8th Maths : Chapter 2 : Measurements : Try these, Recap Exercise, Student Activities, Think and answer - Numerical Example solved problems and Numerical problems Questions with Solution

__Think__

1. 22/7 and 3.14 are rational numbers. Is ‘π’ a rational number? Why?

Solution:

22/7 and 3.14 are rational numbers π has non−terminating and non−repoating decimal expansion. So it is not a rational number. It is an irrational number.

2. When is the ‘ *π* ’ day celebrated? Why?

Solution:

March 14th is the π day celebrated for every year. Approximately value of ‘π’ is 3.14.

__Think__

The given circular figure is divided into six equal parts. Can we call the equal parts as sectors? Why?

Answer:

No, the equal parts are not sectors. Because a sector is a plane surface that is enclosed between two radii and the circular arc of the circle.

Here the boundaries are not radii.

__Try these__

Fill the central angle of the shaded sector (each circle is divided into equal sectors)

__Think__

Instead of multiplying by 1/2, 1/3 and 1/4 , we shall multiply by 180°/360°, 120°/360° and 90°/360° respectively. Why?

Solution:

So, 180°/ 360° = 1/2

120°/360° = 1/3

90°/360° = 1/4

__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*12 = 4π*r*12

Area = 4 × old area.

__Think__

All the sides of a rhombus are equal. Is it a regular polygon?

Solution:

For a regular polygon all sides and all the angles must be equal. But in a rhombus all the sides are equal. But all the angles are not equal

∴ It is not a regular polygon.

__Try these__

In the above example split the given mat into two trapeziums and verify your answer.

Solution:

Area of the mat = Area of I trapezium + Area of II trapezium

= [ 1/2 × *h*1 × (*a*1 + *b*1 )] + [ 1/2 × *h*2 × (*a*2 + *b*2 )] sq. units

= [ 1/2 × 2 × (7 + 5)] + 1/2 × 2 × (9 + 7) sq. feet

= 12 + 16 = 28 sq.feet

∴ Cost per sq.feet = ₹ 20

Cost for 28 sq. feet = ₹20 × 28 = ₹560

∴ Total cost for the entire mat = ₹560

Both the answers are the same.

__Try these__

Tabulate the number of faces (F), vertices (V) and edges (E) for the following polyhedrons. Also find F+V–E

What do you observe from the above table? We observe that, F+V–E = 2 in all the cases. This is true for any polyhedron and this relation F+V–E = 2 is known as Euler’s formula.

From the table F + V – E = 2 for all the solid shapes.

__Activity__

Draw a line to match the following shapes to their relevant nets.

__Activity__

1. Draw each of the given solid figures on an isometric dot sheet

2. Draw each of the given solid figures on a grid paper

__Activity__

Draw and name the two dimensional shapes (2-D) which you get in the cross section of the following solid shapes.

__Activity__

• Using a bangle, draw a circle on the paper and cut it. Then mark any two points A and B on it and fold the circle so that the fold has A and B on it. Now, this line segment represents a chord.

• By paper folding, find two diameters and hence the centre of a circle.

• Check whether the diameter of a circle is twice its radius.

__Note__

The circumference of a circle is 2πr units, which can also be written as πd units. As π =3.14 (approximately) and it is slightly more than 3, for some quick guess, we shall say that a circle whose diameter is *d units* shall have its circumference slightly more than three times its diameter.

For example,

If a round table of diameter of 3*m* is to be decorated by flower strings around it, it will require a little more than 9*m* of flower strings.

__Note__

The part which has a smaller arc is called as the ‘minor segment’ and the part which has a larger arc is called as the ‘major segment’.

__Note__

1. If a circle of radius *r* units divided into *n* equal sectors, then the

Length of the arc of each of the sectors = [1/*n*] × 2*π**r units*

and

Area of each of the sectors = [ 1/*n* ] × *π*r2 *sq.units*

2. Also, the area of the sector is derived as = [θº/360°] × πr2

__Note__

1. The perimeter of a semi-circle

*P = l+2r units*

= πr + 2r = (π + 2)*r* *units*

2. The perimeter of a circular quadrant

P = *l* + 2r = πr / 2 + 2*r* = ( π/2 + 2 ) *r* *units*

__Do you Know:__

1. The area of the unshaded regions in each of the squares of side *a* units are the same in all the cases given below.

2. If the biggest circle is cut from a square of side ‘*a*’ *units*, then the remaining area in the square is approximately 3/14 *a*2 *sq.units*. ( π = 22/7)

3. The area of the biggest circle cut out from the square of ‘a’ units *= *11/14 *a*2 *sq. units* (approximately)

4. In the given figure if *π =* 22/7 , the area of the unshaded part of a square of side *a* units is approximately 3/7 *a*2 *sq.units* and that of the shaded part is approximately 4/7 *a*2 *sq.units*.

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8th Maths : Chapter 2 : Measurements : Try these, Student Activities, Think and answer | Questions with Answers, Solution | Measurements | Chapter 2 | 8th Maths

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