GSEB Class 10 Maths Solutions Chapter 12 Areas Related to Circles Exercise 12.1

Question 1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.
Answer: Let \( r_1 \) and \( r_2 \) be the radii of two circles, and \( R \) be the radius of the third circle whose circumference is equal to the sum of the circumferences of the first two circles.
We are given:
\( r_1 = 19 \) cm
\( r_2 = 9 \) cm
Let \( C_1 \) and \( C_2 \) be the circumferences of the two given circles, and \( C \) be the circumference of the third circle.
According to the problem, \( C = C_1 + C_2 \).
We know that the circumference of a circle is given by \( 2\pi r \).
So, \( 2\pi R = 2\pi r_1 + 2\pi r_2 \)
We can factor out \( 2\pi \) from the right side:
\( 2\pi R = 2\pi (r_1 + r_2) \)
Now, we can divide both sides by \( 2\pi \):
\( R = r_1 + r_2 \)
Substitute the given values for \( r_1 \) and \( r_2 \):
\( R = 19 + 9 \)
\( R = 28 \) cm
Thus, the radius of the circle whose circumference equals the sum of the circumferences of the two given circles is 28 cm.
In simple words: When you add the outside lengths (circumferences) of two circles, the new circle's outside length will simply be found by adding the radii of the first two circles together. So, a 19 cm radius plus a 9 cm radius makes a 28 cm radius.

Exam Tip: Remember that circumference is directly proportional to the radius, so when circumferences add, radii also add. This is a common shortcut for such problems.

Question 2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having an area equal to the sum of the areas of the two circles.
Answer: Let \( r_1 \) and \( r_2 \) be the radii of the two given circles, and \( R \) be the radius of the third circle.
We are given:
\( r_1 = 8 \) cm
\( r_2 = 6 \) cm
Let \( A_1 \) and \( A_2 \) be the areas of the two given circles, and \( A \) be the area of the third circle.
According to the problem, \( A = A_1 + A_2 \).
We know that the area of a circle is given by \( \pi r^2 \).
So, \( \pi R^2 = \pi r_1^2 + \pi r_2^2 \)
We can factor out \( \pi \) from the right side:
\( \pi R^2 = \pi (r_1^2 + r_2^2) \)
Now, we can divide both sides by \( \pi \):
\( R^2 = r_1^2 + r_2^2 \)
Substitute the given values for \( r_1 \) and \( r_2 \):
\( R^2 = 8^2 + 6^2 \)
\( R^2 = 64 + 36 \)
\( R^2 = 100 \)
To find \( R \), we take the square root of both sides:
\( R = \sqrt{100} \)
\( R = 10 \) cm
Hence, the radius of the circle whose area equals the sum of the areas of the two given circles is 10 cm.
In simple words: To find the radius of a new circle whose area is the sum of two smaller circles' areas, you square the radii of the small circles, add them up, and then take the square root of that sum. For 8 cm and 6 cm, it gives 10 cm.

Exam Tip: For problems involving areas, remember that areas are proportional to the square of the radius. So, while circumferences add directly, areas require squaring the radii first.

Question 3. The figure depicts an archery target marked with its five scoring regions from the center outwards as Gold, Red, Blue, Black, and White. The diameter of the region representing the Gold score 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
Answer: Let's find the area of each scoring region step-by-step.

**1. Gold Scoring Region**
Diameter of Gold region = 21 cm
Radius of Gold region, \( R_1 = \frac{21}{2} \) cm \( = 10.5 \) cm
Area of Gold scoring region \( = \pi R_1^2 \)
\( = \frac{22}{7} \times (10.5)^2 \)
\( = \frac{22}{7} \times 10.5 \times 10.5 \)
\( = 22 \times 1.5 \times 10.5 \)
\( = 346.5 \) cm\(^2\)

**2. Red Scoring Region**
The red band is 10.5 cm wide.
Radius to the outer edge of Red region, \( R_2 = R_1 + 10.5 = 10.5 + 10.5 = 21 \) cm
Area of Red scoring region \( = \pi (R_2^2 - R_1^2) \)
\( = \frac{22}{7} [(21)^2 - (10.5)^2] \)
Using the identity \( a^2 - b^2 = (a-b)(a+b) \):
\( = \frac{22}{7} (21 - 10.5)(21 + 10.5) \)
\( = \frac{22}{7} \times 10.5 \times 31.5 \)
\( = 22 \times 1.5 \times 31.5 \)
\( = 1039.5 \) cm\(^2\)

**3. Blue Scoring Region**
The blue band is 10.5 cm wide.
Radius to the outer edge of Blue region, \( R_3 = R_2 + 10.5 = 21 + 10.5 = 31.5 \) cm
Area of Blue scoring region \( = \pi (R_3^2 - R_2^2) \)
\( = \frac{22}{7} [(31.5)^2 - (21)^2] \)
\( = \frac{22}{7} (31.5 - 21)(31.5 + 21) \)
\( = \frac{22}{7} \times 10.5 \times 52.5 \)
\( = 22 \times 1.5 \times 52.5 \)
\( = 1732.5 \) cm\(^2\)

