GSEB Class 7 Maths Solutions Chapter 11 Perimeter and Area Exercise 11.3

Question 1. Find the circumference of the circles with the following radius: ( Taken \( \pi = \frac {22}{7} \) )
(a) 14 cm
(b) 28 mm
(c) 21 cm
Answer:
(a) Here, the radius (r) is 14 cm.
Therefore, the Circumference (C) is calculated as \( 2\pi r \).
So, \( C = 2 \times \frac { 22 }{ 7 } \times 14 \) cm
\( = 2 \times 22 \times 2 \) cm
\( = 88 \) cm
(b) Here, the radius (r) is 28 mm.
Therefore, the Circumference (C) is calculated as \( 2\pi r \).
So, \( C = 2 \times \frac { 22 }{ 7 } \times 28 \) mm
\( = 2 \times 22 \times 4 \) mm
\( = 176 \) mm
(c) Here, the radius (r) is 21 cm.
Therefore, the Circumference (C) is calculated as \( 2\pi r \).
So, \( C = 2 \times \frac { 22 }{ 7 } \times 21 \) cm
\( = 2 \times 22 \times 3 \) cm
\( = 132 \) cm
In simple words: To find how far around a circle it is, you use a formula. Just put the given radius into the formula with the value of pi, and then do the math to get the distance around.

Exam Tip: Remember the formula for circumference is \( C = 2\pi r \). Ensure you use the correct units (cm, mm, m) in your final answer.

Question 2. Find the area of the following circles, given that: ( Taken \( \pi = \frac { 22 }{ 7 } \) )
(a) radius = 14 mm
(b) diameter = 49 m
(c) radius = 5 cm
Answer:
(a) Here, the radius (r) is 14 mm.
Therefore, the Area of the circle is calculated as \( \pi r^2 \).
So, Area \( = \frac { 22 }{ 7 } \times (14)^2 \) mm\( ^2 \)
\( = \frac { 22 }{ 7 } \times 14 \times 14 \) mm\( ^2 \)
\( = 22 \times 2 \times 14 \) mm\( ^2 \)
\( = 616 \) mm\( ^2 \)
(b) Here, the diameter is 49 m.
Therefore, the Radius (r) is \( \frac { 49 }{ 2 } \) m.
Now, the area of the circle is calculated as \( \pi r^2 \).
So, Area \( = \frac { 22 }{ 7 } \times \left( \frac { 49 }{ 2 } \right)^2 \) m\( ^2 \)
\( = \frac { 22 }{ 7 } \times \frac { 49 }{ 2 } \times \frac { 49 }{ 2 } \) m\( ^2 \)
\( = 22 \times \frac { 7 }{ 2 } \times \frac { 49 }{ 2 } \) m\( ^2 \)
\( = \frac { 11 \times 7 \times 49 }{ 2 } \) m\( ^2 \)
\( = \frac { 3773 }{ 2 } \) m\( ^2 \)
\( = 1886.5 \) m\( ^2 \)
(c) Here, the radius (r) is 5 cm.
Therefore, the Area of the circle is calculated as \( \pi r^2 \).
So, Area \( = \frac { 22 }{ 7 } \times (5)^2 \) cm\( ^2 \)
\( = \frac { 22 }{ 7 } \times 5 \times 5 \) cm\( ^2 \)
\( = \frac { 550 }{ 7 } \) cm\( ^2 \)
In simple words: To find the area, you use another formula. If you are given the diameter, first halve it to get the radius. Then, multiply pi by the radius squared to get the space inside the circle.

Exam Tip: Pay attention to whether radius or diameter is given. Always calculate radius first if diameter is provided, then apply the area formula \( A = \pi r^2 \).

Question 3. If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Taken \( \pi = \frac { 22 }{7} \) )
Answer: Here, the circumference of the circle is 154 m.
Let's consider the radius to be r m.
Therefore, \( 2\pi r = 154 \)
Or \( 2 \times \frac { 22 }{ 7 } \times r = 154 \)
Or \( r = \frac { 154 \times 7 }{ 2 \times 22 } \) m
\( = \frac { 7 \times 7 }{ 2 } \) m
Or \( r = \frac { 49 }{ 2 } \) m
\( = 24.5 \) m
Now, the Area of a circle is \( \pi r^2 \).
Area \( = \frac { 22 }{ 7 } \times (24.5)^2 \) m\( ^2 \)
\( = \frac { 22 }{ 7 } \times \left( \frac { 49 }{ 2 } \right)^2 \) m\( ^2 \)
\( = \frac { 22 }{ 7 } \times \frac { 49 }{ 2 } \times \frac { 49 }{ 2 } \) m\( ^2 \)
\( = \frac { 11 \times 7 \times 49 }{ 2 } \) m\( ^2 \)
\( = \frac { 3773 }{ 2 } \) m\( ^2 \)
\( = 1886.5 \) m\( ^2 \)
In simple words: First, use the circumference to find the radius of the circle. Once you have the radius, you can then use it to calculate the total area of that circular sheet.

