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Geometry H Chapter 8 Review Worksheet answers 1. A convex polygon has 17 sides. (a) Find the sum of its interior angles. (b) Find the sum of its exterior angles. (a) Formula for sum of the interior angles : S I 180(n 2) S17 = 180(17-2) = 180(15) = 2700 (b) The sum of the exterior angle is always 360. 2. The sum of the interior angles of a complex polygon is 3,600. How many sides does it have? S I 180(n 2) 3,600 = 180(n-2) 20 = n – 2 22 = n The polygon has 22 sides. 3. A regular polygon has 12 sides. (a) Find the measure of each interior angle. (b) Find the measure of each exterior angle. (a) Formula for int. angle of regular polygon : 180(n 2) mi = n 180(12 2) 1800 mi = 150 12 12 (b) Formula for ext. angle of regular polygon : 360 me = n 360 me = 30 12 4. In a regular polygon, the measure of each interior angle is 120. How many sides does it have? Method 1 180(n 2) mi = n 180(n 2) 120 = n 120n = 180(n-2) 120n = 180n – 360 -60n = -360 n = 6 The polygon has 6 sides. Method 2 If each interior angle has a measure of 120, each exterior angle measures 60. 360 me = n 360 60 = n 60n = 360 n = 6 5. In a regular polygon, the measure of each exterior angle is 15. How many sides does it have? 360 n 360 15 = n 15n = 360 n = 24 The polygon has 24 sides. me = The figure below is for questions #6-9. Quad. ABCD is a parallelogram. 6. AB = 8x – 7 DC = 2x + 11 x = ___ Since opposite sides of a parallelogram are congruent, AB = DC . 8x – 7 = 2x + 11 6x = 18 x = 3 Geometry H Chapter 8 Review Worksheet answers 7. mB = 25 – y mD = 3y – 11 y = ___ Since opposite angles of a parallelogram are congruent, mB = mD. 25 – y = 3y – 11 36 = 4y 9 = y 8. mA = 5z + 10 mD = 12z z = ___ 11. RS UT and RU || ST No. Theorem 8-9 requires that the sides that are parallel also be the sides that are congruent. 12. URS STU and RST TUR Yes, by Theorem 8-8 Since consecutive angles of a parallelogram are supplementary, mA + mD = 180. 5z + 10 + 12z = 180 17z + 10 = 180 17z = 170 z = 10 13. RT SU No. Theorem 8-10 requires that the diagonals bisect each other, but not that they be congruent to each other. A quad with congruent diagonals might be an isosceles trapezoid. 9. Draw the diagonals of ABCD. Call the point of intersection Q. AQ = w2 QC = 2w + 3 w = ____ Use this figure for questions #14-15 Since the diagonals of a parallelogram bisect each other, AQ = QC. w2 = 2w + 3 w2 - 2w – 3 = 0 (w – 3)(w + 1) = 0 w – 3 = 0 or w + 1 = 0 w=3 or w = -1 If w = 3, AQ = 9 and QC = 9 If w = -1, AQ = 1 and QC = 1 So, both 3 and -1 are possibilities for w. Use this figure for questions #10-13. With the given information, does quad. RSTU have to be a parallelogram? 10. RS UT and RU ST Yes, by Theorem 8-7 ABCD is a rectangle. 14. AC = 3x – 20 and BD = x + 30. x = ___ The diagonals of a rectangle are congruent. So, 3x – 20 = x + 30 2x = 50 x = 25 15. BC = 8 and mBDC = 30. QC = ___ DCB is a right triangle. BC is the leg opposite BDC. BD is the hypotenuse. BC sinBDC = BD 8 sin30 = BD 8 16 BD = sin 30 Since BD = 16, AC = 16. Since Q is the midpoint of AC , QC = 8. Geometry H Chapter 8 Review Worksheet answers Use this figure for questions #16-18. GHIJ is a rhombus 16. GH = 4y+120 and HI = 6y-20. y = ___ Consecutive sides of a rhombus are congruent. So, 4y + 120 = 6y - 20 140 = 2y 70 = y 17. GI = 24 and HJ = 10. GH = ___ The diagonals of a rhombus bisect each other. So, GK = 12 and KH = 5. The diagonals of a rhombus are . Using the Pythagorean Theorem, (GH)2 = (GK)2 + (KH)2 (GH)2 = 122 + 52 (GH)2 = 169 GH = 13 10 cos 25 10 40 Perimeter = 4∙GH = 4∙ = cos 25 cos 25 = 44.13511676 = 44.1 Area HK In right GKH, tanHGK = GK HK tan 25 = 10 10∙tan 25 = HK Area of GHI = ½ ∙ GI ∙ HK = ½ ∙ 20 ∙ 10∙tan 25 = 100∙tan 25 Since GJI GHI, Area of GJI = 100∙tan 25 Area GHIJ = Area of GHI + Area of GJI = 100∙tan 25 + 100∙tan 25 = 200∙tan 25 = 93.26153163 = 93.3 GH = Use this figure for questions #19-20 WXYZ is a trapezoid with WX || ZY . 18. mHGJ = 50. GI = 20. Find the perimeter and area of GHIJ. (Round to the nearest tenth.) Since mHGJ = 50, mHGK = 25 (Diagonals of rhombi bisects opposite angles.) Perimeter 19. mZWX = 110. mWZY = ___ Since they are same side interior angle of parallel lines, the are spupplementary. So, mWZY = 70. Since GI = 20, GK = 10. (Diagonals of rhombi bisects each other.) GK In right GKH, cosHGK = GH 10 cos 25 = GH 20. Is it possible for WY and XZ to bisect each other? Explain. No. That would make WXYZ a parallelogram by Theorem 8-10. Geometry H Chapter 8 Review Worksheet answers Use this figure for questions #21-24 Use this figure for questions #25-27 ABCD is a kite with AB = AD 21. BC = 2x + 10 and DC = 6x – 30 x = _____ Since BC = DC, 2x + 10 = 6x – 30 40 = 4x 10 = x 22. BE = 6 and AE = 8 AB = ___ Since the diagonals of a kite are perpendicular AEB is a rt. . By Pythagorean Theorem, (AB)2 = (AE)2 + (BE)2 (AB)2 = (8)2 + (6)2 (AB)2 = 100 AB = 10 23. mABC = 8y and m ADC = 10y – 30 y = ___ Since ABC and ADC are concruent, mABC = mADC 8y = 10y – 30 -2y = -30 y = 15 24. BD = 4z + 10 and BE = 3z – 8 z = ___ Since E is the midpoint of BD , BD = 2∙BE 4z + 10 = 2(3z – 8) 4z + 10 = 6z – 16 26 = 2z 13 = z ABCD is a trapezoid with AB || DC 25. AB = 20 and DC = 56 JK = ___ Since JK is the midsegment of the trapezoid, AB DC JK 2 20 56 76 JK 38 2 2 26. AB = 12 and JK = 22 DC = ___ Since JK is the midsegment of the trapezoid, AB DC JK 2 12 DC 22 2 44 = 12 + DC 32 = DC 27. A is 20 larger than AJK. mA = ___ Since JK is the midsegment of the trapezoid, AB || JK Since A and AJK are same-side interior angles of parallel lines, mA + mAJK = 180 Since mA = mAJK + 20 mAJK + 20 + mAJK = 180 2mAJK + 20 = 180 mAJK = 80 So, mA = 80 + 20 = 100