8th Maths : Chapter 2 : Measurements : Parts of a Circle: Numerical Example solved problems

__Example 2.1__

A circular shaped gymnasium ring of radius 35*cm* is divided into 5 equal arcs shaded with different colours. Find the length of each of the arcs.

*Solution:*

Radius, *r* = 35 *cm* and *n* = 5.

Length of each of the arcs, *l* = 1/*n* × 2πr units

= 1/5 × 2 × π × 35 = 14 π *cm*.

__Example 2.2__

A spinner of radius 7.5 *cm* is divided into 6 equal sectors. Find the area of each of the sectors.

*Solution:*

Radius, *r* = 7.5 *cm* and *n* = 6.

Area of each of the sectors, *A* = 1/*n* × π*r*2 *sq. units*

= 1/6 × π × 7.5 × 7.5

= 9.375π *sq. cm*

__Example 2.3__

Kamalesh has a dining table, circular in shape of radius 70 *cm* whereas Tharun has a circular quadrant dining table of radius 140 *cm*. Whose dining table has a greater area? [π = 22/ 7]

*Solution:*

Area of the dining table with Kamalesh = π*r*2 *sq. units*

= 22/7 × 70 × 70

*A *= 15400* sq.cm *(approximately.)

Area of the circular quadrant dining table with Tharun

= 1/4* *π*r*2* *=* *1/4* *×* *22/7* *×140* *×140

* A *= 15400* sq.cm *(approximately.)

We find that, the area of the dining tables of both of them have the same area.

Think

If the radius of a circle is doubled, what will happen to the area of the new circle so formed?

**Solution:**

If *r* = 2*r*_{1}

⇒ Area of the circle = π*r*^{2}
= π (2*r*_{1})^{2} = π4*r*_{1}^{2} =
4π*r*_{1}^{2}

Area = 4 × old area.

__Example 2.4__

Four identical medals, each of diameter 7*cm* are placed as shown in Fig. 2.18. Find the area of the shaded region between the medals. [π = 22/7]

*Solution:*

Diameter, *d* = 7 *cm,* therefore *r* = 7/2 *cm*.

Area of the shaded region = Area of the square – 4 × Area of the circular quadrant

*= a*2* *−* *4* *×* *1/4* *π*r*2

= (7×7) – ( 4 × 1/4 × 22/7 × 7/2 ×7/2 )

= 49 –38.5 = 10.5 *sq.cm.* (approximately)

Tags : Measurements | Chapter 2 | 8th Maths , 8th Maths : Chapter 2 : Measurements

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