Question

The area of the shaded segment is 100cm^2. Calculate the value of r.

Answer 1

Hello, 

The formula for finding the area of a circular region is: $A= \frac{ \alpha *r^{2} }{2}$

then:
$A_{1} = \frac{80*r^{2} }{2}$

With the two radius it is formed an isosceles triangle, so, we must obtain its area, but first we obtain the height and the base.

$cos(40)= \frac{h}{r} \\ \\ h= r*cos(40)\\ \\ \\ sen(40)= \frac{b}{r} \\ \\ b=r*sen(40)$

Now we can find its area:
$A_{2}=2* \frac{b*h}{2} \\ \\ A_{2}= [r*sen(40)][r*cos(40)]\\ \\A_{2}= r^{2}*sen(40)*cos(40)$

The subtraction of the two areas is 100cm^2, then:

$A_{1}-A_{2}=100cm^{2} \\ (40*r^{2})-(r^{2}*sen(40)*cos(40) )=100cm^{2} \\ 39.51r^{2}=100cm^{2} \\ r^{2}=2.53cm^{2} \\ r=1.59cm$

Answer: r= 1.59cm

Answer 2

Ok so we need to subtract the area of the triangle from the area of the segment and this will equal 100.
We know that the area of the segment is:
$\frac{80}{360} * \pi r^{2}$
And that the area of the triangle is:
$\frac{1}{2} r^{2} sin(80)$
Therefore:
$\frac{80}{360} * \pi r^{2} - \frac{1}{2} r^{2} sin(80)=100$
We can simplify it through these steps:
$\frac{80}{360} * \pi r^{2} - \frac{1}{2} r^{2} sin(80)=100$
$4 \pi r^{2} - 9 r^{2} sin(80)=1800$
$r^{2}(4 \pi -9sin(80))=1800$
$r^{2} = \frac{1800}{4 \pi -9sin(80)}$
$r= \sqrt{\frac{1800}{4 \pi -9sin(80)} }$
Therefore r=22.04cm (4sf)