OP Malhotra Class 9 Maths Solutions Chapter 17 Circle Circumference and Area Exercise 17 (B)

S Chand Class 9 ICSE Maths Solutions Chapter 17 Circle: Circumference and Area Ex 17(B)

Take \( \pi = \frac {22}{7} \) unless mentioned otherwise.

Question 1. Find the area of the circle whose :
(ii) radius = 14 cm.
(iii) diameter = 2.8 cm.
Answer:
(i) Diameter = 7 cm
The radius \( (r) = \frac { 7 }{ 2 } \) cm.
\( \implies \) Area of the circle \( = \pi r^2 = \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac {7}{ 2 } \text{ cm}^2 \)
\( \implies = \frac { 77 }{ 2 } = 38.5 \text{ cm}^2 \)
(ii) Radius \( (r) = 14 \) cm
\( \implies \) Area \( = \pi r^2 = \frac { 22 }{ 7 } \times 14 \times 14 \text{ cm}^2 = 616 \text{ cm}^2 \)
(iii) Diameter \( = 2.8 \) cm
\( \implies \) Radius \( (r) = \frac { 2.8 }{ 2 } = 1.4 \) cm
\( \implies \) Area \( = \pi r^2 = \frac { 22 }{ 7 } \times 1.4 \times 1.4 \text{ cm}^2 \)
\( \implies = 6.16 \text{ cm}^2 \)
In simple words: To find the area of a circle, first find the radius. If you have the diameter, divide it by 2 to get the radius. Then, use the formula \( \pi r^2 \), where \( \pi \) is approximately \( \frac{22}{7} \). Remember to always include the correct units in your final answer.

🎯 Exam Tip: Pay close attention to whether the question provides the radius or the diameter, as using the wrong one directly will lead to an incorrect area calculation.

Question 2. A horse is tied to a pole with 28 m long string. Find out the area which the horse can graze.
Answer: The length of the string is 28 m, which acts as the radius for the horse's grazing area.
\( \implies \) Radius \( (r) = 28 \) m
The area the horse can graze is the area of a circle.
\( \implies \) Area \( = \pi r^2 = \frac { 22 }{ 7 } \times 28 \times 28 \text{ m}^2 = 2464 \text{ m}^2 \)
In simple words: The string acts like the radius of a circle, and the horse can graze anywhere inside that circle. So, we find the area of a circle using the string's length as the radius.

🎯 Exam Tip: When a problem describes something tied to a point and moving freely, imagine it sweeping out a circle. The length of the string or tether will always be the radius of that circle.

Question 3. Find the radius of a circular field whose area is
(ii) 1386 cm²
Answer:
(i) Area of a circular field \( = 154 \text{ cm}^2 \)
Let \( r \) be the radius.
\( \implies \pi r^2 = 154 \)
\( \implies \frac { 22 }{ 7 } r^2 = 154 \)
\( \implies r^2 = \frac{154 \times 7}{22} = 49 \)
\( \implies r^2 = (7)^2 \)
\( \implies r = 7 \)
Therefore, the radius \( = 7 \) cm.
(ii) Area of the circular field \( = 1386 \text{ cm}^2 \)
Let \( r \) be the radius of the field.
\( \implies \pi r^2 = 1386 \)
\( \implies \frac { 22 }{ 7 } r^2 = 1386 \)
\( \implies r^2 = \frac{1386 \times 7}{22} = 63 \times 7 = 441 = (21)^2 \)
\( \implies r = 21 \)
Therefore, the radius \( = 21 \) cm.
In simple words: To find the radius when you know the area, you work backward from the area formula. Divide the area by \( \pi \) and then find the square root of that number to get the radius. This is the opposite of finding the area from the radius.

🎯 Exam Tip: Remember that \( r^2 \) means \( r \times r \). To find \( r \), you need to take the square root of \( r^2 \). Make sure to perform inverse operations correctly when rearranging formulas.

Question 4. The area of a circle is 24.64 cm². Find its circumference.
Answer: The area of the circle is \( 24.64 \text{ cm}^2 \). Let \( r \) be the radius.
\( \implies \pi r^2 = 24.64 \)
\( \implies \frac { 22 }{ 7 } r^2 = 24.64 \)
\( \implies r^2 = \frac{24.64 \times 7}{22} = 1.12 \times 7 = 7.84 \)
\( \implies r^2 = (2.8)^2 \)
\( \implies r = 2.8 \) cm.
Now, the circumference of the circle is \( 2\pi r \).
\( \implies \text{Circumference} = 2 \times \frac { 22 }{ 7 } \times 2.8 \text{ cm} = 17.6 \text{ cm} \)
In simple words: First, use the given area to find the circle's radius. Once you have the radius, you can easily calculate the circumference using the formula \( 2 \pi r \). This involves two steps: finding the radius, then finding the circumference.

🎯 Exam Tip: This is a two-step problem. First, correctly find the radius from the area, and then use that radius to find the circumference. Avoid rounding until the final answer to maintain accuracy.

