Download GH , KJ , AB b. m is the angle bisector of ∠BCD. By the Angle

Self-Test
(pp. 371–372)
___ ___ ___
1. a. GH, KJ, AB
b. m is the angle bisector of ∠BCD. By the Angle
Symmetry Theorem, the symmetry line of an
angle contains the bisector of the angle.
2.
square
6. a. Since ∠RFT is a straight angle, (4x + 50) +
(5x + 40) = 180, so then 9x = 90, and x = 10.
b. No. m∠RBF = 38 and m∠TBF = 43, so
m∠R ≠ m∠T and BRT is not isosceles.
Thus, BR ≠ BT.
3. Answers vary. Sample:
7. The largest side is the side opposite the largest
angle. m∠R = 180___
- 40 - 80 = 60. Because ∠J is
the largest angle, GR is the longest side.
8. m∠P + m∠X = 180 by the Trapezoid Angle
Theorem. Thus, m∠X = 70. Since FPXL is
isosceles, this means that m∠L = 70 as well.
9. 18
10. a. 8-fold rotational symmetry
4. In order for a kite to have rotation symmetry,
both its diagonals must be symmetry lines.
Any rhombus or square is a kite with rotation
symmetry.
b. 4-fold rotational symmetry
11. ∠W intercepts a semicircle, so by Thales’
Theorem, it is a right angle, that is, m∠W = 90.
12. Because RAN is equilateral, m∠R and
m∠RNA are both 60. Because the measure of
an intercepted arc is twice the measure of the
= 120° = mAN
. Because
inscribed angle, mRBA
= mRBA
+ m AN
, mRAN
= 240°.
mRAN
13. a. Sometimes but not always. All rectangles
are trapezoids, but not all trapezoids are
rectangles.
5. 1, 2, or 4
isosceles trapezoid
rectangle
A107
Geometry
b. Sometimes but not always. In kites, one
diagonal is a symmetry line and the
perpendicular bisector of the other diagonal.
In rhombuses, both diagonals are symmetry
lines and perpendicular bisectors of each
other.
14. a. The measure of one of the interior angles of a
(n - 2) · 180
, so with n = 8, the
regular n-gon is __________
n
measure of one of the interior angles is 135.
m∠MTN = 180 - 135 = 45.
b. In ∠BFS, m∠B = 135. The triangle is
isosceles, so the base angles are congruent.
180 - 135 =
22.5.
Thus, m∠BFS = ________
2
15. ∠R ∠V and ∠T ∠P because alternate
interior angles formed when parallel lines are cut
by a transversal are congruent (Parallel Lines
Theorem), and ∠R ∠P is given, so by the
Transitive
of Congruence, ∠V ∠T.
___ Property
___
Thus, SV ST by the Converse of the Isosceles
Triangle Base Angles Theorem.
16. Diagonals of a rhombus are angle bisectors, so
FOA is equilateral. Therefore, FO = OA =
FA = 18 cm, so each side of rhombus FOAM is
18 cm, and the perimeter is 72 cm.
17.
polygon
equiangular
polygon
equilateral
polygon
regular
polygon
rhombus
square
equilateral
triangle
18. Because MANDC is a regular pentagon,
(5 - 2) · 180
= 108. Because AN = ND,
m∠AND = __________
5
AND is isosceles with base angles measuring
(180 - 108) ÷ 2 = 36. So m∠ADN = 36.
Because m∠CDN = m∠CDA + m∠ADN,
108 = m∠CDA + 36, so m∠CDA = 72.
ACD is isosceles, so m∠DCA = m∠CDA = 72.
Thus m∠CAD = 180 - 72 - 72 = 36.
Because MC = CD, MCD is isosceles, and
by a calculation similar to a previous one,
m∠MDC = 36.
A108
Geometry
19. Quadrilateral II also has congruent diagonals
because it is lower in the hierarchy than
type G, but nothing can be concluded about
Quadrilateral I because it is higher in the
hierarchy than type G.
20. II and V both have type 5 symmetry. III and IV
both have type 2 symmetry.