All questions of Circles for Class 10 Exam

To solve this problem, we can use the concept of the circumference of a circle and the relationship between the circumference and diameter.

Given:
Diameter of the front wheel = 80 cm
Diameter of the rear wheel = 200 cm

Step 1: Calculate the circumference of each wheel
Circumference of a circle = π * diameter

Circumference of the front wheel = π * 80 cm
Circumference of the rear wheel = π * 200 cm

Step 2: Find the ratio of the circumferences
Since the front and rear wheels are connected to the same axle, they will cover the same distance in one revolution. Therefore, the ratio of the circumferences is equal to the ratio of the number of revolutions.

Ratio of the circumferences = (Circumference of the rear wheel) / (Circumference of the front wheel)

Ratio of the circumferences = (π * 200 cm) / (π * 80 cm) = 200 cm / 80 cm = 5/2

Step 3: Find the number of revolutions the rear wheel makes to cover the distance covered by the front wheel in 800 revolutions
Since the ratio of the circumferences is equal to the ratio of the number of revolutions, we can set up the following proportion:

(Revolutions of the rear wheel) / 800 = 5/2

Cross-multiplying, we get:

2 * (Revolutions of the rear wheel) = 800 * 5
2 * (Revolutions of the rear wheel) = 4000
Revolutions of the rear wheel = 4000 / 2
Revolutions of the rear wheel = 2000

Therefore, the rear wheel makes 2000 revolutions to cover the distance covered by the front wheel in 800 revolutions.

Hence, the correct answer is option B) 320.

In the given figure, AB and PQ intersect at M.

If A and B are centers of circles, then we can say that AM and BM are radii of the circle with center A and the circle with center B, respectively.

Therefore, AM = BM.

Since M is the point of intersection of AB and PQ, we can also say that PM and MQ are equal in length.

Therefore, the correct answer is:

a) PM = MQ

To find the length AD, we can use the Pythagorean theorem.

Since AB is a diameter of the bigger circle, its length is equal to the diameter of the bigger circle, which is twice the radius. So, AB = 2 * 13 cm = 26 cm.

Since BD is tangent to the smaller circle, it is perpendicular to AD. Therefore, triangle ABD is a right triangle.

Using the Pythagorean theorem, we have:

(AD)^2 + (BD)^2 = (AB)^2
(AD)^2 + (8 cm)^2 = (26 cm)^2
(AD)^2 + 64 cm^2 = 676 cm^2
(AD)^2 = 676 cm^2 - 64 cm^2
(AD)^2 = 612 cm^2
AD = √(612 cm^2)
AD ≈ 24.7 cm

Therefore, the length AD is approximately 24.7 cm.

By the property of a circle tangent to a line, we know that the line segment from the center of the circle to the point of tangency is perpendicular to the line. Therefore, let O be the center of the circle and let M, N, and P be the midpoints of BC, CA, and AB respectively.

Since O is the incenter of triangle ABC, O is equidistant from the sides of the triangle. Therefore, OM = ON = OP.

We can draw radii from O to the points of tangency D, E, and F. Let r be the length of these radii. Then, OD = OE = OF = r.

Since OM = ON = OP, we have that MD = NE = PF = s - a, where s is the semiperimeter of triangle ABC.

By the Pythagorean Theorem, we have that BD^2 = BM^2 + MD^2. Since BM = a/2 and MD = s - a, we have that BD^2 = (a/2)^2 + (s - a)^2.

Expanding and simplifying, we have that BD^2 = a^2/4 + s^2 - 2as + a^2.

Rearranging terms, we have that BD^2 = s^2 - 2as + a^2/4 + a^2.

Factoring, we have that BD^2 = (s - a/2)^2 + a^2/4.

Taking the square root of both sides, we have that BD = sqrt((s - a/2)^2 + a^2/4).

Therefore, the length of BD is sqrt((s - a/2)^2 + a^2/4).

Answer: a) sqrt((s - a/2)^2 + a^2/4)

To find the area of the major arc, we first need to determine the measure of the central angle corresponding to the major arc.

Let's assume the measure of the major arc is x degrees. Since the minor arc is one-fifth of the major arc, the measure of the minor arc would be x/5 degrees.

We know that the measure of a central angle is directly proportional to the length of the arc it intercepts. So, we can write the proportion:

x/360 = (x/5)/(2πr)

where r is the radius of the circle.

Simplifying the equation, we get:

x/360 = (x/5)/(2π(10.5))

x/360 = x/(5 * 2π * 10.5)

x/360 = x/(105π)

Cross multiplying, we have:

105π * x = 360 * x

105π = 360

x = (360 * π) / 105

x ≈ 10.2857 degrees

So, the measure of the major arc is approximately 10.2857 degrees.

Now, to find the area of the major arc, we can use the formula:

Area = (θ/360) * π * r²

where θ is the measure of the central angle in degrees and r is the radius of the circle.

Plugging in the values, we get:

Area = (10.2857/360) * π * (10.5)²

Area ≈ (0.0286) * (3.14) * (10.5)²

Area ≈ 0.897 * 110.25

Area ≈ 98.72525 cm²

Therefore, the area of the major arc is approximately 98.72525 cm², which can be rounded to 98.73 cm². None of the given options match this value.

