Review Sections 6.1,6.2,6.4,6.6 Section 6.1 Polygon or not? 2 3 4 5 6 7 Section 6.1 Regular, irregular, convex, concave 8 9 10 11 12 Sum of interior angles and exterior angles 13 14 15 16 Parallelograms 17 18 19 20 21 Special Parallelograms 22 23 25 27 28 29 30 31 Kites 32 34 36 39 40 Trapezoids 41 42 43 44 45 46 47 48 49 50 51

Example 1A: Identifying Polygons Tell whether the figure is a polygon. If it is a polygon, name it by the number of sides. polygon, hexagon

Example 1B: Identifying Polygons Tell whether the figure is a polygon. If it is a polygon, name it by the number of sides. polygon, heptagon

Example 1C: Identifying Polygons Tell whether the figure is a polygon. If it is a polygon, name it by the number of sides. not a polygon

Check It Out! Example 1a Tell whether each figure is a polygon. If it is a polygon, name it by the number of its sides. not a polygon

Check It Out! Example 1b Tell whether the figure is a polygon. If it is a polygon, name it by the number of its sides. polygon, nonagon

Check It Out! Example 1c Tell whether the figure is a polygon. If it is a polygon, name it by the number of its sides. not a polygon

Example 2A: Classifying Polygons Tell whether the polygon is regular or irregular. Tell whether it is concave or convex. irregular, convex

Example 2B: Classifying Polygons Tell whether the polygon is regular or irregular. Tell whether it is concave or convex. irregular, concave

Example 2C: Classifying Polygons Tell whether the polygon is regular or irregular. Tell whether it is concave or convex. regular, convex

Check It Out! Example 2a Tell whether the polygon is regular or irregular. Tell whether it is concave or convex. regular, convex

Check It Out! Example 2b Tell whether the polygon is regular or irregular. Tell whether it is concave or convex. irregular, concave

Check It Out! Example 3a Find the sum of the interior angle measures of a convex 15-gon. (n – 2)180° Polygon  Sum Thm. (15 – 2)180° A 15-gon has 15 sides, so substitute 15 for n. 2340° Simplify.

Example 3B: Finding Interior Angle Measures and Sums in Polygons Find the measure of each interior angle of a regular 16-gon. Step 1 Find the sum of the interior angle measures. (n – 2)180° Polygon  Sum Thm. Substitute 16 for n and simplify. (16 – 2)180° = 2520° Step 2 Find the measure of one interior angle. The int. s are , so divide by 16.

Check It Out! Example 4a Find the measure of each exterior angle of a regular dodecagon. A dodecagon has 12 sides and 12 vertices. sum of ext. s = 360°. Polygon  Sum Thm. A regular dodecagon has 12  ext. s, so divide the sum by 12. measure of one ext. The measure of each exterior angle of a regular dodecagon is 30°.

Example 4A: Finding Interior Angle Measures and Sums in Polygons Find the measure of each exterior angle of a regular 20-gon. A 20-gon has 20 sides and 20 vertices. sum of ext. s = 360°. Polygon  Sum Thm. A regular 20-gon has 20  ext. s, so divide the sum by 20. measure of one ext.  = The measure of each exterior angle of a regular 20-gon is 18°.

Example 2A: Using Properties of Parallelograms to Find Measures WXYZ is a parallelogram. Find YZ.  opp. s  YZ = XW Def. of  segs. 8a – 4 = 6a + 10 Substitute the given values. Subtract 6a from both sides and add 4 to both sides. 2a = 14 a = 7 Divide both sides by 2. YZ = 8a – 4 = 8(7) – 4 = 52

Example 2B: Using Properties of Parallelograms to Find Measures WXYZ is a parallelogram. Find mZ . mZ + mW = 180°  cons. s supp. (9b + 2) + (18b – 11) = 180 Substitute the given values. 27b – 9 = 180 Combine like terms. 27b = 189 Add 9 to both sides. b = 7 Divide by 27. mZ = (9b + 2)° = [9(7) + 2]° = 65°