Exam Tip: This question requires two steps: first finding the radius from circumference, and then using that radius to calculate the area. Make sure to clearly show both calculations.

Question 4. A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs Rs. 4 per metre. ( Taken \( \pi = \frac { 22 }{ 7 } \) )
Answer: The diameter of the circular garden is 21 m.
Therefore, the Radius (r) is \( \frac { 21 }{ 2 } \) m.
The Circumference (C) is calculated as \( 2\pi r \).
So, the Circumference of the circular garden \( = 2 \times \frac { 22 }{ 7 } \times \frac { 21 }{ 2 } \) m
\( = 22 \times 3 \) m
\( = 66 \) m
Since the rope makes 2 rounds of fence, the total rope length will be twice the garden's circumference.
Rope length \( = 2 \times [\text{Circumference of the garden}] \)
\( = 2 \times 66 \) m
\( = 132 \) m
Thus, the length of the required rope is 132 m.
Now, the cost of the rope \( = \text{Rs. } 4 \times 132 \)
\( = \text{Rs. } 528 \)
In simple words: First, find the distance around the garden. Then, double that distance because the gardener wants two layers of fence. Finally, multiply this total length by the price per meter to find the total cost.

Exam Tip: Be careful to read the question fully to ensure all parts (length and cost, and multiple rounds) are addressed in your solution. Convert diameter to radius if necessary.

Question 5. From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take \( \pi = 3.14 \))
Answer: The radius of the larger circular sheet (R) is 4 cm.
The radius of the removed circular sheet (which is the inner circle) (r) is 3 cm.
We know that the Area of a circle is calculated as \( \pi \times (\text{Radius})^2 \).
Therefore, the Area of the remaining sheet is found by subtracting the area of the inner circle from the area of the outer circle.
Area of remaining sheet \( = [\text{Area of outer circle}] - [\text{Area of inner circle}] \)
\( = \pi R^2 - \pi r^2 \)
\( = \pi [R^2 - r^2] \)
\( = \pi (R - r)(R + r) \)
\( = 3.14(4 - 3)(4 + 3) \) cm\( ^2 \)
\( = 3.14(1)(7) \) cm\( ^2 \)
\( = 21.98 \) cm\( ^2 \)
In simple words: First, figure out the area of the big circular sheet. Then, calculate the area of the smaller circle that's being cut out. Subtract the smaller area from the larger area to find how much sheet is left over.

Exam Tip: When finding the area of a remaining region after a cut-out, remember to subtract the area of the removed portion from the original total area. Using the difference of squares identity \( (R^2 - r^2) = (R-r)(R+r) \) can simplify calculations.

Question 6. Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one metre of the lace costs Rs. 15. (Take \( \pi = 3.14 \))
Answer: The diameter of the circular table cover is 1.5 m.
Therefore, the Radius (r) is \( \frac { 1.5 }{ 2 } \) m.
The Circumference (C) is calculated as \( 2\pi r \).
So, \( C = 2 \times 3.14 \times \frac { 1.5 }{ 2 } \) m
\( = 3.14 \times 1.5 \) m
\( = 4.71 \) m
Thus, the length of the lace required is 4.71 m.
Now, the cost of the lace \( = \text{Rs. } 15 \times 4.71 \)
\( = \text{Rs. } 70.65 \)
In simple words: First, calculate the distance around the table cover, which is its circumference. That's how much lace Saima needs. Then, multiply the lace's total length by its price per meter to figure out the total cost.

Exam Tip: Ensure you use the specified value of \( \pi \) (here, 3.14) for calculations. Calculate the circumference first, and then use that length to find the total cost.

