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### 12/02/2014 Conditions for Parallelograms

Grade 9 ASP

Justify each statement.

1.

2.

Evaluate each expression for x = 12 and

y = 8.5.

3. 2x + 7

4. 16x –9

5.(8y + 5)°

Reflex Prop. of

Conv. of Alt. Int. s Thm.

31

183

73°

Prove that a given quadrilateral is a parallelogram.

You have learned to identify the properties of a parallelogram. Now you will be given the properties of a quadrilateral and will have to tell if the quadrilateral is a parallelogram. To do this, you can

use the definition of a parallelogram or the conditions below.

The two theorems below can also be used to show that a given quadrilateral is a parallelogram.

Example 1A: Verifying Figures are Parallelograms

Show that JKLM is a parallelogram for a = 3 and b = 9.

Step 1 Find JK and LM.

Given

LM = 10a + 4

JK = 15a – 11

Substitute and simplify.

LM = 10(3)+ 4 = 34

JK = 15(3) – 11 = 34

Step 2 Find KL and JM.

Given

KL = 5b + 6

JM = 8b – 21

Substitute and simplify.

KL = 5(9) + 6 = 51

JM = 8(9) – 21 = 51

Since JK = LM and KL = JM, JKLM is a parallelogram by Theorem 6-3-2.

Example 1B: Verifying Figures are Parallelograms

Show that PQRS is a parallelogram for x = 10 and y = 6.5.

mQ = (6y + 7)°

Given

Substitute 6.5 for y and simplify.

mQ = [(6(6.5) + 7)]° = 46°

mS = (8y – 6)°

Given

Substitute 6.5 for y and simplify.

mS = [(8(6.5) – 6)]° = 46°

mR = (15x – 16)°

Given

Substitute 10 for x and simplify.

mR = [(15(10) – 16)]° = 134°

Since 46° + 134° = 180°, R is supplementary to both Q and S. PQRS is a parallelogram by Theorem 6-3-4.

Therefore,

Check It Out! Example 1

Show that PQRS is a parallelogram for a = 2.4 and b = 9.

mQ = 74°, and mR = 106°, so Q and R are supplementary.

So one pair of opposite sides of PQRS are || and .

By Theorem 6-3-1, PQRS is a parallelogram.

Example 2A: Applying Conditions for Parallelograms

Determine if the quadrilateral must be a parallelogram. Justify your answer.

Yes. The 73° angle is supplementary to both its corresponding angles. By Theorem 6-3-4, the quadrilateral is a parallelogram.

Example 2B: Applying Conditions for Parallelograms

Determine if the quadrilateral must be a parallelogram. Justify your answer.

No. One pair of opposite angles are congruent. The other pair is not. The conditions for a parallelogram are not met.

Determine if the quadrilateral must be a parallelogram. Justify your answer.

Yes

The diagonal of the quadrilateral forms 2 triangles.

Two angles of one triangle are congruent to two angles of the other triangle, so the third pair of angles are congruent by the Third Angles Theorem.

So both pairs of opposite angles of the quadrilateral are congruent .

By Theorem 6-3-3, the quadrilateral is a parallelogram.

Determine if each quadrilateral must be a parallelogram. Justify your answer.

No. Two pairs of consective sides are congruent.

None of the sets of conditions for a parallelogram are met.

To say that a quadrilateral is a parallelogram by

definition, you must show that both pairs of opposite sides are parallel.

Example 3A: Proving Parallelograms in the Coordinate Plane

Show that quadrilateral JKLM is a parallelogram by using the definition of parallelogram. J(–1, –6), K(–4, –1), L(4, 5), M(7, 0).

Find the slopes of both pairs of opposite sides.

Since both pairs of opposite sides are parallel, JKLM is a parallelogram by definition.

AB and CD have the same slope, so . Since AB = CD, . So by Theorem 6-3-1, ABCD is a parallelogram.

Example 3B: Proving Parallelograms in the Coordinate Plane

Show that quadrilateral ABCD is a parallelogram by using Theorem 6-3-1.A(2, 3), B(6, 2), C(5, 0), D(1, 1).

Find the slopes and lengths of one pair of opposite sides.

Both pairs of opposite sides have the same slope so and by definition, KLMN is a parallelogram.

Check It Out! Example 3

Use the definition of a parallelogram to show that the quadrilateral with vertices K(–3, 0), L(–5, 7), M(3, 5), and N(5, –2) is a parallelogram.

