CBSE Class 10 Mathematics Area Relates to Circles Assignment Set 08

X - Mathematics Assignment No-01 - Areas Related to Circles

Q1. Find the area of a circle whose circumference is equal to the perimeter of a square of side 11 cm.
Answer: Side of the square = 11 cm
Perimeter of the square = \(4 \times \text{side} = 4 \times 11 = 44\text{ cm}\)
Circumference of the circle = Perimeter of the square = \(44\text{ cm}\)
Let \(r\) be the radius of the circle.
Circumference = \(2\pi r = 44\)
\(2 \times \frac{22}{7} \times r = 44\)
\(\frac{44}{7} \times r = 44\)
\(r = \frac{44 \times 7}{44} = 7\text{ cm}\)
Area of the circle = \(\pi r^2 = \frac{22}{7} \times (7)^2 = \frac{22}{7} \times 49 = 154\text{ cm}^2\)
In simple words: First find the perimeter of the square, set it equal to the circle's circumference to find the radius, and then calculate the area using \(\pi r^2\).

Exam Tip: Always use \(\pi = \frac{22}{7}\) when dimensions are multiples of 7 for easier calculations.

Q2. If the perimeter of a semi-circular protractor is 54 cm, find the diameter and radius of the protractor.
Answer: Perimeter of a semi-circular protractor = \(\pi r + 2r = r(\pi + 2)\)
Given perimeter = 54 cm
\(r \left( \frac{22}{7} + 2 \right) = 54\)
\(r \left( \frac{22 + 14}{7} \right) = 54\)
\(r \left( \frac{36}{7} \right) = 54\)
\(r = \frac{54 \times 7}{36} = \frac{3 \times 7}{2} = \frac{21}{2} = 10.5\text{ cm}\)
Radius \(r = 10.5\text{ cm}\)
Diameter \(d = 2r = 2 \times 10.5 = 21\text{ cm}\)
In simple words: The perimeter of a semicircle includes both the curved boundary (\(\pi r\)) and the flat diameter (\(2r\)). Use this formula to solve for the radius.

Exam Tip: Do not confuse the perimeter of a semicircle with half of a circle's circumference; remember to add the diameter.

Q3. A wire is looped in the form of a circle of radius 28 cm. It is rebent into a square form. Find the diagonal of the square.
Answer: Radius of the circle \(r = 28\text{ cm}\)
Circumference of the circle = \(2\pi r = 2 \times \frac{22}{7} \times 28 = 2 \times 22 \times 4 = 176\text{ cm}\)
Length of the wire (Perimeter of the square) = 176 cm
Let \(a\) be the side of the square.
\(4a = 176\)
\(a = \frac{176}{4} = 44\text{ cm}\)
Diagonal of the square = \(a\sqrt{2} = 44\sqrt{2}\text{ cm}\)
In simple words: The length of the wire remains constant, so the circle's circumference equals the square's perimeter. Find the side length and then use the diagonal formula.

Exam Tip: The diagonal of a square with side \(a\) is always \(a\sqrt{2}\).

Q4. The area of a square is 484 cm\(^2\). The wire of the square is bent to form a circle. Find the area of circle.
Answer: Area of the square = 484 cm\(^2\)
Let \(a\) be the side of the square.
\(a^2 = 484\)
\(a = \sqrt{484} = 22\text{ cm}\)
Perimeter of the square = \(4a = 4 \times 22 = 88\text{ cm}\)
Circumference of the circle = Perimeter of the square = 88 cm
\(2\pi r = 88\)
\(2 \times \frac{22}{7} \times r = 88\)
\(\frac{44}{7} \times r = 88\)
\(r = \frac{88 \times 7}{44} = 2 \times 7 = 14\text{ cm}\)
Area of the circle = \(\pi r^2 = \frac{22}{7} \times (14)^2 = \frac{22}{7} \times 196 = 22 \times 28 = 616\text{ cm}^2\)
In simple words: Find the square's side from its area, calculate the perimeter to get the circle's circumference, find the radius, and finally calculate the circle's area.

Exam Tip: Memorize squares of numbers up to 30 (like \(22^2 = 484\)) to save time in exams.

