18/02/2014 CH.6.5 Conditions for Special Parallelogram

18/02/2014 CH.6.5Conditions for Special Parallelogram

Classwork

(Pages 434 to 436 ) Exercises

1, 4, 5, 6, 9 to 16, 17, 18, 20, 22, 24 to 26, 27, 28, 33(a, b, c), 34, 39, 40, 41, 43.

HomeworkHomework Booklet Chapter: 6.5

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°

Warm Up

1. Find AB for A (–3, 5) and B (1, 2).

2. Find the slope of JK for J(–4, 4) and K(3, –3).

ABCD is a parallelogram. Justify each statement.

3. ABC CDA

4. AEB CED

5

–1

Vert. s Thm.

opp. s

Prove that a given quadrilateral is a rectangle, rhombus, or square.

Objective

When you are given a parallelogram with certainproperties, you can use the theorems below to determine whether the parallelogram is a rectangle.

Example 1: Carpentry Application

A manufacture builds a mold for a desktop so that , , and mABC = 90°. Why must ABCD be a rectangle?

Both pairs of opposites sides of ABCD are congruent, so ABCD is a . Since mABC = 90°, one angle ABCD is a right angle. ABCD is a rectangle by Theorem 6-5-1.

Check It Out! Example 1

A carpenter’s square can be used to test that an angle is a right angle. How could the contractor use a carpenter’s square to check that the frame is a rectangle?

Both pairs of opp. sides of WXYZ are , so WXYZ is a parallelogram. The contractor can use the carpenter’s square to see if one of WXYZ is a right . If one angle is a right , then by Theorem 6-5-1 the frame is a rectangle.

Below are some conditions you can use to determine whether a parallelogram is a rhombus.

In order to apply Theorems 6-5-1 through 6-5-5, the quadrilateral must be a parallelogram.

Caution

To prove that a given quadrilateral is a square, it is sufficient to show that the figure is both a rectangle and a rhombus. You will explain why this is true in Exercise 43.

You can also prove that a given quadrilateral is arectangle, rhombus, or square by using the definitions of the special quadrilaterals.

Remember!

Example 2A: Applying Conditions for Special ParallelogramsDetermine if the conclusion is valid. If not, tell what additional information is needed to make it valid.

Given:Conclusion: EFGH is a rhombus.

The conclusion is not valid. By Theorem 6-5-3, if one pair of consecutive sides of a parallelogram are congruent, then the parallelogram is a rhombus. By Theorem 6-5-4, if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. To apply either theorem, you must first know that ABCD is a parallelogram.

Example 2B: Applying Conditions for Special ParallelogramsDetermine if the conclusion is valid. If not, tell what additional information is needed to make it valid.

Given:

Conclusion: EFGH is a square.

Step 1 Determine if EFGH is a parallelogram.

Given

EFGH is a parallelogram. Quad. with diags. bisecting each other

Example 2B Continued

Step 2 Determine if EFGH is a rectangle.

Given.

EFGH is a rectangle.

Step 3 Determine if EFGH is a rhombus.

EFGH is a rhombus.

with diags. rect.

with one pair of cons. sides rhombus


Step 4 Determine is EFGH is a square.

Since EFGH is a rectangle and a rhombus, it has four right angles and four congruent sides. So EFGH is a square by definition.

The conclusion is valid.


Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid.

Given: ABC is a right angle.

Conclusion: ABCD is a rectangle.

The conclusion is not valid. By Theorem 6-5-1, if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle. To apply this theorem, you need to know that ABCD is a parallelogram .

Example 3A: Identifying Special Parallelograms in the Coordinate Plane

Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply.

P(–1, 4), Q(2, 6), R(4, 3), S(1, 1)

Example 3A Continued

Step 1 Graph PQRS.

Step 2 Find PR and QS to determine if PQRS is a rectangle.


Since , the diagonals are congruent. PQRS is a rectangle.

Step 3 Determine if PQRS is a rhombus.

Step 4 Determine if PQRS is a square.

Since PQRS is a rectangle and a rhombus, it has four right angles and four congruent sides. So PQRS is a square by definition.


Since , PQRS is a rhombus.

Example 3B: Identifying Special Parallelograms in the Coordinate Plane

W(0, 1), X(4, 2), Y(3, –2), Z(–1, –3)

Step 1 Graph WXYZ.


Step 2 Find WY and XZ to determine if WXYZ is a rectangle.

