Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 19 Mensuration I

Exercise 19.1

Answer 4.
Answer:
Given: Circumference = 396 m

Step 1: Find the radius using circumference formula.
\( 2\pi r = 396 \)
\( 2 \times \frac{22}{7} \times r = 396 \)
\( r = 396 \times \frac{7}{22} \times \frac{1}{2} \)
\( r = 63 \text{ m} \)

Step 2: Find the area using area formula.
\( \text{Area} = \pi r^2 \)
\( = \frac{22}{7} \times 63 \times 63 \)
\( = 12474 \text{ m}^2 \)

Area of the circular field = 12474 m²
In simple words: We found the radius first by using the circumference formula. Then we used that radius to find the area of the circle.

📝 Teacher's Note: Show students that we use the same radius value in both formulas. Make them write the circumference formula first, then the area formula.

🎯 Exam Tip: Always find radius first when circumference is given. Write both formulas clearly and substitute values step by step.

Answer 5.
Answer:
Given: Area = \( \pi r^2 = 81\pi \)

Step 1: Find the radius.
\( \pi r^2 = 81\pi \)
\( r^2 = 81 \)
\( r = 9 \text{ cm} \)

Step 2: Find the circumference.
\( \text{Circumference} = 2\pi r \)
\( = 2 \times \pi \times 9 \)
\( = 18\pi \text{ cm} \)

Circumference = 18π cm
In simple words: We found the radius by taking the square root of the area value. Then we used that radius to find the circumference.

📝 Teacher's Note: When area has π in it, the radius calculation becomes very easy. The π cancels out on both sides.

🎯 Exam Tip: When area is given as "number × π", just take square root of that number to get radius. Keep π in the final answer.

Answer 6.
Answer:
Given: Diameter of wheel = 1.4 m, Distance travelled = 2.2 km = 2200 m

Step 1: Find circumference of wheel.
\( \text{Circumference} = 2\pi r = \pi d \)
\( = \frac{22}{7} \times 1.4 = 4.4 \text{ m} \)

Step 2: Find number of revolutions.
\( \text{No. of revolutions} = \frac{\text{Distance}}{\text{Circumference}} \)
\( = \frac{2200}{4.4} = 500 \)

Wheel makes 500 revolutions in travelling 2.2 km
In simple words: One revolution means the wheel goes around once and covers a distance equal to its circumference. We divide total distance by circumference to get number of revolutions.

📝 Teacher's Note: Use a coin to demonstrate - one roll of the coin covers a distance equal to the coin's circumference. This makes the concept clear.

🎯 Exam Tip: Always convert km to m when working with wheel problems. Formula is: Revolutions = Distance ÷ Circumference.

Answer 7.
Answer:
Given: Diameter = 70 cm, Speed = 10 revolutions per second

Step 1: Find circumference of wheel.
\( \text{Circumference} = 2\pi r = \pi d \)
\( = \frac{22}{7} \times 70 = 220 \text{ cm} = 2.2 \text{ m} \)

Step 2: Find revolutions per hour.
\( \text{Revolutions per hour} = 10 \times 60 \times 60 = 36000 \)

Step 3: Find distance covered in one hour.
\( \text{Distance} = 36000 \times 2.2 = 79200 \text{ m/hour} \)

Step 4: Convert to speed in km/hour.
\( \text{Speed} = \frac{79200}{1000} = 79.2 \text{ km/hour} \)

Speed = 79.2 km/hr
In simple words: We found how far the wheel travels in one hour by multiplying revolutions per hour with circumference. Then we converted meters to kilometers.

📝 Teacher's Note: Break this into small steps - first find circumference, then revolutions per hour, then distance, then speed. Students get confused if you try to do everything together.

🎯 Exam Tip: Convert cm to m early in the solution. Remember: 1 hour = 3600 seconds. Always write final answer in km/hr as asked.

Answer 8.
Answer:
Given: Diameter = 140 cm, Speed = 66 km/hour

Step 1: Find circumference of wheel.
\( \text{Circumference} = 2\pi r = \pi d \)
\( = \frac{22}{7} \times 140 = 440 \text{ cm} = 4.4 \text{ m} \)

Step 2: Find distance travelled in one hour.
\( \text{Distance} = 66 \text{ km} = 66000 \text{ m} \)

Step 3: Find distance travelled in one minute.
\( \text{Distance in 1 minute} = \frac{66000}{60} = 1100 \text{ m} \)

Step 4: Find revolutions per minute.
\( \text{Revolutions per minute} = \frac{\text{Distance}}{\text{Circumference}} = \frac{1100}{4.4} = 250 \)

Number of revolutions = 250 rpm
In simple words: We found how far the vehicle travels in one minute. Then we divided this by the wheel's circumference to get revolutions per minute.

