Solutions to Review for Exam 2 on Chapters 4, 5, 6

Solutions to Review for Exam 2 on Chapters 4, 5, 6 — MATH 310 1 (a) In any triangle, the sum of the interior angles is 180 degrees. GIVEN: Triangle ABC PROVE: a + b + c = 180 degrees Construction: Draw line L through B parallel to segment AC. Mark angles x and y. Proof: a = x and c = y (Alternate interior angles, segment AC parallel to L) x + b + y = 180 degrees (Angles on a line) Therefore, a + b + c = 180 degrees. 1 (b) Each exterior angle of a triangle is the sum of the opposite interior angles. GIVEN: Triangle ABC PROVE: e = a + b Proof: c + e = 180 degrees (Angles on a line) e = 180 - c a + b + c = 180 degrees (Sum of angles in a triangle) a + b = 180 - c Therefore, e = a + b. 1 (c) Opposite angles in a parallelogram are equal. GIVEN: ABCD is a parallelogram PROVE: a = c Construction: Extend sides AB, DC. Mark angle X. Proof: a = x (Corresponding angles, segment AD parallel to BC) x = c (Alternate interior angles, segment AB parallel to DC) Therefore, a = c. Page 2 2. Congruence shortcuts: (a) RHL (Right angle, Hypotenuse, Leg) (b) AAS (or ASA) — showing 28 deg and 62 deg angles with sides 17 cm. (c) SSS (Side, Side, Side) (d) SAS (Side, Angle, Side) — All radii in a circle are the same length. 3. BISECT ANGLE CONSTRUCTION Why does vector BX bisect Angle ABC? Key: Draw circles: center B (any radius), centers R, S with same radius. Given: BR = BS, RX = SX Prove: a = b Proof: BR = BS (given) RX = SX (given) BX = BX (common side) Therefore, Triangle BRX is congruent to Triangle BSX by SSS. a = b (Corresponding angles of congruent triangles). 4 (a) Given: AB = BC, BD = BE Prove: AD = EC Proof: a = c (base angles of isosceles triangle ABC) x = y (base angles of isosceles triangle DBE) d = 180 - x and e = 180 - y (Angles on a line) Since x = y, then d = e. Triangle ABD is congruent to Triangle CBE by AAS. AD = EC (Corresponding sides of congruent triangles). Page 3 4 (b) Given: AC = BD Prove: Angle ACB = Angle DBC and AB = DC Proof: AC = BD (given) Angle ABC = Angle DCB = 90 degrees (given) BC = BC (common side) Therefore, Triangle ABC is congruent to Triangle DCB by RHL. Angle ACB = Angle DBC (corresponding parts of congruent triangles). AB = DC (corresponding parts of congruent triangles). 5. ABCD is a rectangle. Find x and y. AC = BD (diagonals of a rectangle are equal) AE = EC (1/2 AC) and BE = ED (1/2 BD) (diagonals of a parallelogram bisect each other) Therefore, EC = ED. Triangle DEC is an isosceles triangle. Angle DEC = 130 degrees (vertical angles) Angle EDC = y (base angles of isosceles triangle) y + y + 130 = 180 (Sum of angles in a triangle) 2y = 50 -> y = 25. x + y = 90 (definition of a rectangle) x + 25 = 90 -> x = 65. Alternatively (using exterior angle of triangle): 130 = x + x -> 2x = 130 -> x = 65. 