Question 5. A copper wire, when bent in the form of a square, encloses an area of 121 cm². If the same wire is bent into the form of a circle, find the area of the circle.
Answer: The area of the square is \( 121 \text{ cm}^2 \).
\( \implies \) Side of the square \( (a) = \sqrt{\text{Area}} = \sqrt{121} = 11 \) cm.
The length of the wire is the perimeter of the square.
\( \implies \text{Perimeter of the wire} = 4a = 4 \times 11 = 44 \) cm.
When the same wire is bent into a circle, its length becomes the circumference of the circle.
\( \implies \text{Circumference} = 44 \) cm.
Now, we find the radius of this circle. The circumference formula is \( 2\pi r \).
\( \implies 2\pi r = 44 \)
\( \implies 2 \times \frac{22}{7} \times r = 44 \)
\( \implies r = \frac{44 \times 7}{2 \times 22} = 7 \) cm.
Finally, we find the area of the circle.
\( \implies \text{Area of the circle} = \pi r^2 = \frac { 22 }{ 7 } \times 7 \times 7 = 154 \text{ cm}^2 \)
In simple words: The wire has a fixed length. First, calculate this length by finding the perimeter of the square. Then, use this length as the circumference of the circle to find its radius. Finally, use the radius to find the area of the circle.

🎯 Exam Tip: The key to solving this problem is understanding that the length of the wire remains constant. Therefore, the perimeter of the square is equal to the circumference of the circle.

Question 6. The circumference of a circle is equal to the perimeter of a square. The area of the square is 484 sq m. Find the area of the circle.
Answer: The area of the square is \( 484 \text{ sq m} \).
\( \implies \) Side of the square \( (a) = \sqrt{\text{Area}} = \sqrt{484} \text{ m} = 22 \text{ m} \).
The perimeter of the square \( = 4a = 4 \times 22 \text{ m} = 88 \text{ m} \).
We are told that the circumference of the circle is equal to the perimeter of the square.
\( \implies \text{Circumference of the circle} = 88 \text{ m} \).
Now we find the radius \( (r) \) of the circle using the circumference formula \( 2\pi r \).
\( \implies 2\pi r = 88 \)
\( \implies r = \frac{\text{circumference}}{2\pi} = \frac{88 \times 7}{2 \times 22} \text{ m} = 14 \text{ m} \).
Finally, we calculate the area of the circle.
\( \implies \text{Area of the circle} = \pi r^2 = \frac { 22 }{ 7 } \times 14 \times 14 \text{ sq. m} = 616 \text{ sq. m} \)
In simple words: First, figure out the side of the square using its area, then find the square's perimeter. This perimeter is the same as the circle's circumference. Use the circumference to find the circle's radius, and then use that radius to get the circle's area.

🎯 Exam Tip: Remember to carry out each step carefully: square root for side from area, multiply by 4 for perimeter, use perimeter as circumference to find radius, then use radius for circle area.

Question 7. A wire is in the form of a circle of radius 42 cm. It is bent into a square. Determine the side of the square and compare the areas of the regions enclosed in the two cases.
Answer: The radius of the circular wire \( (r) = 42 \) cm.
The circumference of the circle \( = 2\pi r \).
\( \implies 2 \times \frac { 22 }{ 7 } \times 42 = 264 \) cm.
(i) When the wire is bent into a square, its length becomes the perimeter of the square.
\( \implies \text{Perimeter of the square} = 264 \) cm.
The side of the square \( = \frac { 264 }{ 4 } = 66 \) cm.
(ii) Now, we calculate the area of the circle and the square.
Area of the circle \( = \pi r^2 = \frac { 22 }{ 7 } \times 42 \times 42 \text{ cm}^2 = 5544 \text{ cm}^2 \).
Area of the square \( = (\text{side})^2 = (66)^2 = 4356 \text{ cm}^2 \).
To compare the areas, we find the ratio:
\( \implies \text{Ratio of areas} = 5544 : 4356 \)
We can simplify this ratio by dividing both numbers by common factors.
\( \implies \frac{5544}{4356} = \frac{56}{44} = \frac{14}{11} \).
So the ratio is \( 14 : 11 \).
In simple words: First, find the total length of the wire by calculating the circle's circumference. This length is then used as the perimeter of the square to find its side. After that, calculate the area of both the circle and the square, and compare them by finding their ratio.

🎯 Exam Tip: Always remember that when a wire is reshaped, its total length (perimeter or circumference) stays the same. The area, however, changes depending on the shape, so always calculate and compare both areas.

Question 8. From a copper plate, which is a square of side 12.5 cm, a circular disc of diameter 7 cm is cut off. Find the weight of the remaining part, if 1 sq. cm of the plate weighs 0.8 gm.
Answer: The side of the square plate is \( 12.5 \) cm.
The diameter of the circular disc cut off is \( 7 \) cm.