You have learned several ways to determine whether a quadrilateral is a parallelogram. You can use the given information about a figure to decide which condition is best to apply.

To show that a quadrilateral is a parallelogram, you only have to show that it satisfies one of these sets of conditions.

The legs of a keyboard tray are connected by a bolt at their midpoints, which allows the tray to be raised or lowered. Why is PQRS always a parallelogram?

Since the bolt is at the midpoint of both legs, PE = ER and SE = EQ. So the diagonals of PQRS bisect each other, and by Theorem 6-3-5, PQRS is always a parallelogram.

Since ABRS is a parallelogram, it is always true that .

Since AB stays vertical, RS also remains vertical no matter how the frame is adjusted.

Check It Out! Example 4

The frame is attached to the tripod at points A and B such that AB = RS and BR = SA. So ABRS is also a parallelogram. How does this ensure that the angle of the binoculars stays the same?

Therefore the viewing never changes.

1. Show that JKLM is a parallelogram for a = 4 and b = 5.

2. Determine if QWRT must be a parallelogram. Justify your answer.

JN = LN = 22; KN = MN = 10; so JKLM is a parallelogram by Theorem 6-3-5.

No; One pair of consecutive s are , and one pair of opposite sides are ||. The conditions for a parallelogram are not met.

3. Show that the quadrilateral with vertices E(–1, 5), F(2, 4), G(0, –3), and H(–3, –2) is a parallelogram.

Since one pair of opposite sides are || and , EFGH is a parallelogram by Theorem 6-3-1.

L.O.

- Application in Racing
- Using Properties in Parallelograms to find Measures.
- Parallelograms in Coordinate Planes.
- Using Properties of Parallelograms in Proof.

1. Name the polygon by the number of its sides. Then tell whether the polygon is regular or irregular, concave or convex.

nonagon; irregular; concave

2. Find the sum of the interior angle measures of a convex 11-gon.

1620°

3. Find the measure of each interior angle of a regular 18-gon.

4. Find the measure of each exterior angle of a regular 15-gon.

160°

24°

CLASSWORK AND HOMEWORK

CLASSWORK

HOMEWORK

See Homework booklet. Week 3

- (Pages 407 to 409 )

1, 2, 3 to 13, 14, 21 to 24, 25, 26, 27 to 30, 32 to 43, 46, 47, 50, 51, 52, 53.

Prove and apply properties of parallelograms.

Use properties of parallelograms to solve problems.

Any polygon with four sides is a quadrilateral. However, some quadrilaterals have special properties. These special quadrilaterals are given their own names.

Opposite sides of a quadrilateral do not share a vertex. Opposite angles do not share a side.

A quadrilateral with two pairs of parallel sides is a parallelogram. To write the name of a parallelogram, you use the symbol .

DG = 31 mm, and mFCD = 42°. Find CF.

opp. sides

Example 1A: Properties of Parallelograms

CF = DE

Def. of segs.

CF = 74 mm

Substitute 74 for DE.

DG = 31 mm, and mFCD = 42°. Find mEFC.

cons. s supp.

Example 1B: Properties of Parallelograms

mEFC + mFCD = 180°

mEFC + 42= 180

Substitute 42 for mFCD.

mEFC = 138°

Subtract 42 from both sides.

DG = 31 mm, and mFCD = 42°. Find DF.

diags. bisect each other.

Example 1C: Properties of Parallelograms

DF = 2DG

DF = 2(31)

Substitute 31 for DG.

DF = 62

Simplify.

Check It Out! Example 1a

In KLMN, LM = 28 in.,

LN = 26 in., and mLKN = 74°. Find KN.

LM = KN

Def. of segs.

LM = 28 in.

Substitute 28 for DE.

Check It Out! Example 1b

In KLMN, LM = 28 in.,

LN = 26 in., and mLKN = 74°. Find mNML.

NML LKN

mNML = mLKN

Def. of s.

mNML = 74°

Substitute 74° for mLKN.

Def. of angles.

Check It Out! Example 1c

In KLMN, LM = 28 in.,

LN = 26 in., and mLKN = 74°. Find LO.

LN = 2LO

26 = 2LO

Substitute 26 for LN.

LO = 13 in.

Simplify.

Example 2A: Using Properties of Parallelograms to Find Measures

WXYZ is a parallelogram. Find YZ.

YZ = XW

Def. of segs.