Question 7. Find the perimeter of the figure, which is a semicircle including its diameter.
Answer: The diameter of the semicircle is 10 cm.
Therefore, the Radius (r) is \( \frac { 10 }{ 2 } = 5 \) cm.
Now, the circumference of the semicircle is \( \frac { 2\pi r }{ 2 } = \pi r \).
So, Circumference \( = \frac { 22 }{ 7 } \times 5 \) cm
\( = \frac { 110 }{7} \) cm
\( = 15.71 \) cm (approximately)
Therefore, the Perimeter of the semicircle \( = \text{Circumference} + \text{Diameter} \)
\( = 15.71 \) cm \( + 10 \) cm
\( = 25.71 \) cm
In simple words: The perimeter of a semicircle is found by adding the curved part (half the circumference of a full circle) to the straight part (the diameter).

Exam Tip: For the perimeter of a semicircle, do not forget to add the length of the diameter to the length of the arc (half circumference) to get the complete boundary.

Question 8. Find the cost of polishing a circular tabletop of diameter 1.6 m, if the rate of polishing is Rs. 15/m². (Take \( \pi = 3.14 \))
Answer: The diameter of the circular table-top is 1.6 m.
Therefore, the Radius (r) is \( \frac { 1.6 }{ 2 } \) m.
Now, the area of the circular table-top is \( \pi r^2 \).
Area \( = 3.14 \times \left( \frac { 1.6 }{ 2 } \right)^2 \) m\( ^2 \)
\( = 3.14 \times (0.8)^2 \) m\( ^2 \)
\( = 3.14 \times 0.64 \) m\( ^2 \)
\( = 2.0096 \) m\( ^2 \)
Since the rate of polishing is Rs. 15/m\( ^2 \).
Therefore, the Total cost of polishing the circular table-top \( = \text{Rs. } 15 \times 2.0096 \)
\( = \text{Rs. } 30.144 \)
Thus, the required cost of polishing the circular table-top is Rs. 30.144.
In simple words: First, find the radius from the diameter. Then, calculate the total surface area of the tabletop. Finally, multiply this area by the given polishing rate to determine the overall cost.

Exam Tip: When dealing with costs based on area, ensure you calculate the area accurately first. Double-check your calculations, especially for decimal values and squared terms.

Question 9. Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure enclose more area, the circle or the square? ( Taken \( \pi = \frac { 22 }{7} \))
Answer: Here, the total length of the wire is 44 cm.
The wire is bent to form a circle.
Therefore, the Circumference of the circle formed is 44 cm.
Let the radius of the circle be r.
So, \( 2\pi r = 44 \)
Or \( 2 \times \frac { 22 }{ 7 } \times r = 44 \)
Or \( r = \frac { 44 \times 7 }{ 2 \times 22 } \) cm
\( = 7 \) cm.
Thus, the required radius of the circle is 7 cm.
Now, the area of the circle formed is \( \pi r^2 \).
Area \( = \frac { 22 }{ 7 } \times (7)^2 \) cm\( ^2 \)
\( = \frac { 22 }{ 7 } \times 7 \times 7 \) cm\( ^2 \)
\( = 154 \) cm\( ^2 \)
Next, the same wire is rebent to form a square.
Therefore, the Perimeter of the square formed is equal to the Circumference of the circle, which is 44 cm.
Let the side of the square be x cm.
So, \( 4 \times x = 44 \) cm
Since the Perimeter of a square \( = 4 \times \text{Side} \).
Or \( x = \frac { 44 }{ 4 } \) cm
\( = 11 \) cm
Now, the area of the square \( = \text{Side} \times \text{Side} \)
\( = 11 \) cm \( \times 11 \) cm
\( = 121 \) cm\( ^2 \)
Since 154 cm\( ^2 \) is greater than 121 cm\( ^2 \).
Therefore, the circle encloses a greater area.
In simple words: First, use the wire's length as the circle's circumference to find its radius and area. Then, use the same wire length as the square's perimeter to find its side length and area. Finally, compare the two areas to see which shape covers more space.

Exam Tip: Remember that the length of the wire remains constant whether it's shaped into a circle (circumference) or a square (perimeter). This allows you to find the dimensions of each shape and then compare their areas.