The radius of the circular disc \( (r) = \frac { 7 }{ 2 } \) cm.
Area of the square plate \( = \text{side}^2 = (12.5)^2 \text{ cm}^2 = 156.25 \text{ cm}^2 \).
Area of the circular disc \( = \pi r^2 = \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \text{ cm}^2 = \frac { 77 }{ 2 } = 38.5 \text{ cm}^2 \).
Area of the remaining part \( = \text{Area of square} - \text{Area of circular disc} \).
\( \implies = 156.25 - 38.50 = 117.75 \text{ cm}^2 \).
The weight of one sq. cm of the plate is \( 0.8 \) gm.
Total weight of the remaining portion \( = 117.75 \times 0.8 \text{ gm} = 94.2 \text{ gm} \).
In simple words: First, find the total area of the square plate. Then, calculate the area of the circle that was cut out. Subtract the circle's area from the square's area to find the remaining area. Finally, multiply this remaining area by the weight per square centimeter to get the total weight of the remaining copper plate.

🎯 Exam Tip: When a portion is "cut off," it means you need to subtract its area. Ensure accurate calculations for both square and circle areas, and then multiply by the given weight per unit area.

Question 9. The circumference of two circles are in the ratio 2: 3. Find the ratio of their areas.
Answer: Let the circumferences of the two circles be \( C_1 \) and \( C_2 \).
The ratio of their circumferences is given as \( 2:3 \).
\( \implies \frac{C_1}{C_2} = \frac{2}{3} \).
Let the radius of the first circle be \( r_1 \) and the second circle be \( r_2 \).
\( C_1 = 2\pi r_1 \) and \( C_2 = 2\pi r_2 \).
\( \implies \frac{2\pi r_1}{2\pi r_2} = \frac{2}{3} \)
\( \implies \frac{r_1}{r_2} = \frac{2}{3} \).
So, the ratio of their radii is also \( 2:3 \).
Now, we need to find the ratio of their areas. Let \( A_1 \) and \( A_2 \) be their areas.
\( A_1 = \pi r_1^2 \) and \( A_2 = \pi r_2^2 \).
\( \implies \frac{A_1}{A_2} = \frac{\pi r_1^2}{\pi r_2^2} = \frac{r_1^2}{r_2^2} = \left(\frac{r_1}{r_2}\right)^2 \).
Since \( \frac{r_1}{r_2} = \frac{2}{3} \).
\( \implies \frac{A_1}{A_2} = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \).
The ratio of their areas is \( 4:9 \).
In simple words: If the ratio of the circumferences of two circles is known, then the ratio of their radii is the same. To find the ratio of their areas, simply square the ratio of their radii. This means if the radii are in a ratio of 2:3, the areas will be in a ratio of \( 2^2 : 3^2 \).

🎯 Exam Tip: Remember that circumference is directly proportional to radius, while area is proportional to the square of the radius. This relationship is crucial for quickly solving ratio problems involving circles.

Question 10. A square park has each side of 100 m. At each corner of the park, there is a flower bed in the form of a quadrant of radius 14 m as shown in the figure. Find the area of the remaining part of the park. (Take \( \pi = \frac{22}{7} \))
Answer: The side of the square park is \( 100 \) m.
Area of the square park \( = (\text{side})^2 = (100)^2 = 10000 \text{ m}^2 \).
At each corner, there is a flower bed in the form of a quadrant (a quarter circle) with radius \( 14 \) m.

Area of one quadrant \( = \frac { 1 }{ 4 }\pi r^2 \).
\( \implies \frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times (14) \times (14) \text{ m}^2 = 154 \text{ m}^2 \).
Since there are 4 such quadrants (one at each corner), the total area of the 4 quadrants \( = 154 \times 4 = 616 \text{ m}^2 \).
The area of the remaining portion of the park \( = \text{Area of square park} - \text{Total area of 4 quadrants} \).
\( \implies = 10000 - 616 = 9384 \text{ m}^2 \).
In simple words: First, find the total area of the square park. Then, calculate the area of one flower bed (which is a quarter of a circle). Since there are four such beds, multiply that area by four to get the total area of all flower beds. Finally, subtract this total from the square park's area to find what's left.

🎯 Exam Tip: Remember that four quadrants (quarter circles) of the same radius combine to form one full circle. So, you could also calculate the area of one full circle with the given radius and subtract it from the square's area.

Question 11. The radius of a circular field is 20 m. Inside it runs a path 5 m wide all around. Find the area of the path.
Answer: The radius of the circular field \( (R) = 20 \) m.
The width of the inside path is \( 5 \) m.

The inner radius \( (r) = \text{Outer radius} - \text{Width of path} \).
\( \implies r = 20 - 5 = 15 \) m.
The area of the path is the difference between the outer circle's area and the inner circle's area.
\( \implies \text{Area of path} = \pi R^2 - \pi r^2 = \pi (R^2 - r^2) \).
We can use the difference of squares formula, \( a^2 - b^2 = (a-b)(a+b) \).
\( \implies \text{Area of path} = \pi (R - r)(R + r) \).
\( \implies = \frac { 22 }{ 7 } [(20)^2 - (15)^2] \text{ m}^2 \)
\( \implies = \frac { 22 }{ 7 } (20 + 15) (20 - 15) \text{ m}^2 \)
\( \implies = \frac {22}{7} \times 35 \times 5 = 550 \text{ m}^2 \).
In simple words: When there's a path around or inside a circular area, its area is found by subtracting the area of the smaller circle from the area of the larger circle. First, find both radii, then calculate their respective areas, and finally subtract.

