# Answers to Review for Quiz 10 ```Geo I
Review for Semester 1 Quiz 10
Name ________________
Date ___________
Q (-2, 1)
U (2, 3)
A (3, 1)
D (-1, -1)
1. What is the most specific name
actual WORDS) each of your steps.
Step 1: Find the slopes of all 4 sides. This will allow me
To determine if the quadrilateral is a trapezoid, some
Kind of a parallelogram, or neither. I got slopes of
2/4 = &frac12; , -2/1, 2/4 = &frac12;, and -2/1.
2 pairs of opposite sides tells me this is a parallelogram
Step 2: Look at the slopes of a pair of consecutive sides to determine if they are perpendicular. This will tell
me whether or not the parallelogram is a rectangle.
2/4 = &frac12; and -2/1
Opposite reciprocal slopes tell me that this IS a rectangle.
Step 3: Check to see if it is also a rhombus by determining if the diagonals are perpendicular. I found the
diagonals to have slopes of 0 and 4/3
The slopes are not opposite reciprocals of each other, so this is NOT a rhombus. The most specific
name for the quadrilateral is a rectangle.
2. If QUAD is reflected over the y-axis, give the image points.
Q’ (2, 1)
U’ (-2, 3)
A’(-3, 1)
D’ (1, -1)
3. If QUAD rotated 180&deg;, give the image points.
Q’ (2, -1)
U’ (-2, -3)
A’(-3, -1)
D’ (1, 1)
4. If QUAD is translated 3 units to the right, and 1 unit down, give the image
points.
Q’ (1, 0)
U’ (5, 2)
A’(6, 0)
D’ (2, -2)
Q (-2, 1)
U (2, 3)
A (3, 1)
D (-1, -1)
5. Find the length of QU.
42 + 22 = d2
From -2 to 2 is 4 spaces, from 1 to 3 is 2 spaces.
The length of QU is the square
root of 20.
6. Find the midpoint of UA
(Average!)
3 + 2 = 5 5 / 2 = 2.5
1+3=4
(2.5, 2)
4/2=2
7. Find the length of the diagonal QA.
8. Find the midpoint of diagonal UD
From -2 to 3 is 5 spaces, from 1 to 1 is 0 spaces.
(Average!)
-1 + 2 = 1 1 / 2 = .5
52 + 02 = d2
-1 + 3 = 2
The length of QU is the 5
2/2=1
(.5, 1)
.
For # 9 &amp; 10, ΔABC = ΔDEF.
9. Which is the largest &lt; in ΔABC?
E
A
2x + 3
B
D
(2x – 10)&deg;
4x – 1
3x – 2
C
10. Which is the shortest side in ΔDEF?
6x – 6
F
(x + 15)&deg;
B
D
A
(x + 25)&deg;
(2x)&deg;
C
F
E
2x + 3 = 3x – 2
5=x
DE =3(5) – 2 = 13
EF = 4(5) – 1 = 19
DF = 6(5) – 6 = 24
2x – 10 = x + 25
x = 35
m&lt;A = 2(35) – 10 = 60
m&lt;B = 35 + 15 = 50
m&lt;C = 2(35) = 70
DF is the longest side, so E is the largest &lt; in ΔDEF.
(Flashlight) &lt;B corresponds to &lt;E, so
&lt;B is the smallest &lt; in ΔABC, AC is the shortest
side. (Flashlight) DF corresponds to AC, so
&lt;B is the largest &lt; in ΔABC
AC is the shortest side in ΔDEF
C
Complete the proof.
1
AB // CD
B
4
AB // CD
Given
Given
&lt;1 = &lt;3
&lt;2 = &lt;4
// lines  alt int &lt;s =
// lines  alt int &lt;s =
∆ABC = ∆CDA
ASA