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°] × 2πr 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 (π = 3.14) .
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 cm2
= (6 × 6) + ( 8 × 6) + (1/2 × 6 × 4) + (8 × 4) cm2
= 36 + 48 + 12 + 32 cm2
= 128 cm2
Required Area = 128 cm2
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°] × πr2 = [120° / 360°] × π × 3
× 3
= 3π feet2
(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] feet2 = 3π / 2
feet2
= (1/2 )(3π) feet2
∴ The double door requires the minimum area.
6. In a rectangular field which measures
15 m × 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 no cows can graze. (π = 3.14)
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 units2
= 15 × 8 m2 = 120 m2
Area of 4 quadrant circles = 4 × (1/4) πr2
units
Radius of the circle = 3m
Area of 4 quadrant circles = 4 × (1/4) × 3.14 × 3 × 3 = 28.26m2
Area of the circle at the middle = πr2 units
= 3.14 × 3 × 3 m2 = 28.26m2
∴ Area where none of the
cows can graze
= [120 − 28.26 − 28.26]m2 = 120 − 56.52 m2
= 63.48m2 (approx)
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)
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 a2 units2 = √3/4 × 6 × 6 cm2
= [1.732 / 4] × 6 × 6 cm2 = 15.588 cm2
Area of 3 sectors = 3 × [θ/360°] × [πr2]
sq.units
= 3 × [60°/360°] × 3.14 × 3 × 3 cm2 = 14.13 cm2
∴ Area of ther shaded
region = 15.588 – 14.13 cm2 = 1.458 cm2
Required area = 1.458 cm2 (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 cm2
4.
Challenging Problems
5. double door requires
the minimum area
6. 63.48 m2 (approximately)
7. 1.46 cm2 (approximately)
8. (i) F = 10 (ii) V =
4 (iii) E= 28
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