# KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1

KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.1.

## Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.1

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.
Solution:
Radius of 1st Circle, r1 = 19 cm.
Circumference of 1st circle, C1 = 2πr1
$$=2 \times \frac{22}{7} \times 19$$
Radius of 2nd circle, r2 = 9 cm.
Circumference of 2nd circle, C2 = 2πr2
$$=2 \times \frac{22}{7} \times 9$$
Sum of the circumferences of the two circles
$$=2 \times \frac{22}{7} \times 19+2 \times \frac{22}{7} \times 9$$
$$C_{3}=2 \times \frac{22}{7}(19+9)$$
$$=2 \times \frac{22}{7} \times 28$$
= 2 × 22 × 4
∴ C3 = 176 cm.
Circumference of 3rd circle, C3 = 176 cm. then, r3 = ?
2πr3 = C3
$$2 \times \frac{22}{7} \times r_{3}=176$$
$$r_{3}=176 \times \frac{7}{22} \times \frac{1}{2}$$
∴ r3 = 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.
Solution:
We have,
Radius of circle – I, r1 = 8 cm
Radius of circle – II, r2 = 6 cm
∴ Area of circle – I = πr12 = π(8)2 cm2
Area of circle-II = πr22 = π(6)2 cm2
Let the radius of the circle – III be R cm.
∴ Area of circle-III = πR2
Now, according to the condition,
πr12 + πr22 = πR2
⇒ π(8)2 + π(6)2 = πR2
⇒ π(82 + 62) = πR2
⇒ 82 + 62 = R2
⇒ 64 + 36 = R2
⇒ 100 = R2
⇒ 102 = R2 ⇒R = 10
Thus, the radius of the new circle = 10 cm. Question 3.
The figure given below 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 the 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:
i) The diameter of the region representing Gold Score :
21
d = 21 cm. ∴ Radius, $$r=\frac{21}{2} \mathrm{cm}$$
∴ Area of the region representing Gold score
= πr2 A = 346.5 sq. cm. ii) Radius of the region representing Gold score and Red score:
r = 10.5 + 10.5 = 21 cm.
∴ Area of the region representing Red score: Area of the region representing Gold and Red Score — Area of the region representing Gold score.
= πr2 – πr2 ∴ A = 1039.5 sq.cm.

iii) Total radii of the region representing Gold, Red and Blue region :
r =10.5 + 10.5 + 10.5 = 31.5 cm
$$=\frac{63}{2} \mathrm{cm}$$
∴ Area of the region representing Blue region:
[ Area of the region representing Gold, Red and Blue regions ] — Area of the region representing Gold and Red region
∴ A = πr2 – πr2  iv) Similarly, to find Area of the region representing Black region :
∴ A = πr2 – πr2 ∴ A = 2425.5 sq. cm.

v) Area of the region representing white:
∴ A = πr2 – πr2 ∴ A = 3118.5 sq. cm

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?
Solution:
Diameter of a wheel = 80 cm
∴ Radius of the wheel = $$\frac{80}{2}$$ cm = 40 cm
So, circumference of the wheel
= 2πr = 2 × $$\frac{22}{7}$$ × 40 cm
⇒ Distance covered by a wheel in one revolution = $$\frac{2 \times 22 \times 40}{7}$$ cm
Distance travelled by the car in 1 hour (i.e., in 60 mins)
= 66 km = 66 × 1000 × 100 cm
∴ Distance travelled in 10 minutes Thus, the required number of revolutions = 4375 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) n units
C) 4 units
D) 7 units
Solution:
A) 2 units
Circumference of a circle = Area of the circle.
2πr = πr2
2 × π × r = π × r × r
∴ 2 = r
∴ r = 2 units.

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