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Properties of Kites and Trapezoids. 6-6. Holt Geometry. Warm Up Solve for x. 1. x 2 + 38 = 3 x 2 – 12 2. 137 + x = 180 3. 4. Find FE. 5 or –5. 43. 156. A kite is a quadrilateral with exactly two pairs of congruent consecutive sides. Example 1: Problem-Solving Application.
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Properties of Kites and Trapezoids 6-6 Holt Geometry
Warm Up Solve for x. 1.x2 + 38 = 3x2 – 12 2. 137 + x = 180 3. 4. Find FE. 5 or –5 43 156
A kiteis a quadrilateral with exactly two pairs of congruent consecutive sides.
Example 1: 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 . She has a dowel that is 36 cm long. About how much wood will she have left after cutting the last dowel?
Example 2A: Using 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°
Example 2A Continued 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°
Example 2B: Using 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°
Example 2B Continued mABC + mBCD + mABC + mDAB = 360° mABC + 76°+ mABC + 54° = 360° Substitute. 2mABC = 230° Simplify. mABC = 115° Solve.
Example 2C: Using 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 trapezoidis a quadrilateral with exactly one pair of parallel sides. Each of the parallel sides is called a base. The nonparallel sides are called legs. Base anglesof 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.
Reading Math Theorem 6-6-5 is a biconditional statement. So it is true both “forward” and “backward.”
Example 3A: Using Properties of Isosceles Trapezoids Find mA. mC + mB = 180° Same-Side Int. s Thm. 100 + mB = 180 Substitute 100 for mC. mB = 80° Subtract 100 from both sides. A B Isos. trap. s base mA = mB Def. of s mA = 80° Substitute 80 for mB
Example 3B: Using Properties of Isosceles Trapezoids KB = 21.9m and MF = 32.7. Find FB. Isos. trap. s base 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.
Example 3B Continued Same line. KFJ MJF Isos. trap. s base Isos. trap. legs SAS ∆FKJ ∆JMF CPCTC BKF BMJ Vert. s FBK JBM
Example 3B Continued Isos. trap. legs AAS ∆FBK ∆JBM CPCTC FB = JB Def. of segs. FB = 10.8 Substitute 10.8 for JB.
Check It Out! Example 3a Find mF. mF + mE = 180° Same-Side Int. s Thm. E H Isos. trap. s base mE = mH Def. of s mF + 49°= 180° Substitute 49 for mE. mF = 131° Simplify.
Check It Out! Example 3b JN = 10.6, and NL = 14.8. Find KM. Isos. trap. s base Def. of segs. KM = JL JL = JN + NL Segment Add Postulate KM = JN + NL Substitute. KM = 10.6 + 14.8 = 25.4 Substitute and simplify.
Example 4A: Applying 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.
Example 4B: Applying Conditions for Isosceles Trapezoids AD = 12x – 11, and BC = 9x – 2. Find the value of x so that ABCD is isosceles. Diags. isosc. trap. Def. of segs. AD = BC Substitute 12x – 11 for AD and 9x – 2 for BC. 12x – 11 = 9x – 2 Subtract 9x from both sides and add 11 to both sides. 3x = 9 x = 3 Divide both sides by 3.
Check It Out! Example 4 Find the value of x so that PQST is isosceles. Trap. with pair base s isosc. trap. Q S mQ = mS Def. of s Substitute 2x2 + 19 for mQ and 4x2 – 13 for mS. 2x2 + 19 = 4x2 – 13 Subtract 2x2 and add 13 to both sides. 32 = 2x2 Divide by 2 and simplify. x = 4 or x = –4
The midsegment of a trapezoidis 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.
Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. Solve. EF = 10.75
1 16.5 = (25 + EH) 2 Check It Out! Example 5 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