Holt Geometry 8-3 Conditions for Parallelograms 8-3 Conditions for Parallelograms Holt Geometry Warm...

Holt Geometry

8-3 Conditions for Parallelograms8-3 Conditions for Parallelograms

Holt Geometry

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Geometry

8-3 Conditions for Parallelograms

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°

Holt Geometry

8-3 Conditions for Parallelograms

Prove that a given quadrilateral is a parallelogram.

Objective

Holt Geometry

8-3 Conditions for Parallelograms

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.

Holt Geometry

8-3 Conditions for Parallelograms

Holt Geometry

8-3 Conditions for Parallelograms

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

Holt Geometry

8-3 Conditions for Parallelograms

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

Holt Geometry

8-3 Conditions for Parallelograms

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

Holt Geometry

8-3 Conditions for Parallelograms

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.

Holt Geometry

8-3 Conditions for Parallelograms

Example 1B Continued

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

Holt Geometry

8-3 Conditions for Parallelograms

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,

Holt Geometry

8-3 Conditions for Parallelograms

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.

Holt Geometry

8-3 Conditions for Parallelograms

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.

Holt Geometry

8-3 Conditions for Parallelograms

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.

Holt Geometry

8-3 Conditions for Parallelograms

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.

Holt Geometry

8-3 Conditions for Parallelograms

To say that a quadrilateral is a parallelogram bydefinition, you must show that both pairs of opposite sides are parallel.

Helpful Hint

Holt Geometry

8-3 Conditions for Parallelograms

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.

Holt Geometry

8-3 Conditions for Parallelograms

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.

Holt Geometry

8-3 Conditions for Parallelograms

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.

Holt Geometry

8-4 Properties of Special Parallelograms8-4 Properties of Special Parallelograms

Holt Geometry

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Geometry

8-4 Properties of Special Parallelograms

Warm UpSolve for x.

1. 16x – 3 = 12x + 13

2. 2x – 4 = 90

ABCD is a parallelogram. Find each measure.

3. CD 4. mC

4

47

14 104°

Holt Geometry

8-4 Properties of Special Parallelograms

Prove and apply properties of rectangles, rhombuses, and squares.

Use properties of rectangles, rhombuses, and squares to solve problems.

Objectives

Holt Geometry

8-4 Properties of Special Parallelograms

rectanglerhombussquare

Vocabulary

Holt Geometry

8-4 Properties of Special Parallelograms

A second type of special quadrilateral is a rectangle. A rectangle is a quadrilateral with four right angles.

Holt Geometry

8-4 Properties of Special Parallelograms

Since a rectangle is a parallelogram by Theorem 6-4-1, a rectangle “inherits” all the properties of parallelograms that you learned in Lesson 6-2.

Holt Geometry

8-4 Properties of Special Parallelograms

Example 1: Craft Application

A woodworker constructs a rectangular picture frame so that JK = 50 cm and JL = 86 cm. Find HM.

Rect. diags.

Def. of segs.

Substitute and simplify.

KM = JL = 86

diags. bisect each other

Holt Geometry

8-4 Properties of Special Parallelograms

Check It Out! Example 1a

Carpentry The rectangular gate has diagonal braces. Find HJ.

Def. of segs.

Rect. diags.

HJ = GK = 48

Holt Geometry

8-4 Properties of Special Parallelograms

Check It Out! Example 1b

Carpentry The rectangular gate has diagonal braces. Find HK.

Def. of segs.

Rect. diags.

JL = LG

JG = 2JL = 2(30.8) = 61.6 Substitute and simplify.

Rect. diagonals bisect each other

Holt Geometry

8-4 Properties of Special Parallelograms

A rhombus is another special quadrilateral. A rhombus is a quadrilateral with four congruent sides.

Holt Geometry

8-4 Properties of Special Parallelograms

Holt Geometry

8-4 Properties of Special Parallelograms

Like a rectangle, a rhombus is a parallelogram. So you can apply the properties of parallelograms to rhombuses.