Q5. A wheel makes 5000 revolutions and covers a distance of 11 km. Find the area of the wheel.
Answer: Total distance covered = \(11\text{ km} = 11 \times 1000 \times 100\text{ cm} = 1,100,000\text{ cm}\)
Number of revolutions = 5000
Distance covered in 1 revolution (Circumference) = \(\frac{\text{Total Distance}}{\text{Number of revolutions}}\)
Circumference = \(\frac{1,100,000}{5000} = \frac{1100}{5} = 220\text{ cm}\)
Let \(r\) be the radius of the wheel.
\(2\pi r = 220\)
\(2 \times \frac{22}{7} \times r = 220\)
\(\frac{44}{7} \times r = 220\)
\(r = \frac{220 \times 7}{44} = 5 \times 7 = 35\text{ cm}\)
Area of the wheel = \(\pi r^2 = \frac{22}{7} \times (35)^2 = \frac{22}{7} \times 35 \times 35 = 22 \times 5 \times 35 = 3850\text{ cm}^2\)
In simple words: Divide the total distance by the number of revolutions to find the circumference, find the radius, and then compute the area.

Q6. The circumference of a circular wheel exceeds the diameter by 16.8 cm. Find the diameter of the wheel.
Answer: Let the diameter of the wheel be \(d\) and radius be \(r\) (where \(d = 2r\)).
According to the question, Circumference - Diameter = 16.8 cm
\(2\pi r - d = 16.8\)
\(\pi d - d = 16.8 \quad (\text{since } 2r = d)\)
\(d(\pi - 1) = 16.8\)
\(d \left( \frac{22}{7} - 1 \right) = 16.8\)
\(d \left( \frac{22 - 7}{7} \right) = 16.8\)
\(d \left( \frac{15}{7} \right) = 16.8\)
\(d = \frac{16.8 \times 7}{15} = \frac{117.6}{15} = 7.84\text{ cm}\)
In simple words: Set up an equation where circumference minus diameter equals 16.8 cm, substitute \(\pi d\), and solve for diameter \(d\).

Exam Tip: Expressing circumference directly in terms of diameter (\(\pi d\)) simplifies problems involving both quantities.

Q7. The circumference of a circular park is 352 m. A road 7 m wide is around the park. Find the area of the road.
Answer: Let the radius of the circular park be \(r\).
Circumference = \(2\pi r = 352\)
\(2 \times \frac{22}{7} \times r = 352\)
\(\frac{44}{7} \times r = 352\)
\(r = \frac{352 \times 7}{44} = 8 \times 7 = 56\text{ m}\)
Width of the road = 7 m
Radius of the outer circle (\(R\)) = \(r + 7 = 56 + 7 = 63\text{ m}\)
Area of the road = Area of outer circle - Area of inner park
\(= \pi R^2 - \pi r^2 = \pi (R^2 - r^2) = \pi (R - r)(R + r)\)
\(= \frac{22}{7} \times (63 - 56) \times (63 + 56)\)
\(= \frac{22}{7} \times 7 \times 119\)
\(= 22 \times 119 = 2618\text{ m}^2\)
In simple words: Find the inner radius from the circumference, add the road width to get the outer radius, and subtract the inner circle's area from the outer circle's area.

Exam Tip: Use the algebraic identity \(R^2 - r^2 = (R-r)(R+r)\) to make area calculations of concentric circles much faster.

Q8. Two circles touch externally. The sum of their areas is \(130\pi\text{ cm}^2\) and the distance between their centres is 14 cm. Find the radii of the circles.
Answer: Let the radii of the two circles be \(r_1\) and \(r_2\).
Since they touch externally, the distance between their centres is equal to the sum of their radii:
\(r_1 + r_2 = 14 \implies r_2 = 14 - r_1\) --- (1)
Given the sum of their areas is \(130\pi\text{ cm}^2\):
\(\pi r_1^2 + \pi r_2^2 = 130\pi\)
\(r_1^2 + r_2^2 = 130\) --- (2)
Substitute equation (1) into equation (2):
\(r_1^2 + (14 - r_1)^2 = 130\)
\(r_1^2 + 196 - 28r_1 + r_1^2 = 130\)
\(2r_1^2 - 28r_1 + 196 - 130 = 0\)
\(2r_1^2 - 28r_1 + 66 = 0\)
Divide by 2:
\(r_1^2 - 14r_1 + 33 = 0\)
Factorize the quadratic equation:
\(r_1^2 - 11r_1 - 3r_1 + 33 = 0\)
\(r_1(r_1 - 11) - 3(r_1 - 11) = 0\)
\((r_1 - 11)(r_1 - 3) = 0\)
So, \(r_1 = 11\) or \(r_1 = 3\).
If \(r_1 = 11\text{ cm}\), then \(r_2 = 14 - 11 = 3\text{ cm}\).
If \(r_1 = 3\text{ cm}\), then \(r_2 = 14 - 3 = 11\text{ cm}\).
Thus, the radii of the two circles are 3 cm and 11 cm.
In simple words: Set up two equations: one for the sum of radii equaling the center distance, and another for the sum of areas. Solve the resulting quadratic equation.

