**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, r_{1} = 19 cm.

Circumference of 1st circle, C_{1} = 2πr_{1}

\(=2 \times \frac{22}{7} \times 19\)

Radius of 2nd circle, r_{2} = 9 cm.

Circumference of 2nd circle, C_{2} = 2πr_{2}

\(=2 \times \frac{22}{7} \times 9\)

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

∴ C_{3} = 176 cm.

Circumference of 3rd circle, C_{3} = 176 cm. then, r_{3 }= ?

2πr_{3} = C_{3}

\(2 \times \frac{22}{7} \times r_{3}=176\)

\(r_{3}=176 \times \frac{7}{22} \times \frac{1}{2}\)

∴ r_{3} = 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, r_{1} = 8 cm

Radius of circle – II, r_{2} = 6 cm

∴ Area of circle – I = πr_{1}^{2} = π(8)^{2} cm^{2}

Area of circle-II = πr_{2}^{2} = π(6)^{2} cm^{2}

Let the radius of the circle – III be R cm.

∴ Area of circle-III = πR^{2}

Now, according to the condition,

πr_{1}^{2} + πr_{2}^{2} = πR^{2}

⇒ π(8)^{2} + π(6)^{2} = πR^{2}

⇒ π(8^{2} + 6^{2}) = πR^{2}

⇒ 8^{2} + 6^{2} = R^{2}

⇒ 64 + 36 = R^{2}

⇒ 100 = R^{2}

⇒ 10^{2} = R^{2} ⇒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

= πr^{2}

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.

= πr^{2} – πr^{2}

∴ 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 = πr^{2} – πr^{2}

iv) Similarly, to find Area of the region representing Black region :

∴ A = πr^{2} – πr^{2}

∴ A = 2425.5 sq. cm.

v) Area of the region representing white:

∴ A = πr^{2} – πr^{2}

∴ 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 = πr^{2}

2 × π × r = π × r × r

∴ 2 = r

∴ r = 2 units.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.1, drop a comment below and we will get back to you at the earliest.