Holt Geometry

8-4 Properties of Special Parallelograms

Example 2A: Using Properties of Rhombuses to Find Measures

TVWX is a rhombus. Find TV.

Def. of rhombus

Substitute given values.Subtract 3b from both sides and add 9 to both sides.

Divide both sides by 10.

WV = XT

13b – 9 = 3b + 410b = 13

b = 1.3

Holt Geometry

8-4 Properties of Special Parallelograms

Example 2A Continued

Def. of rhombus

Substitute 3b + 4 for XT.

Substitute 1.3 for b and simplify.

TV = XT

TV = 3b + 4

TV = 3(1.3) + 4 = 7.9

Holt Geometry

8-4 Properties of Special Parallelograms

Rhombus diag.

Example 2B: Using Properties of Rhombuses to Find Measures

TVWX is a rhombus. Find mVTZ.

Substitute 14a + 20 for mVTZ.

Subtract 20 from both sides and divide both sides by 14.

mVZT = 90°

14a + 20 = 90°

a = 5

Holt Geometry

8-4 Properties of Special Parallelograms

Example 2B Continued

Rhombus each diag. bisects opp. s

Substitute 5a – 5 for mVTZ.

Substitute 5 for a and simplify.

mVTZ = mZTX

mVTZ = (5a – 5)°

mVTZ = [5(5) – 5)]° = 20°

Holt Geometry

8-4 Properties of Special Parallelograms

Check It Out! Example 2a

CDFG is a rhombus. Find CD.

Def. of rhombus

Substitute

Simplify

Substitute

Def. of rhombus

Substitute

CG = GF

5a = 3a + 17

a = 8.5

GF = 3a + 17 = 42.5

CD = GF

CD = 42.5

Holt Geometry

8-4 Properties of Special Parallelograms

Check It Out! Example 2b

CDFG is a rhombus. Find the measure.

mGCH if mGCD = (b + 3)°and mCDF = (6b – 40)°

mGCD + mCDF = 180°

b + 3 + 6b – 40 = 180°

7b = 217°

b = 31°

Def. of rhombus

Substitute.

Simplify.

Divide both sides by 7.

Holt Geometry

8-4 Properties of Special Parallelograms

Check It Out! Example 2b Continued

mGCH + mHCD = mGCD

2mGCH = mGCDRhombus each diag. bisects opp. s

2mGCH = (b + 3)

2mGCH = (31 + 3)

mGCH = 17°

Substitute.

Substitute.

Simplify and divide both sides by 2.

Holt Geometry

8-4 Properties of Special Parallelograms

A square is a quadrilateral with four right angles and four congruent sides. In the exercises, you will show that a square is a parallelogram, a rectangle, and a rhombus. So a square has the properties of all three.

Holt Geometry

8-4 Properties of Special Parallelograms

Rectangles, rhombuses, and squares are sometimes referred to as special parallelograms.

Helpful Hint

Holt Geometry

8-4 Properties of Special Parallelograms

Lesson Quiz: Part I

A slab of concrete is poured with diagonal spacers. In rectangle CNRT, CN = 35 ft, and NT = 58 ft. Find each length.

1. TR 2. CE

35 ft 29 ft

Holt Geometry

8-4 Properties of Special Parallelograms

Lesson Quiz: Part II

PQRS is a rhombus. Find each measure.

3. QP 4. mQRP

42 51°

Holt Geometry

8-4 Properties of Special Parallelograms

Lesson Quiz: Part III

5. The vertices of square ABCD are A(1, 3), B(3, 2), C(4, 4), and D(2, 5). Show that its diagonals are congruent perpendicular bisectors of each other.

Holt Geometry

8-4 Properties of Special Parallelograms

Lesson Quiz: Part IV

ABE CDF

6. Given: ABCD is a rhombus. Prove: