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Lesson 6–6 Objectives. Be able to use properties of kites to solve problems Be able to use properties of trapezoids to solve problems. State HSCE:. A kite is a quadrilateral with exactly two pairs of congruent consecutive sides. Problem–Solving Application.
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Lesson 6–6 Objectives • Be able to use properties of kites to solve problems • Be able to use properties of trapezoids to solve problems
A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.
Problem–Solving Application Lucy is framing a kite with wooden dowels. She uses two dowels that measure 18 cm, one dowel that measures 30 cm, and two dowels that measure 27 cm. To complete the kite, she needs a dowel to place along KL. She has a dowel that is 36 cm long. About how much wood will she have left after cutting the last dowel?
1 Make a Plan Understand the Problem The diagonals of a kite are perpendicular, so the four triangles are right triangles. Let N represent the intersection of the diagonals. Use the Pythagorean Theorem and the properties of kites to find , and . Add these lengths to find the length of . 2 Problem–Solving Application The answer will be the amount of wood Lucy has left after cutting the dowel.
3 Solve Problem–Solving Application N bisects JM. Pythagorean Thm. Pythagorean Thm.
Problem–Solving Application Lucy needs to cut the dowel to be 32.4 cm long. The amount of wood that will remain after the cut is, 36 – 32.4 3.6 cm Lucy will have 3.6 cm of wood left over after the cut.
To estimate the length of the diagonal, change the side length into decimals and round. , and . The length of the diagonal is approximately 10 + 22 = 32. So the wood remaining is approximately 36 – 32 = 4. So 3.6 is a reasonable answer. 4 Problem–Solving Application Look Back
Properties of Kites • In kite ABCD, mDAB = 54°, and mCDF = 52°. • Find mBCD Kite cons. sides ∆BCD is isos. 2 sides isos. ∆ isos. ∆base s CBF CDF mCBF = mCDF Def. of s Polygon Sum Thm. mBCD + mCBF + mCDF = 180°
mBCD + mCBF + mCDF = 180° Substitute mCDF for mCBF. mBCD + mCBF+ mCDF= 180° Substitute 52 for mCBF. mBCD + 52°+ 52° = 180° Subtract 104 from both sides. mBCD = 76°
Properties of Kites • In kite ABCD, mDAB = 54°, and mCDF = 52°. • Find mABC ADC ABC Kite one pair opp. s Def. of s mADC = mABC Polygon Sum Thm. mABC + mBCD + mADC + mDAB = 360° Substitute mABC for mADC. mABC + mBCD + mABC+ mDAB = 360°
mABC + mBCD + mABC + mDAB = 360° mABC + 76°+ mABC + 54° = 360° Substitute. 2mABC = 230° Simplify. mABC = 115° Solve.
Properties of Kites • In kite ABCD, mDAB = 54°, and mCDF = 52°. • Find mFDA CDA ABC Kite one pair opp. s mCDA = mABC Def. of s mCDF + mFDA = mABC Add. Post. 52° + mFDA = 115° Substitute. mFDA = 63° Solve.
A trapezoid is a quadrilateral with exactly one pair of parallel sides. • Each of the parallel sides is called a base. • The nonparallel sides are called legs. • Baseangles of a trapezoid are two consecutive angles whose common side is a base.
If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid.
Properties of Isosceles Trapezoids Find the measures of the other angles. mD= 100° Thm 6-6-3 mC + mB = 180° Same-Side Int. s Thm. 100 + mB = 180 Substitute 100 for mC. mB = 80° Subtract 100 from both sides. Thm 6-6-3 A B mA = mB Def. of s mA = 80° Substitute 80 for mB
Properties of Isosceles Trapezoids • KB =21.9 and MF = 32.7 • Find FB Thm 6-6-3 KJ = FM Def. of segs. KJ = 32.7 Substitute 32.7 for FM. Seg. Add. Post. KB + BJ = KJ 21.9 + BJ = 32.7 Substitute 21.9 for KB and 32.7 for KJ. BJ = 10.8 Subtract 21.9 from both sides.
Same line. KFJ MJF Thm 6-6-3 Def of Isos. trap. SAS ∆FKJ ∆JMF CPCTC BKF BMJ Vert. s FBK JBM
Def of Isos. trap. AAS ∆FBK ∆JBM CPCTC FB = JB Def. of segs. FB = 10.8 Substitute 10.8 for JB.
Properties of Isosceles Trapezoids • KB =21.9 and MF = 32.7 • Find JB Thm 6-6-3 KJ = FM Def. of segs. KJ = 32.7 Substitute 32.7 for FM. Seg. Add. Post. KB + BJ = KJ 21.9 + BJ = 32.7 Substitute 21.9 for KB and 32.7 for KJ. BJ = 10.8 Subtract 21.9 from both sides.
Properties of Isosceles Trapezoids • KB =21.9 and MF = 32.7 • Find MB Seg. Add. Post. MB + FB = MF MB + 10.8 = 32.7 Substitute 10.8 for FB and 32.7 for MF. MB = 21.9 Subtract 10.8 from both sides.
Conditions for Isosceles Trapezoids Find the value of a so that PQRS is isosceles. Trap. with pair base s isosc. trap. S P mS = mP Def. of s Substitute 2a2 – 54 for mS and a2 + 27 for mP. 2a2 – 54 = a2 + 27 Subtract a2 from both sides and add 54 to both sides. a2 = 81 a = 9 or a = –9 Find the square root of both sides.
The midsegment of a trapezoid is the segment whose endpoints are the midpoints of the legs. In Lesson 5-1, you studied the Triangle Midsegment Theorem. The Trapezoid Midsegment Theorem is similar to it.
Trapezoid Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. Solve. EF = 10.75
1 16.5 = (25 + EH) 2 Trapezoid Midsegments Find EH. Trap. Midsegment Thm. Substitute the given values. Simplify. Multiply both sides by 2. 33= 25 + EH Subtract 25 from both sides. 13= EH