Special Parallelograms
Mar 18, 2016
Special Parallelograms
Warm Up
1.What is the sum of the interior angles of 11-gon.
2.Given Parallelogram ABCD, what is the value of y?
3.Explain why the quadrilateral, JKLM, is a Parallelogram
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 ApplicationA 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
When you are given a parallelogram with certainproperties, you can use the theorems below to determine whether the parallelogram is a rectangle.
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.
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 rhombusSubstitute given values.Subtract 3b from both sides and add 9 to both sides.
Divide both sides by 10.
WV = XT13b – 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 + 4TV = 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
SubstituteSimplifySubstitute
Def. of rhombusSubstitute
CG = GF
5a = 3a + 17a = 8.5
GF = 3a + 17 = 42.5CD = GFCD = 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.
Below are some conditions you can use to determine whether a parallelogram is a rhombus.
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.
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.
One special characteristic of a square is that the diagonals are•Congruent•Perpendicular•Bisect one another
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.
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.
111slope 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.
Check It Out! Example 3 Continued
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.
Check It Out! Example 3 Continued
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.
Example 2B: Applying Conditions for Special Parallelograms
Determine 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)
Example 3A Continued
Step 1 Graph PQRS.
Step 2 Find PR and QS to determine is 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.
Check It Out! Example 4
Given: PQTS is a rhombus with diagonalProve:
Check It Out! Example 4 Continued
Statements Reasons1. 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. SAS7. 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: