Geometry B U2 L3 Proving That a Quadrilateral Is a Parallelogram Answers
1. Complete this statement: A polygon with all sides the same length is said to be ______.
Answer - B. Equilateral
2. A road sign is in the shape of a regular heptagon. What is the measure of each angle on the sign? Round to the nearest tenth.
Answer - A. 128.6°
3. In the figure, the horizontal lines are parallel and AB=BC=CD. Find the measure of JM. The diagram is not drawn to scale.
Answer - B. 21
4. Find the sum of the measures of the interior angles of the figure.
Answer - B. 720
5. Find the value of x. The diagram is not drawn to scale.
Answer - C. x = 99
6. The sum of the measures of two exterior angles of a triangle is 264°. What is the measure of the third exterior angle?
Answer - B. 96
7. How many sides does a regular polygon have if each exterior angle measures 72°?
Answer - C. 5
8. What are the missing reasons in the proof?
Answers - 3. definition of parallelogram
4. alternate interior angles theroem
5. reflexive property of congruence
6. ASA
9.Based on the information in the diagram, determine whether the figure is a parallelogram. If so, select the answer with the correct justification.
Answer - D. Yes, both pairs of opposite sides are congruent.
10. Complete the two-column proof.
Answers - 2. SV // TU
3. SV // TU
7. Corresponding part of congruent triangle
8. Definition of parallelogram.
100% Hope this helps! Btw, this is only the first half of the quiz. The second half is different for everyone.
9 answers
1. Find the values of the variables x,y, and z in the parallelogram. The diagram is not drawn to scale. (diagram shows a tilted square with 33 in top left corner, 109 in top right, z in bottom left, and x and y in bottom right.)
Answer - x=33, y=38, z=109
Explanation -
x=33 (alternate angles)
33+109+y=180
142+y=180
y=180-142
y=38
z=109
2. In parallelogram DEFG, DH = x + 1, HF = 3y, GH = 3x − 4, and HE = 5y + 1. Find the values of x and y. The diagram is not drawn to scale.
Answer -
DH = HF
x + 1 = 3y
y = (x + 1) ⁄ 3
GH = HE
3x – 4 = 5y + 1
y = (3x – 5) ⁄ 5
y = y
(x + 1) ⁄ 3 = (3x – 5) ⁄ 5
5x + 5 = 9x – 15
4x = 20
x = 5
y = (x + 1) ⁄ 3
y = (5 + 1) ⁄ 3
y = 2
3. If ON = 8x − 8, LM = 7x + 4, NM = x − 5, and OL = 3y − 6, find the values of x and y for which LMNO must be a parallelogram. The diagram is not drawn to scale.
Answer -
ON = LM
8x-8 = 7x+4
x = 12
OL = NM
3y-6 = x-5
3y-6 = 12-5
3y = 13
y = 13/3