📝 Teacher's Note: rpm means "revolutions per minute". This is commonly used for wheels and engines. Make students understand this unit.

🎯 Exam Tip: Convert speed from km/hr to m/min by dividing by 60 and multiplying by 1000. Write "rpm" clearly in the answer.

Answer 9.
Answer:
Given: Diameter = 42 cm, Speed = 9 revolutions per second

Step 1: Find circumference of wheel.
\( \text{Circumference} = 2\pi r = \pi d \)
\( = \frac{22}{7} \times 42 = 132 \text{ cm} = 1.32 \text{ m} \)

Step 2: Find revolutions per hour.
\( \text{Revolutions per hour} = 9 \times 60 \times 60 = 32400 \)

Step 3: Find distance covered in one hour.
\( \text{Distance} = 32400 \times 1.32 = 42768 \text{ m/hour} \)

Step 4: Convert to speed in km/hour.
\( \text{Speed} = \frac{42768}{1000} = 42.768 \text{ km/hour} = 43 \text{ km/hr} \)

Speed = 43 km/hr
In simple words: We calculated how many times the wheel turns in one hour, then multiplied by circumference to get distance, then converted to km/hr.

📝 Teacher's Note: This is the reverse of the previous problem. Here we go from revolutions to speed, there we went from speed to revolutions.

🎯 Exam Tip: When rounding the final answer, round to the nearest whole number unless the question asks for decimal places.

Answer 10.
Answer:
Given: Diameter = 35 cm, Time = 2 minutes = 120 seconds, Speed of rope = 1.1 m/s

Step 1: Find circumference of wheel.
\( \text{Circumference} = 2\pi r = \pi d \)
\( = \frac{22}{7} \times 35 = 110 \text{ cm} = 1.1 \text{ m} \)

Step 2: Find length of rope pulled.
\( \text{Length} = \text{Speed} \times \text{Time} = 1.1 \times 120 = 132 \text{ m} \)

Step 3: Find number of revolutions.
\( \text{Revolutions} = \frac{\text{Length of rope}}{\text{Circumference}} = \frac{132}{1.1} = 120 \)

Number of revolutions = 120
In simple words: The rope moves at a fixed speed for 2 minutes, so we found total rope length. Each revolution pulls rope equal to wheel's circumference.

📝 Teacher's Note: Use a simple pulley system to show this - as the wheel turns, the rope moves. One full turn moves rope equal to circumference.

🎯 Exam Tip: Convert minutes to seconds when speed is given in m/s. Formula: Distance = Speed × Time.

Answer 11.
Answer:
Given: Circumference = 280 cm = 2.8 m, Distance travelled = 490 m

Step 1: Find number of revolutions.
\( \text{Number of revolutions} = \frac{\text{Distance}}{\text{Circumference}} = \frac{490}{2.8} = 175 \)

Roller takes 175 revolutions in moving 490 m
In simple words: We divided the total distance by the circumference to find how many complete turns the roller made.

📝 Teacher's Note: This is a direct application of the revolution formula. No complicated steps needed when circumference is already given.

🎯 Exam Tip: When circumference is given directly, just divide distance by circumference. Convert units if needed.

Answer 12.
Answer:
Given: Diameter = \( 4\frac{5}{11} = \frac{49}{11} \) cm, Distance = 6.3 km = 6300 m

Step 1: Find circumference.
\( \text{Circumference} = 2\pi r = \pi d \)
\( = \frac{22}{7} \times \frac{49}{11} = 14 \text{ cm} = 0.14 \text{ m} \)

Step 2: Find number of revolutions.
\( \text{Number of revolutions} = \frac{6300}{0.14} = 45000 \)

Wheel takes 45000 revolutions in moving 6.3 km
In simple words: We converted the mixed number to improper fraction, found circumference, then divided total distance by circumference.

📝 Teacher's Note: When diameter is a mixed number, convert to improper fraction first. This makes calculation easier.

