, 0) and (x, 0). Since the second endpoint lies on the x-axis, its y-coordinate is 0.
Using the distance formula for the diameter:
Distance = √((x - (-4))² + (0 - 0)²) = 6
√((x + 4)²) = 6 ⟹ x + 4 = ±6
Solving for x:
x + 4 = 6 ⟹ x = 2 or x + 4 = -6 ⟹ x = -10
Thus, the valid point is (2, 0).
Therefore, the other endpoint is (2, 0).

Step 1: For the pair of equations to not have a solution, the lines must be parallel. This happens when:
(a₁/a₂) = (b₁/b₂) ≠ (c₁/c₂)
Here:
a₁ = 5, b₁ = 2, c₁ = -7, a₂ = 2, b₂ = k, c₂ = 1
Step 2: Use the condition (a₁/a₂) = (b₁/b₂):
5/2 = 2/k ⟹ 5k = 4 ⟹ k = 4/5.
Final Answer: The value of k is 4/5.

We know that for a quadratic equation ax² + bx + c, the product of the zeroes (roots) is given by:
Product of the zeroes = c/a
In our case, the polynomial is kx² - 4x - 7, where:
a = k, b = -4, c = -7.
We are given that the product of the zeroes is 2. Therefore, we can set up the equation:
c/a = 2
Substituting the values of c and a:
-7/k = 2 ⟹ k = -7/2.
Thus, the correct answer is: (b) -7/2

In an arithmetic progression (A.P.), the nth term is given by the formula:
aₙ = a + (n - 1)d
Where:
aₙ is the nth term, a is the first term, d is the common difference.
We are given:
a = 8 (the first term),
a₁₀ = -19 (the 10th term),
We need to find d (the common difference).
Substitute the known values into the formula for the 10th term:
a₁₀ = a + (10 - 1)d ⟹ -19 = 8 + 9d
Now, solve for d:
-19 - 8 = 9d ⟹ -27 = 9d ⟹ d = -3
Thus, the correct answer is: (d) -3

The formula for the mid-point of a line segment joining the points (x₁, y₁) and (x₂, y₂) is given by:
((x₁ + x₂)/2, (y₁ + y₂)/2)
Substitute the coordinates of the points (-1, 3) and (8, 3/2) into this formula:
(((-1) + 8)/2, (3 + 3/2)/2)
Simplifying:
(7/2, (6 + 3)/4) = (7/2, 9/4)
Thus, the mid-point of the line segment is (7/2, 9/4), which corresponds to option (d).

Step 1: Analyze Assertion (A):
PA and PB are tangents drawn to the circle. The angle between the tangents at the external point P and the angles subtended by the points of tangency form a quadrilateral OAPB.
A quadrilateral is cyclic if all its vertices lie on the circumference of a single circle.
Since OAPB satisfies this condition, Assertion (A) is true.
Step 2: Analyze Reason (R):
The given Reason states that in a cyclic quadrilateral, opposite angles are equal. This is incorrect. The correct property of a cyclic quadrilateral is that the sum of opposite angles is 180°. Thus, Reason (R) is false.
Conclusion: Assertion (A) is true, but Reason (R) is false.

The given polynomial is p(x) = x² - 2x - 3. To find the zeroes, we solve x² - 2x - 3 = 0 by factoring:
x² - 2x - 3 = (x - 3)(x + 1) = 0
Thus, the zeroes of the polynomial are x = 3 and x = -1.
The graph of a quadratic polynomial intersects the x-axis at its zeroes. Therefore, the points where the graph intersects the x-axis are (-1, 0) and (3, 0), as given in the reason.
Since both the assertion and the reason are true, and the reason explains the assertion, the correct answer is (a).

We are given that √5 is an irrational number. We need to prove that 6 - 4√5 is also an irrational number.

Proof:

Assume, for the sake of contradiction, that 6 - 4√5 is a rational number. This means that 6 - 4√5 = p/q, where p and q are integers and q ≠ 0.

Now, let's isolate √5:

6 - 4√5 = p/q

4√5 = 6 - p/q

√5 = (6q - p)/(4q)

Thus, √5 is expressed as a ratio of two integers, which implies that √5 is a rational number.

But this contradicts the given information that √5 is irrational.

Therefore, our assumption that 6 - 4√5 is rational must be false. Hence, 6 - 4√5 is irrational.

Conclusion:

Thus, we have proved that 6 - 4√5 is an irrational number.

