a) Solve by writing the area as a function of h.

b) Solve by writing the area as a function of a.

c) Identify the type of triangle of maximum area.

Gregory Emery
2022-01-10
Answered

Find the area of the largest isosceles triangle that can be inscribed in a circle of radius 6.

a) Solve by writing the area as a function of h.

b) Solve by writing the area as a function of a.

c) Identify the type of triangle of maximum area.

a) Solve by writing the area as a function of h.

b) Solve by writing the area as a function of a.

c) Identify the type of triangle of maximum area.

You can still ask an expert for help

temnimam2

Answered 2022-01-11
Author has **36** answers

See the figure above

Using Pythagoras theorem, we can write

Length of the Base

In the given problem, radius of the circle is 6. Therefore

Length of the Base

Area of a triangle

Area of a triangle

Differentiate

Equate A’(h) to 0

Multiply throughout by

Area cannot be minimum when h = 3, because minimum area is 0.

deginasiba

Answered 2022-01-12
Author has **31** answers

For solving (b), we can find the total height $\left({h}_{T}\right)$ of the isosceles triangle by adding the value for h found in (a) to the other height 6:

${h}_{T}=6+3=9$ .

So, we can say that the total height of the triangle is 9 units. Then we can find the base:

$b=\sqrt{36-{3}^{2}}=\sqrt{36-9}=\sqrt{27}=3\sqrt{3}$ ,

We can then divide the base by 2 to get the length of the base for half the triangle (by the AAS.

Similar Triangles Theorem) to get the base to be$\frac{3\sqrt{3}}{2}$ , We can now focus on the right half of the

triangle divided by the dotted line going through the vertex point of the circle, We can tell that the

dotted line through the vertex makes a right angle. Thus, we ean use Trigonometry to solve for a.

So,

$a=\mathrm{tan}\frac{opposite}{adjacent}=\mathrm{tan}\frac{\frac{3\sqrt{3}}{2}}{9}=\mathrm{tan}\frac{27\sqrt{3}}{2}$ .

Since we have found what a is, we can find the function of it with respect to a:

$$\phantom{\rule{0ex}{0ex}}tan(\begin{array}{c}\frac{(6+h)(\sqrt{36+{h}^{2})}}{2}\end{array})$$ .

or as an incorporated function:

$\frac{1}{2}bh=\frac{1}{2}\left(6\mathrm{sin}a\right)\left(12\mathrm{cos}a\right)=36\mathrm{sin}a\mathrm{cos}a$ .

And we are done with (b)

For (c), By the Isoperimetric Theorem, it states

Theorem: Among all triangles inscribed in a given circle, the equilateral one has the largest area.

Therefore, the equilateral triangle has the maximum area.

So, we can say that the total height of the triangle is 9 units. Then we can find the base:

We can then divide the base by 2 to get the length of the base for half the triangle (by the AAS.

Similar Triangles Theorem) to get the base to be

triangle divided by the dotted line going through the vertex point of the circle, We can tell that the

dotted line through the vertex makes a right angle. Thus, we ean use Trigonometry to solve for a.

So,

Since we have found what a is, we can find the function of it with respect to a:

or as an incorporated function:

And we are done with (b)

For (c), By the Isoperimetric Theorem, it states

Theorem: Among all triangles inscribed in a given circle, the equilateral one has the largest area.

Therefore, the equilateral triangle has the maximum area.

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