🎯 Exam Tip: Be careful with fraction multiplication. Convert mixed numbers to improper fractions before calculating.

Answer 13.
Answer:
Given: Area of circular ring = 88 cm², Outer radius = (r + 1) cm, Inner radius = r cm

Step 1: Set up the equation.
\( \text{Area of ring} = \text{Area of outer circle} - \text{Area of inner circle} \)
\( \pi(r + 1)^2 - \pi r^2 = 88 \)

Step 2: Expand and simplify.
\( \pi(r^2 + 2r + 1) - \pi r^2 = 88 \)
\( \pi r^2 + 2\pi r + \pi - \pi r^2 = 88 \)
\( 2\pi r + \pi = 88 \)
\( \frac{22}{7}(2r + 1) = 88 \)

Step 3: Solve for r.
\( 2r + 1 = 88 \times \frac{7}{22} = 28 \)
\( 2r = 27 \)
\( r = 13.5 \text{ cm} \)

Step 4: Find outer radius.
\( r + 1 = 13.5 + 1 = 14.5 \text{ cm} \)

Therefore, radii are 13.5 cm and 14.5 cm
In simple words: A ring is like a circle with a hole in it. We found the difference between the big circle area and small circle area equals 88.

📝 Teacher's Note: Use two circles drawn on paper - one inside the other. The shaded area between them is the ring. This visual helps students understand.

🎯 Exam Tip: For ring problems, always write "Area of ring = Area of outer circle - Area of inner circle". Expand brackets carefully.

Answer 14.
Answer:
Step 1: Find the area between two concentric circles.
Area between circles = Area of larger circle - Area of smaller circle

Step 2: Use the circle area formula.
Area of circle = \( \pi r^2 \)
Radius of bigger circle = r1 = 13 cm
Radius of smaller circle = r2 = 6 cm

Step 3: Calculate the required area.
Required Area = \( \pi r_1^2 - \pi r_2^2 \)
\( = \frac{22}{7} \times 13 \times 13 - \frac{22}{7} \times 6 \times 6 \)
\( = 531.1429 - 113.1429 \)
\( = 418 \text{ cm}^2 \)

Hence, required area = 418 cm²
In simple words: We found the area of the big circle and the small circle. Then we subtracted the small area from the big area to get the area between them.

📝 Teacher's Note: Draw two circles on the board — one inside the other. Show students that the shaded region is like a ring. The area of this ring is what we need to find.

🎯 Exam Tip: Always write the formula first. Then substitute values carefully. Make sure to subtract the smaller area from the larger area — not the other way around.

Answer 16.
Answer:
Step 1: Find the side of the square from its area.
Area of the square = 484 cm²
Side of the square = \( \sqrt{484} = 22 \text{ cm} \)

Step 2: Find the perimeter of the square.
Perimeter of the square = \( 4 \times 22 = 88 \text{ cm} \)

Step 3: Find the radius of the circle.
Perimeter of square = perimeter of circle
\( 2\pi r = 88 \)
\( r = 88 \times \frac{7}{22} \times \frac{1}{2} \)
\( r = 14 \text{ cm} \)

Step 4: Calculate the area of the circle.
Radius of circle = 14 cm
Area of circle = \( \pi r^2 \)
\( = \frac{22}{7} \times 14 \times 14 \)
\( = 616 \text{ cm}^2 \)

Hence, area of the circle = 616 cm²
In simple words: We found how long each side of the square is. Then we found the distance around the square. This same distance goes around the circle. From this, we found the circle's area.

📝 Teacher's Note: Show students a square and a circle with the same perimeter. Help them see that both shapes have the same distance around their edges, but different areas inside.

🎯 Exam Tip: Remember that perimeter of square = 4 × side and circumference of circle = 2πr. When they are equal, you can find one shape's measurements from the other.