🎯 Exam Tip: For problems involving concentric circles (a circle within a circle), the area of the region between them is best found using the formula \( \pi(R^2 - r^2) \), where R is the outer radius and r is the inner radius. The algebraic identity \( (R^2 - r^2) = (R-r)(R+r) \) often simplifies calculations.

Question 12. A road 3.5 m wide surrounds a circular plot whose circumference is 44 m. Find the cost of paving the road of Rs. 10 per m².
Answer: The width of the road is \( 3.5 \) m.
The circumference of the circular plot is \( 44 \) m.

First, find the radius of the inner circular plot \( (r) \) from its circumference.
\( \implies \text{Circumference} = 2\pi r = 44 \).
\( \implies 2 \times \frac{22}{7} \times r = 44 \).
\( \implies r = \frac{44 \times 7}{2 \times 22} = 7 \) m.
The outer radius \( (R) = \text{Inner radius} + \text{Width of road} \).
\( \implies R = 7 + 3.5 = 10.5 \) m.
Area of the road \( = \text{Area of outer circle} - \text{Area of inner circle} = \pi R^2 - \pi r^2 \).
\( \implies = \pi (R^2 - r^2) = \pi [(10.5)^2 - (7)^2] \text{ m}^2 \).
\( \implies = \frac { 22 }{ 7 } (10.5 + 7) (10.5 - 7) \text{ m}^2 \).
\( \implies = \frac { 22 }{ 7 } \times 17.5 \times 3.5 \text{ m}^2 = 192.5 \text{ m}^2 \).
The cost of paving the road is Rs. \( 10 \) per m².
Total cost \( = 192.5 \times 10 = \text{Rs. } 1925 \).
In simple words: First, use the circumference to find the radius of the circular plot. Then, add the road's width to this radius to get the outer radius. Calculate the area of the road by subtracting the inner circle's area from the outer circle's area. Finally, multiply the road's area by the cost per square meter to find the total cost.

🎯 Exam Tip: Always draw a quick sketch to visualize the inner and outer radii. Remember to correctly apply the circumference formula to find the inner radius first, as this is a common point of error.

Question 13. A lawn is in the shape of a semi-circle of diameter 35 dm. The lawn is surrounded by a flower bed of width 3.5 dm all round. Find the area of the flower bed in dm².
Answer: The inner diameter of the semi-circular lawn is \( 35 \) dm.
Inner radius \( (r) = \frac { 35 }{ 2 } = 17.5 \) dm.
The width of the flower bed is \( 3.5 \) dm.

The outer radius \( (R) = \text{Inner radius} + \text{Width of flower bed} \).
\( \implies R = 17.5 + 3.5 = 21 \) dm.
The area of the flower bed is the difference between the outer semi-circle's area and the inner semi-circle's area.
\( \implies \text{Area of flower bed} = \frac { 1 }{ 2 } \pi R^2 - \frac { 1 }{ 2 } \pi r^2 = \frac { 1 }{ 2 } \pi (R^2 - r^2) \).
\( \implies = \frac { 1 }{ 2 } \times \frac { 22 }{ 7 } [(21)^2 - (17.5)^2] \text{ dm}^2 \).
\( \implies = \frac { 11 }{ 7 } (21 + 17.5) (21 - 17.5) \text{ dm}^2 \).
\( \implies = \frac { 11 }{ 7 } \times 38.5 \times 3.5 \text{ dm}^2 \).
\( \implies = 11 \times 5.5 \times 3.5 \text{ dm}^2 \).
\( \implies = 211.75 \text{ dm}^2 \).
In simple words: First, calculate the inner radius of the semi-circular lawn from its diameter. Add the flower bed's width to get the outer radius. Then, find the area of the flower bed by subtracting the inner semi-circle's area from the outer semi-circle's area. Remember that for a semi-circle, you use half the area formula of a full circle.

🎯 Exam Tip: When dealing with semi-circles, always remember to divide the full circle's area formula by two. Also, be careful to add the width of the path to the radius for the outer shape, not the diameter.

Question 14. A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the surface area of the road.
Answer: The width of the road is \( 7 \) m.
The circumference of the circular park is \( 352 \) m.

First, find the radius of the inner circular park \( (r) \) from its circumference.
\( \implies \text{Circumference} = 2\pi r = 352 \).
\( \implies r = \frac{\text{circumference}}{2\pi} = \frac{352 \times 7}{2 \times 22} = 56 \) m.
The outer radius \( (R) = \text{Inner radius} + \text{Width of road} \).
\( \implies R = 56 + 7 = 63 \) m.
The area of the road \( = \text{Area of outer circle} - \text{Area of inner circle} = \pi R^2 - \pi r^2 \).
\( \implies = \pi (R^2 - r^2) = \pi [(63)^2 - (56)^2] \text{ m}^2 \).
\( \implies = \frac { 22 }{ 7 } (63 + 56) (63 - 56) \text{ m}^2 \).
\( \implies = \frac { 22 }{ 7 } \times 119 \times 7 \text{ m}^2 = 2618 \text{ m}^2 \).
In simple words: First, use the park's circumference to find its radius. Add the road's width to this radius to find the outer radius. Then, subtract the area of the park from the area of the larger circle (park plus road) to get the road's area.

