MHB Find the radius of the sector adjoining a triangle

[mathlearn]

https://i.imgsafe.org/5172eee94d.jpg

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We know that the area of the sector should be $\frac{40}{360}$*$\frac{22}{7}$*$r$*r

Any ideas on how to begin?

Many Thanks:)

 

[MarkFL]

We know the area of the triangle and the area of the sector are equal, so using the respective formula for those areas we may state:

$$\frac{1}{2}(2\pi)(r)=\frac{1}{2}\left(40^{\circ}\cdot\frac{\pi}{180^{\circ}}\right)r^2$$

What do you get when solving for $r$?

 

[mathlearn]

We know the area of the triangle and the area of the sector are equal, so using the respective formula for those areas we may state:

$$\frac{1}{2}(2\pi)(r)=\frac{1}{2}\left(40^{\circ}\cdot\frac{\pi}{180^{\circ}}\right)r^2$$

What do you get when solving for $r$?

$\displaystyle \frac{1}{2}(2\pi)(r)=\frac{1}{2}\left(40^{\circ}\cdot\frac{\pi}{180^{\circ}}\right)r^2$

$\displaystyle (\pi)(r)=\left(\frac{\pi}{9^{\circ}}\right)r^2$

$9 \displaystyle (\pi)(r)=\pi r^2$

Now Let's use factorization to find r ,

$9 \displaystyle (\pi)(r)=\pi r^2$

$9 \displaystyle (\pi r)=(\pi r) * r $

$9 cm =r$

Now to check whether It is correct,

We know the area of the triangle and the area of the sector are equal

$ \displaystyle \frac{1}{2} * 2 * \frac{22}{7} * 9 = \frac{22}{7} * \frac{40}{360}* 9^2$

$ \displaystyle \frac{22}{7} * 9 = \frac{22}{7} * \frac{1}{9}* 9^2$

$ \displaystyle \frac{22}{7} * 9 = \frac{22}{7} * \frac{1}{9}* 9 * 9$

$ \displaystyle \frac{22}{7} * 9 = \frac{22}{7} * 9 $

Correct I guess ? :)

Many Thanks :)

 

[MarkFL]

Yes, I also got:

$$r=9\text{ cm}$$

In your second line, the degrees would have "cancelled" and so you would just have:

$$(\pi)(r)=\left(\frac{\pi}{9}\right)r^2$$

When checking the answer, I would simply use the $\pi$ symbol rather than a rational approximation for $\pi$. :)

 

[mathlearn]

:) Thanks For the advice.