1. Megan,
Sorry for the delay. I have to wrestle to use the computer. Anyway, I was looking at the questions. The
answer key only provides “answers.” There are no explanations or whatsoever. Go figure.
Here are my solutions.
For number 5-2 # 2a
We did not have to solve for x. I looked at the diagram again and saw that a side length measurement is
provided for segment KH, which is 10. Since there are similar markings (double dashes) on both
segments KH and KD, they are congruent sides. Hence, segment KD also has 10. From this information,
point K is 10 units from ray EH, in the same way point K is 10 units from ray ED. This is supported by
the statement at the end of the “Vocabulary and Key Concepts” that “the distance from a point to a line is
the length of the perpendicular segment from the point to the line.
For numbers 5-3 # 1a
We have to follow exactly the steps from Example 1 except that we need to plot the points given: (0,0),
(-8,0), and (0,6). Once plotted, connect the three points with lines. You should have a triangle on
Quadrant II (the quadrant where x and y are negative).
Here are the steps to find the center of the circle:
Let’s go vertical (down/up on the y-axis). Our triangle vertically has endpoints (0,0) and (0,6). To get its
perpendicular bisector, we need to compute for the midpoint. So, ((0+0)/2, (0+6)/2)) = (0, 3). Draw a
straight horizontal line through y=3 until this line crosses the hypotenuse of the triangle.
Let’s go horizontal (right/left on the x-axis). Our triangle horizontally has endpoints (0,0) and (-8,0). To
get its perpendicular bisector, we need to compute for the midpoint. So, ((0+-8)/2, (0+0)/2)) = (-4, 0).
Draw a straight vertical line through x=-4 until this line crosses the hypotenuse of the triangle.
Where the lines, y=3 and x=-4, intersect, it is the point that is the center of the circle that circumscribes
the triangle. The coordinates of this point should be (-4,3).
For numbers 5-3 # 1b
Theorem 5-6, all the perpendicular bisectors of the sides of a triangle are concurrent. As it relates to the
question, we did not have to find a third perpendicular bisector because, obviously, once we found the
perpendicular bisectors of the two sides of the triangle, didn’t the lines meet up at a point on the third
side, which basically became its perpendicular bisector, too? (Look at your two lines drawn, which are
your perpendicular bisectors from 5-3 1a.) Hence, these bisectors overlap at one point – concurrent.
For numbers 5-3 #3
Going back to Example 3 and its drawing, we know from our computations that WX = 24, which is the
measure of the whole line segment. There are two ways to answer these. By common sense, we know
WM = 16, which is 2/3 of WX (24). So, to get the remaining 1/3 which is segment MX, WX-WM=MX.
Thus, we have 24-16 = 8. By algebraic manipulation MX = (1/3)WX, hence, MX = (1/3)24=8.
As required, we have to check WM + MX = WX. So, there’s 16 + 8 = 24. It checksJ
Hope this helps. Come after school if you have questions. This time, I’ll makes sure, I have my notes with
me.
Mrs. Crespo