Question-1 :- The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles (Take π = 22/7).Solution :-
Radius (r₁) of 1st circle = 19 cm Radius (r₂) or 2nd circle = 9 cm Let the radius of 3rd circle be r. Circumference of 1st circle = 2πr₁ = 2π (19) = 38π Circumference of 2nd circle = 2πr₂ = 2π (9) = 18π Circumference of 3rd circle = 2πr Given that : Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle 2πr = 38π + 18π 2πr = 56π r = 28 cm Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm.
Question-2 :- The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles (Take π = 22/7).Solution :-
Radius (r₁) of 1st circle = 8 cm Radius (r₂) or 2nd circle = 6 cm Let the radius of 3rd circle be r. Area of 1st circle = πr₁² = 64π Area of 2nd circle = πr₂² = 36π Area of 3rd circle = πr² Given that : Area of 3rd circle = Area of 1st circle + Area of 2nd circle πr² = 64π + 36π πr² = 100π r² = 100 r = ± 10cm However, the radius cannot be negative. Therefore, the radius of the circle having area equal to the sum of the areas of the two circles is 10 cm.
Question-3 :- In Figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions (Take π = 22/7).Solution :-
Radius (r₁) of gold region (i.e., 1st circle) = 21/2 = 10.5cm Given That : Each circle is 10.5 cm wider than the previous circle. Therefore, radius (r₂) of 2nd circle = 10.5 + 10.5 = 21 cm Radius (r₃) of 3rd circle = 21 + 10.5 = 31.5 cm Radius (r₄) of 4th circle = 31.5 + 10.5 = 42 cm Radius (r₅) of 5th circle = 42 + 10.5 = 52.5 cm Area of gold region = Area of 1st circle = πr₁² = π x (10.5)² = 346.5 cm² Area of red region = Area of 2nd circle − Area of 1st circle = πr₂² - πr₁² = π x (21)² - π x (10.5)² = 441π - 110.25π = 330.75π = 1039.5 cm² Area of blue region = Area of 3rd circle − Area of 2nd circle = πr₃² - πr₂² = π x (31.5)² - π x (21)² = 992.25π - 441π = 551.25π = 1732.5 cm² Area of black region = Area of 4th circle − Area of 3rd circle = πr₄² - πr₃² = π x (42)² - π x (31.5)² = 1764π - 992.25π = 771.75π = 2425.5 cm² Area of white region = Area of 5th circle − Area of 4th circle = πr₅² - πr₄² = π x (52.5)² - π x (42)² = 2756.25π - 1764π = 992.25π = 3118.5 cm² Therefore, areas of gold, red, blue, black, and white regions are 346.5 cm², 1039.5 cm², 1732.5 cm², 2425.5 cm², and 3118.5 cm² respectively.
Question-4 :- The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour? (Take π = 22/7).Solution :-
Diameter of the wheel of the car = 80 cm Radius (r) of the wheel of the car = 40 cm Circumference of wheel = 2πr = 2π (40) = 80π cm Speed of car = 66 km/hour = (66 x 100000)/60 cm/hour = 110000 cm/hour Distance travelled by the car in 10 minutes = 110000 × 10 = 1100000 cm Let the number of revolutions of the wheel of the car be n. n × Distance travelled in 1 revolution (i.e., circumference) = Distance travelled in 10 minutes n x 80π = 1100000 n = 110000/80π n = (110000 x 7)/(80 x 35) n = 35000/8 n = 4375 Therefore, each wheel of the car will make 4375 revolutions.
Question-5 :- Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units (B) π units (C) 4 units (D) 7 units .
Let the radius of the circle be r. Circumference of circle = 2πr Area of circle = πr² Given that : The circumference of the circle and the area of the circle are equal. 2πr = πr² 2 = r r = 2cm Therefore, the radius of the circle is 2 units. Hence, the correct answer is A.