18/02/2014 CH.6.5 Conditions for Special Parallelogram
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Transcript of 18/02/2014 CH.6.5 Conditions 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
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.
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
Example 2B Continued
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.
Check It Out! Example 2
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)
Step 2 Find PR and QS to determine if PQRS is a rectangle.
Example 3A Continued
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.
Example 3A Continued
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.
Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply.
Step 2 Find WY and XZ to determine if WXYZ is a rectangle.
Thus WXYZ is not a square.
Example 3B Continued
Since , WXYZ is not a rectangle.
Step 3 Determine if WXYZ is a rhombus.
Example 3B Continued
Since (–1)(1) = –1, , WXYZ is a rhombus.
Check It Out! Example 3A
Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply.
K(–5, –1), L(–2, 4), M(3, 1), N(0, –4)
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.
Check It Out! Example 3A Continued
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 3A Continued
Check It Out! Example 3B
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(–4, 6) , Q(2, 5) , R(3, –1) , S(–3, 0)
Step 2 Find PR and QS to determine if PQRS is a rectangle.
Check It Out! Example 3B Continued
Since , PQRS is not a rectangle. Thus PQRS is not a square.
Step 3 Determine if PQRS is a rhombus.
Check It Out! Example 3B Continued
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
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
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
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
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
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°
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.
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.
Check It Out! Example 3
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 .
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.
Check It Out! Example 3 Continued
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 .
Check It Out! Example 4 Continued
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 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.