Question

Find the maximum area of an isosceles triangle which can be inscribed in a circle of radius 6cm.​

Answer 1

Solution ............

Answer 2

Answer:

Explanation:

TO FIND : The maximum area of a triangle inscribed in a circle of radius ‘a'

I've calculated the maximum area by taking radius a=3. Then in the end 3 can be replaced by a

Area ( triangle) = 1/2 * base * altitude

So, to get maximum area, i started with the largest base, which will be the diameter= 6, and as shown in the figure, the largest altitude = AO=3. Hence, we concluded that to get max area, at this level the triangle has to be isosceles. ie AB = AC. SO, area = 1/2*6*3=9 unit²

NOW, We increase its altitude, so that Base will become <6

(1)● first step we increase altitude by OM= 1.4 . Because if we increase by 1.5. Then the triangle becomes equilateral. As in that condition centroid O divides median AM in the ratio AO:ON = 2:1.

SO , if we increase altitude by 1.4 = OM.

=> BM = √(BO² - OM²) = √(9–1.4²) = √(9–1.96) = √7.04= 2.653

So, area ( triangleABC) = 1/2 * BC * AM

= 2.653* 4.4 = 11.6732 ●●●●●●●●●(1)

NOW, (2)● in 2nd step, we increase altitude by 3/2 or 1.5. So, OM = 3/2, BO = 3 Hence BM = √(9–9/4)=√(27/4) = 3√3/2 .

So, AB = √( AM² + BM²) = √(81/4+27/4) =3√3

Hence we concluded that AB = 2BM .

SO, triangle ( ABC) is an equilateral triangle ●…

And area of this equilateral triangle ABC = 3√3/2 * 9/2 = 27√3/4 = (27 * 1.732)/4 = 46.764 /4 = 11.691.●●●●●●●●●●(2)

Now in (3)●rd step we increase altitude by 1.6. ie OM = 1.6 , SO BM = √(9–1.6²) = √(9–2.56)= √6.44 = 2.538 ,

So, AM = 3+1.6= 4.6

SO, area( triangle ABC) = 1/2 * BC * AM = BM * AM = 2.538* 4.6 = 11.6748 ●●●●●●●●●(3)

Now by comparing all 3 above areas , we notice that as we increase the altitude , area increases up to a limit, then it starts decreasing…

And that limit, where the area is maximum is when all the 3 sides of the triangle are equal…

ie, when the triangle is an equilateral triangle..

So, maximum area of a triangle inscribed in a circle of radius a = we calculate AB first x² = 9a²/4 +x² /4 => x² - x²/4 = 9a² /4 => x² = 3a² => x= √3a = BC

1/2 * BC * AM = 1/2 * √3a * 3a/2

= 3√3a² /4 unit² ………..ANS