Answer 17.
Answer:
Step 1: Find the side of the equilateral triangle.
Area of equilateral triangle = \( 121\sqrt{3} \text{ cm}^2 \)
\( \frac{s^2\sqrt{3}}{4} = 121\sqrt{3} \)
\( s^2 = 484 \)
\( s = 22 \text{ cm} \)

Step 2: Find the perimeter of the triangle.
Side of the triangle = 22 cm
Perimeter of the triangle = \( 3 \times 22 = 66 \text{ cm} \)

Step 3: Find the radius of the circle.
Perimeter of the circle = perimeter of the triangle
\( 2\pi r = 66 \)
\( r = 66 \times \frac{7}{22} \times \frac{1}{2} \)
\( r = 10.5 \text{ cm} \)

Step 4: Calculate the area of the circle.
Radius of circle = 10.5 cm
Area of circle = \( \pi r^2 \)
\( = \frac{22}{7} \times 10.5 \times 10.5 \)
\( = 346.5 \text{ cm}^2 \)

Hence, area of the circle = 346.5 cm²
In simple words: We found the side length of the triangle from its area. Then we found the distance around the triangle. A circle with the same distance around it has this area.

📝 Teacher's Note: Remind students that an equilateral triangle has all three sides equal. So perimeter = 3 × side. The formula for area includes √3, which makes it special.

🎯 Exam Tip: Write the equilateral triangle area formula clearly: Area = (s²√3)/4. Remember that perimeter of triangle = perimeter of circle when they are equal.

Answer 18.
Answer:
Step 1: Find the area of the original circle.
Area of the circle with radius 7 cm = \( \pi r^2 \)
\( = \frac{22}{7} \times 7 \times 7 \)
\( = 154 \text{ cm}^2 \)

Step 2: Find the area of the new circle.
Area of new circle = 25 × area of the circle with radius 7 cm
\( = 25 \times 154 \text{ cm}^2 \)
\( = 3850 \text{ cm}^2 \)

Step 3: Find the radius of the new circle.
\( \pi R^2 = 3850 \)
\( R^2 = 3850 \times \frac{7}{22} \)
\( R^2 = 1225 \)
\( R = 35 \text{ cm} \)

Step 4: Calculate the circumference.
Radius of new circle = 35 cm
Circumference of new circle = \( 2\pi R \)
\( = 2 \times \frac{22}{7} \times 35 \)
\( = 220 \text{ cm} \)

Hence, circumference of new circle = 220 cm
In simple words: The new circle has 25 times more area than the small circle. We found how big the radius must be for this much area. Then we found the distance around this big circle.

📝 Teacher's Note: Show students that when area becomes 25 times bigger, the radius becomes 5 times bigger (since 5² = 25). This helps them understand the relationship between area and radius.

🎯 Exam Tip: When area increases by a factor, the radius increases by the square root of that factor. So if area × 25, then radius × 5. Always check your answer this way.

Answer 19.
Answer:
Step 1: Find the side of the square from its area.
Area of the square = 484 m²
Side of the square = \( \sqrt{484} = 22 \text{ m} \)

Step 2: Find the perimeter of the square.
Perimeter of the square = \( 4 \times 22 = 88 \text{ m} \)

Step 3: Find the radius of the circle.
Perimeter of square = circumference of circle
\( 2\pi r = 88 \)
\( r = 88 \times \frac{7}{22} \times \frac{1}{2} \)
\( r = 14 \text{ m} \)

Step 4: Calculate the area of the circle.
Radius of circle = 14 m
Area of circle = \( \pi r^2 \)
\( = \frac{22}{7} \times 14 \times 14 \)
\( = 616 \text{ m}^2 \)

Hence, area of the circle = 616 m²
In simple words: We found the side of the square first. Then we found the distance around the square. A circle with the same distance around it has an area of 616 m².

📝 Teacher's Note: This is the same type as Answer 16, but with different units (meters instead of centimeters). Show students that the method stays the same regardless of units.

🎯 Exam Tip: Always check your units. If the question gives area in m², your final answer should also be in m². Don't mix centimeters and meters.

Answer 20.
Answer:
Step 1: Find the area of the circle.
Radius of the circle = 42 cm
Area of the circle = \( \pi r^2 \)
\( = \frac{22}{7} \times 42 \times 42 \)
\( = 5544 \text{ cm}^2 \)

Step 2: Find the circumference of the circle.
Circumference of the circle = \( 2\pi r \)
\( = 2 \times \frac{22}{7} \times 42 \)
\( = 264 \text{ cm} \)

Step 3: Find the side of the square.
Perimeter of the square = Circumference of the circle = 264 cm
\( 4 \times \text{side} = 264 \)
\( \text{side} = \frac{264}{4} = 66 \text{ cm} \)

Step 4: Find the area of the square and the ratio.
Area of square = side² = \( 66 \times 66 = 4356 \text{ cm}^2 \)

Area of the circle : Area of square = \( \frac{5544}{4356} = \frac{14}{11} \)

Area of the circle : Area of square = 14 : 11
In simple words: We found the areas of both shapes when they have the same perimeter. The circle's area is bigger than the square's area. The ratio tells us how many times bigger.

