<p>This one I don't have scanner its on page 516 test 3 section 2</p>
<p>7)In the circle above,pentagon ABCDE is equilateral. What is the ratio of arc ABC to the length of arc AEC?</p>
<p>1) 1 to 2
2) 2 to 3
3) 2 to 5
4) 3 to 5
5) 4 to 5</p>
<p>I have made the figure into triangles in order to know the ratio and I choosed 1:2 but the answer is 2:3 please tell me why.</p>
<p>18)Any 2 points determine a line. If there are 6 points in a plane,no 3 of which lie on the same line,how many lines are determined by pairs of these 6 points?</p>
<p>1)15
2)18
3)20
4)30
5)36</p>
<p>Okay since there are 2 points which determine a line and 6 points on a plane,then we can make three parrallel lines and draw them,so I computed 20 lines this is how I pictured it in my mind.</p>
<p>Suppose that "|" mark the points.</p>
<p><<-|-------|->></p>
<p><<-|-------|->></p>
<p><<-|-------|->></p>
<p>So the bottom point | can connect with the 8 points I computed 13 lines,but the answer is wrong.</p>
<p>7) Given: Equilateral pentagon ABCDE is inscribed in a circle.
Problem: What is the ratio of the length of arc ABC to the length of arc AEC?</p>
<p>Solution: By symmetry, the arclengths between adjacent vertices of the pentagon are all the same. For arc ABC, you should count two arcs between adjacent vertices(A to B, B to C). For arc AEC, you should count three arcs between adjacent vertices (A to E, E to D, D to C).<br>
So the ratio is 2 arclengths to 3 arclengths (2 to 3)</p>
<p>18) Once you recognize the problem as a combination problem, you should be able to solve the problem. Some examples of combinations include:</p>
<p>a) Six people in a room shake hands with each other. How many handshakes are there?
b) Six dots form the vertices of an equilateral hexagon. What is the maximum number of line segments you can draw between the dots, including edges and diagonals?</p>
<p>Answer: Use the formula for combinations.
Combinations: nCr = n!/[(r!)(n-r)!] How many ways can you choose r things out of a group of size n, where the order doesn’t matter.</p>
<p>Here n = 6, r = 2</p>
<p>nCr = 6C2 = 6!/[2! (6-2)!] = 6!/[2! (4!)] = 6<em>5/ (1</em>2) = 15</p>
<p>Alternate method: Draw the six dots which form the equilateral hexagon. Next, draw as many edges and diagonals as possible, while counting the line segments.</p>
<p>If you are not familiar with the formula for combinations then an alternate way to solve 18 is to count as follows.</p>
<p>Number the points 1 through 6. This is just for convenience, and to allow me to explain the approach.</p>
<p>Start with point 1. How many distinct lines can you draw where one of the lines starts at point 1? Well 5: 1–2, 1–3, 1–4, etc.</p>
<p>Now start with point 2. Same question. But we musn’t include the line to point 1 (because we’ve already taken that into account). So you get 4: 2–3, 2–4, etc.</p>
<p>Then with point 3. We musn’t include the line to point 1 or 2, because we’ve already counted those. You get 3:</p>
<p>Etc.</p>
<p>So the answer is: 5 + 4 + 3 + 2 + 1 = 15</p>
<p>Thanks alot guys that helped alot.</p>