Check It Out! Example 2a EFGH is a parallelogram. Find JG.  diags. bisect each other. EJ = JG Def. of  segs. 3w = w + 8 Substitute. 2w = 8 Simplify. w = 4 Divide both sides by 2. JG = w + 8 = 4 + 8 = 12

Check It Out! Example 2b EFGH is a parallelogram. Find FH.  diags. bisect each other. FJ = JH Def. of  segs. 4z – 9 = 2z Substitute. 2z = 9 Simplify. z = 4.5 Divide both sides by 2. FH = (4z – 9) + (2z) = 4(4.5) – 9 + 2(4.5) = 18

Lesson Quiz: Part II QRST is a parallelogram. Find each measure. 2. TQ 3. mT 71° 28

Example 1: Craft Application A woodworker constructs a rectangular picture frame so that JK = 50 cm and JL = 86 cm. Find HM. Rect.  diags.  KM = JL = 86 Def. of  segs.  diags. bisect each other Substitute and simplify.

Example 2A: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find TV. WV = XT Def. of rhombus 13b – 9 = 3b + 4 Substitute given values. 10b = 13 Subtract 3b from both sides and add 9 to both sides. b = 1.3 Divide both sides by 10.

Example 2A Continued TV = XT Def. of rhombus Substitute 3b + 4 for XT. TV = 3b + 4 TV = 3(1.3) + 4 = 7.9 Substitute 1.3 for b and simplify.

Example 2B: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find mVTZ. mVZT = 90° Rhombus  diag.  14a + 20 = 90° Substitute 14a + 20 for mVTZ. Subtract 20 from both sides and divide both sides by 14. a = 5

Example 2B Continued Rhombus  each diag. bisects opp. s mVTZ = mZTX mVTZ = (5a – 5)° Substitute 5a – 5 for mVTZ. mVTZ = [5(5) – 5)]° = 20° Substitute 5 for a and simplify.

Check It Out! Example 2a CDFG is a rhombus. Find CD. CG = GF Def. of rhombus 5a = 3a + 17 Substitute a = 8.5 Simplify GF = 3a + 17 = 42.5 Substitute CD = GF Def. of rhombus CD = 42.5 Substitute

Check It Out! Example 2b CDFG is a rhombus. Find the measure. mGCH if mGCD = (b + 3)° and mCDF = (6b – 40)° mGCD + mCDF = 180° Def. of rhombus b + 3 + 6b – 40 = 180° Substitute. 7b = 217° Simplify. b = 31° Divide both sides by 7.

Check It Out! Example 2b Continued mGCH + mHCD = mGCD Rhombus  each diag. bisects opp. s 2mGCH = mGCD 2mGCH = (b + 3) Substitute. 2mGCH = (31 + 3) Substitute. mGCH = 17° Simplify and divide both sides by 2.

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. mQRP 42 51°

Example 2A: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD. Kite  cons. sides  ∆BCD is isos. 2  sides isos. ∆ CBF  CDF isos. ∆ base s  mCBF = mCDF Def. of   s mBCD + mCBF + mCDF = 180° Polygon  Sum Thm.

Example 2A Continued mBCD + mCBF + mCDF = 180° Substitute mCDF for mCBF. mBCD + mCDF + mCDF = 180° Substitute 52 for mCDF. mBCD + 52° + 52° = 180° Subtract 104 from both sides. mBCD = 76°

Example 2B: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC. ADC  ABC Kite  one pair opp. s  mADC = mABC Def. of  s Polygon  Sum Thm. mABC + mBCD + mADC + mDAB = 360° Substitute mABC for mADC. mABC + mBCD + mABC + mDAB = 360°

Example 2B Continued mABC + mBCD + mABC + mDAB = 360° mABC + 76° + mABC + 54° = 360° Substitute. 2mABC = 230° Simplify. mABC = 115° Solve.

Example 2C: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mFDA. CDA  ABC Kite  one pair opp. s  mCDA = mABC Def. of  s mCDF + mFDA = mABC  Add. Post. 52° + mFDA = 115° Substitute. mFDA = 63° Solve.

