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Review Sections 6.1,6.2,6.4,6.6 Section 6.1 Polygon or not? 2 3 4 5 6 7234567 Section 6.1 Regular, irregular, convex, concave 8 9 10 11 1289101112 Sum of interior angles and exterior angles 13 14 15 1613141516 Parallelograms 17 18 19 20 211718192021 Special Parallelograms 22 23 25 27 28 29 30 312223252728293031 Kites 32 34 36 39 403234363940 Trapezoids 41 42 43 44 45 46 47 48 49 50 514142434445464748495051 1

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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 2

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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 3

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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 4

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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 5

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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 6

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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 7

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Example 2A: Classifying Polygons Tell whether the polygon is regular or irregular. Tell whether it is concave or convex. irregular, convex 8

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Example 2B: Classifying Polygons Tell whether the polygon is regular or irregular. Tell whether it is concave or convex. irregular, concave 9

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Example 2C: Classifying Polygons Tell whether the polygon is regular or irregular. Tell whether it is concave or convex. regular, convex 10

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Check It Out! Example 2a Tell whether the polygon is regular or irregular. Tell whether it is concave or convex. regular, convex 11

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Check It Out! Example 2b Tell whether the polygon is regular or irregular. Tell whether it is concave or convex. irregular, concave 12

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Check It Out! Example 3a Find the sum of the interior angle measures of a convex 15-gon. (n – 2)180° (15 – 2)180° 2340° Polygon Sum Thm. A 15-gon has 15 sides, so substitute 15 for n. Simplify. 13

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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. Step 2 Find the measure of one interior angle. (n – 2)180° (16 – 2)180° = 2520° Polygon Sum Thm. Substitute 16 for n and simplify. The int. s are , so divide by 16. 14

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Find the measure of each exterior angle of a regular dodecagon. Check It Out! Example 4a A dodecagon has 12 sides and 12 vertices. sum of ext. s = 360°. A regular dodecagon has 12 ext. s, so divide the sum by 12. The measure of each exterior angle of a regular dodecagon is 30°. Polygon Sum Thm. measure of one ext. 15

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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°. A regular 20-gon has 20 ext. s, so divide the sum by 20. The measure of each exterior angle of a regular 20-gon is 18°. Polygon Sum Thm. measure of one ext. = 16

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Example 2A: Using Properties of Parallelograms to Find Measures WXYZ is a parallelogram. Find YZ. Def. of segs. Substitute the given values. Subtract 6a from both sides and add 4 to both sides. Divide both sides by 2. YZ = XW 8a – 4 = 6a + 10 2a = 14 a = 7 YZ = 8a – 4 = 8(7) – 4 = 52 opp. s 17

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

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Check It Out! Example 2a EFGH is a parallelogram. Find JG. Substitute. Simplify. EJ = JG 3w = w + 8 2w = 8 w = 4Divide both sides by 2. JG = w + 8 = 4 + 8 = 12 Def. of segs. diags. bisect each other. 19

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Check It Out! Example 2b EFGH is a parallelogram. Find FH. Substitute. Simplify. FJ = JH 4z – 9 = 2z 2z = 9 z = 4.5Divide both sides by 2. Def. of segs. FH = (4z – 9) + (2z) = 4(4.5) – 9 + 2(4.5) = 18 diags. bisect each other. 20

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Lesson Quiz: Part II QRST is a parallelogram. Find each measure. 2. TQ 3. mT 28 71° 21

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Example 1: Craft Application A 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 22

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Example 2A: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find TV. Def. of rhombus Substitute given values. Subtract 3b from both sides and add 9 to both sides. Divide both sides by 10. WV = XT 13b – 9 = 3b + 4 10b = 13 b = 1.3 23

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Example 2A Continued Def. of rhombus Substitute 3b + 4 for XT. Substitute 1.3 for b and simplify. TV = XT TV = 3b + 4 TV = 3(1.3) + 4 = 7.9 24

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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 = 5a = 5 25

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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° 26

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Check It Out! Example 2a CDFG is a rhombus. Find CD. Def. of rhombus Substitute Simplify Substitute Def. of rhombus Substitute CG = GF 5a = 3a + 17 a = 8.5 GF = 3a + 17 = 42.5 CD = GF CD = 42.5 27

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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. 28

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Check It Out! Example 2b Continued mGCH + mHCD = mGCD 2mGCH = mGCD Rhombus each diag. bisects opp. s 2mGCH = (b + 3) 2mGCH = (31 + 3) mGCH = 17° Substitute. Simplify and divide both sides by 2. 29

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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. TR2. CE 35 ft29 ft 30

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Lesson Quiz: Part II PQRS is a rhombus. Find each measure. 3. QP4. mQRP 4251° 31

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Kite cons. sides Example 2A: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD. ∆BCD is isos. 2 sides isos. ∆ isos. ∆ base s Def. of s Polygon Sum Thm. CBF CDF mCBF = mCDF mBCD + mCBF + mCDF = 180° 32

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Example 2A Continued Substitute mCDF for mCBF. Substitute 52 for mCDF. Subtract 104 from both sides. mBCD + mCDF + mCDF = 180° mBCD + 52° + 52° = 180° mBCD = 76° mBCD + mCBF + mCDF = 180° 33