First, simplify the given expression:

11 × 19 × 23 + 3 × 11

Factor out 11 from both terms:

11 (19 × 23 + 3)

Now, calculate inside the parentheses:

19 × 23 = 437

437 + 3 = 440

Thus, the expression becomes:

11 × 440 = 4840

Since 4840 is divisible by 11, it is not a prime number.

Therefore, 11 × 19 × 23 + 3 × 11 = 4840 is not a prime number.

In the given right-angled triangle ΔABC, we have ∠A = 90°, so ∠B + ∠C = 90°. This implies that:

∠B = 90° - ∠C

We are given that tan C = √3. From the definition of tangent, we know:

tan C = opposite/adjacent = AB/AC

Thus, AB/AC = √3, implying that:

AB = √3 × AC

Let AC = x, then AB = √3x.

Next, we use the Pythagorean theorem in ΔABC:

BC² = AB² + AC²

BC² = (√3x)² + x² = 3x² + x² = 4x²

BC = 2x

Now, let's calculate the required expression sin B + cos C - cos² B.

1. Finding sin B:

Since ∠B = 90° - ∠C, we know:

sin B = cos C

2. Finding cos C:

Using the definition of cosine:

cos C = AC/BC = x/2x = 1/2

3. Finding cos B:

Since sin B = cos C = 1/2, we use the Pythagorean identity to find cos B:

cos² B = 1 - sin² B = 1 - (1/2)² = 1 - 1/4 = 3/4

cos B = √3/2

Now, substitute these values into the expression sin B + cos C - cos² B:

sin B + cos C - cos² B = 1/2 + 1/2 - 3/4

= 1 - 3/4 = 1/4

Thus, the value of sin B + cos C - cos² B is 1/4.

Since ΔOAP ~ ΔOBQ (AA criterion), we can use the property of similar triangles:

OA/OB = AP/BQ

Substitute the given values:

15/12 = 10/BQ

Now solve for BQ:

BQ = (10 × 12)/15 = 8 cm

Thus, the length of BQ is 8 cm.

From the first equation:

x + 2y = 9 → x = 9 - 2y

Substitute this expression for x into the second equation:

y - 2(9 - 2y) = 2

Simplifying:

y - 18 + 4y = 2 → 5y = 20 → y = 4

Now substitute y = 4 into x = 9 - 2y:

x = 9 - 2(4) = 9 - 8 = 1

Thus, the solution is x = 1 and y = 4.

For the first equation x + y + 1 = 0: Substitute x = -4 and y = 3:

(-4) + 3 + 1 = 0 This is true, so the point lies on the first line.

For the second equation x - y = 1: Substitute x = -4 and y = 3:

(-4) - 3 = -7 This is false, so the point does not lie on the second line.

Thus, the point (-4, 3) lies on the first line but not on the second line.

Step 1: Start with the left-hand side (LHS): LHS = (√(sec A - 1))/(√(sec A + 1)) + (√(sec A + 1))/(√(sec A - 1))

Step 2: Simplify by taking a common denominator: LHS = ((sec A - 1) + (sec A + 1))/(√(sec A + 1) * √(sec A - 1))

Step 3: Simplify the numerator: Numerator = (sec A - 1) + (sec A + 1) = 2 sec A

Step 4: Simplify the denominator using the identity (√a * √b = √(a * b)): Denominator = √((sec A + 1)(sec A - 1)) = √(sec² A - 1)

Step 5: Use the trigonometric identity sec² A - 1 = tan² A: LHS = 2 sec A / √(tan² A) = 2 sec A / tan A

Step 6: Simplify using the definition tan A = sin A / cos A and sec A = 1 / cos A: LHS = (2 * (1 / cos A)) / (sin A / cos A) = 2 / sin A = 2 csc A

Step 7: Conclude: LHS = RHS

Final Answer: (√(sec A - 1))/(√(sec A + 1)) + (√(sec A + 1))/(√(sec A - 1)) = 2 csc A

In this problem, we are given two concentric circles. The radius of the smaller circle is OQ = 6 cm, and the radius of the larger circle is OA = r cm. The chord CD of the larger circle is tangent to the smaller circle at point Q, and we are asked to find the length of CD.

We are given the following information:
PA = 16 cm, where PA is the tangent to the larger circle at point A.
OP = 20 cm, where O is the center of both circles, and P is a point outside the larger circle.

Now, let’s use the property that the tangent to a circle from an external point is perpendicular to the radius at the point of tangency. Therefore, the length of PA is perpendicular to the radius OA.

We can now apply the Pythagorean theorem to the right triangle OPA, where OP = 20 cm and PA = 16 cm:

OP² = OA² + PA²

Substituting the given values:

20² = r² + 16²
400 = r² + 256
r² = 144
r = 12 cm

Now, we know the radius OA = 12 cm.