📝 Teacher's Note: This shows students that for the same perimeter, a circle always has more area than a square. This is an important property in geometry.

🎯 Exam Tip: When finding ratios, always simplify the fraction to its lowest terms. Here 5544:4356 simplifies to 14:11. Show your simplification steps clearly.

Answer 21.
Answer:
Given:
Area of garden = 5544 m²

Step 1: Find radius of circular garden.
Area = \( \pi r^2 = 5544 \)
\( r^2 = 5544 \times \frac{7}{22} \)
\( r^2 = 1764 \)
\( r = 42 \) m

Radius of circular garden = 42 m

Step 2: Find radius including road.
Radius of garden with surrounding road = R = 42 + 7 m = 49 m

Step 3: Find area including road.
Area of garden with surrounding road = \( \pi R^2 \)
\( = \frac{22}{7} \times 49 \times 49 \)
\( = 7546 \) m²

Area of garden with surrounding road = 7546 m²

Step 4: Find area of road only.
Area of road = Area of garden with surrounding road - area of garden
\( = 7546 - 5544 \) m²
\( = 2002 \) m²

Area of road = 2002 m²

Step 5: Calculate cost.
Cost of tarring one sq m = Rs. 150
Cost of tarring area of road = Rs.(150 × 2002) = Rs 3,00,300

Cost of tarring the road = Rs. 3,00,300
In simple words: We found the radius of the garden first. Then we added the road width to get the total radius. We found both areas and subtracted to get only the road area. Then we multiplied by the cost per square meter.

📝 Teacher's Note: Draw two circles on the board - one small for garden, one big for garden plus road. The ring between them is the road area. Students can see this clearly.

🎯 Exam Tip: Always write "Given" first. Show each step clearly. Don't forget to subtract the garden area from total area to get only road area. Write units in final answer.

Answer 22.
Answer:
Given:
Circumference of the plot = 176 m

Step 1: Find radius of circular plot.
\( 2\pi r = 176 \)
\( r = 176 \times \frac{7}{22} \times \frac{1}{2} \)
\( r = 28 \) m

Radius of circular plot = 28 m

Step 2: Find area of circular plot.
Area of circular plot = \( \pi r^2 \)
\( = \frac{22}{7} \times 28 \times 28 \)
\( = 2464 \) m²

Area of circular plot = 2464 m²

Step 3: Find radius including road.
Radius of circular plot with surrounding road = R = (28 + 4.2) m = 32.2 m

Step 4: Find area including road.
Area of circular plot with surrounding road = \( \pi R^2 \)
\( = \frac{22}{7} \times 32.2 \times 32.2 \)
\( = 3258.64 \) m²

Area of circular plot with surrounding road = 3258.64 m²

Step 5: Find area of road only.
Area of road = Area of circular plot with surrounding road - Area of circular plot
\( = (3258.64 - 2464) \) m²
\( = 794.64 \) m²

Area of road = 794.64 m²

Step 6: Calculate cost.
Cost of paving one sq m = Rs. 75
Cost of paving the road = Rs (75 × 794.64) = Rs. 59,598

Cost of paving the road = Rs. 59,598
In simple words: First we found the radius from circumference. Then we calculated the plot area. We added road width to get total radius and found total area. Road area is the difference. Finally we multiplied by cost per square meter.

📝 Teacher's Note: Use the formula circumference = 2πr to find radius first. Then area = πr². Show students that road is like a ring around the plot.

🎯 Exam Tip: From circumference, always find radius first using 2πr formula. Show all steps clearly. Don't forget to subtract original plot area from total area to get road area only.

Answer 23.
Answer:

[Diagram: Two circles touching each other. Left circle has radius r cm, right circle has radius (10-r) cm.]