Question 10. From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed {as shown in the adjoining figure). Find the area of the remaining sheet. (Taken \( \pi = \frac { 22 }{ 7 } \))
Answer: The radius of the main circular sheet is 14 cm.
Therefore, the Area of the sheet \( = \pi r^2 \)
\( = \frac { 22 }{ 7 } \times (14)^2 \) cm\( ^2 \)
\( = \frac { 22 }{ 7 } \times 14 \times 14 \) cm\( ^2 \)
\( = 22 \times 2 \times 14 \) cm\( ^2 \)
\( = 616 \) cm\( ^2 \)
Now, the radius of each small circle is 3.5 cm.
Therefore, the Area of one small circle \( = \frac { 22 }{ 7 } \times 3.5 \times 3.5 \) cm\( ^2 \)
\( = 22 \times 0.5 \times 3.5 \) cm\( ^2 \)
\( = 38.5 \) cm\( ^2 \)
So, the area of two small circles \( = 2 \times 38.5 \) cm\( ^2 \)
\( = 77 \) cm\( ^2 \)
Again, the length of the small rectangle is 3 cm, and its breadth is 1 cm.
The Area of the small rectangle \( = 3 \times 1 \) cm\( ^2 \)
\( = 3 \) cm\( ^2 \)
Therefore, the Total area removed from the sheet \( = 77 \) cm\( ^2 \) \( + 3 \) cm\( ^2 \)
\( = 80 \) cm\( ^2 \)
The Area of the remaining sheet \( = 616 \) cm\( ^2 \) \( - 80 \) cm\( ^2 \)
\( = 536 \) cm\( ^2 \)
Thus, the required area of the remaining sheet is 536 cm\( ^2 \).
In simple words: First, calculate the area of the large circular sheet. Then, calculate the areas of the two small circles and the rectangle that are cut out. Add these removed areas together. Finally, subtract the total removed area from the large sheet's area to find what's left.

Exam Tip: Break down complex area problems into simpler shapes. Calculate the area of each component, then use addition and subtraction to find the area of the composite or remaining figure.

Question 11. A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take \( \pi = 3.14 \))
Answer: The side of the square is 6 cm.
The Area of the square \( = \text{Side} \times \text{Side} \)
\( = 6 \) cm \( \times 6 \) cm
\( = 36 \) cm\( ^2 \)
The radius of the circle cut out from the sheet, r, is 2 cm.
Therefore, the Area of the circle \( = \pi r^2 \)
\( = 3.14 \times (2)^2 \) cm\( ^2 \)
\( = 3.14 \times 2 \times 2 \) cm\( ^2 \)
\( = 12.56 \) cm\( ^2 \)
The area of the remaining sheet \( = 36 \) cm\( ^2 \) \( - 12.56 \) cm\( ^2 \)
\( = 23.44 \) cm\( ^2 \)
In simple words: First, find the area of the square sheet. Then, calculate the area of the circle that is removed. Subtract the circle's area from the square's area to find how much aluminium sheet is remaining.

Exam Tip: When a shape is cut out from a larger one, the area of the remaining part is always the area of the larger shape minus the area of the cut-out. Ensure accurate area calculations for both shapes.

Question 12. The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. (Take \( \pi = 3.14 \))
Answer: Here, the circumference of the circle is 31.4 cm.
Let's assume the radius of the circle is r.
Therefore, \( 2\pi r = 31.4 \)
Or \( 2 \times 3.14 \times r = 31.4 \)
Or \( r = \frac {31.4 }{ 2 \times 3.14 } \) cm
\( = 5 \) cm
Thus, the radius of the circle (r) is 5 cm.
Now, the area of the circle is \( \pi r^2 \).
Area \( = 3.14 \times (5)^2 \) cm\( ^2 \)
\( = 3.14 \times 5 \times 5 \) cm\( ^2 \)
\( = 78.5 \) cm\( ^2 \)
In simple words: First, use the given circumference and the formula to find the circle's radius. After that, use the newly found radius to calculate the total area that the circle covers.

Exam Tip: This question is a two-part problem. Start by finding the radius from the given circumference, and then use that radius to compute the area. Always follow the specified value of \( \pi \).

Question 13. A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? ( \( \pi = 3.14 \) )
Answer: The diameter of the flower bed is 66 m.
Therefore, the Radius (r) of the flower bed is \( \frac { 66 }{ 2 } \) m \( = 33 \) m.
Since the width of the surrounding path is 4 m.
Therefore, the Radius of the outer circle (R) is \( 33 \) m \( + 4 \) m \( = 37 \) m.
Now, the area of the outer circle is \( \pi R^2 \).
And the area of the inner circle is \( \pi r^2 \).
Therefore, the Area of the path \( = \pi R^2 - \pi r^2 \)
\( = \pi [R^2 - r^2] \)
\( = \pi [(R + r)(R - r)] \)
\( = 3.14[(37 + 33)(37 - 33)] \) m\( ^2 \)
\( = 3.14[70 \times 4] \) m\( ^2 \)
\( = 3.14 \times 280 \) m\( ^2 \)
\( = 879.2 \) m\( ^2 \)
Thus, the area of the path is 879.2 m\( ^2 \).
In simple words: Calculate the radius of the flower bed, then add the path's width to find the outer radius. Find the area of both the inner and outer circles. Subtract the inner area from the outer area to discover the area covered by the path.