🎯 Exam Tip: This question involves a common pattern of finding the area of a circular ring (annulus). Make sure to correctly identify the inner and outer radii before applying the area difference formula.

Question 15. The figure between the circumferences of two concentric circles is 346.5 cm². The circumference of the inner circle is 88 cm. Calculate the radius of the outer circle.
Answer: The area of the shaded portion (the region between the two concentric circles) is \( 346.5 \text{ cm}^2 \).
The circumference of the inner circle is \( 88 \) cm.

First, find the radius of the inner circle \( (r) \) from its circumference.
\( \implies \text{Circumference} = 2\pi r = 88 \).
\( \implies r = \frac{88 \times 7}{2 \times 22} = 14 \) cm.
Let the outer radius be \( R \).
The area of the shaded portion \( = \pi R^2 - \pi r^2 = \pi (R^2 - r^2) \).
\( \implies 346.5 = \pi [R^2 - (14)^2] \).
\( \implies 346.5 = \frac{22}{7} [R^2 - 196] \).
\( \implies R^2 - 196 = \frac{346.5 \times 7}{22} \).
\( \implies R^2 - 196 = 110.25 \).
\( \implies R^2 = 110.25 + 196 = 306.25 \).
\( \implies R = \sqrt{306.25} = 17.5 \) cm.
The radius of the outer circle is \( 17.5 \) cm.
In simple words: First, use the inner circle's circumference to find its radius. Then, set up an equation where the area of the shaded region (the big circle's area minus the small circle's area) equals 346.5. Solve this equation for the unknown outer radius.

🎯 Exam Tip: Always solve for the inner radius first if its circumference is given. Then, substitute this value into the formula for the area between two concentric circles to find the unknown outer radius.

Question 16. Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm. A third concentric circle is drawn outside the 7 cm circle so that the area enclosed between it and the 7 cm circle is the same as that between the two inner circles. Find the radius of the third circle, correct to one decimal place.
Answer: The radius of the first inner circle \( (r_1) = 3.5 \) cm.
The radius of the second circle \( (r_2) = 7 \) cm.

Area between the two inner circles \( = \pi (r_2^2 - r_1^2) \).
\( \implies = \frac { 22 }{ 7 } [(7)^2 - (3.5)^2] \text{ cm}^2 \).
\( \implies = \frac { 22 }{ 7 } (7 + 3.5) (7 - 3.5) \text{ cm}^2 \).
\( \implies = \frac { 22 }{ 7 } \times 10.5 \times 3.5 \text{ cm}^2 = 115.5 \text{ cm}^2 \).
Let the radius of the third circle be \( R \).
The area between the second circle (radius \( r_2 = 7 \) cm) and the third circle (radius \( R \)) is the same as the area between the two inner circles, which is \( 115.5 \text{ cm}^2 \).
\( \implies \pi (R^2 - r_2^2) = 115.5 \).
\( \implies \frac { 22 }{ 7 } [R^2 - (7)^2] = 115.5 \).
\( \implies \frac { 22 }{ 7 } [R^2 - 49] = 115.5 \).
\( \implies R^2 - 49 = \frac{115.5 \times 7}{22} \).
\( \implies R^2 - 49 = 36.75 \).
\( \implies R^2 = 36.75 + 49 = 85.75 \text{ cm}^2 \).
\( \implies R = \sqrt{85.75} \approx 9.26 \) cm.
The radius of the third circle, correct to one decimal place, is \( 9.3 \) cm.
In simple words: First, calculate the area between the two given inner circles. Since the area between the second and third circle is the same, use this area to find the radius of the third circle. You'll subtract the known area of the second circle from the total area and then solve for the unknown radius.

🎯 Exam Tip: When dealing with multiple concentric circles, identify each radius correctly and keep track of which area corresponds to which pair of circles. Rounding should only be done at the very final step, as requested.

Question 17. A circular field has a perimeter of 650 m. A plot, in the shape of a square having its vertices on the circumference of the field, is marked in the field. Calculate the area of the square plot.
Answer: The perimeter of the circular field (its circumference) is \( 650 \) m.
The diameter of the field \( = \frac{\text{circumference}}{\pi} \).
\( \implies = \frac{650}{\frac{22}{7}} = \frac{650 \times 7}{22} \) m.

The square plot has its vertices on the circumference of the circle. This means the diagonal of the square is equal to the diameter of the circle.
\( \implies \text{Diagonal of square} = \text{Diameter of the circle} = \frac{650 \times 7}{22} \) m.
The area of a square can also be calculated using its diagonal \( d \) with the formula \( \text{Area} = \frac{d^2}{2} \).
\( \implies \text{Area of square plot} = \frac{1}{2} \left(\frac{650 \times 7}{22}\right)^2 \)
\( \implies = \frac{1}{2} \left(\frac{4550}{22}\right)^2 \)
\( \implies = \frac{1}{2} (206.818)^2 \)
\( \implies = \frac{1}{2} (42773.76) \text{ m}^2 = 21386.88 \text{ m}^2 \).
Rounded to the nearest whole number, the area of the square plot is \( 21387 \text{ m}^2 \).
In simple words: First, use the circular field's perimeter (circumference) to find its diameter. Because the square's corners touch the circle, the circle's diameter is the same as the square's diagonal. Then, use the formula for a square's area based on its diagonal to get the final answer.

