+2446 To figure out this problem, we must know a few formulas that relate to a circle. Of course, you must know the area formula for a circle:
Let A= Area of a circle
Let r= radius of the circle:
\(A_{\circ}=\pi r^2\)
However, there is a problem currently, because we do not know the circumference of this circle! However, we can solve for the radius if we know the circumference. The formula for finding the circumference is:
Let C = circumference of a circle
Let r = radius
\(C_{\circ}=2\pi r\)
We know the circumference of the circle because it is given; the circumference is 9.42 inches. Let's solve for r:
| \(C_{\circ}=2\pi r\) | This is the formula for finding the circumference of a circle. We already know the circumference, so plug it into this equation |
| \(9.42=2\pi r\) | Divide by 2 on both sides of the equation |
| \(4.71=\pi r\) | Divide by \(\pi\) on both sides. |
| \(\frac{4.71}{\pi}=r\) | I am going to leave the r in this form because I want my final answer to be as exact as possible. |
Now that we know what r equals, we can substitute this into the area formula for a circle:
| \(A_{\circ}=\pi r^2\) | Substitute \(r\hspace{1mm}\text{for}\hspace{1mm}\frac{4.71}{\pi}\) |
| \(A_{\circ}=\pi (\frac{4.71}{\pi})^2\) | We could simply input this into the calculator, but we can actually simplify this further. Let's do it! First, do \((\frac{4.71}{\pi})^2\). Remember that the exponent is distributed to both the numerator and denominator |
| \(A_{\circ}=\frac{\pi}{1}*\frac{22.1841}{\pi^2}\) | Multiply the fractions together. |
| \(A_{\circ}=\frac{22.1841\pi}{\pi^2}\) | Now, we will utilize a fraction rule stating that \(\frac{a^b}{a^c}=a^{b-c}\) |
| \(A_{\circ}=22.1841\pi^{-1}\) | Now, we will use a power rule that says that \(a^b=\frac{1}{a^b}\hspace{3mm},b<0\) |
| \(A_{\circ}=\frac{22.1841}{\pi}\) | You cannot simplify further, so evaluate with a calculator now. |
| \(A_{\circ}=\frac{22.1841}{\pi}\approx7.0614in^2\) | Of course, keep units in your answer. |
+2446 I'll also answer your next question about making the segments perpendicular. Find the slope of \(\overline{SR}\) by using the slope formula:
| \(m_{\overline{SR}}=\frac{1-(-1)}{2-(-1)}\) | Subtracting a negative number is the same as adding a positive number |
| \(m_{\overline{SR}}=\frac{1+1}{2+1}\) | Simplify the numerator and denominator. |
| \(m_{\overline{SR}}=\frac{2}{3}\) | The fraction is already in simplest terms |
Let's find the slope of \(\overline{PQ}\):
| \(m_{\overline{PQ}}=\frac{1-(-a)}{a-0}\) | Subtracting a negative is the same as adding a positive |
| \(m_{\overline{PQ}}=\frac{1+a}{a-0}\) | Subtracting a number by 0 is itself. |
| \(m_{\overline{PQ}}=\frac{1+a}{a}\) | |
For \(\overline{PQ}\hspace{1mm}||\hspace{1mm}\overline{SR}\), the slopes must be the same. Therefore, set their slopes equal to each other:
| \(\frac{1+a}{a}=\frac{2}{3}\) | Multiply by a on both sides on the equation to get rid of one of the fractions |
| \(1+a=\frac{2}{3}a\) | Multiply by 3 on both sides to get rid of the other fraction |
| \(3(1+a)=2a\) | Distribute the 3 onto both terms in the parentheses |
| \(3+3a=2a\) | Subtract 3a from both sides |
| \(3=-a\) | Multiply by -1 on both sides to isolate a. |
| \(a=-3\) | |
Therefore, the coordinates are now \(P\hspace{1mm}(-3,1)\hspace{1mm}\text{and}\hspace{1mm}Q\hspace{1mm}(0,3)\).
+2446 Oh, I am sorry. I did not anwer the other questions that you asked. Here we go. Our ultimate goal is to find what value for n makes \(\overline{AB}\hspace{1mm}\perp\hspace{1mm}\overline{CD}\). Just like in the previous problem, find the slope of \(\overline{CD}\). Since I have already found the slope in the previous problem that I answered, I don't need to do that calculation again:
\(m\hspace{1mm}\text{of}\hspace{1mm}\overline{CD}=\frac{1}{2}\)
We already know the slope of \(\overline{AB}\) because we calculated it in the previous problem as well:
\(m\hspace{1mm}\text{of}\hspace{1mm}\overline{AB}=\frac{n-5}{-4}\)
The difference with this problen, however, is that we have to find the value for n that makes the segment perpendicular! For a segment to be perpendicular, its slope must be the opposite reciprocal of the to another segment. In other words, its slope, when multiplied to the other segment must be -1. Of course, we are comparing the slope of \(\overline{AB}\) to the slope of \(\overline{CD}\):
| \(\frac{1}{2}m_{\overline{AB}}=-1\) | Multiply by 2 on both sides to get rid of the pesky fraction |
| \(m_{\overline{AB}}=-2\) | |
Now we know that the slope of \(\overline{AB}=-2\), we can solve for n:
| \(-2=\frac{n-5}{-4}\) | Solve for n. Multiply by -4 on both sides to get rid of the fraction |
| \(8=n-5\) | Add 5 to both sides to isolate n. |
| \(13=n\) | |
Therefore, in order for \(\overline{AB}\perp\overline{CD}\), the coordinate is \(A\hspace{1mm}(-1,13)\)
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