1. ## semi circle question

Let the interval [-r,r] be the base of a semicircle. If a point is selected at random from this interval, assign a probability to the event that the length of the perpendicular segment from the point to the semicircle is less than r/2.

2. Originally Posted by calculuskid1
Let the interval [-r,r] be the base of a semicircle. If a point is selected at random from this interval, assign a probability to the event that the length of the perpendicular segment from the point to the semicircle is less than r/2.
You can use Pythagoras' theorem to calculate the fraction of the diameter for which the vertical line
to the semicircle is under half the radius.

$\displaystyle\left(\frac{r}{2}\right)^2+x^2=r^2\Ri ghtarrow\ x^2=r^2-\frac{r^2}{4}=\frac{4r^2-r^2}{4}=\frac{3r^2}{4}$

$\displaystyle\Rightarrow\ x=\frac{\sqrt{3}r}{2}$

This is the base length of two right-angled triangles both sides of the y-axis.

You only need to work with one of them.

To calculate the length of the line segment for which the perpendicular is under half the radius...

$\displaystyle\ r-\frac{\sqrt{3}r}{2}=r\left(1-\frac{\sqrt{3}}{2}\right)$

The probability is the ratio of that length to the radius.

3. The probability is the ratio of that length to the radius.
Do I have to show that.
The radius is r so how would I make that a ratio?

4. Originally Posted by calculuskid1
The probability is the ratio of that length to the radius.
Do I have to show that.
The radius is r so how would I make that a ratio?
The probability is the ratio of 2(r-x) to 2r,

because if the point is on either line segment of length r-x (on the diameter,
touching the circumference), then the perpendicular height to the semicircle
is less than half the radius.

You can also use trigonometry to find the ratio.

$\displaystyle\ probability=\frac{2(r-x)}{diameter}=\frac{2(r-x)}{2r}=\frac{r-x}{r}=\frac{r\left(1-\frac{\sqrt{3}}{2}\right)}{r}=\left(1-\frac{\sqrt{3}}{2}\right)$

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### what is the i base of a semicircle

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