Preparing for the Regents Examination Geometry, AK

6. AP ES (Distances between
parallel lines are
7.
equal.)
−−
AP −−
ES
8. PAQ SER
9.
(Leg–Acute Angle)
−−
QP −−
RS (CPCTC)
10-8 Kites
(page 239)
1 A kite is a quadrilateral with only two pairs
of adjacent congruent sides. A rhombus has
four congruent sides.
2 True
3 True
4 False
5 True
6 True
7 False
8 False
9 True
10 a −−
AC
b 1 8; 2 7; 3 6; 4 5
11 a 90 b 45 c 17
d m4 25 and mKLM 50
Note: Since there are many variations of proofs,
the following is simply one set of acceptable
statements to complete each proof. Depending
on the textbook used, the wording and format
of reasons may differ, so they have not been
supplied for the method of congruence applied
in each problem. (These solutions are intended
to be used as a guide—other possible solutions
may vary.)
12 1. −−
QS is the perpendicular bisector of −−
PR .
2. −−
PT −−
TR
3. QTP QTR
4. −−
QT −−
QT
5. QPT QRT (SAS SAS)
6. STP STR
7. −−
TS −−
TS
8. STP STR (SAS SAS)
9. −−
QP −−−
QR (CPCTC)
10. −−
SP −−
SR (CPCTC)
11. −−
QT −−
ST
12. PQRS is not ( Definition of
a rhombus. a rhombus)
13. PQRS is a kite. (Definition of a
kite)
10-9 Areas of Polygons
(pages 243–245)
1 4
2 16
3 9, 12, 15
4 Length is 4 ft and width is 5 ft.
5 a h 10 x
b A x(10 x)
6 a x 2 b 9 x 2 c x 2 4x 4
d x 2 4x 4 e 4 x 2 4x 1
7 a 25 _
2
b 18
49
c _
2
d 1
8 11
e 9
9 a 12 b 45 c 20
d 18 √ 3 e 45 _
2
10 √ 36 144 6 √ 5
11 a BD 10
12 72 √ 3
b 120
13 s 6
14 a 13 b 120 c 120 _
13
15 a 21x b 10 x 2 c 4x 12
d 18x 12 e 2 x 2 16 a x 4
b 4(4) 4 12
17 8
18 27 √ 3
19 30 √ 2
20 10 and 20
6x
21 a 96
22 72
23 100
24 a 6
b 260 c 60
b SMP and AML are similar. Let x be the
perpendicular drawn from M to −−
AP .
Then x _ 20
6 x _ , and x 5.
4
25 a _ x 10
12 x _
20
b x 4
c Area of RCS 20. Area of QCT 80.
4(8 11)
26 a A _
2 38 b 9 ( 5, _
2 )
27 a Slope of −−
JE 3 _ , slope of
4 −−
EN 4 _ . The
3
slopes of the two legs are negative reciprocals,
therefore perpendicular, forming a
right angle.
b A 1 _ (JE)(EN) 25
2
10-9 Areas of Polygons 65