Question

*50 points*
On a circular table cover of radius 42cm, a design is formed by a girl leaving a equilateral triangle ABC in the middle as shown in the figure. find the covered area of design.

Answer 1

Heya...!!!!

............✍️

Solution.....

Construction: Draw OD _|_ BC and join OB and OC

In ∆BOD and ∆COD

OB = OC (radiii)

OD = OD (common)

And ∆ODB = ∆ODC = 90°

Therefore

by using R.H.S congruent condition

∆ODB and ∆ODC

=> BD = DC. (by CPCT)

or BC = 2BD......(i)

∆ODB and ∆COD. (by CPCT)

=> ∆BOD and ∆COD

$\frac{120}{2}$

=> 60°

Now In ∆BOD,we have

$\sin(60 {}^0{} ) =$

$\frac{BD}{OB}$

$\sqrt{ \frac{3}{2} } = \frac{BD}{32}$

$2BD \: = 32\sqrt{3}$

$BD = 16 \sqrt{3}$

From 1 we get,

BC = 2BD

$2 \times 16 \sqrt{3} \: = 32 \sqrt{3}$

Now Area of Shaded Part

Area of Circle - Area of ∆ABC

$({ \pi \: r - \frac{ \sqrt{3} }{4} (side) {}^{2} } \: cm {}^{2} )$

$( \frac{22}{7} \times 32 \times 32 - \sqrt{} \frac{3}{4} \times 32 \sqrt{3 {}^{2} } ) \: m \\ cm {}^{2}$

$\frac{22582}{7} - 782 \sqrt{3} )cm {}^{2}$

.........✍️.....Hope it helps

Answer 2

I hope it is help full for you