EXERCISE
1. The diameter of two circles are 38 cm and 18 cm respectively.
(i) Find the radius of the circle which has circumference equal to the sum of the circumferences
of the two circles.
Solution : Diamter of first circle = 38 cm.
therefore, Radius of first circle $r_1$= 19 cm.
Therefore perimeter $p_1$= 2π × $r_1$ = 2 × $22/7 × 19$
Diameter of second radius = 18 cm
therefore radius of secong circle $r_2$= 9 cm.
Therefore perimeter $p_2$= 2π × $r_2$ = 2 × $22/7 × 9$
Given: Circumference of new circle= Sum of circumference of first circle and second circle
2π × $r$ = 2 × $22/7 × 19$ + 2 × $22/7 × 9$
Therefore radius of new Circle= 28 cm
(ii) Find the cost of fencing and ploughing the new circular field if the rate of fencing
and ploughing are ₹ 20 per meter and ₹ 10 per $m^2$
Solution
Radius of new field = 28 cm.
Circumference = 2π × $r$ = 2×$22/7$ × $28$ = 176 m
Therefore total cost of fencing : 176 × 20 = ₹ 3520
Total area of new field = $πr^2$ = $22/7 × 28×28 $ = 2464 $m^2$
Total cost for ploughing : rate × total area = 10 × 2464 = ₹ 24640
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.
Solution
Radius of first circle= 8 cm
Area of first circle = $π {r_1}^2$ = $π×8×8 $ $cm^2$
Radius of second circle= 6 cm
Area of first circle = $π {r_2}^2$ = $π×6×6 $ $cm^2$
Area of new circle = sum of the areas of the two circles
$π {r}^2$ = $π×64 + π×36 $ =$π×(64+36)$
$π {r}^2$ = $π×100$
${r}$ = $10$
3. 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.
Solution
Diameter of gold band= 21 cm
Width of other band = 10.5 cm
Therefore radius of gold band = 10.5 cm.
Area of concentric circle is given as = π $({r_1}^2 -{r_2}^2)$ = π $({r_1} -{r_2})({r_1}+{r_2})$
Area of gold region = $π {r}^2$ = ${22/7} ×{10.5}^2$ =346.5 $cm^2$
Area of red region = π $({r_1} -{r_2})({r_1}+{r_2})$
= ${22/7}× (10.5)×({10.5}+{21})$
= ${22/7}× (10.5)×(31.5)$
= $1039.5$ $cm^2$
Area of blue region = π $({r_1} -{r_2})({r_1}+{r_2})$
= ${22/7}× (10.5)×({31.5}+{21})$
= ${22/7}× (10.5)×({31.5}+{21})$
= ${22/7}× (10.5)×(52.5)$
= $1732.5$ $cm^2$
Area of black region = π $({r_1} -{r_2})({r_1}+{r_2})$
= ${22/7}× (10.5)×(43+31.5) $
= ${22/7}× (10.5)×(43+31.5) $
= $ 2458.5 $ $cm^2$
Area of white region = π $({r_1} -{r_2})({r_1}+{r_2})$
= ${22/7}× (10.5)×(54.5+43) $
= ${22/7}× (10.5)×(54.5+43) $
= $3217.5 $ $cm^2$
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?
Solution:
Formula of speed = ${distance}/{Time}$ km/h
(Always check the unit of measurement of various parameters before solving. All the unit must be in same.)
(You will notice the unit of speed is given in km/hour and time in min. So convert the time unit to hour)
So, time taken = 10 min = $10/60$ hours
So, distance travel by the car travelling in $10/60$ hours at a speed of 66 km/hour =
= $66 {km}/{hour} × 10/60 hour$
= $11$ $km$
Circumference of wheel of diameter 80 cm = π ×d
= $22/7$ × 80 $cm$
So wheel makes one full revolution when it covers a distance equal to its circumference.
So for 1 revolution , distance covered = $22/7$ × 80 $cm$
Let $n$ be the number of revolution for a distance covered = 11 $km$
= 1100000 $cm$ (1 km = 1000 m and 1 m = 100 cm)
Therefore $n$ = $1100000/{22/7 × 80}$
= 4375 revolutions
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
Solution
Let $r$ be the radius of he circle.
Given Perimeter of circle= area of circle
$2πr $ = $π r^2$
$r=2$ unit