Check It Out! Example 2a In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQRT. Kite  cons. sides  ∆PQR is isos. 2  sides  isos. ∆ RPQ  PRQ isos. ∆  base s  mQPT = mQRT Def. of  s

Check It Out! Example 2a Continued mPQR + mQRP + mQPR = 180° Polygon  Sum Thm. Substitute 78 for mPQR. 78° + mQRT + mQPT = 180° 78° + mQRT + mQRT = 180° Substitute. 78° + 2mQRT = 180° Substitute. Subtract 78 from both sides. 2mQRT = 102° mQRT = 51° Divide by 2.

Check It Out! Example 2b In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQPS. QPS  QRS Kite  one pair opp. s  mQPS = mQRT + mTRS  Add. Post. mQPS = mQRT + 59° Substitute. mQPS = 51° + 59° Substitute. mQPS = 110°

Check It Out! Example 2c In kite PQRS, mPQR = 78°, and mTRS = 59°. Find each mPSR. mSPT + mTRS + mRSP = 180° Polygon  Sum Thm. mSPT = mTRS Def. of  s mTRS + mTRS + mRSP = 180° Substitute. 59° + 59° + mRSP = 180° Substitute. Simplify. mRSP = 62°

Example 3A: Using Properties of Isosceles Trapezoids Find mA. mC + mB = 180° Same-Side Int. s Thm. 100 + mB = 180 Substitute 100 for mC. mB = 80° Subtract 100 from both sides. A  B Isos. trap. s base  mA = mB Def. of  s mA = 80° Substitute 80 for mB

Example 3B: Using Properties of Isosceles Trapezoids KB = 21.9 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.

Check It Out! Example 3a Find mF. mF + mE = 180° Same-Side Int. s Thm. E  H Isos. trap. s base  mE = mH Def. of  s mF + 49° = 180° Substitute 49 for mE. mF = 131° Simplify.

Check It Out! Example 3b JN = 10.6, and NL = 14.8. Find KM. Isos. trap. s base  KM = JL Def. of  segs. 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 mS = mP Def. of  s Substitute 2a2 – 54 for mS and a2 + 27 for mP. 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 mQ = mS Def. of  s Substitute 2x2 + 19 for mQ and 4x2 – 13 for mS. 2x2 + 19 = 4x2 – 13 Subtract 2x2 and add 13 to both sides. 32 = 2x2 Divide by 2 and simplify. x = 4 or x = –4

Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. EF = 10.75 Solve.

Substitute the given values. Check It Out! Example 5 Find EH. Trap. Midsegment Thm. 1 16.5 = (25 + EH) 2 Substitute the given values. Simplify. 33 = 25 + EH Multiply both sides by 2. 13 = EH Subtract 25 from both sides.

Lesson Quiz: Part II A. mWZY = 61°. Find mWXY. B. XV = 4.6, and WY = 14.2. Find VZ. 119° 9.6

Lesson Quiz: Part II Find LP. 18

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?

Understand the Problem 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 .

Example 1 Continued Solve 3 N bisects JM. Pythagorean Thm. Pythagorean Thm.

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.

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.

Check It Out! Example 1 What if...? Daryl is going to make a kite by doubling all the measures in the kite. What is the total amount of binding needed to cover the edges of his kite? How many packages of binding must Daryl buy?

Understand the Problem Check It Out! Example 1 Continued 1 Understand the Problem The answer has two parts. • the total length of binding Daryl needs • the number of packages of binding Daryl must buy

Check It Out! Example 1 Continued 2 Make a Plan The diagonals of a kite are perpendicular, so the four triangles are right triangles. Use the Pythagorean Theorem and the properties of kites to find the unknown side lengths. Add these lengths to find the perimeter of the kite.

Check It Out! Example 1 Continued Solve 3 Pyth. Thm. Pyth. Thm. perimeter of PQRS =

Check It Out! Example 1 Continued Daryl needs approximately 191.3 inches of binding. One package of binding contains 2 yards, or 72 inches. packages of binding In order to have enough, Daryl must buy 3 packages of binding.

Check It Out! Example 1 Continued 4 Look Back To estimate the perimeter, change the side lengths into decimals and round. , and . The perimeter of the kite is approximately 2(54) + 2 (41) = 190. So 191.3 is a reasonable answer.