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Preparing for the Regents Examination Geometry, AK

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3 4x 2 3x 3<br />

x 5<br />

RS 18<br />

4 Perimeter of MABC 8<br />

5 mADC 110<br />

6 2x 2 x 8<br />

x 10<br />

CD 18<br />

7 mADC 94<br />

8 a Midpoint of −−<br />

AC midpoint of −−<br />

BD (6, 5)<br />

b Slope of −−<br />

AC 1, slope of −−<br />

BD 1.<br />

Slopes are negative reciprocals, diagonals<br />

are perpendicular.<br />

9 a (9, 5)<br />

b PQ PS 5 √ 2<br />

c Slope of −−<br />

PR 1 _ , slope of<br />

3 −−<br />

QS 3.<br />

Slopes are negative reciprocals, diagonals<br />

are perpendicular.<br />

10 a Slope of −−−<br />

AD slope of −−<br />

BC k _<br />

x<br />

.<br />

AD BC √ x 2 k 2 . ABCD is a parallelogram.<br />

(One pair of opposite sides have<br />

<strong>the</strong> same length and are parallel.)<br />

b Slope of −−<br />

CA k _<br />

−−<br />

, slope of BD <br />

x k k _<br />

x k ,<br />

not negative reciprocals. The diagonals are<br />

not perpendicular.<br />

Note: Since <strong>the</strong>re are many variations of proofs,<br />

<strong>the</strong> following is simply one set of acceptable<br />

statements to complete each proof. Depending<br />

on <strong>the</strong> textbook used, <strong>the</strong> wording and <strong>for</strong>mat<br />

of reasons may differ, so <strong>the</strong>y have not been<br />

supplied <strong>for</strong> <strong>the</strong> method of congruence applied<br />

in each problem. (These solutions are intended<br />

to be used as a guide—o<strong>the</strong>r possible solutions<br />

may vary.)<br />

11 −−−<br />

QR −−<br />

AC , −−<br />

PS −−<br />

AC , −−−<br />

QR −−<br />

PS . Likewise,<br />

−−<br />

QP −−<br />

BD , −−<br />

RS −−<br />

BD , and −−<br />

QP −−<br />

RS . Thus, PQRS<br />

is a parallelogram. Since −−<br />

AC −−<br />

BD , −−−<br />

QR<br />

and −−<br />

PS are perpendicular to −−<br />

QP and −−<br />

RS .<br />

Thus, PQRS has at least one right angle, and<br />

PQRS is a rectangle.<br />

12 1. Rhombus ABCD<br />

2. −−<br />

ED<br />

3. −−<br />

BF −−<br />

AC<br />

4. −−<br />

EF −−<br />

AC<br />

5. F bisects −−<br />

AC .<br />

6. EA EC<br />

62 Chapter 10: Quadrilaterals<br />

7. −−<br />

EA −−<br />

EC<br />

8. ACE is isosceles. (Definition of an<br />

isosceles triangle)<br />

13 Prove that −−<br />

DF −−<br />

AB and −−<br />

DE −−<br />

BF by using<br />

<strong>the</strong> midpoints and alternate interior angles.<br />

Thus, EBFD is a parallelogram. By <strong>the</strong> division<br />

postulate, −−<br />

EB −−<br />

FB . EBFD is a rhombus<br />

because a rhombus is a parallelogram with<br />

two congruent consecutive sides.<br />

14 1. Parallelogram ABCD<br />

2. −−<br />

AB −−−<br />

CD<br />

3. BAD 2<br />

4. mBAD m2<br />

(Corresponding angles<br />

are congruent.)<br />

5. BAD 1 CAD<br />

6. mBAD m1 (Whole is greater<br />

than a part.)<br />

7. m2 m1 (Substitution<br />

postulate)<br />

15 1. Rhombus PQRS<br />

2. −−<br />

QS −−<br />

PR<br />

3. −−−<br />

QC −−<br />

CS<br />

4. B is <strong>the</strong> midpoint of −−−<br />

QC .<br />

5. D is <strong>the</strong> midpoint of −−<br />

CS .<br />

6. −−<br />

BC −−−<br />

CD<br />

7. ___<br />

AC ___<br />

AC<br />

8. ABC ADC<br />

9.<br />

(SAS SAS)<br />

−−<br />

AB −−−<br />

AD<br />

(CPCTC)<br />

10. BAD is isosceles. (Definition of an<br />

isosceles triangle)<br />

16 If a quadrilateral is equilateral, it is a parallelogram<br />

with a pair of consecutive congruent<br />

sides.<br />

17 The diagonal creates two congruent triangles<br />

(ASA ASA) and corresponding congruent<br />

sides are consecutive.<br />

18 1. −−−<br />

CD BE <br />

2. −−−<br />

CD −−<br />

BA<br />

3. −−−<br />

AD BF <br />

4. −−−<br />

AD −−<br />

BC<br />

5. ABCD is a ( Definition of a<br />

parallelogram. parallelogram)<br />

6. BAD BCD (Opposite angles<br />

7. BG bisects FBE.<br />

are congruent.)<br />

8. CBD ABD<br />

9. ABCD is a rhombus. (Diagonal −−<br />

BD<br />

bisects opposite<br />

angles.)

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