Next, let’s use the property of tangents and the fact that the chord CD of the larger circle is tangent to the smaller circle at Q. From geometry, the length of the chord CD can be found using the following formula:

CD = 2√(OP² - OQ²)

Substituting the known values OP = 20 cm and OQ = 6 cm:

CD = 2√(20² - 6²)
CD = 2√(400 - 36)
CD = 2√364
CD = 12√3 cm

Final Answer: CD = 12√3 cm

- Let PT and QT be two tangents drawn from the external point T to the circle with center O.
- Since PT and QT are tangents, the angle between a tangent and the radius is always 90°, so:
∠OTP = ∠OTQ = 90°
- Also, we know that ∠PTQ = ∠OTP + ∠OTQ, which means:
∠PTQ = 90° + 90° = 180°

Therefore, ∠PTQ = 2 ∠OPQ.

- The solid consists of a cylindrical part and two hemispherical ends. The total height of the solid is the sum of the height of the cylinder and the height of the two hemispheres.
- Let the radius of the cylinder be r = 14/2 = 7 cm.
- The height of the cylinder is h = 20 - 2r = 20 - 2(7) = 6 cm.
- Surface area of the solid is the sum of the curved surface area of the cylinder and the surface area of the two hemispheres:

Surface Area = 2πr² + 2πrh + 2πr²

Substituting values:

Surface Area = 2π(7)² + 2π(7)(6) + 2π(7)² = 2π(49) + 2π(42) + 2π(49)
Surface Area = 2π(49 + 42 + 49) = 2π(140) = 280π cm²
Therefore, the surface area is:
Surface Area = 280π cm² ≈ 880 cm²

Final Answer: Surface Area = 880 cm²

Radius of the glass r = 10/2 = 5 cm

Capacity of glass = volume of cylinder - volume of hemisphere

Volume of cylinder = πr²h
Volume of hemisphere = (2/3)πr³

Capacity of glass = π × 5² × 14 - (2/3)π × 5³

= 3.14 × 5 × 5 × 14 - (2/3) × 3.14 × 5 × 5 × 5

= 2512 cm³ or 837.33 cm³ (approx)

Final Answer: 837.33 cm³

The alarm clocks will beep together again at the Least Common Multiple (LCM) of 20 and 25 minutes.

LCM of 20 and 25 = 100 minutes

Thus, the next time they beep together will be 100 minutes after 12:00 PM, which is 1 hour and 40 minutes later, or 1:40 PM.

Final Answer: 1:40 PM

Step 1: Assume the coordinates of A and B: - Since A lies on the x-axis, its coordinates are (x, 0). - Since B lies on the y-axis, its coordinates are (0, y). Step 2: Use the section formula: The coordinates of P(2, -3) divide AB in the ratio AP : PB = 3 : 1. By the section formula: P(x, y) = ((m2x1 + m1x2) / (m1 + m2), (m2y1 + m1y2) / (m1 + m2)) Here: - m1 = 3, m2 = 1, - A(x, 0), B(0, y), - P(2, -3). Step 3: Substitute into the section formula: For the x-coordinate: 2 = (1 × x + 3 × 0) / (3 + 1) = x / 4 → x = 8 For the y-coordinate: -3 = (1 × 0 + 3 × y) / (3 + 1) = 3y / 4 → y = -4 Step 4: Write the coordinates of A and B: - Coordinates of A are (x, 0) = (8, 0), - Coordinates of B are (0, y) = (0, -4). Final Answer: The coordinates of A are (8, 0), and the coordinates of B are (0, -4).

Let the smaller angle be x. Then, the larger angle is x + 18°.

Since the angles are supplementary, their sum is 180°: x + (x + 18°) = 180° Simplifying: 2x + 18° = 180° 2x = 162° x = 81°

Thus, the smaller angle is 81°, and the larger angle is: 81° + 18° = 99°

Final Answer: The two angles are 99° and 81°.

Assuming AOB to be a straight line and hence the diameter of the circle:

∠ACB = 90°
Then in ∆ACB, we have:

AC² + BC² = 28² + 21² = (35)² = AB²
Thus, AB = 35 cm is the diameter, and
r = 35 / 2 = 17.5 cm.

Now, for the area of the shaded region:

Area of shaded region = area of quadrant + (½ × π r² − area of ∆ACB)

Substitute the values:

= (3/4 × (22/7) × (35/2) × (35/2)) − (½ × 28 × 21)

Simplifying this:

= 721.9 − 294 = 427.9 (approx.)

