Question

The segment in the smaller circle is made by a line equal to the radius of the smaller circle. The circle can fit exactly in the semi-ci cle as shown. If the area of shaded region is given by + Zv3), what is the value of x-Y+Z given that X and Y are coprime? R a​

Answer 1

Step-by-step explanation:

The segment in the smaller circle is made by a line equal to the radius of the smaller circle. The circle can fit exactly in the semi-ci cle as shown. If the area of shaded region is given by + Zv3), what is the value of x-Y+Z given that X and Y are coprime? R a

Answer 2

Assuming the question you are asking is this, so reframing the question:

The segment in the smaller circle is made by a line equal to the radius of the smaller circle. The circle can fit exactly in the semi-circle as shown. If the area of shaded region is given by $\frac{R^{2}}{8} \ (\frac{X\pi }{Y} +Z\sqrt{3} )$, what is the value of X-Y+Z given that X and Y are coprime?

Answer:

The value of $X-Y+Z$ is 3.5.

Step-by-step explanation:

Given the radius of bigger circle is R.  

Shaded portion is a semicircle, so the area of the semicircle is  $\frac{\pi R^{2}}{2}$  

Given a segment in the smaller circle is made by a line equal to the radius of the smaller circle.

From the figure, the radius of the smaller circle is $\frac{R}{2}$

Therefore, the shaded portion in small circle is an equilateral triangle, with length of side, $s=\frac{R}{2}$

Area of equilateral triangle is $\frac{\sqrt{3} }{4} s^{2}$

                                                $=\frac{\sqrt{3} }{4} (\frac{R}{2} )^{2}=\frac{\sqrt{3} }{4} \frac{R} {4}^{2}$

Therefore, area of the shaded region is  

                              $\frac{\pi R^{2}}{2}+\frac{\sqrt{3} }{4} \frac{R} {4}^{2}=\frac{ R^{2}}{8}(4\pi+\frac{\sqrt{3} }{2} })$

Comparing this with the given area, $\frac{R^{2}}{8} \ (\frac{X\pi }{Y} +Z\sqrt{3} )$

we get, $X=4, Y=1$ and $Z=\frac{1}{2}$

Therefore, $X-Y+Z=4-1+\frac{1}{2} =3.5$

And $X=4, Y=1$, the common factor of 4 and 1 is 1.

So, they are co-prime.