12/02/2014 Conditions for Parallelograms

Grade 9 ASP

Warm UpJustify 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.

Objective

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 canuse 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

Substitute and simplify.

JK = 15a – 11

JK = 15(3) – 11 = 34

LM = 10a + 4

LM = 10(3)+ 4 = 34

Example 1A Continued

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

Step 2 Find KL and JM.

Given

Substitute and simplify.

KL = 5b + 6

KL = 5(9) + 6 = 51

JM = 8b – 21

JM = 8(9) – 21 = 51

Example 1B: Verifying Figures are Parallelograms

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

Given

Substitute 6.5 for y and simplify.

Given

Substitute 6.5 for y and simplify.

mQ = (6y + 7)°

mQ = [(6(6.5) + 7)]° = 46°

mS = (8y – 6)°

mS = [(8(6.5) – 6)]° = 46°

mR = (15x – 16)°

mR = [(15(10) – 16)]° = 134°

Given Substitute 10 for x and

simplify.

Example 1B Continued

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

Check It Out! Example 1

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

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

PQ = RS = 16.8, so

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

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

Therefore,

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.

Check It Out! Example 2a

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

The diagonal of the quadrilateral forms 2 triangles.

Yes

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.

Check It Out! Example 2b

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

Helpful Hint

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.

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.

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

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

Helpful Hint

Example 4: Application

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.

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?

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

Therefore the viewing never changes.

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

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

Lesson Quiz: Part I

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.

Lesson Quiz: Part II

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.1. Application in Racing

2. Using Properties in Parallelograms to find Measures.

3. Parallelograms in Coordinate Planes.

4. Using Properties of Parallelograms in Proof.

Lesson Quiz

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

nonagon; irregular; concave2. 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

(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.

HOMEWORK

See Homework booklet. Week 3

Prove and apply properties of parallelograms.

Use properties of parallelograms to solve problems.

Objectives

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.

Helpful Hint

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

Example 1A: Properties of Parallelograms

Def. of segs.

Substitute 74 for DE.

In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find CF.

CF = DE

CF = 74 mm

opp. sides

Substitute 42 for mFCD.

Example 1B: Properties of Parallelograms

Subtract 42 from both sides.

mEFC + mFCD = 180°

mEFC + 42 = 180

mEFC = 138°

In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find mEFC.

cons. s supp.

Example 1C: Properties of Parallelograms

Substitute 31 for DG.

Simplify.

In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find DF.

DF = 2DG

DF = 2(31)

DF = 62

diags. bisect each other.

Check It Out! Example 1a

In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find KN.

Def. of segs.

Substitute 28 for DE.

LM = KN

LM = 28 in.

opp. sides

Check It Out! Example 1b

Def. of angles.

In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find mNML.

Def. of s.

Substitute 74° for mLKN.

NML LKN

mNML = mLKN

mNML = 74°

opp. s

Check It Out! Example 1c

In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find LO.

Substitute 26 for LN.

Simplify.

LN = 2LO

26 = 2LO

LO = 13 in.

diags. bisect each other.

Example 2A: Using Properties of Parallelograms to Find Measures

WXYZ is a parallelogram. Find YZ.

Def. of segs.

Substitute the given values.

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

Divide both sides by 2.

YZ = XW

8a – 4 = 6a + 10

2a = 14

a = 7

YZ = 8a – 4 = 8(7) – 4 = 52

opp. s

Example 2B: Using Properties of Parallelograms to Find Measures

WXYZ is a parallelogram. Find mZ .

Divide by 27.

Add 9 to both sides.

Combine like terms.

Substitute the given values.

mZ + mW = 180°

(9b + 2) + (18b – 11) = 180

27b – 9 = 180

27b = 189

b = 7

mZ = (9b + 2)° = [9(7) + 2]° = 65°

cons. s supp.

Check It Out! Example 2a

EFGH is a parallelogram.Find JG.

Substitute.

Simplify.

EJ = JG

3w = w + 8

2w = 8

w = 4 Divide both sides by 2.

JG = w + 8 = 4 + 8 = 12

Def. of segs.

diags. bisect each other.

Check It Out! Example 2b

EFGH is a parallelogram.Find FH.

Substitute.

Simplify.

FJ = JH

4z – 9 = 2z

2z = 9

z = 4.5 Divide both sides by 2.

Def. of segs.

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

diags. bisect each other.

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

Remember!

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.

Step 1 Graph the given points.

J

K

L

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

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.

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.

J

K

L

M

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

Example 3 Continued

Step 4 Use the slope formula to verify that

J

K

L

M

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.

Step 1 Graph the given points.

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

P

Q

S

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.

Check It Out! Example 3 Continued

P

Q

S

R

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.

Check It Out! Example 3 Continued

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

Step 4 Use the slope formula to verify that

P

Q

S

R

Example 4A: Using Properties of Parallelograms in a Proof

Write a two-column proof.

Given: ABCD is a parallelogram.

Prove: ∆AEB ∆CED

Example 4A Continued

Proof:

Statements Reasons

1. ABCD is a parallelogram 1. Given

4. SSS Steps 2, 3

2. opp. sides

3. diags. bisect each other

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

Example 4B Continued

Proof:

Statements Reasons

1. GHJN and JKLM are parallelograms. 1. Given

2. cons. s supp. 2. H and HJN are supp.

M and MJK are supp.

3. Vert. s Thm.3. HJN MJK

4. H M 4. Supps. Thm.

Check It Out! Example 4

Write a two-column proof.

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

Prove: N K

Proof:

Statements Reasons

1. GHJN and JKLMare parallelograms. 1. Given

2. N and HJN are supp.

K and MJK are supp.

Check It Out! Example 4 Continued

2. cons. s supp.

3. Vert. s Thm.

4. Supps. Thm.4. N K

3. HJN MJK

Lesson Quiz: Part I

In PNWL, NW = 12, PM = 9, and mWLP = 144°. Find each measure.

1. PW 2. mPNW

18 144°

Real World Problems

Lesson Quiz: Part II

QRST is a parallelogram. Find each measure.

2. TQ 3. mT

28 71°

Lesson Quiz: Part III

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

coordinates of vertex D.

(8, –5)

Lesson Quiz: Part IV

6. Write a two-column proof.Given: RSTU is a parallelogram.Prove: ∆RSU ∆TUS

Statements Reasons

1. RSTU is a parallelogram. 1. Given

4. SAS4. ∆RSU ∆TUS

3. R T

2. cons. s

3. opp. s