Final Answer: The area of the shaded region is 427.9 sq. cm.

Let h be the height of the tower and x be the length of the original shadow.
From the first situation (altitude 30°):

tan 30° = h / (x + 40)
(1/√3) = h / (x + 40)
h = (x + 40) / √3

From the second situation (altitude 60°):

tan 60° = h / x
√3 = h / x
h = √3 x

Equating the two expressions for h:

(x + 40) / √3 = √3 x

Solving:
x + 40 = 3x
40 = 2x
x = 20

Therefore, the height of the tower is:
h = √3 × 20 = 20√3 ≈ 34.64 m
The length of the original shadow is x = 20 m.

Final Answer: Height of the tower = 34.64 m, Length of the shadow = 20 m.

Let the height of the multi-storeyed building be h, and the distance between the two buildings be d.

1. Using the angle of depression of 45°:

tan(45°) = 8 / d₁
Since tan(45°) = 1, we have:
d₁ = 8 m
This is the horizontal distance from the base of the multi-storeyed building to the base of the 8 m tall building.

2. Using the angle of depression of 30°:

tan(30°) = (h - 8) / (d₁ + 14)
Substituting tan(30°) = 1 / √3 and d₁ = 8:

(1/√3) = (h - 8) / 22
h - 8 = 22 / √3 ≈ 12.67
Thus:
h = 8 + 12.67 = 20.67 m

Final Answer: Height of the multi-storeyed building = 20.67 m, Distance between the two buildings = 14 m.

Given: AB / PQ = AC / PR = AD / PM

We have:
CE / RN = AC / PR = 2 / PN → ∆AEC ~ ∆PNR

Thus, ∠1 = ∠2, and similarly ∠3 = ∠4

Therefore, ∠BAC = ∠QPR
Also, AB / PQ = AC / PR
Thus, ∆ABC ~ ∆PQR

We are given that the n-th term sum of the A.P. is Sn = 4n² - n.

(i) Find the first term and common difference:
The first term of the A.P., a₁, is simply S₁, the sum of the first term. Thus:
S₁ = 4(1)² - 1 = 4 - 1 = 3.
So, the first term a₁ = 3.

To find the common difference, we use the fact that the difference between consecutive sums gives the n-th term:

aₙ = Sₙ - Sₙ₋₁.
Let's find a₂ and a₃ to identify the common difference.
S₂ = 4(2)² - 2 = 16 - 2 = 14, S₁ = 3.
Thus:
a₂ = S₂ - S₁ = 14 - 3 = 11.
The common difference d is the difference between consecutive terms:
d = a₂ - a₁ = 11 - 3 = 8.
Thus, the common difference is d = 8.

(ii) Write the A.P.:
The first term is a₁ = 3 and the common difference is d = 8. So, the A.P. is:
3, 11, 19, 27, 35, ...

(iii) Which term of the A.P. is 107?
To find which term is 107, we use the formula for the n-th term of an A.P.:

aₙ = a₁ + (n-1) * d.
Substitute the known values a₁ = 3, d = 8, and aₙ = 107:

107 = 3 + (n-1) * 8.
Simplify and solve for n:
107 - 3 = (n-1) * 8, 104 = (n-1) * 8,
n - 1 = 104 / 8 = 13, n = 14.

Thus, the 14th term of the A.P. is 107.

Given: In ∆ABC, DE || BC To Prove: AD / DB = AE / EC

Construction: Join BE, DC. Draw DM ⊥ AC and EN ⊥ AB.

Proof:

(1) ar(∆ADE) / ar(∆DBE) = (1/2) × AD × EN / (1/2) × DB × EN = AD / DB (i)

(2) ar(∆ADE) / ar(∆DCE) = (1/2) × AE × DM / (1/2) × EC × DM = AE / EC (ii)

From (i), (ii), and the fact that ar(∆BDE) = ar(∆CDE) (iii), we get:

AD / DB = AE / EC

Hence proved.

(i) Median Class
The cumulative frequency is calculated as follows:

The total number of leaves is 40, so the median will lie at the 20th position. From the cumulative frequency, the 20th leaf lies in the class 100-110, so the median class is 100-110.

(ii) Leaves of length equal to or more than 10 cm (100 mm):
The relevant classes are:
- 100-110: 12 leaves
- 110-120: 5 leaves
- 120-130: 4 leaves
- 130-140: 2 leaves
Total = 12 + 5 + 4 + 2 = 23 leaves of length ≥ 10 cm.

