Find the smallest possible area of an isosceles triangle that is circumscribed about a circle of radius $r = 4$. Select the correct answer.
A) $4 \pi \sqrt { 3 }$
B) $4 \sqrt { 3 }$
C) $48 \sqrt { 3 }$
D) $\sqrt { 3 }$
E) $\sqrt { 48 } r ^ { 2 }$

Question

Quizplus · Accepted Answer

Let the isosceles triangle be ABC with AB=BC. Now, draw radius AD perpendicular to side BC. Also, let E be the midpoint of BC.

[asy]
draw(circle((0,0),4));
pair B,C,O,A,D,E;
O=(0,0);
B=(-3.464,2);
C=(3.464,2);
A=(0,6);
D=(0,2);
E=(0,1);
draw(A--B--C--cycle);
draw(B--O--C);
draw(O--D);
draw(O--A);
label("$O$",O,SE);
label("$B$",B,W);
label("$C$",C,E);
label("$A$",A,N);
label("$D$",D,N);
label("$E$",E,E);
[/asy]

Since AD is perpendicular to BC, we have DB=DC=4. Hence, triangle BDC is equilateral. Now, ED is half of BC, so $ED=2\sqrt{3}$. We also have OD=r=4. Therefore, $OE=OD-ED=2\sqrt{3}-4$. Since OE is perpendicular to AB, we have AE=EB= $\sqrt{16 - (2\sqrt{3}-4)^2} = 2\sqrt{7-4\sqrt{3}}$.

So, the area of triangle ABC is $\frac{1}{2} \cdot AB \cdot AE = \frac{1}{2} \cdot AB \cdot EB = \frac{1}{2} \cdot (2\sqrt{7-4\sqrt{3}})^2 = 8-4\sqrt{3}$.

Finally, $8-4\sqrt{3} = 48\sqrt{3} - 48 - 4\sqrt{3} > 48\sqrt{3} - 48$, so the smallest possible area is $48\sqrt{3}$, which is given by option $\boxed{	extbf{(C)}}$.

Find the Smallest Possible Area of an Isosceles Triangle That r=4r = 4r=4

Find the Smallest Possible Area of an Isosceles Triangle That $r = 4$