The Shape-Shifting Triangle- Part 2

Let the angle between the two equal sides be $\theta$ . The area of the triangle is thus $\dfrac{1}{2}\cdot 3\cdot 3\cdot \sin \theta$ . This is maximized when $\sin\theta=1$ or $\theta=90^{\circ}$ . Thus, the area $A=\dfrac{1}{2}\cdot3\cdot 3\cdot 1=4.5$ , and $A^2=\boxed{20.25}$ .

Well I did a lot of tough way to solve this! I used Calculas by formatting a function of A and a. By differentiating twice, I got A= 4.5

Then A^2 = 20.25

Area of triangle = $\frac{1}{2} \times$ a b $\sin$ c

$a$ and $b$ are fixed as $3$ , so is $\frac{1}{2}$

all that is left is finding the maximum $\sin$ c , and we know $\sin 90^ \circ = 1$

The maximum area $A$ would be $\frac{1}{2} \times$ a b $\sin 90^ \circ = 4.5$

$A^2 = 4.5^2 = 20.25$

We can find the area of a triangle using this formula:1/2•length of one side •length of the other side•sin of the angle between them.Here,1/2•3•3•sin(angle).Now we know that the maximum value of sin(angle) is when that angle is 90.