50 = y + y -> 2y = 50 -> y = 25. Page 4 6. Properties Table (True/False): (a) Pair of parallel sides: Trapezoid(T), Kite(F), Parallelogram(T), Rectangle(T), Rhombus(T), Square(T) (b) Two pairs of equal adjacent sides: Trapezoid(F), Kite(T), Parallelogram(F), Rectangle(F), Rhombus(T), Square(T) (c) Diagonals are perpendicular: Trapezoid(F), Kite(T), Parallelogram(F), Rectangle(F), Rhombus(T), Square(T) (d) Pair of equal opposite angles: Trapezoid(F), Kite(T), Parallelogram(T), Rectangle(T), Rhombus(T), Square(T) (e) Diagonal bisects pair of opposite angles: Trapezoid(F), Kite(T), Parallelogram(F), Rectangle(F), Rhombus(T), Square(T) (f) Opposite sides are parallel: Parallelogram(T), Rectangle(T), Rhombus(T), Square(T) (g) Opposite sides are equal: Parallelogram(T), Rectangle(T), Rhombus(T), Square(T) (h) Opposite angles are equal: Parallelogram(T), Rectangle(T), Rhombus(T), Square(T) (i) Diagonals bisect each other: Parallelogram(T), Rectangle(T), Rhombus(T), Square(T) (j) Four right angles: Rectangle(T), Square(T) (k) Diagonals are equal: Rectangle(T), Square(T) (l) Sides are all equal: Rhombus(T), Square(T) (m) Diagonals are bisectors of opposite angles: Rhombus(T), Square(T) 7 (a) Opposite sides of a parallelogram are equal. GIVEN: ABCD is a parallelogram PROVE: AB = DC, BC = AD Proof: In Triangle ABC and Triangle CDA, s = x (alt int angles, AB parallel to DC) y = r (alt int angles, AD parallel to BC) AC = AC (common side) Therefore, Triangle ABC is congruent to Triangle CDA by ASA. AB = DC and BC = AD (Corresponding sides of congruent triangles). Page 5 7 (b) If two opposite sides of a quadrilateral are equal and parallel, then the quadrilateral is a parallelogram. Given: AB = DC, AB parallel to DC Prove: AD parallel to BC Proof: AB = DC (given) Angle BAC = Angle DCA (alt int angles, AB parallel to DC) AC = AC (common side) Therefore, Triangle CAB is congruent to Triangle ACD by SAS. Angle BCA = Angle DAC (corresponding angles of congruent triangles). AD parallel to BC (alternate interior angles, converse). ABCD is a parallelogram. 7 (c) GIVEN: ABCD is a Kite (i) PROVE: Angle B = Angle D Proof: AB = AD and BC = CD (definition of Kite) AC = AC (common side) Therefore, Triangle ABC is congruent to Triangle ADC by SSS. Angle B = Angle D (corresponding angles of congruent triangles). This proves: in a kite, at least one pair of opposite angles are equal. (ii) PROVE: a = b and c = d By (i), Triangle ABC congruent to Triangle ADC. a = b and c = d (corresponding angles). This proves: in a kite, at least one of its diagonals bisects a pair of opposite angles. Page 6 7 (iii) PROVE: Segment BD is perpendicular to segment AC Proof: By (i), BC = CD By (ii), c = d CE = CE (common side) Therefore, Triangle BCE is congruent to Triangle DCE by SAS. x = y (corresponding angles) x + y = 180 degrees (angles on a line) x + x = 180 -> 2x = 180 -> x = 90 degrees. Therefore, BD is perpendicular to AC. This proves: in a kite, the diagonals are perpendicular to each other. 