Question

In circular table cover of radius 32cm design is formed leaving an equilateral triangle KBC in the middle as shown in ( figure 12.24 ) find the area of the design ?

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Answer 1

Answer :

We have to find the area of the design.

Hence,

We can say that,

Area of the design = Area of the circle - Area of ∆ ABC

So,
Let's find the area of circle

So,

r = 32 cm

We know that,

$\sf{Area \: of \: circle = πr^2}$

$\sf{ = \frac{22}{7} \times 32 \times 32}$

$\sf{ = \frac{22}{7} \times 1024}$

$\sf{ = \frac{22528}{7} \: cm {}^{2}}$

Now,
Let's find the area of ∆ ABC

Here
We have to do some construction !

Firstly,

Let the centre of circle be 'T'

Join TA , TB, TC

Hence,

TA = TB = TC ( Radius of same circle )

AB = AC = BC ( Sides of equi. Triangle )

Hence,
We can say that ,

Angle ATB = Angle BTC = Angle ATC ... (1)

But,

Sum of angles around any point equals to 360°

Hence,

Angle ATB + Angle BTC + Angle ATC = 360°

Angle BTC + Angle BTC + Angle BTC = 360°

3 ( Angle BTC ) = 360°

Angle BTC = 360°/3

Angle BTC = 120°

Now,
We know that

∆ ATB , ∆ ATC and ∆ BTC are congruent.

Hence,

Area ∆ ATB = Area ∆ ATC = Area ∆ BTC

Hence,

$\sf{area \: of \triangle \: BTC = \frac{1}{3} \times Area \: of \: \triangle ABC}$

Now,

Let's find Area of ∆ BTC

Area of ∆ BTC = 1/2 × Base × Height

Now,
Again, We have to do some construction !

Draw TM perpendicular to BC

So, Angle TMB = Angle TMC = 90°

In ∆ TMB and ∆ TMC ,

Angle TMB = Angle TMC

TB = TC ( Radius )

TM = TM ( Common side )

Hence,

∆ TMB = ∆ TMC

Hence,

By c. p. c. t,

Angle BTM = Angle CTM

Angle BTM = Angle CTM = 1/2 Angle BTC

Angle BTM = Angle CTM = 60°

Now,

We know that,

∆ TMB = ∆ TMC

By c. p. c. t

TM = CM

Hence,

TM = CM = 1/2 × BC ... (2)

Now,

In ∆ TMB,

sin T = opposite / Hypotenuse

sin 60° = BM / TB

√3/2 = BM / 32

32√3 = 2 MB

16 √3 = MB

Now,

In ∆ TMB,

cos T = adjacent / Hypotenuse

cos 60° = TM / TB

1/2 = TM / 32

32 = 2 TM

16 = TM

Now,

From (2) ,

BM = 1/2 × BC

BC = 2 BM

BC = 2 × 16 √3

BC = 32 √3

Now,

Area of ∆ BTC = 1/2 × Base × Height

= 1/2 × BC × TM

= 1/2 × 32 √3 × 16

= 32√3 × 8

= 256√3

Now,

We know that,

Area (∆BTC) = 1/3 × Area of ∆ ABC

256√3 = 1/3 × Area of ∆ ABC

3 ( 256 √3 ) = Area of ∆ ABC

Area of ∆ ABC = 768 √3 sq.cm

Now,

Finally ,

Area of the design = Area of the circle - Area of ∆ ABC

$\sf{ = \frac{22528}{7} - 768 \sqrt{3}}$

$\sf{ = 3218.2 - 768 \sqrt{3}}$
We know that,
√3 = 1.73

$\sf{ = 3128.2 -( 768 \times 1.73)}$

$\sf{ = 3218.2 - 1328.6}$

$\sf{ = 1,889.6 \: cm {}^{2} }$

Hence,

Area of the design is 1889.6 cm^2 or 1890 cm^2.

_____________________________

If any doubt, write in comments box.

Answer 2

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