Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 2 Measurements Ex 2.4 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 2 Measurements Ex 2.4

Question 1.

Two gates are fitted at the entrance of a library. To open the gates easily, a wheel is fixed at 6 feet distance from the wall to which the gate is fixed. If one of the gates is opened to 90°, find the distance moved by the wheel (π = 3.14).

Amswer:

Let A be the position of the wall AC be the gate in initial position and AB be position when it is moved 90°.

Now the arc length BC gives the distance moved by the wheel.

Length of the arc = \(\frac{\theta}{360^{\circ}}\) × 2πr units

= \(\frac{90^{\circ}}{360^{\circ}}\) × 2 × 3.14 × 6 feets

= 3.14 × 3 feets

= 9.42 feets

∴ Distance moved by the wheel = 9.42 feets.

Question 2.

With his usual speed, if a person covers a circular track of radius 150 m in 9 minutes, find the distance that he covers in 3 minutes (π = 3.14).

Answer:

Radius of the circular track = 150m

Distance covers in 9 minutes = Perimeter of the circle = 2 × π × r units

Distance covered in 9 min = 2 × 3.14 × 150m

Distance covered in 1 min = \(\frac{2 \times 3.14 \times 150}{9} \mathrm{m}\)

Distance he covers in 3min = 314m

Question 3.

Find the area of the house drawing given in the figure.

Answer:

Area of the house = Area of a square of side 6 cm + Area of a rectangle with l = 8cm, b = 6 cm + Area of a ∆ with b = 6 cm and h = 4 cm + Area of a parallelogram with b = 8 cm, h = 4 cm

= (side × side) + (l × b) + (\(\frac { 1 }{ 2 }\) × b × h) + bh cm^{2}

= (6 × 6) + (8 × 6) + (\(\frac { 1 }{ 2 }\) × 6 × 4) + (8 × 4) cm^{2}

= 36 + 48 + 12 + 32 cm^{2}

= 128 cm^{2}

Required Area = 128 cm^{2}

Question 4.

Draw the top, front and side view of the following solid shapes

(i)

Answer:

(a) Top view

(b) Front view

(c) Side view

(ii)

Answer:

(a) Top view

(b) Front view

(c) Side view

Challenging Problems

Question 5.

Guna has fixed a single door of width 3 feet in his room where as Nathan has fixed a double door, each of width 1 \(\frac { 1 }{ 2 }\) feet in his room. From the closed position, if each of the single and double doors can open up to 120°, whose door takes a minimum area?

Answer:

Width of the door that Guna fixed = 3 feet.

When the door is open the radius of the sector = 3 feet

Angle covered = 120°

∴ Area required to open the door

= 3π feet^{2}

(b) Width of the double doors that Nathan fixed = 1 \(\frac { 1 }{ 2 }\) feet.

Angle described to open = 120°

Area required to open = 2 × Area of the sector

z

= \(\frac { 1 }{ 2 }\) (3π) feet^{2}

∴ The double door requires the minimum area.

Question 6.

In a rectangular field which measures 15 m x 8m, cows are tied with a rope of length 3m at four corners of the field and also at the centre. Find the area of the field where none of the cow can graze. (π = 3.14).

Answer:

Area of the field where none of the cow can graze = Area of the rectangle – [Area of 4 quadrant circles] – Area of a circle

Area of the rectangle = l × b units^{2}

= 15 × 8m^{2} = 120m^{2}

Area of 4 quadrant circles = 4 × \(\frac { 1 }{ 4 }\) πr^{2} units

Radius ofthe circle = 3m

Area of 4 quadrant circles = 4 × \(\frac { 1 }{ 4 }\) × 3.14 × 3 × 3 = 28.26 m^{2}

Area of the circle at the middle = πr^{2} units

= 3.14 × 3 × 3m^{2} = 28.26m^{2}

∴ Area where none of the cows can graze

= [120 – 28.26 – 28.26]m^{2} = 120 – 56.52 m^{2} = 63.48 m^{2}

Question 7.

Three identical coins, each of diameter 6 cm are placed as shown. Find the area of the shaded region between the coins. (π = 3.14) (√3 = 1.732)

Answer:

Given diameter of the coins = 6 cm

∴ Radius of the coins = \(\frac { 6 }{ 2 }\) = 3 cm

Area of the shaded region = Area of equilateral triangle – Area of 3 sectors of angle 60°

Area of the equilateral triangle = \(\frac{\sqrt{3}}{4}\) a^{2} units^{2} = \(\frac{\sqrt{3}}{4}\) × 6 × 6 cm^{2}

= \(\frac{1.732}{4}\) × 6 × 6 cm^{2} = 15.588 cm^{2}

Area of 3 sectors = 3 × \(\frac{\theta}{360^{\circ}}\) × πr^{2} sq.units

= 3 × \(\frac{60^{\circ}}{360^{\circ}}\) × 3.14 × 3 × 3 cm^{2} = 1.458 cm^{2}

∴ Area of the shaded region = 15.588 – 14.13 cm^{2} = 1.458 cm^{2}

Required area = 1.458 cm^{2} (approximately)

Question 8.

Using Euler’s formula, find the unknowns.

Answer:

Euler’s formula is given by F + V – E = 2

(i) V = 6, E = 14

By Euler’s formula = F + 6 – 14 = 2

F = 2 + 14 – 6

F = 10

(ii) F = 8,E = 10

By Euler’s formula = 8 + V – 10 = 2

V = 2 – 8 + 10

V = 4

(iii) F = 20, V = 10

By Euler’s formula = 20 + 10 — E = 2

30 – E = 2

E = 30 – 2.

E = 28

Tabulating the required unknowns