8a – 4 = 6a + 10

Substitute the given values.

Subtract 6a from both sides and add 4 to both sides.

2a = 14

a = 7

Divide both sides by 2.

YZ = 8a – 4 = 8(7) – 4 = 52

Example 2B: Using Properties of Parallelograms to Find Measures

WXYZ is a parallelogram. Find mZ.

mZ + mW = 180°

(9b + 2)+ (18b –11) = 180

Substitute the given values.

Combine like terms.

27b – 9 = 180

27b = 189

b = 7

Divide by 27.

mZ = (9b + 2)° = [9(7) + 2]° = 65°

Check It Out! Example 2a

EFGH is a parallelogram.

Find JG.

EJ = JG

Def. of segs.

3w = w + 8

Substitute.

2w = 8

Simplify.

w = 4

Divide both sides by 2.

JG = w + 8 = 4 + 8 = 12

Check It Out! Example 2b

EFGH is a parallelogram.

Find FH.

FJ = JH

Def. of segs.

4z – 9 = 2z

Substitute.

2z = 9

Simplify.

z = 4.5

Divide both sides by 2.

FH = (4z – 9) + (2z) = 4(4.5) – 9 + 2(4.5) = 18

When you are drawing a figure in the coordinate plane, the name ABCD gives the order of the vertices.

K

J

Example 3: Parallelograms in the Coordinate Plane

Three vertices of JKLM are J(3, –8), K(–2, 2), and L(2, 6). Find the coordinates of vertex M.

Since JKLM is a parallelogram, both pairs of opposite sides must be parallel.

Step 1 Graph the given points.

Step 2 Find the slope of by counting the units from K to L.

- The rise from 2 to 6 is 4.
- The run of –2 to 2 is 4.

L

K

J

Example 3 Continued

- Step 3 Start at J and count the
- same number of units.
- A rise of 4 from –8 is –4.
- A run of 4 from 3 is 7. Label (7, –4) as vertex M.

M

Step 4 Use the slope formula to verify that

L

K

M

J

Example 3 Continued

The coordinates of vertex M are (7, –4).

S

P

Check It Out! Example 3

Three vertices of PQRS are P(–3, –2), Q(–1, 4), and S(5, 0). Find the coordinates of vertex R.

Since PQRS is a parallelogram, both pairs of opposite sides must be parallel.

Step 1 Graph the given points.

Step 2 Find the slope of by counting the units from P to Q.

- The rise from –2 to 4 is 6.
- The run of –3 to –1 is 2.

Q

S

P

Check It Out! Example 3 Continued

R

- Step 3 Start at S and count the
- same number of units.
- A rise of 6 from 0 is 6.
- A run of 2 from 5 is 7. Label (7, 6) as vertex R.

Step 4 Use the slope formula to verify that

R

Q

S

P

Check It Out! Example 3 Continued

The coordinates of vertex R are (7, 6).

Example 4A: Using Properties of Parallelograms in a Proof

Write a two-column proof.

Given: ABCD is a parallelogram.

Prove:∆AEB∆CED

each other

2. opp. sides

Example 4A Continued

Proof:

1. ABCD is a parallelogram

1. Given

4. SSS Steps 2, 3

Example 4B: Using Properties of Parallelograms in a Proof

Write a two-column proof.

Given: GHJN and JKLM are parallelograms. H and M are collinear. N and K are collinear.

Prove:H M

M and MJK are supp.

2. cons. s supp.

Example 4B Continued

Proof:

1.GHJN and JKLM are

parallelograms.

1. Given

3.HJN MJK

3. Vert. s Thm.

4.H M

4. Supps. Thm.

Write a two-column proof.

Given: GHJN and JKLM are parallelograms.

H and M are collinear. N and K are collinear.

Prove: N K

K and MJK are supp.

2. cons. s supp.

Check It Out! Example 4 Continued

Proof:

1.GHJN and JKLM

are parallelograms.

1. Given

3.HJN MJK

3. Vert. s Thm.

4.N K

4. Supps. Thm.

5. Three vertices of ABCD are A (2, –6), B (–1, 2), and C(5, 3). Find the coordinates of vertex D.

(8, –5)

1. Given

3.R T

4. ∆RSU∆TUS

4. SAS

2. cons. s

3. opp. s

Lesson Quiz: Part IV

6. Write a two-column proof.

Given:RSTU is a parallelogram.

Prove: ∆RSU∆TUS

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