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I've a right triangle that is inscribed in a circle with radius $r$ the hypotunese of the triangle is equal to the diameter of the circle and the two other sides of the triangle are equal to eachother.

Prove that when you divide the area of the circle by the area of the triangle that you will get $\pi$.

This is what I did:

The area of a triangle is $\frac{height\times width}{2}$ and the area of a circle is $\pi r^2$. Now I do not know how to continue.

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  • $\begingroup$ Have you drawn a picture? Have you determined any special properties about this triangle. Any ideas how to calculate height and width? $\endgroup$ – Doug M May 14 '18 at 18:03
  • $\begingroup$ @DougM No the problem is given without a picture. $\endgroup$ – Looper May 14 '18 at 18:03
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    $\begingroup$ I didn't ask if the problem was given with a picture. I asked if you have tried to draw a picture. $\endgroup$ – Doug M May 14 '18 at 18:06
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Well, first of all let's summarise the things we can say about this problem and after that we can set up a system of equations.

  1. By the Pythagorean theorem, we can write: $$\left|\text{AB}\right|^2=\left|\text{AC}\right|^2+\left|\text{BC}\right|^2\tag1$$  2. The area enclosed by a circle is given by: $$\mathcal{A}_{\space\circ}=\pi\cdot\text{r}_{\space\circ}^2\tag2$$

Where $\text{r}_{\space\circ}$ is the radius of the circle.  3. The connection between the radius and the diameter of a circle is given by: $$\text{d}_{\space\circ}=2\cdot\text{r}_{\space\circ}\tag3$$  4. The area of a triangle is given by: $$\mathcal{A}_{\space\triangle}=\frac{1}{2}\cdot\text{b}_{\space\triangle}\cdot\text{h}_{\space\triangle}\tag4$$ Where $\text{b}_{\space\triangle}$ is the length of the base of the triangle, and $\text{h}_{\space\triangle}$ is the height or altitude of the triangle.


Now, from your question we know that:

 - The Hypotenuse of the triangle is equal to the diameter of the circle: $$\left|\text{AB}\right|=\text{d}_{\space\circ}\tag5$$  - The two other sides of the right triangle, $\triangle\text{ABC}$, are equal to each other: $$\left|\text{AC}\right|=\left|\text{BC}\right|\tag6$$  - The $\text{b}_{\space\triangle}$ and $\text{h}_{\space\triangle}$ of $\triangle\text{ABC}$ are equal to each other and equal to the two other sides of the right triangle: $$\text{b}_{\space\triangle}=\text{h}_{\space\triangle}=\left|\text{AC}\right|=\left|\text{BC}\right|\tag7$$

So, we can write:

$$ \begin{cases} \left|\text{AB}\right|=\text{d}_{\space\circ}\\ \\ \text{d}_{\space\circ}=2\cdot\text{r}_{\space\circ}\\ \\ \mathcal{A}_{\space\circ}=\pi\cdot\text{r}_{\space\circ}^2\\ \\ \left|\text{AB}\right|^2=\left|\text{AC}\right|^2+\left|\text{AC}\right|^2=2\cdot\left|\text{AC}\right|^2\\ \\ \mathcal{A}_{\space\triangle}=\frac{1}{2}\cdot\text{b}_{\space\triangle}\cdot\text{h}_{\space\triangle}\\ \\ \text{b}_{\space\triangle}=\text{h}_{\space\triangle}=\left|\text{AC}\right|=\left|\text{BC}\right| \end{cases}\tag8 $$

Which give for $\mathcal{A}_{\space\triangle}$:

$$\mathcal{A}_{\space\triangle}=\frac{\mathcal{A}_{\space\circ}}{\pi}\space\Longleftrightarrow\space\frac{\mathcal{A}_{\space\circ}}{\mathcal{A}_{\space\triangle}}=\pi\tag9$$

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Hints:

  • You need to calculate the height and width in terms of $r$: there are two ways of doing this which will give the same area

  • You need to divide the area of the circle by the area of the triangle

![enter image description here

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Hint: Find the side of the triangle using pythagoras theorem.
$hypotenuse=2r$

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The task given "the right triangle" and "two other sides of the triangle are equal to each other". This means this is a right iscoceles triangle or $45-45-90$ triangle.

Assume that this is the $45-45-90$ triangle $ABC$ right and isoceles at $A$, draw the altitude $AH$ (perpendicular to $BC$), then we can prove that $HA=HB=HC=\dfrac{BC}{2}=r$. So the area of $\Delta{ABC}$ is:

$\dfrac{\text{base}\times \text{height}}{2}=\dfrac{2r\times r}{2}=r^2$

Alternatively, you can also use the Pythagorean theorem to calculate $AB$ and $AC$, then use the special formula to calculate the area of the right triangle (in this case, it is also isoceles).

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Hint: The area of this triangle is given by $$A_1=\frac{2r\cdot r}{2}=r^2$$ and the area of the circle is $$A_2=\pi r^2$$

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    $\begingroup$ The OP seems aware of these formulas already. $\endgroup$ – Matthew Leingang May 14 '18 at 18:10
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Consider the triangle $\triangle ABC$ with angles $45^\circ-90^\circ -45^\circ$ and hypotenuse $BC = 2r$

$AB = BC\cdot \cos(45^\circ) = \sqrt2r$

$AC = BC\cdot \sin(45^\circ) = \sqrt2r$

Area of triangle = $\frac12\cdot AB\cdot AC = r^2$

ratio = $\frac{\pi r^2}{r^2} = \pi$

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