🎯 Exam Tip: Recognize that when a square is inscribed in a circle (vertices on the circumference), the diagonal of the square is equal to the diameter of the circle. This relationship simplifies finding the square's area.

Question 18. The inside perimeter of a practice running track with semicircular ends and straight parallel sides is 312 m. The length of the straight portions of the track is 90 m. If the track has uniform width of 2 m throughout, find its area.
Answer: The inside perimeter of the running track is \( 312 \) m.
The length of the straight portions is \( 90 \) m. There are two straight portions.

The perimeter of the two inner semi-circular ends \( = \text{Total perimeter} - (\text{2} \times \text{length of straight portion}) \).
\( \implies = 312 - (2 \times 90) = 312 - 180 = 132 \) m.
Since there are two semi-circular ends, the perimeter of one hemisphere end \( = \frac{132}{2} = 66 \) m.
This perimeter is half the circumference of a circle \( (2\pi r) \). So, \( \pi r = 66 \).
\( \implies r = \frac{66}{\pi} = \frac{66 \times 7}{22} = 3 \times 7 = 21 \) m.
So, the inner radius of the semi-circular ends is \( 21 \) m.
The track has a uniform width of \( 2 \) m.
The outer radius \( (R) = \text{Inner radius} + \text{Width} = 21 + 2 = 23 \) m.
The area of the track is the sum of the area of the two rectangular portions and the area of the two semi-circular ends (which form a circular ring).
Area of two rectangular portions \( = 2 \times (\text{length} \times \text{width}) = 2 \times (90 \times 2) = 360 \text{ m}^2 \).
Area of the two semi-circular ends \( = \pi R^2 - \pi r^2 \) (because two semi-circular rings form one full circular ring).
\( \implies = \pi (R^2 - r^2) = \pi [(23)^2 - (21)^2] \).
\( \implies = \pi (23 + 21) (23 - 21) \).
\( \implies = \pi \times 44 \times 2 = 88\pi \text{ m}^2 \).
Total area of the track \( = 88\pi + 360 \text{ m}^2 \).
If \( \pi = 3.14 \), then \( = (88 \times 3.14) + 360 = 276.32 + 360 = 636.32 \text{ m}^2 \).
In simple words: First, find the length of the curved parts of the inner track by subtracting the straight parts from the total perimeter. Use this to find the inner radius of the semi-circles. Then, add the track's width to get the outer radius. Calculate the area of the two straight rectangular parts. Calculate the area of the two curved parts by subtracting the inner semi-circular areas from the outer ones. Add these two areas to get the total area of the track.

🎯 Exam Tip: Break down complex shapes like running tracks into simpler geometric figures (rectangles and circular rings). Calculate each component's area separately and then sum them up. Ensure you use the correct radii for inner and outer semi-circles.

Question 19. The boundary of the shaded region in the figure consists of three semi-circular arcs, the smaller ones being equal. If the diameter of the larger arc is 10 cm. Calculate
(i) the length of the boundary
(ii) the area of the shaded region.
(Take \( \pi \) to be 3.14)
Answer: The diameter of the larger semi-circular arc is \( 10 \) cm.
So, the radius of the larger semi-circle \( (R) = \frac { 10 }{ 2 } = 5 \) cm.
The diameter of each of the two smaller semi-circles is \( 5 \) cm (half of the larger diameter).
So, the radius of each smaller semi-circle \( (r) = \frac { 5 }{ 2 } = 2.5 \) cm.

(i) Length of the boundary of the shaded portion:
This includes the arc of the larger semi-circle and the arcs of the two smaller semi-circles.
Length of arc of larger semi-circle \( = \pi R \).
Length of arc of one smaller semi-circle \( = \pi r \).
Length of boundary \( = \pi R + 2 \times \pi r = \pi (R + 2r) \).
\( \implies = 3.14 (5 + 2 \times 2.5) \).
\( \implies = 3.14 (5 + 5) = 3.14 \times 10 = 31.4 \) cm.
(ii) Area of the shaded region:
Area of shaded region \( = \text{Area of larger semi-circle} - \text{Area of one smaller semi-circle} + \text{Area of other smaller semi-circle} \).
\( \implies = \frac { 1 }{ 2 } \pi R^2 - \frac { 1 }{ 2 } \pi r^2 + \frac { 1 }{ 2 } \pi r^2 \).
Notice that the areas of the two smaller semi-circles cancel each other out in the calculation, or rather, the shaded area is simply the larger semicircle minus the unshaded small semicircle, plus the shaded small semicircle, which simplifies to the area of the large semicircle.
So, Area of shaded region \( = \frac { 1 }{ 2 } \pi R^2 \).
\( \implies = \frac { 1 }{ 2 } \times 3.14 \times 5 \times 5 \text{ cm}^2 \).
\( \implies = \frac { 1 }{ 2 } \times 3.14 \times 25 \text{ cm}^2 \).
\( \implies = 39.25 \text{ cm}^2 \).
Rounded to one decimal place, the area is \( 39.3 \text{ cm}^2 \).
In simple words: For the boundary, add the curved length of the big half-circle and the curved lengths of the two small half-circles. For the shaded area, calculate the area of the big half-circle. The areas of the two small half-circles cancel each other out in the calculation since one is subtracted and the other is added.