Given:
Let r be radius of one circle, so the radius of other circle is (10-r)

Step 1: Set up equation for sum of areas.
Sum of the areas = \( \pi r^2 + \pi(10 - r)^2 \)

Step 2: Use given total area.
\( \pi r^2 + \pi(10 - r)^2 = 58\pi \)
\( \pi r^2 + \pi(100 - 20r + r^2) = 58\pi \)
\( r^2 + 100 - 20r + r^2 = 58 \)
\( 2r^2 + 100 - 20r - 58 = 0 \)
\( 2r^2 - 20r + 42 = 0 \)
\( r^2 - 10r + 21 = 0 \)
\( (r - 7)(r - 3) = 0 \)
\( r = 7, r = 3 \)

Therefore, radii of the two circles is 7cm and 3 cm
In simple words: We said one circle has radius r and the other has radius (10-r) because they add up to 10. We made an equation using their total area and solved it to find both radii.

📝 Teacher's Note: Draw two circles and show students that if one radius is r, the other must be (10-r) since they add to 10. This makes the algebra easy.

🎯 Exam Tip: Always define variables clearly. Write "Let r be radius of one circle" first. Show the quadratic equation and factorization clearly. Check that both radii add to 10.

Answer 24.
Answer:
Given:
Sum of the diameters = 112 cm

Step 1: Find sum of radii.
Therefore, sum of the radii = \( \frac{112}{2} \) cm = 56cm

Step 2: Set up variables.
If r is the radius of one circle, radius of other circle is (56-r)

Step 3: Use given total area.
Sum of the areas = \( \pi r^2 + \pi(56 - r)^2 \)
\( \pi r^2 + \pi(56 - r)^2 = 5236 \)
\( \pi r^2 + \pi(3136 - 112r + r^2) = 5236 \)
\( (r^2 + 3136 - 112r + r^2) \times \frac{22}{7} = 5236 \)
\( 2r^2 - 112r + 3136 = 1666 \)
\( 2r^2 - 112r + 3136 - 1666 = 0 \)
\( 2r^2 - 112r + 1470 = 0 \)
\( r^2 - 56r + 735 = 0 \)
\( (r - 35)(r - 21) = 0 \)
\( r = 35, r = 21 \)

Therefore, radii of the two circles is 35 cm and 21 cm
In simple words: We found that the sum of radii is half of sum of diameters. Then we used the same method as before - one radius is r, other is (56-r). We solved the equation to get both radii.

📝 Teacher's Note: Remind students that radius = diameter ÷ 2, so sum of radii = sum of diameters ÷ 2. This is a common mistake to forget.

🎯 Exam Tip: First convert sum of diameters to sum of radii by dividing by 2. Then follow the same method as previous question. Always check your answer by adding the two radii.

Answer 25.
Answer:
Given:
The sum of the radii = 10.5 cm
\( r_1 + r_2 = 10.5 \) cm ...............(i)

The difference of circumferences = 13.2 cm
\( 2\pi(r_1 - r_2) = 13.2 \)
\( r_1 - r_2 = \frac{13.2}{2\pi} \)
\( r_1 - r_2 = \frac{13.2 \times 7}{2 \times 22} \)
\( r_1 - r_2 = 2.1 \) ...........(ii)

Step 1: Solve the system of equations.
Adding (i) and (ii)
\( 2r_1 = 12.6 \)
\( r_1 = 6.3 \) cm

Therefore, \( r_2 = (10.5 - 6.3) \) cm = 4.2 cm

Hence, the two radii are 6.3 cm and 4.2 cm
In simple words: We have two equations - one for sum of radii and one for difference of radii. We solved them together by adding the equations to find each radius.

📝 Teacher's Note: This is a system of linear equations. Show students that adding the equations eliminates one variable. Teach them to check by substituting back.

🎯 Exam Tip: Convert circumference difference to radius difference first by dividing by 2π. Then solve the two equations by addition and subtraction method. Always verify your answer.

Answer 26.
Answer:
[Diagram: A semicircular flower bed with an inner semicircle of 2m width and outer edge of 7m width, showing a ring-shaped garden area.]

Given:
Diameter of the inner semi-circle = 42 m
Hence, radius of the inner semi-circle = r = 21 m

Width of the flower bed = 7m

Diameter of the outer semi-circle = (42 + 2 × 7) m = 56 m
Hence, radius of the outer semi-circle = R = 28 m

Step 1: Find area of the semicircular flower bed
Area of the semicircular flower bed = Area of outer semi-circle − area of inner semi-circle

Step 2: Apply the formula
\[ = \frac{1}{2}\pi R^2 - \frac{1}{2}\pi r^2 \]
\[ = \frac{1}{2}\pi(R^2 - r^2) \]
\[ = \frac{1}{2} \times \frac{22}{7} \times (28^2 - 21^2) \]
\[ = \frac{1}{2} \times \frac{22}{7} \times (784 - 441) \]
\[ = \frac{1}{2} \times \frac{22}{7} \times 343 \]
\[ = 539m^2 \]

Area of semicircular flower bed = 539 m²
In simple words: We found the area by subtracting the inner semicircle from the outer semicircle. This gives us just the flower bed part in the middle.

