Triangle ABO is situated within the Circle with center O so

Since OA = OB = r , Perimeter (AOB) = AB + OB + OA = 2OA + AB = 8 + \(\sqrt{32}\)

Hence, \(AB = \sqrt{32}\)
and OA = OB = r = 4

\(Lenght of arc AB = diameter * Pi * 45/360\)

--> \(Lenght of arc AB = Pi * \sqrt{2} / 2\)

Finally, the area of the portion bounded by line AB and arc AB \(Area = AB* ArcAB - 2r\)

\(Area = (Pi \sqrt{2} * \sqrt{32} / 2) - 8\)

Thus, Answer A : 4Pi-8

Did not understand this one. Probably, I am missing something

Hi

First of all, Points A and B lie on the circle with center O : OA = OB = r hence triangle OAB is iscocele triangle . (1)

Second, given that perimeter of triangle AOB is equal to \(AB + OB + OA = 8 + \sqrt{32}\)

From (1) , \(2OA + AB = 8 + \sqrt{32}\) Thus, \(OA = 4\) AND \(AB = \sqrt{32}\) because \(\sqrt{32}\)cannot equal 2OA ( otherwise the triangle OAB wouldn't stand as isocele )

Now, we know that OA = OB = r = 4 and \(AB = \sqrt{32}\) and OAB is isocele

and AB^2 = 32 and OA^2 = OB^2 = 16 Hence OAB is a right isocele triangle. from pythagor (AB^2 = OA^2 + OB^2)

Area of shaded region = area of sector AOB - area of triangle AOB

area of sector AOB = Lenght of Arc(AB) * AB
Lenght of Arc (AB) = diameter * Pi * 45/360 = Pi \(\sqrt{2}\) / 2

area of sector AOB = 4Pi
area of triangle = 4*4 /2 = 8

Finally: area of shaed region = 4Pi - 8

Hope that helps

Thanks!. This surely helped me.

I have a problem with this part..

\(\sqrt{32} = 4 \sqrt{2}\)

Therefore, why couldn't the two equal sides be \(2\sqrt{2}\) each and the third side be 8?

And even then, how are you getting the angle of the triangle? I feel like there are some number properties of triangles i'm missing that I need to memorize here...

Hi dave785,

Answer to the 1st question: The lenght of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides; Therefore OA and OB must be equal to 4 and AB must be equal to \(\sqrt{32}\)

Now , for the second question, which angle are you talking about ?

Lenght of Arc (AB) = diameter * Pi * 45/360 = Pi \(\sqrt{2}\) / 2


How is this operation not equal to just pi.
Diameter is 8, so 8*45/360 cancels out and the only thing left is Pi.
Where do you get sqrt(2)/2 from?

Small correction. I believe length of arc is found always from centre angle which should be 90 degrees and not 45 degrees.

If an angle in the triangle is 90 degress, shouldn't the hypotenuse be the diameter. How is this scenario possible?

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