Exam Tip: For problems involving concentric circles (like a path around a circular bed), always calculate both the inner and outer radii accurately. The area of the path is the difference between the area of the larger circle and the smaller circle.

Question 14. A circular flower garden has an area of 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Take \( \pi = 3.14 \))
Answer: The area of the circular garden is 314 m\( ^2 \).
Let's suppose the radius of the garden is r m.
Therefore, \( \pi r^2 = 314 \)
Or \( 3.14 \times r^2 = 314 \)
Or \( \frac { 314 }{ 100 } \times r^2 = 314 \)
Or \( r^2 = 314 \times \frac { 100 }{ 314 } \)
\( = 100 \)
\( r^2 = 10^2 \)
\( \implies r = 10 \) m
Now, the radius of the area covered by the sprinkler is 12 m.
Since 12 m is greater than 10 m.
Therefore, the sprinkler covers an area beyond the garden, which means, yes, the entire garden is covered by the sprinkler.
In simple words: First, use the garden's area to find its radius. Then, compare this garden radius to the sprinkler's reach. If the sprinkler's reach is longer than the garden's radius, it will water the whole garden.

Exam Tip: To determine if a sprinkler covers an area, always compare the sprinkler's reach (radius) with the radius of the area to be covered. If the sprinkler's radius is equal to or greater than the garden's radius, it will cover it fully.

Question 15. Find the circumference of the inner and the outer circles, shown in the figure? (Take \( \pi = 3.14 \))
Answer: The Radius of the outer circle (R) is 19 m.
Therefore, the Circumference of the outer circle \( = 2\pi R \)
\( = 2 \times 3.14 \times 19 \) m
\( = 119.32 \) m
Again, the Radius of the inner circle (r) \( = 19 \) m \( - 10 \) m
\( = 9 \) m
The Circumference of the inner circle \( = 2\pi r \)
\( = 2 \times 3.14 \times 9 \) m
\( = 56.52 \) m
In simple words: To find the circumference of the outer circle, use its given radius. For the inner circle, subtract the space between the circles from the outer radius to find the inner radius. Then, calculate its circumference.

Exam Tip: Carefully read the labels on diagrams to determine radii. For concentric circles, the distance between them allows you to deduce the inner radius if only the outer radius and the gap are provided.

Question 16. How many times a wheel of radius 28 cm must rotate to go 352 m? ( Taken \( \pi = \frac { 22 }{ 7 } \))
Answer: The radius of the wheel (r) is 28 cm.
Therefore, the Circumference of the wheel \( = 2\pi r \)
\( = 2 \times \frac { 22 }{ 7 } \times 28 \) cm
\( = 2 \times 22 \times 4 \) cm
\( = 176 \) cm
So, the wheel can cover 176 cm of distance in one rotation.
The total distance to be covered is 352 m.
Let's convert this to centimeters: \( 352 \times 100 \) cm \( = 35200 \) cm.
Therefore, the Number of rotations \( = \frac { \text{Total distance} }{ \text{Circumference per rotation} } \)
\( = \frac { 35200 }{ 176 } \)
\( = 200 \)
Thus, a distance of 352 m will be covered in 200 turns.
In simple words: First, find the distance the wheel travels in one full spin (its circumference). Convert the total distance needed into the same units. Then, divide the total distance by the distance per spin to find out how many times the wheel needs to turn.

Exam Tip: Ensure consistent units throughout your calculations. Convert all measurements to a single unit (e.g., cm or m) before performing division to find the number of rotations.

Question 17. The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take \( \pi = 3.14 \))
Answer: The length of the minute hand is 15 cm.
Therefore, the Radius of the circle made by the tip of the minute hand (r) is 15 cm.
The Circumference of the circle formed \( = 2\pi r \)
\( = 2 \times 3.14 \times 15 \) cm
\( = 94.2 \) cm
Since the minute hand completes one full revolution in one hour.
Thus, the distance covered by the tip of the minute hand in 1 hour is 94.2 cm.
In simple words: The minute hand's length is the circle's radius. In one hour, the minute hand makes a full circle. So, you need to find the distance around that circle (the circumference) to know how far the tip moves.

Exam Tip: Understand that one full revolution of a minute hand represents the circumference of the circle it traces. The length of the hand is the radius for this calculation.