You do not need the area of a semi circle. You just want the area of a sector angle 120 - area of the isos triangle with angle 120. 1/3.pi.3^2 - 1/2.3^2.sin120 = 3pi -9rt3/4

@marcelowanderleycorreia8876

Ай бұрын

It´s another way to solve the question.

@skwest

Ай бұрын

Yep, just the sector...

@hanswust6972

Ай бұрын

Faster, so did I.

@phungcanhngo

24 күн бұрын

good solution.

Notice that AC is the side of an equilateral triangle inscribed in a circle with radius R=3. We know AC=R•√3=3•√3 and the heigh of triangle AOC is the abscess of equilateral triangle , so height = R/2=3/2. Shaded area = Area of the circular sector OAC - area of triangle AOC =( π•3²•120)360 - (3√3 •3/2)/2 = …….. = 3π- 9√3/4

As the 30° angle ∠CAB is on the circumference, that means that minor arc BC covers twice that, or 60°. Thus minor arc CA covers 120°. Let O be the center of the semicircle. Draw radius OC. The shaded area is the area of the sector covered by minor arc CA minus the area of triangle ∆AOC. A = (θ/360)πr² - (1/2)r²sin(θ) A = (120/360)π3² - 3²/2sin(120) A = 9π/3 - (9/2)(√3/2) A = 3π - (9√3)/4 ≈ 5.53

Complete the circle and the equilateral triangle inscribed in the circle. Subtract the area of the equilateral triangle from the area of the circle. Then divide by three to get the area = ¾ (4π − 3√3) ≈ 5.53 square units.

@PS-mh8ts

Ай бұрын

Excellent! 💯💯

شكرا لكم على المجهودات S=1/2 [3^2 (2pi)/3 -3^2 sin2pi/3]

That was done the hard way. Answer = 3*Pi - sqrt(3)*9/4 Easy way: Draw radius perpendicular to AC through O to meet semicircle at D and bisect AC at M. Angle AOM is 60 degrees. 30,60,90 triangle has sides in ratio of 1,sqrt(3),2 therefore our triangle has sides 1.5,1.5*sqrt(3),3 and area sqrt(3)*9/8. 60 degree segment of a circle has an area of (Pi * r^2)/6. Half shaded area equals Segment - triangle therefore shaded area = 2*((Pi*3*3/6)-sqrt(3)*9/8) = 3*Pi - sqrt(3)*9/4

Sector AOC - triangle AOC

But why so complicated?! 😮 You have the area of sector AOC... Now subtract the area of triangle AOC... Et voilà... You got the value of shaded area. 😂😅👍🏼

@User-jr7vf

Ай бұрын

Yea, it was surprisingly easy. I was like "is that all we have to do?" because this channel typically uploads much harder problems.

What the name you are using in writing and drawing

The correct answer is 3pi - 16sqrt(3)/8

Connect O to C and B to C Angle ABC=90° Angle BAC=30° So Angle ABC=60° OA=OB=OC=3 Triangle OBC is equilateral So angle BOC=OBC=OBC=60° Shaded area=1/2(π)3^2-(1)2(3)2sin(120°)-π(3)^2(60/360) So : area shaded=3π-(9√3)/4

A somewhat strange convoluted method shown in the video, and if Thales's theorem is applied to determine ∠ACB = 90° then no hint of any trigonometry is needed.

Instead of calculating the sector cob if u had calculated d area of sector aoc.. and then subtracted the area of triangle aoc..would have served d purpose as well, wouldnt it?

S=3(4π-3√3)/4≈5,49

Sir, could you solve this transcendental equation: x^x + x =30 without guessing

@DB-lg5sq

Ай бұрын

Yes

Yellow Area=(12π-9√3)/4

Area equals 5.5278

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