Exam Tip: For externally touching circles, always remember distance between centers = \(r_1 + r_2\).

Q9. Two circles touch internally. The sum of their areas is \(116\pi\text{ cm}^2\) and the distance between their centres is 6 cm. Find the radii of the circles.
Answer: Let the radii of the two circles be \(r_1\) and \(r_2\) (where \(r_1 > r_2\)).
Since they touch internally, the distance between their centres is equal to the difference of their radii:
\(r_1 - r_2 = 6 \implies r_1 = r_2 + 6\) --- (1)
Given the sum of their areas is \(116\pi\text{ cm}^2\):
\(\pi r_1^2 + \pi r_2^2 = 116\pi\)
\(r_1^2 + r_2^2 = 116\) --- (2)
Substitute equation (1) into equation (2):
\((r_2 + 6)^2 + r_2^2 = 116\)
\(r_2^2 + 12r_2 + 36 + r_2^2 = 116\)
\(2r_2^2 + 12r_2 + 36 - 116 = 0\)
\(2r_2^2 + 12r_2 - 80 = 0\)
Divide by 2:
\(r_2^2 + 6r_2 - 40 = 0\)
Factorize the quadratic equation:
\(r_2^2 + 10r_2 - 4r_2 - 40 = 0\)
\(r_2(r_2 + 10) - 4(r_2 + 10) = 0\)
\((r_2 + 10)(r_2 - 4) = 0\)
Since radius cannot be negative, \(r_2 = 4\text{ cm}\).
Substitute \(r_2 = 4\) into equation (1):
\(r_1 = 4 + 6 = 10\text{ cm}\)
Thus, the radii of the circles are 4 cm and 10 cm.
In simple words: For internally touching circles, the distance between centers is the difference of radii (\(r_1 - r_2 = 6\)). Combine this with the area sum equation to solve for radii.

Exam Tip: Be careful with signs: internal touch means difference of radii, external touch means sum of radii.

Q10. Three horses are tethered at three corners of a triangular plot having sides 20 m, 30 m and 40 m with ropes of 7 m length each. Find the area of this plot which can be grazed by the horses.
Answer: Let the three corners of the triangular plot have interior angles \(A\), \(B\), and \(C\).
Each horse grazes a sector of a circle of radius \(r = 7\text{ m}\).
The area grazed by the three horses represents the sum of the areas of the three sectors at the three vertices:
Total grazed area = \(\frac{\angle A}{360^{\circ}} \pi r^2 + \frac{\angle B}{360^{\circ}} \pi r^2 + \frac{\angle C}{360^{\circ}} \pi r^2\)
\(= \frac{\pi r^2}{360^{\circ}} (\angle A + \angle B + \angle C)\)
Since the sum of angles in a triangle is \(180^{\circ}\) (\(\angle A + \angle B + \angle C = 180^{\circ}\)):
Total grazed area = \(\frac{180^{\circ}}{360^{\circ}} \times \pi r^2 = \frac{1}{2} \times \pi r^2\)
Substitute \(r = 7\text{ m}\) and \(\pi = \frac{22}{7}\):
\(= \frac{1}{2} \times \frac{22}{7} \times (7)^2\)
\(= \frac{1}{2} \times \frac{22}{7} \times 49 = 11 \times 7 = 77\text{ m}^2\)
In simple words: Instead of finding individual angles, use the fact that the sum of all three corner angles in any triangle is 180 degrees. This combines the three grazed sectors into a single semicircle of radius 7 m.

Exam Tip: This clever shortcut (\(\frac{\text{sum of angles}}{360^{\circ}} \times \pi r^2\)) works for any polygon where horses are tethered at all corners.