Ex 11.1, 13
A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm2. (Use √3 = 1.7)
In the question a round table cover
is given which has six equal designs.
Let us join ends of the design
Since the designs are equal,
the angle made by the designs at point O will be equal.
We know that
Sum of angles at a point = 360°
6 × Angle made by one design = 360°
Angle made by one design = 360/6
Angle made by one design = 60°
Let us look at one design
We see that
Area of one design = Area of segment APB
And,
Total Area of design = 6 × Area of one design
= 6 × Area of segment APB
Finding Area of segment APB
Now,
Area of segment APB = Area of sector OAPB – Area of ΔOAB
Area of sector OAPB
Area of sector OAPB = 𝜃/360×𝜋𝑟2
= 𝟔𝟎/𝟑𝟔𝟎×𝟐𝟐/𝟕×(𝟐𝟖)𝟐
= 1/6×22/7 × 28 × 28
= 1/6 × 22 × 4 × 28
= 410.67 cm2
Finding area of Δ AOB
We draw OM ⊥ AB
∴ ∠ OMB = ∠ OMA = 90°
And, by symmetry
M is the mid-point of AB
∴ BM = AM = 1/2 AB
In right triangle Δ OMA
sin O = (side opposite to angle O)/Hypotenuse
sin 𝟑𝟎° = 𝐀𝑴/𝑨𝑶
1/2=𝐴𝑀/28
28/2 = AM
AM = 14
In right triangle Δ OMA
cos O = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑂)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝟑𝟎° = 𝑶𝑴/𝑨𝑶
√3/2=𝑂𝑀/21
√3/2 × 28 = OM
OM = 14 √𝟑
From (1)
AM = 𝟏/𝟐AB
2AM = AB
AB = 2AM
Putting value of AM
AB = 2 × 1/2 × 14
AB = 28 cm
Now,
Area of Δ AOB = 1/2 × Base × Height
= 𝟏/𝟐 × AB × OM
= 1/2 × 28 × 14√3
= 196√3
= 196 × 1.7
= 333.2 cm2
Therefore,
Area of segment APB
= Area of sector OAPB – Area of ΔOAB
= 410.67 – 333.2 cm2
= 77.47 cm2
Now,
Area of design = 6 × Area of segment ACB
= 6 × 77.47
= 464.82 cm2
We need to find cost of the design
Finding Cost of the design
Given that find the cost of making the designs at the rate of Rs 0.35 per cm2
Now,
Cost of making 1 cm2 design = Rs 0.35
Cost of making 464.82 cm2 design = Rs 0.35 × 464.82
= Rs 162.68
Made by
Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo
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