📝 Teacher's Note: Draw two semicircles on the board - one inside the other. Show students that the flower bed is the space between them. Use simple numbers like 5 and 10 first to make it easy.

🎯 Exam Tip: Always write the formula for area of semicircle first. Show both semicircles clearly in your working. Write the final answer with correct units (m²).

Answer 27.
Answer:

Given:
Radius of innermost circle = r1 = 6.3 cm
Radius of central circle = r2 = 8.4 cm

Step 1: Find area between two inner circles
Area between two inner circles = \( \pi r2^2 - \pi r1^2 \)
= \( \pi(8.4)^2 - \pi(6.3)^2 \)
= 70.56π - 39.69π ...........(i)
= 221.76 - 124.74
= 97.02cm²

Area between two inner circles = 97.02 cm²

Step 2: Find radius of third circle
Let radius of third circle be r

Area between next two circles = \( \pi r^2 - \pi r2^2 \)
= \( \pi r^2 - \pi(8.4)^2 \)
= \( \pi r^2 - 70.56π \) ...........(II)

Step 3: Solve for equal areas
Given that (i) and (ii) are equal
\[ \pi r^2 - 70.56π = 70.56π - 39.69π \]
\[ r^2 - 70.56 = 70.56 - 39.69 \]
\[ r^2 = 70.56 - 39.69 + 70.56 \]
\[ r^2 = 101.43 \]
\[ r = 10.07cm \]

Therefore, radius of third circle is 10.07 cm
In simple words: We made the ring areas equal. The first ring area helped us find what the second ring area should be. Then we worked backwards to find the third circle size.

📝 Teacher's Note: Draw three circles like a target board. Explain that each ring has the same area. Students can see this pattern easily with colored rings.

🎯 Exam Tip: Set up the equation clearly by making both ring areas equal. Show all steps when solving the quadratic. Round your final answer to 2 decimal places.

Answer 28.
Answer:
[Diagram: A rectangle of 44 cm × 28 cm with a circle of diameter 28 cm inscribed inside it.]

Given:
Area of piece of paper = 44 × 28 = 1232 cm²

The biggest circle that can be cut from rectangular piece of paper is of diameter 28 cm (i.e. Radius 14 cm).

Step 1: Find area of circle
Area of circle = \( \pi r^2 \)
= \( \frac{22}{7} \times 14 \times 14 \)
= 616cm²

Area of the circle = 616 cm²

Step 2: Find remaining paper area
Area of paper left = area of paper − area of circle
= (1232 − 616) cm²
= 616 cm²

Therefore, area of paper left = 616 cm²
In simple words: We cut out the biggest circle possible from the rectangle. Half the paper was used for the circle, half was left over.

📝 Teacher's Note: Take a rectangular paper and draw the biggest circle that fits inside. Show students that the circle diameter equals the smaller side of rectangle.

🎯 Exam Tip: The biggest circle diameter equals the smaller dimension of the rectangle. Always subtract circle area from total area to get remaining area.

Answer 30.
Answer:

Given:
An equilateral triangle has all angles of 60 degrees
The horse will be able to graze over a sector of a circle of radius 21cm and angle 60°

Step 1: Apply sector area formula
Area = \( \frac{60°}{360°} \pi r^2 \)
= \( \frac{60°}{360°} \times \frac{22}{7} \times 21 \times 21 \)
= 231m²

Horse can graze in 231 m²
In simple words: The horse is tied at the corner of a triangle. It can move in a fan shape (sector) that covers 1/6 of a full circle because 60° is 1/6 of 360°.

📝 Teacher's Note: Draw an equilateral triangle and show the horse at one corner. The rope creates a 60° fan shape. This is 1/6 of a full circle.

🎯 Exam Tip: Remember that equilateral triangle has 60° angles. Use the sector formula with angle 60°. Convert angle to fraction of 360° first.