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Kite one pair opp. s Example 2B: Using Properties of Kites Def. of s Polygon Sum Thm. In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC. ADC ABC mADC = mABC mABC + mBCD + mADC + mDAB = 360° mABC + mBCD + mABC + mDAB = 360° Substitute mABC for mADC. 34

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Example 2B Continued Substitute. Simplify. mABC + mBCD + mABC + mDAB = 360° mABC + 76° + mABC + 54° = 360° 2mABC = 230° mABC = 115° Solve. 35

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Kite one pair opp. s Example 2C: Using Properties of Kites Def. of s Add. Post. Substitute. Solve. In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mFDA. CDA ABC mCDA = mABC mCDF + mFDA = mABC 52° + mFDA = 115° mFDA = 63° 36

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Check It Out! Example 2a In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQRT. Kite cons. sides ∆PQR is isos. 2 sides isos. ∆ isos. ∆ base s Def. of s RPQ PRQ mQPT = mQRT 37

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Check It Out! Example 2a Continued Polygon Sum Thm. Substitute 78 for mPQR. mPQR + mQRP + mQPR = 180° 78° + mQRT + mQPT = 180° Substitute. 78° + mQRT + mQRT = 180° 78° + 2mQRT = 180° 2mQRT = 102° mQRT = 51° Substitute. Subtract 78 from both sides. Divide by 2. 38

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Check It Out! Example 2b In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQPS. Kite one pair opp. s Add. Post. Substitute. QPS QRS mQPS = mQRT + mTRS mQPS = mQRT + 59° mQPS = 51° + 59° mQPS = 110° 39

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Check It Out! Example 2c Polygon Sum Thm. Def. of s Substitute. Simplify. In kite PQRS, mPQR = 78°, and mTRS = 59°. Find each mPSR. mSPT + mTRS + mRSP = 180° mSPT = mTRS mTRS + mTRS + mRSP = 180° 59° + 59° + mRSP = 180° mRSP = 62° 40

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Isos. trap. s base Example 3A: Using Properties of Isosceles Trapezoids Find mA. Same-Side Int. s Thm. Substitute 100 for mC. Subtract 100 from both sides. Def. of s Substitute 80 for mB mC + mB = 180° 100 + mB = 180 mB = 80° A B mA = mB mA = 80° 41

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Example 3B: Using Properties of Isosceles Trapezoids KB = 21.9 and MF = 32.7. Find FB. Isos. trap. s base Def. of segs. Substitute 32.7 for FM. Seg. Add. Post. Substitute 21.9 for KB and 32.7 for KJ. Subtract 21.9 from both sides. KJ = FM KJ = 32.7 KB + BJ = KJ 21.9 + BJ = 32.7 BJ = 10.8 42

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Isos. trap. s base Same-Side Int. s Thm. Def. of s Substitute 49 for mE. mF + mE = 180° E H mE = mH mF = 131° mF + 49° = 180° Simplify. Check It Out! Example 3a Find mF. 43

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Check It Out! Example 3b JN = 10.6, and NL = 14.8. Find KM. Def. of segs. Segment Add Postulate Substitute. Substitute and simplify. Isos. trap. s base KM = JL JL = JN + NL KM = JN + NL KM = 10.6 + 14.8 = 25.4 44

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Example 4A: Applying Conditions for Isosceles Trapezoids Find the value of a so that PQRS is isosceles. a = 9 or a = –9 Trap. with pair base s isosc. trap. Def. of s Substitute 2a 2 – 54 for mS and a 2 + 27 for mP. Subtract a 2 from both sides and add 54 to both sides. Find the square root of both sides. S P mS = mP 2a 2 – 54 = a 2 + 27 a 2 = 81 45

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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. Substitute 12x – 11 for AD and 9x – 2 for BC. Subtract 9x from both sides and add 11 to both sides. Divide both sides by 3. AD = BC 12x – 11 = 9x – 2 3x = 9 x = 3 46

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Check It Out! Example 4 Find the value of x so that PQST is isosceles. Subtract 2x 2 and add 13 to both sides. x = 4 or x = –4 Divide by 2 and simplify. Trap. with pair base s isosc. trap. Q S Def. of s Substitute 2x 2 + 19 for mQ and 4x 2 – 13 for mS. mQ = mS 2x 2 + 19 = 4x 2 – 13 32 = 2x 2 47

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Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. Solve. EF = 10.75 48

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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 1 16.5 = ( 25 + EH ) 2 49

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Lesson Quiz: Part II A. mWZY = 61°. Find mWXY. B. XV = 4.6, and WY = 14.2. Find VZ. 119° 9.6 50

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Lesson Quiz: Part II Find LP. 18 51

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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? 52

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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. 53

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Solve 3 N bisects JM. Pythagorean Thm. Example 1 Continued 54

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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 55

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Look Back 4 Example 1 Continued 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. 56

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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? 57

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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 58

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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 59

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Solve 3 Pyth. Thm. Check It Out! Example 1 Continued perimeter of PQRS = 60

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Daryl needs approximately 191.3 inches of binding. One package of binding contains 2 yards, or 72 inches. In order to have enough, Daryl must buy 3 packages of binding. Check It Out! Example 1 Continued packages of binding 61

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Look Back 4 Check It Out! Example 1 Continued 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. 62

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