Find the dimensions of the isosceles triangle of largest area that can be inscri

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Find the dimensions of the isosceles triangle of largest area that can be inscri

IMLOG10ct 2021-11-20 Answered

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

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Expert Answer

Wasither1957

Answered 2021-11-21 Author has 17 answers

Step 1
From the figure, by Pythagoras theorem we can write
$\frac{b}{2} = \sqrt{r^{2} - x^{2}}$
$b = 2 \sqrt{r^{2} - x^{2}} \to (1)$

Step 2
Area of Triangle

= \frac{1}{2}

(Base)(Height)

A = \frac{1}{2} (b) (r + x)

Use Eqn(1) to substitute the value of b

A = \frac{1}{2} (2 \sqrt{r^{2} - x^{2}}) (r + x)

A (x) = (r + x) \sqrt{r^{2} - x^{2}}

Step 3
Differentiate

A (x)

A^{'} (x) = \frac{d [(r + x) \sqrt{r^{2} - x^{2}}]}{d x}

Use Product Rule

A^{'} (x) = \sqrt{r^{2} - x^{2}} \cdot \frac{d [r + x]}{d x} + (r + x) \cdot \frac{d [\sqrt{r^{2} - x^{2}}]}{d x}

A^{'} (x) = \sqrt{r^{2} - x^{2}} \cdot 1 + (r + x) \cdot \frac{d [\sqrt{r^{2} - x^{2}}]}{d (r^{2} - x^{2})} \times \frac{d (r^{2} - x^{2})}{d x}

A^{'} (x) = \sqrt{r^{2} - x^{2}} + (r + x) \cdot \frac{1}{2 \sqrt{r^{2} - x^{2}}} \times (- 2 x)

A^{'} (x) = \sqrt{r^{2} - x^{2}} - \frac{r x + x^{2}}{\sqrt{r^{2} - x^{2}}}

Step 4
Solve for

A^{'} (x) = 0

\sqrt{r^{2} - x^{2}} - \frac{r x + x^{2}}{\sqrt{r^{2} - x^{2}}} = 0

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Elizabeth Witte

Answered 2021-11-22 Author has 24 answers

Step 1
The equation of circle will be
$x^{2} + y^{2} = r^{2}$

Step 2
Now, area of triangle

A = \frac{1}{2} (2 x) (r + y)

= x (r + y)

A (y) = \sqrt{r^{2} - y^{2}} (r + y)

triangle derivative

A^{'} (y) = \sqrt{r^{2} - y^{2}} + (r + y) \cdot \frac{- 2 y}{2 \sqrt{r^{2} - y^{2}}}

= \frac{r^{2} - r y - 2 y^{2}}{\sqrt{r^{2} - y^{2}}}

equating to zero

\frac{r^{2} - r y - 2 y^{2}}{\sqrt{r^{2} - y^{2}}} = 0

r^{2} - r y - 2 y^{2} = 0

y = \frac{r \pm \sqrt{9 r^{2}}}{- 4} \Rightarrow y = - r or y = \frac{r}{2}

rejecting,

y = - r

Therefore, we check to see

y = \frac{r}{2}

is maximum

A^{'} (y) > 0

for

y < \frac{r}{2}

and

A^{'} (y) < 0

for

y > \frac{r}{2}

∴ y = \frac{r}{2}

is maximum

x^{2} + {(\frac{r}{2})}^{2} = r^{2} \Rightarrow x = \frac{\sqrt{3}}{2} r

Dimension of triangle with maximum area is base

= 2 x = 2 \cdot \frac{\sqrt{3}}{2} r = \sqrt{3} r

h = r + y = r + \frac{r}{2} = \frac{3}{2} r

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