Preparing for the Regents Examination Geometry, AK

31 a 20 √ 3 35
b 20
c 200 √ 3 346.4
32 a AB BC 5 √ 2 . ABC is isosceles because
there is one pair of congruent sides.
b Slope of −−
AB 1. Slope of −−
BC 1. Slopes
are negative reciprocals, thus −−
AB −−
BC .
c Area of ABC 25
33 −−
DE −−
AV because the slope of −−
DE slope
of −−
AV 1 _ .
3 −−−
DA is not parallel to −−
VE because
the slope of −−−
DA is not equal to the slope
of −−
VE . DA VE 5; nonparallel sides are
congruent. Therefore, DAVE is an isosceles
trapezoid.
34 Slope of −−
PQ slope of −−
RS 3 _ . Slope of
4 −−
PS slope of −−−
RQ 4 _ . Slope of
3 −−
PR 1 _ .
7
Slope of −−
QS 7. PQRS is a parallelogram
because opposite sides have the same slope,
and a rhombus because the diagonals are
perpendicular—slopes are negative reciprocals,
and a square because consecutive sides
are perpendicular—slopes are negative reciprocals.
35 Slope of −−
AB slope of −−−
CD 0. Slope of
−−
BC slope of −−−
AD 4 _ . Slope of
3 −−
AC 2.
Slope of −−
BD 1 _ . ABCD is a parallelogram
2
because opposite sides have the same slope,
and a rhombus because the diagonals are
perpendicular—slopes are negative
reciprocals.
36 Slope of −−
LE slope of −−
AF 4 _ . Slope
5
of −−
EA slope of −−
FL 4 _ . Slope of
5 −−
LA is
udefined. Slope of −−
EF 0. LEAF is a parallelogram
because opposite sides have the same
slope, and a rhombus because the diagonals
are perpendicular.
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.)
37 1. Quadrilateral TRIP
2. −−
TR −−
RI
3. m1 m2
4. 1 2
5. −−
RA −−
RA
6. a RAT RAI (SAS SAS)
7. −−
TA −−
AI (CPCTC)
8. TAP IAP (Supplements of
congruent angles
are congruent.)
9. −−
AP −−
AP
10. TAP IAP (SAS SAS)
11. b 3 4 (CPCTC)
38 1. Parallelogram PQRS
2. Diagonal −−
QS bisects PQR.
3. PQS RQS
4. RQS QSP (Alternate interior
angles are congruent.)
5. PQS QSR (Alternate interior
angles are congruent.)
6. RQS QSR (Transitive postulate)
7. QSP QSR
8. −−
PS −−
PQ
9. PS PQ
10. PQRS is a rhombus. (A rhombus is a
parallelogram with
two congruent
consecutive sides.)
39 1. ABED is a rhombus.
2. −−
BD intersects −−−−
AOEC at O.
3. BOE and DOE are right angles.
4. BOE and DOE are right triangles.
5. −−
OE −−
OE
6. −−
BE −−
DE
7. BOE DOE (HL HL)
8. −−
BO −−−
OD (CPCTC)
9. BOC and DOC are right triangles.
10. −−−
OC −−−
OC
11. BOC DOC (Leg-leg)
12. a −−
BC −−−
DC (CPCTC)
13. −−
AB −−−
AD
14. −−
AC −−
AC
15. ABC ADC (SSS SSS)
16. b ABC ADC (CPCTC)
40 Mark the point of intersection of the diagonals
E. The diagonals bisect each other,
thus AE DE. By the triangle inequality
theorem, mADE mDCE and
mADC mDCB by the multiplication
postulate of inequality.
Chapter Review 67