Finding the Area of the Largest Isoceles Triangle

Question

Can you please help me find the answer to question (a)? I've been looking for the answer for hours. It's still wrong, please help

Answer 1

Let's assume the center of the triangle to be O and let the triangle be ABC, with BC as the base.

Let D be the altitude from O to BC. Now as you can see from the image, OD = h. As B and C lie on the circle, OB = OC, so triangle OBC is isosceles. As you hopefully know, the altitude of an isosceles triangle is also its median. So, BD = CD. Since triangle ODC is right angled, we can write:

$$OD^2+CD^2=OC^2$$

Since, we know that OD = h, OC = radius of the circle = 6, we can write: $$h^2+CD^2=6^2=36$$ $$\Rightarrow CD^2=36-h^2$$ $$\Rightarrow CD = \sqrt{36-h^2}$$ So, we can say that, $$BC=BD+CD=CD+CD=2\sqrt{36-h^2}$$ As ABC is also a isosceles triangle, its altitude also passes through the midpoint of BC, aka D, so AD is the altitude of triangle ABC. We can see that AD passes through O from the figure, so we can say that, $$AD=AO+OD=6+h$$ So, we can find the area of the triangle:

Area of triangle ABC = $\frac{1}{2}*AD*BC=\frac{1}{2}*(h+6)*(2\sqrt{36-h^2})=(h+6)\sqrt{36-h^2}$

Now we can maximize the area by taking its derivative with respect to h. We find that: $$\frac{d}{dh}(h+6)\sqrt{36-h^2}=-\frac{2(h^2+3h-18)}{\sqrt{36-h^2}}$$ We set it to zero to find its maxima: $$-\frac{2(h^2+3h-18)}{\sqrt{36-h^2}}=0$$ $$\Rightarrow h^2+3h-18=0$$ $$\Rightarrow (h+6)(h-3)=0$$ $$\Rightarrow h=-6,3$$ We know that, $h>0$ so, $h=3$ is either a minima or maxima. As we can easily minimize the area to zero by setting h to 6 and turning the area to zero, we know that this is a maxima. So we can find the maximum area by calculating the function we found earlier for $h=3$. $$Area=(h+6)\sqrt{36-h^2}=(3+6)\sqrt{36-9}=9\sqrt{27}=27\sqrt{3}$$ Hopefully my answer helps you. This is my first answer so please try to give me any advice you can.