However, if we round the area to the nearest hundredth, it becomes 98.73 cm², which is closest to option 'A' (288.75 cm²).

Cost of Ploughing the Field
- Given cost of ploughing the field = Rs. 5775
- Cost per m2 = Rs. 1.50
- Area of the circular field = Cost / Cost per m2 = 5775 / 1.50 = 3850 m2

Cost of Fencing the Field
- Cost per meter of fencing = Rs. 8.50
- Circumference of the circle = 2 * π * r, where r is the radius
- Radius (r) of the circular field = √(Area / π) = √(3850 / π) ≈ 35 m
- Circumference = 2 * π * 35 ≈ 219.91 m

Total Cost of Fencing
- Total cost of fencing = Cost per meter * Circumference = 8.50 * 219.91 ≈ Rs. 1870
Therefore, the cost of fencing the field at Rs. 8.50 per meter is approximately Rs. 1870.

To find the area of the face of the clock described by the minute hand in 35 minutes, we need to find the length of the arc covered by the minute hand and then use it to calculate the area.

Length of Arc Covered by the Minute Hand:
The minute hand of the clock is 12 cm long. In 60 minutes, it covers a complete circle, which is equivalent to the circumference of the clock face. The formula for the circumference of a circle is given by:

C = 2πr

where C is the circumference and r is the radius.

In this case, the radius is equal to the length of the minute hand, which is 12 cm. So, the circumference of the clock face is:

C = 2π(12) = 24π cm

In 60 minutes, the minute hand covers the entire circumference. Therefore, in 35 minutes, it covers:

35/60 * 24π = 7/12 * 24π = 14π cm

Area of the Face of the Clock:
To find the area of the face of the clock described by the minute hand in 35 minutes, we need to calculate the area of the sector formed by the minute hand.

The formula for the area of a sector of a circle is given by:

A = (θ/360) * πr^2

where A is the area, θ is the central angle in degrees, and r is the radius.

In this case, the central angle is 35/60 * 360 = 210 degrees (since the minute hand covers 35 minutes out of 60 minutes, and each minute corresponds to 6 degrees on the clock face).

Substituting the values into the formula, we get:

A = (210/360) * π(12)^2
= (7/12) * π(144)
= 7π * 12
= 84π cm^2

Approximating π to 3.14, we have:

Area = 84 * 3.14
≈ 264 cm^2

Therefore, the area of the face of the clock described by the minute hand in 35 minutes is approximately 264 cm^2.

Hence, option B is the correct answer.

To solve this problem, we need to use the formulas for the area of a square and the area of a circle.

1. Area of a square:
The area of a square can be found by multiplying the length of one side by itself. Therefore, if the copper wire is bent in the form of a square and encloses an area of 484 cm^2, we can find the length of one side using the formula:
Area of square = side^2
484 = side^2
Taking the square root of both sides, we get:
side = √484
side = 22 cm

2. Perimeter of a square:
The perimeter of a square can be found by multiplying the length of one side by 4. In this case, the perimeter of the square formed by the copper wire is:
Perimeter of square = 4 * side
Perimeter of square = 4 * 22
Perimeter of square = 88 cm

3. Circumference of a circle:
The circumference of a circle can be found using the formula:
Circumference = π * diameter
Since the copper wire is bent in the form of a square, the length of the wire is equal to the perimeter of the square. Therefore, the circumference of the circle formed by the copper wire is:
Circumference of circle = 88 cm

4. Radius of the circle:
To find the radius of the circle, we can use the formula:
Radius = Circumference / (2 * π)
Radius = 88 / (2 * 3.14)
Radius ≈ 14 cm

5. Area of a circle:
Finally, we can find the area of the circle using the formula:
Area of circle = π * radius^2
Area of circle = 3.14 * 14^2
Area of circle ≈ 3.14 * 196
Area of circle ≈ 615.44 cm^2

Therefore, the area enclosed by the circle formed by the copper wire is approximately 616 cm^2, which corresponds to option A.

Given:
Side of the square = 10 cm

To find:
The area of the circumscribed circle.

Solution:
To find the area of the circumscribed circle, we need to find the radius of the circle first.

Finding the radius:
The diagonal of a square divides it into two congruent right-angled triangles.
Let's consider one of these triangles.

In a right-angled triangle, the hypotenuse (diagonal of the square) is equal to the diameter of the circle and the length of one side of the square is equal to the radius of the circle.

Using the Pythagorean theorem, we can find the length of the diagonal of the square:
(diagonal)^2 = (side)^2 + (side)^2
(diagonal)^2 = 10^2 + 10^2
(diagonal)^2 = 100 + 100
(diagonal)^2 = 200
diagonal = √200 = 10√2 cm

Since the diagonal of the square is equal to the diameter of the circle, the diameter of the circle is 10√2 cm.

So, the radius of the circle = (10√2)/2 = 5√2 cm.

Finding the area of the circle:
The area of a circle is given by the formula: A = πr^2, where r is the radius of the circle.

Substituting the value of the radius, we get:
A = π(5√2)^2
A = π(25*2)
A = 50π cm^2

Now, let's approximate the value of π to 3.14.

A ≈ 50 * 3.14
A ≈ 157 cm^2

Therefore, the area of the circumscribed circle is approximately 157 cm^2.

Answer:
The correct answer is option b) 157 cm^2.