8th Maths : Chapter 2 : Measurements : Exercise 2.4 : Miscellaneous Practice problems, Challenge Problems, Text Book Back Exercises Questions with Answers, Solution

**Exercise 2.4**

** **

**Miscellaneous
Practice Problems**

** **

**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 90º, find
the distance moved by the wheel (***π =*** 3.14)
.**

**Solution:**

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 = [θ / 360°] × 2π*r* units

= [90° / 360°] × 2 × 3.14 × 6 feets

= 3.14 × 3 feets

= 9.42 feets

∴ Distance moved by the wheel = 9.42 feets.

** **

**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 (**

**Solution:**

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 × 150 m

∴ Distance covered in 1 min = [ 2 × 3.14 × 150 ] / 9 m

Distance covered in 3 min = [ 2 × 3.14 × 150 × 3 ] / 9 = 314 m

Distance he covers in 3 min = 314 m

** **

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

**Solution:**

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 Δ with *b* = 6
cm and *h* = 4 cm + Area of a parallelogram with *b*= 8 cm, *h*
= 4 cm

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

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

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

= 128 cm^{2}

Required Area = 128 cm^{2}

** **

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

**Solution: **

** **

**Challenging
problems**

** **

**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 1/2 feet
in his room. From the closed position, if each of the single and double doors can
open up to 120º, whose door takes a minimum area?**

**Solution: **

**(a)**

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

= [120° / 360°] × π*r*^{2} = [120° / 360°] × π × 3
× 3

= 3π feet^{2}

**(b)**

Width of the double doors that Nathan fixed = 1 (½) feet.

Angle described to open = 120°

Area required to open = 2 × Area of the sector

= 2 × [120°/360°] × π × [3/2] × [3/2] feet^{2} = 3π / 2
feet^{2}

^{}

= (1/2 )(3π) feet^{2}

∴ The double door requires the minimum area.

** **

**6. In a rectangular field which measures
15 m **

**Solution:**

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 × 8 m^{2} = 120 m^{2}

Area of 4 quadrant circles = 4 × (1/4) π*r*^{2}
units

Radius of the circle = 3m

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

Area of the circle at the middle = π*r*^{2} units

= 3.14 × 3 × 3 m^{2} = 28.26m^{2}

∴ Area where none of the
cows can graze

= [120 − 28.26 − 28.26]m^{2} = 120 − 56.52 m^{2}
= 63.48m^{2} (approx)

** **

**7. Three identical coins each of diameter
6 cm are placed as shown. Find the area
of the shaded region between the coins. (**

**Solution: **

Given diameter of the coins = 6 cm

∴ Radius of the coins =
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 = √3/4 *a*^{2} units^{2} = √3/4 × 6 × 6 cm^{2}

= [1.732 / 4] × 6 × 6 cm^{2} = 15.588 cm^{2}

Area of 3 sectors = 3 × [θ/360°] × [π*r*^{2}]
sq.units

= 3 × [60°/360°] × 3.14 × 3 × 3 cm^{2 }= 14.13 cm^{2 }

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

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

** **

**8. Using Euler’s formula, find the unknowns.**

**Solution:**

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

**Answer:**

**Exercise 2.4 **

**Miscellaneous Practice
Problems **

**1. 9.42 feet **

**2. 314 m **

**3. 128 cm^{2} **

**4. **

**Challenging Problems **

**5. double door requires
the minimum area **

**6. 63.48 m^{2} (approximately)**

**7. 1.46 cm^{2} (approximately) **

**8. (i) F = 10 (ii) V =
4 (iii) E= 28**

Tags : Questions with Answers, Solution | Measurements | Chapter 2 | 8th Maths , 8th Maths : Chapter 2 : Measurements

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8th Maths : Chapter 2 : Measurements : Exercise 2.4 | Questions with Answers, Solution | Measurements | Chapter 2 | 8th Maths

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