🎯 Exam Tip: When calculating boundary lengths, sum all arcs that form the outline. For areas of complex shaded regions, break it down into simple geometric shapes and combine their areas (add for overlapping, subtract for cut-outs).

Question 20. A bed of roses is like in the figure. In the centre is a square and on each side there is semi-circle. Side of the square is 21 m. If each rose-plant needs 6 m² of space, find out the number of plants which can be planted in the whole figure.
Answer: The side of the square ABCD is \( 21 \) m.

Area of the square portion \( = (\text{side})^2 = (21)^2 \text{ m}^2 = 441 \text{ m}^2 \).
On each side of the square, there is a semi-circle. The diameter of each semi-circle is equal to the side of the square, which is \( 21 \) m.
So, the radius of each semi-circular portion \( (r) = \frac { 21 }{ 2 } \) m.
Area of one semi-circular portion \( = \frac { 1 }{ 2 } \pi r^2 \).
There are 4 such semi-circular portions, so their total area \( = 4 \times \frac { 1 }{ 2 } \pi r^2 = 2 \pi r^2 \).
\( \implies = 2 \times \frac { 22 }{ 7 } \times \frac { 21 }{ 2 } \times \frac { 21 }{ 2 } \text{ m}^2 = 693 \text{ m}^2 \).
The total area of the figure (the rose bed) \( = \text{Area of square} + \text{Area of 4 semi-circular portions} \).
\( \implies = 441 + 693 = 1134 \text{ m}^2 \).
Each rose plant needs \( 6 \text{ m}^2 \) of space.
Number of plants that can be planted \( = \frac{\text{Total area}}{\text{Space needed per plant}} \).
\( \implies = \frac{1134}{6} = 189 \).
In simple words: First, calculate the area of the central square. Then, find the area of the four semi-circles attached to its sides. Since four half-circles make two full circles, you can find their total area. Add the square's area and the semi-circles' total area to get the total space for plants. Finally, divide this total area by the space each plant needs to find how many plants fit.

🎯 Exam Tip: Break down the complex figure into simpler shapes: a square and four semi-circles. Remember that four semi-circles of the same diameter will form two full circles for easier area calculation. Sum these areas for the total space.

Question 21. Find the area of the region between two concentric circles given in figure, if the length of the chord of the outer circle touching the inner circle is 14 cm.
Answer: Let the radius of the outer circle be \( R \) and the radius of the inner circle be \( r \). The chord of the outer circle that touches the inner circle is 14 cm long. When a radius is drawn to the point where the chord touches the inner circle, it forms a right-angled triangle. The radius of the inner circle is perpendicular to the chord at the point of contact, and it bisects the chord. So, half the chord length is \( \frac{14}{2} = 7 \text{ cm} \). Using the Pythagorean theorem for the right triangle formed by the outer radius \( R \), inner radius \( r \), and half-chord length (7 cm), we get \( R^2 = r^2 + 7^2 \).
\( \implies R^2 - r^2 = 49 \)
The area of the region between the two concentric circles (also called an annulus) is given by \( \text{Area} = \pi R^2 - \pi r^2 = \pi (R^2 - r^2) \).
Substitute the value of \( R^2 - r^2 \):
\( \text{Area} = \frac{22}{7} \times 49 \)
\( \text{Area} = 22 \times 7 = 154 \text{ cm}^2 \)
The area of the region between the circles is 154 cm². This region is often called an annular ring.
In simple words: Imagine two circles inside each other. A line across the bigger circle just touches the smaller one. If this line is 14 cm, the special math rule (Pythagoras theorem) helps us find that the difference between the squares of their radii is 49. Then, we use this to quickly calculate the area of the ring shape between the circles, which comes out to be 154 cm².

🎯 Exam Tip: When dealing with concentric circles and a chord tangent to the inner circle, remember that the square of the outer radius minus the square of the inner radius (\( R^2 - r^2 \)) is always equal to the square of half the chord length. This relationship simplifies calculations for the area of the annulus.