**4. Black Scoring Region**
The black band is 10.5 cm wide.
Radius to the outer edge of Black region, \( R_4 = R_3 + 10.5 = 31.5 + 10.5 = 42 \) cm
Area of Black scoring region \( = \pi (R_4^2 - R_3^2) \)
\( = \frac{22}{7} (42^2 - 31.5^2) \)
\( = \frac{22}{7} (42 - 31.5)(42 + 31.5) \)
\( = \frac{22}{7} \times 10.5 \times 73.5 \)
\( = 22 \times 1.5 \times 73.5 \)
\( = 2425.5 \) cm\(^2\)

**5. White Scoring Region**
The white band is 10.5 cm wide.
Radius to the outer edge of White region, \( R_5 = R_4 + 10.5 = 42 + 10.5 = 52.5 \) cm
Area of White scoring region \( = \pi (R_5^2 - R_4^2) \)
\( = \frac{22}{7} (52.5^2 - 42^2) \)
\( = \frac{22}{7} (52.5 - 42)(52.5 + 42) \)
\( = \frac{22}{7} \times 10.5 \times 94.5 \)
\( = 22 \times 1.5 \times 94.5 \)
\( = 3118.5 \) cm\(^2\)
In simple words: We calculate the area of each colored ring by finding the area of the larger circle up to its outer edge and then subtracting the area of the smaller circle inside it. Each time, the radius grows by 10.5 cm.

Exam Tip: For concentric circles, the area of a ring (annulus) is \( \pi(R^2 - r^2) \). Using the difference of squares formula \( (R-r)(R+r) \) can simplify calculations and reduce potential errors.

Question 4. The wheels of a car are of a diameter of 80 cm each. How many complete revolutions does each wheel make in 10 minutes? When the car is traveling at a speed of 66 km per hour?
Answer: First, let's find the distance a wheel covers in one revolution.
Diameter of wheel = 80 cm
Radius of wheel \( r = \frac{80}{2} = 40 \) cm
Distance covered by wheel in one revolution = Circumference \( = 2\pi r \)
\( = 2 \times \frac{22}{7} \times 40 \) cm
\( = \frac{1760}{7} \) cm
To convert this to meters, divide by 100:
\( = \frac{1760}{7 \times 100} \) m \( = \frac{17.6}{7} \) m

Next, let's find the total distance covered by the car in 10 minutes.
Speed of car = 66 km per hour
Convert speed to meters per minute:
\( \text{Speed in m/min} = \frac{66 \text{ km}}{1 \text{ hour}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ hour}}{60 \text{ minutes}} \)
\( = \frac{66 \times 1000}{60} \) m/minute
\( = \frac{66000}{60} \) m/minute
\( = 1100 \) m/minute
Distance covered in 10 minutes \( = \text{Speed} \times \text{Time} \)
\( = 1100 \text{ m/minute} \times 10 \text{ minutes} \)
\( = 11000 \) m

Finally, calculate the number of complete revolutions.
Number of revolutions \( = \frac{\text{Total distance covered}}{\text{Distance covered in one revolution}} \)
\( = \frac{11000 \text{ m}}{\frac{17.6}{7} \text{ m}} \)
\( = 11000 \times \frac{7}{17.6} \)
\( = 11000 \times \frac{7}{176/10} \)
\( = 11000 \times \frac{70}{176} \)
\( = \frac{770000}{176} \)
\( = 4375 \)
So, each wheel makes 4375 complete revolutions in 10 minutes.
In simple words: First, we find how far the wheel travels in one full spin. Then, we figure out the total distance the car travels in 10 minutes by changing its speed from kilometers per hour to meters per minute. Finally, we divide the total distance by the distance of one spin to learn how many times the wheel turns around.

Exam Tip: Pay close attention to unit conversions (km to m, hours to minutes) to avoid errors. Break down the problem into smaller, manageable steps like distance per revolution, total distance, and then number of revolutions.

Question 5. Tick the correct answer in the following and justify your choice. If the perimeter and area of a circle are numerically equal, then the radius of the circle is -
(a) 2 units
(b) \( \pi \) units
(c) 4 units
(d) 7 units
Answer: (a) 2 units
Let the radius of the circle be \( r \) units.
According to the question, the area of the circle is numerically equal to its perimeter (circumference).
Area of circle \( = \pi r^2 \)
Perimeter (Circumference) of circle \( = 2\pi r \)
Setting them equal:
\( \pi r^2 = 2\pi r \)
To solve for \( r \), we can divide both sides by \( \pi r \) (since \( r \) cannot be zero for a circle).
\( r = 2 \)
Therefore, the radius of the circle is 2 units.
Hence, the correct answer is (a) 2 units.
In simple words: If a circle's area number matches its perimeter number, the radius must be 2. This is because the formulas \( \pi r^2 \) and \( 2\pi r \) become equal only when \( r \) is 2.

Exam Tip: Understand the formulas for area and circumference of a circle. When solving equations involving \( \pi r \), remember that \( \pi \) and \( r \) (for a real circle) are non-zero, allowing you to divide them out safely.