Transcript 6. - Kent
Conditions for Special Parallelograms Entry Task • List the 6 ways to prove a quadrilateral is a parallelogram, show a picture of each. for Special Parallelograms IsConditions it a parallelogram?? Rhombus • 4 congruent sides • Diagonals that Rectangle bisect opp. • 4 rt. Angles Angles • Congruent diagonals • Perpendicular diagonals Square (must have 1 property from rectangle and rhombus) • 4 congruent sides • 4 rt. Angles • Diagonals that bisect opp. Angles • Congruent diagonals • Perpendicular diagonals 6-6 Properties of Kites and Trapezoids Learning targets I know and understand the properties of trapezoids and kites Success Criteria I can use properties of kites and trapezoids to solve problems. Holt Geometry 6-6 Properties of Kites and Trapezoids Vocabulary kite trapezoid base of a trapezoid leg of a trapezoid base angle of a trapezoid isosceles trapezoid midsegment of a trapezoid Holt Geometry 6-6 Properties of Kites and Trapezoids A kite is a quadrilateral with exactly two pairs of congruent consecutive sides. Holt Geometry 6-6 Properties of Kites and Trapezoids *Note – These are NOT parallelograms! Holt Geometry 6-6 Properties of Kites and Trapezoids 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. Base angles 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. Holt Geometry 6-6 Properties of Kites and Trapezoids Holt Geometry 6-6 Properties of Kites and Trapezoids 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. Holt Geometry 6-6 Properties of Kites and Trapezoids Check It Out! Example 2c In kite PQRS, mPQR = 78°, and mTRS = 59°. Find each mPSR. mSPT + mTRS + mRSP = 180° Polygon Sum Thm. mSPT = mTRS Def. of s mTRS + mTRS + mRSP = 180° Substitute. 59° + 59° + mRSP = 180° Substitute. Simplify. mRSP = 62° Holt Geometry 6-6 Properties of Kites and Trapezoids 1. Erin is making a kite based on the pattern below. About how much binding does Erin need to cover the edges of the kite? about 191.2 in. In kite HJKL, mKLP = 72°, and mHJP = 49.5°. Find each measure. 2. mLHJ Holt Geometry 81° 3. mPKL 18° 6-6 Properties of Kites and Trapezoids Example 3A: Using Properties of Isosceles Trapezoids Find mA. mC + mB = 180° 100 + mB = 180 Holt Geometry Same-Side Int. s Thm. Substitute 100 for mC. mB = 80° A B Subtract 100 from both sides. Isos. trap. s base mA = mB Def. of s mA = 80° Substitute 80 for mB 6-6 Properties of Kites and Trapezoids Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. EF = 10.75 Holt Geometry Solve. 6-6 Properties of Kites and Trapezoids Check It Out! Example 5 Find EH. Trap. Midsegment Thm. 1 16.5 = 2 (25 + EH) Substitute the given values. Simplify. 33 = 25 + EH Multiply both sides by 2. 13 = EH Subtract 25 from both sides. Holt Geometry 6-6 Properties of Kites and Trapezoids Holt Geometry 6-6 Properties of Kites and Trapezoids Homework p. 394 #7-23 odds, 26 and 29-35 odds,47-52 Challenge - 63 Holt Geometry 6-6 Properties of Kites and Trapezoids 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? Holt Geometry 6-6 Properties of Kites and Trapezoids Example 1 Continued 1 Understand the Problem The answer will be the amount of wood Lucy has left after cutting the dowel. 2 Make a Plan 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 . Holt Geometry 6-6 Properties of Kites and Trapezoids Example 1 Continued 3 Solve N bisects JM. Pythagorean Thm. Pythagorean Thm. Holt Geometry 6-6 Properties of Kites and Trapezoids Example 1 Continued 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. Holt Geometry 6-6 Properties of Kites and Trapezoids Example 1 Continued 4 Look Back 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. Holt Geometry 6-6 Properties of Kites and Trapezoids 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. AD = BC Def. of segs. Substitute 12x – 11 for AD and 12x – 11 = 9x – 2 9x – 2 for BC. 3x = 9 x=3 Holt Geometry Subtract 9x from both sides and add 11 to both sides. Divide both sides by 3. 6-6 Properties of Kites and Trapezoids Check It Out! Example 4 Find the value of x so that PQST is isosceles. Q S mQ = mS Trap. with pair base s isosc. trap. Def. of s 2 + 19 for mQ Substitute 2x 2x2 + 19 = 4x2 – 13 and 4x2 – 13 for mS. 32 = 2x2 x = 4 or x = –4 Holt Geometry Subtract 2x2 and add 13 to both sides. Divide by 2 and simplify. 6-6 Properties of Kites and Trapezoids 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. KB + BJ = KJ Seg. Add. Post. 21.9 + BJ = 32.7 Substitute 21.9 for KB and 32.7 for KJ. BJ = 10.8 Subtract 21.9 from both sides. Holt Geometry 6-6 Properties of Kites and Trapezoids Example 3B Continued Same line. KFJ MJF Isos. trap. s base Isos. trap. legs ∆FKJ ∆JMF SAS BKF BMJ CPCTC FBK JBM Vert. s Holt Geometry 6-6 Properties of Kites and Trapezoids Example 3B Continued Isos. trap. legs ∆FBK ∆JBM AAS CPCTC Holt Geometry FB = JB Def. of segs. FB = 10.8 Substitute 10.8 for JB. 6-6 Properties of Kites and Trapezoids Lesson Quiz: Part II Use the diagram for Items 4 and 5. 4. mWZY = 61°. Find mWXY. 119° 5. XV = 4.6, and WY = 14.2. Find VZ. 9.6 6. Find LP. 18 Holt Geometry