Thus WXYZ is not a square.


Since , WXYZ is not a rectangle.

Step 3 Determine if WXYZ is a rhombus.


Since (–1)(1) = –1, , WXYZ is a rhombus.

Check It Out! Example 3A


K(–5, –1), L(–2, 4), M(3, 1), N(0, –4)

Step 1 Graph KLMN.

Check It Out! Example 3A Continued


Step 2 Find KM and LN to determine if KLMN is a rectangle.

Since , KMLN is a rectangle.

Step 3 Determine if KLMN is a rhombus.

Since the product of the slopes is –1, the two lines are perpendicular. KLMN is a rhombus.


Step 4 Determine if KLMN is a square.

Since KLMN is a rectangle and a rhombus, it has four right angles and four congruent sides. So KLMN is a square by definition.


Check It Out! Example 3B


P(–4, 6) , Q(2, 5) , R(3, –1) , S(–3, 0)

Check It Out! Example 3B Continued

Step 1 Graph PQRS.

Step 2 Find PR and QS to determine if PQRS is a rectangle.


Since , PQRS is not a rectangle. Thus PQRS is not a square.

Step 3 Determine if PQRS is a rhombus.


Since (–1)(1) = –1, are perpendicular and congruent. PQRS is a rhombus.

Lesson Quiz: Part I

1. Given that AB = BC = CD = DA, what additional

information is needed to conclude that ABCD is a

square?

Lesson Quiz: Part II

2. Determine if the conclusion is valid. If not, tell

what additional information is needed to make it

valid.

Given: PQRS and PQNM are parallelograms.

Conclusion: MNRS is a rhombus.

valid

Lesson Quiz: Part III

3. Use the diagonals to determine whether a parallelogram with vertices A(2, 7), B(7, 9), C(5, 4), and D(0, 2) is a rectangle, rhombus, or square. Give all the names that apply.

AC ≠ BD, so ABCD is not a rect. or a square. The slope of AC = –1, and the slope of BD= 1, so AC BD. ABCD is a rhombus.

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

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

Objectives

rectanglerhombussquare

Vocabulary

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

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.

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

Check It Out! Example 1a

Carpentry The rectangular gate has diagonal braces. Find HJ.

Def. of segs.

Rect. diags.

HJ = GK = 48

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

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

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

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 + 4

10b = 13

b = 1.3


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

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


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°

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

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.

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.

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.

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

Helpful Hint

Example 3: Verifying Properties of Squares

Show that the diagonals of square EFGH are congruent perpendicular bisectors of each other.

Example 3 Continued

Step 1 Show that EG and FH are congruent.

Since EG = FH,

Example 3 Continued

Step 2 Show that EG and FH are perpendicular.

Since ,

The diagonals are congruent perpendicular bisectors of each other.

Example 3 Continued

Step 3 Show that EG and FH are bisect each other.

Since EG and FH have the same midpoint, they bisect each other.


The vertices of square STVW are S(–5, –4), T(0, 2), V(6, –3) , and W(1, –9) . Show that the diagonals of square STVW are congruent perpendicular bisectors of each other.

111

slope of SV =

slope of TW = –11

SV ^ TW

SV = TW = 122 so, SV @ TW .

Step 1 Show that SV and TW are congruent.

Check It Out! Example 3 Continued

Since SV = TW,

Step 2 Show that SV and TW are perpendicular.


Since

The diagonals are congruent perpendicular bisectors of each other.

Step 3 Show that SV and TW bisect each other.

Since SV and TW have the same midpoint, they bisect each other.


Example 4: Using Properties of Special Parallelograms in Proofs

Prove: AEFD is a parallelogram.

Given: ABCD is a rhombus. E is the midpoint of , and F is the midpoint of .

Example 4 Continued

||


Given: PQTS is a rhombus with diagonal

Prove:


Statements Reasons

1. PQTS is a rhombus. 1. Given.

2. Rhombus → eachdiag. bisects opp. s

3. QPR SPR 3. Def. of bisector.

4. Def. of rhombus.

5. Reflex. Prop. of

6. SAS

7. CPCTC

2.

4.

5.

7.

6.

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

Lesson Quiz: Part II

PQRS is a rhombus. Find each measure.

3. QP 4. mQRP

42 51°

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.

Lesson Quiz: Part IV

ABE CDF

6. Given: ABCD is a rhombus. Prove:

18/02/2014 CH.6.5 Conditions for Special Parallelogram

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