(iii) (a) To find the median, use the formula for grouped data:

Median = L + [(N/2 - F) / f] × h

Where:
L = 100 (lower boundary of the median class)
N = 40 (total number of leaves)
F = 17 (cumulative frequency before median class)
f = 12 (frequency of the median class)
h = 10 (class width)

Median = 100 + [(20 - 17) / 12] × 10 = 100 + 2.5 = 102.5 mm
Thus, the Median is 102.5 mm.

(iii) (b) The modal class is the class with the highest frequency, which is 100-110 with 12 leaves. To find the mode, use the formula:

Mode = L + [(f1 - f0) / (2f1 - f0 - f2)] × h

Where:
L = 100 (lower boundary of the modal class)
f1 = 12 (frequency of the modal class)
f0 = 9 (frequency of the class before the modal class)
f2 = 5 (frequency of the class after the modal class)
h = 10 (class width)

Mode = 100 + [(12 - 9) / (2 × 12 - 9 - 5)] × 10 = 100 + (3 / 10) × 10 = 100 + 3 = 103 mm
Thus, the Mode is 103 mm.

(i) Find the length of PQ:
We are given that AP = AQ = 30 cm, and the angle between the tangents, ∠PAQ = 60°. To find PQ, we use the law of cosines in triangle PAQ:

PQ² = AP² + AQ² - 2 × AP × AQ × cos(∠PAQ)

Substituting the given values:

PQ² = 30² + 30² - 2 × 30 × 30 × cos(60°)

Since cos(60°) = 0.5:

PQ² = 900 + 900 - 900 = 900
PQ = √900 = 30 cm

Thus, the length of PQ is 30 cm.

(ii) Find ∠POQ:
Since AP and AQ are tangents to the circle from the point A, the angle between the tangents at P and Q is equal to the angle at the center of the circle subtended by the chord PQ. This means that:

∠POQ = 2 × ∠PAQ = 2 × 60° = 120°

Thus, ∠POQ = 120°.

(iii) (a) Find the length of OA:
To find the length of OA, we use the law of cosines in triangle OAP. Since ∠OAP = 90° (the angle between the radius and the tangent is always 90°), triangle OAP is a right triangle. Using the Pythagorean theorem:

OA² = OP² + AP²

Since OP is the radius and AP = 30 cm, we have:

OA² = r² + 30²
OA = 2r (since ∠PAQ = 60°, OA is double the radius)
Substitute OA = 2r into the Pythagorean theorem:

(2r)² = r² + 30²
4r² = r² + 900
3r² = 900
r² = 300
r = √300 ≈ 17.32 cm

Thus, OA = 2 × 17.32 = 34.64 cm.

OR (b) Find the radius of the mirror:
The radius of the mirror is approximately 17.32 cm.

(i) Obtain a quadratic equation involving R and r:
From the given information, we know the sum of the areas of the two circles is 130 sq m. The area of a circle is given by pi * r^2, where r is the radius. Hence, we can write the equation for the sum of the areas as:

pi * R^2 + pi * r^2 = 130 * pi

Dividing through by pi:

R^2 + r^2 = 130

The distance between the centers of the two circles is given as 14 m, so we have:

R + r = 14

(ii) Write a quadratic equation involving only r:
Using the equation R + r = 14, we substitute R = 14 - r into the equation R^2 + r^2 = 130:

(14 - r)^2 + r^2 = 130

Expanding:

196 - 28r + r^2 + r^2 = 130

Combining like terms:

2r^2 - 28r + 196 = 130

Simplifying:

2r^2 - 28r + 66 = 0

Dividing through by 2:

r^2 - 14r + 33 = 0

(iii) (a) Find the radius r and the corresponding area irrigated:
We solve the quadratic equation r^2 - 14r + 33 = 0. Using factoring or the quadratic formula:

r = (-(-14) ± sqrt((-14)^2 - 4(1)(33)))/(2(1))

r = (14 ± sqrt(196 - 132))/2

r = (14 ± sqrt(64))/2

r = (14 ± 8)/2

Thus, r = 3 or r = 11. Since r < R, we take r = 3.
The area irrigated by the smaller circle is:

pi * r^2 = pi * (3^2) = 9pi sq m

(iii) (b) Find the radius R and the corresponding area irrigated:
Using R + r = 14, we substitute r = 3 to find R:

R = 14 - 3 = 11

The area irrigated by the larger circle is:

pi * R^2 = pi * (11^2) = 121pi sq m

Final Answer:
- Radius r = 3, corresponding area 9pi sq m.
- Radius R = 11, corresponding area 121pi sq m.