Question 22. The figure shows a circle with centre at O and \( \angle \text{AOB} = 90^\circ \). If the radius of the circle is 40 cm, calculate the area of the shaded portion of the circle. (Take \( \pi \) to be 3.14)
Answer: The figure describes a sector of a circle, which is a quadrant because the angle \( \angle \text{AOB} \) is 90°. The radius of the circle is 40 cm. The shaded portion is the area of this quadrant minus the area of the right-angled triangle OAB.
First, calculate the area of the quadrant OAB:
\( \text{Area of quadrant OAB} = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times \pi r^2 \)
\( = \frac{1}{4} \times 3.14 \times (40)^2 \)
\( = \frac{1}{4} \times 3.14 \times 1600 \)
\( = 3.14 \times 400 = 1256 \text{ cm}^2 \)
Next, calculate the area of the triangle OAB. Since \( \angle \text{AOB} = 90^\circ \), it's a right-angled triangle with OA and OB as its perpendicular sides, both equal to the radius.
\( \text{Area of triangle OAB} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{OA} \times \text{OB} \)
\( = \frac{1}{2} \times 40 \times 40 \)
\( = \frac{1}{2} \times 1600 = 800 \text{ cm}^2 \)
Finally, find the area of the shaded portion by subtracting the area of the triangle from the area of the quadrant:
\( \text{Area of shaded portion} = \text{Area of quadrant OAB} - \text{Area of triangle OAB} \)
\( = 1256 \text{ cm}^2 - 800 \text{ cm}^2 = 456 \text{ cm}^2 \)
This shaded region is a circular segment.
In simple words: We have a quarter-circle (quadrant) with a triangle inside it. We first find the area of the whole quarter-circle, then the area of the triangle. By taking away the triangle's area from the quarter-circle's area, we find the area of the shaded part, which is 456 cm².

🎯 Exam Tip: For problems involving sectors and triangles, carefully identify whether the triangle is right-angled. If so, its area can be calculated using half base times height where radii serve as base and height. If not, use the formula \( \frac{1}{2} r^2 \sin \theta \).

Question 23. ABCD is a flower bed. If OA = 21 m and OC = 14 m, find the area of the bed. (Use \( \pi = \frac{22}{7} \))
Answer: The flower bed is shaped like a part of a circular ring (an annular sector). It is the area between two concentric quadrants.
The outer radius \( R \) is given as OA = 21 m.
The inner radius \( r \) is given as OC = 14 m.
Since it's a quadrant (implied by the square corners and context, a quarter of a circle), the angle is 90°.
The area of the flower bed is the area of the larger quadrant minus the area of the smaller quadrant:
\( \text{Area of flower bed} = \text{Area of quadrant OAB} - \text{Area of quadrant OCD} \)
\( = \frac{1}{4} \pi R^2 - \frac{1}{4} \pi r^2 \)
\( = \frac{1}{4} \pi (R^2 - r^2) \)
Substitute the values of \( R \), \( r \), and \( \pi \):
\( = \frac{1}{4} \times \frac{22}{7} ((21)^2 - (14)^2) \)
Using the difference of squares formula, \( a^2 - b^2 = (a-b)(a+b) \):
\( = \frac{1}{4} \times \frac{22}{7} (21 - 14)(21 + 14) \)
\( = \frac{1}{4} \times \frac{22}{7} \times 7 \times 35 \)
\( = \frac{1}{4} \times 22 \times 35 \)
\( = \frac{1}{2} \times 11 \times 35 \)
\( = 11 \times 17.5 = 192.5 \text{ m}^2 \)
This shape is also known as an annular sector.
In simple words: We have a large quarter-circle and a smaller quarter-circle inside it. To find the area of the flower bed between them, we calculate the area of the big quarter-circle and subtract the area of the small quarter-circle. This gives us 192.5 square meters for the flower bed.

🎯 Exam Tip: Always recognize if you can factor out common terms like \( \frac{1}{4} \pi \) when finding the area between concentric sectors. Using the difference of squares formula \( (R^2 - r^2) = (R-r)(R+r) \) can simplify calculations significantly.

Question 24. In the figure, A, B, C and D are centres of equal circles which touch externally in pairs and ABCD is a square of side 14 cm. Find the area of the shaded region.
Answer: Given that A, B, C, and D are the centers of equal circles, and these centers form a square ABCD with side 14 cm. Since the circles touch externally, the radius of each circle is half the side of the square formed by their centers.
Side of square ABCD = 14 cm.
Radius of each circle \( r = \frac{\text{Side}}{2} = \frac{14}{2} = 7 \text{ cm} \).
The problem states the "Area of shaded portions = Area of 3 circles + Area of square". Based on this specific instruction from the provided solution, we calculate:
Area of one circle \( = \pi r^2 = \frac{22}{7} \times (7)^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154 \text{ cm}^2 \)
Area of 3 circles \( = 3 \times 154 = 462 \text{ cm}^2 \)
Area of the square ABCD \( = \text{side}^2 = (14)^2 = 196 \text{ cm}^2 \)
Therefore, the total area of the shaded region according to the given calculation method is:
\( \text{Total area} = \text{Area of 3 circles} + \text{Area of square} \)
\( = 462 \text{ cm}^2 + 196 \text{ cm}^2 = 658 \text{ cm}^2 \)
This implies a specific interpretation of the shaded region, where three full circles and the central square are combined.
In simple words: We have a square made by the centers of four circles, and each circle has a radius of 7 cm. To find the total shaded area, we calculate the area of three full circles and add it to the area of the square. This sum gives us 658 square centimeters.

🎯 Exam Tip: Always clearly identify the radii and dimensions of all shapes involved. For complex diagrams, if a specific calculation method for "shaded area" is implied in the solution steps, follow that method precisely, even if the visual shading might be open to multiple interpretations.