8 (a) Area and Perimeter Area of rectangle = 10cm x 8cm = 80 cm^2 Area of right triangle = 1/2 (4cm x 3cm) = 6 cm^2 Area of shaded region = 80 - 6 = 74 cm^2. By Pythagorean Theorem: hypotenuse is 5 cm (triple 3, 4, 5). Perimeter of shaded region = 5 + 5 + 10 + 8 + 6 = 34 cm. 8 (b) Complex shape area Method 1: Break up into two rectangles and a right triangle. Area (R) = 10cm x 5cm = 50 cm^2 Area (S) = 16cm x 5cm = 80 cm^2 Area (T) = 1/2 (16cm x 12cm) = 96 cm^2 Area of entire region = 50 + 80 + 96 = 226 cm^2. Page 7 8 (b) continued Area of BIG rectangle = 16cm x 22cm = 352 cm^2 Area of small rectangle = 6cm x 5cm = 30 cm^2 Area of right triangle = 1/2 (16cm x 12cm) = 96 cm^2 Area of entire region = 352 - (30 + 96) = 226 cm^2. By Pythagorean Theorem (triple 3,4,5 multiplied by 4), the hypotenuse is 20 cm. Perimeter of shaded region = 20 + 10 + 10 + 5 + 6 + 17 = 68 cm. 9. Swimming pool walkway Area of BIG rectangle = 30ft x 20ft = 600 ft^2 Area of swimming pool = 20ft x 14ft = 280 ft^2 Area of walkway = 600 - 280 = 320 ft^2. 10 (a) Comparison Rectangle 1 (6cm x 4cm): Perimeter = 20cm, Area = 24 cm^2. Rectangle 2 (8cm x 2cm): Perimeter = 20cm, Area = 16 cm^2. The rectangles have the SAME PERIMETER but DIFFERENT AREAS. Page 8 10 (b) Rectangle (4cm x 2cm): Area = 8 cm^2. Triangle (base 4cm, height 4cm): Area = 1/2 (4 x 4) = 8 cm^2. 11 (a) Find x Find the area in two ways: Horizontal base 16cm, corr height 9cm: Area = 1/2 (16 * 9) = 72 cm^2. Tilted base 24cm, corr height x cm: Area = 1/2 (24 * x) = 12x cm^2. 12x = 72 -> x = 6. 11 (b) Area of Parallelogram = Base x Height Find the area in two ways: Horizontal base 27cm, corr height y cm: Area = 27y cm^2. Tilted base 6cm, corr height 18cm: Area = 6 * 18 = 108 cm^2. 27y = 108 -> y = 4. Page 9 11 (c) (i) Area of TRAPEZOID = 1/2 (a + b)h (where a, b are base lengths, h is height). Area = 1/2 (8 + 14)6 = 1/2 (22)6 = 66 in^2. (ii) Area of KITE = 1/2 c*d (where c, d are lengths of diagonals). Area = 1/2 (9 * 5) = 45 / 2 = 22.5 cm^2. (iii) If Area(Triangle ABC) = 42 cm^2, then Area(Triangle ADB) = 42 cm^2. The two triangles are in the same family of SKEWED triangles. They both have same base length (AB) and same height. 12. Study the proof of the Pythagorean Theorem. 13. If a, b, c are lengths of sides of a triangle (c is longest), then: The triangle is a right triangle if and only if a^2 + b^2 = c^2. (a) 15, 20, 25: 15^2 + 20^2 = 225 + 400 = 625. 25^2 = 625. YES! (triple 3,4,5 x 5). (b) 2, 3, 4: 2^2 + 3^2 = 4 + 9 = 13. 4^2 = 16. NO. (c) sqrt(2), sqrt(3), sqrt(5): (sqrt(2))^2 + (sqrt(3))^2 = 2 + 3 = 5. (sqrt(5))^2 = 5. YES! Page 10 (d) 3/4, 4/5, 1: (3/4)^2 + (4/5)^2 = 9/16 + 16/25 = (225 + 256) / 400 = 481/400. 1^2 = 1. NO. Extra example: 3/5, 4/5, 1: (3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 25/25 = 1. YES! (triple 3,4,5 x 1/5). (e) 6, 9, 12: 6^2 + 9^2 = 36 + 81 = 117. 12^2 = 144. NO. 14. Flag pole problem: h^2 + 16^2 = 34^2 h^2 + 256 = 1156 h^2 = 900 -> h = 30. OR recognize the triple (8, 15, 17) multiplied by 2 to get (16, 30, 34). Page 11 15. (a) Triangle ABD is congruent to Triangle ACD by RHL. BD = DC (corresponding sides). So DC = 1/2 (20) = 10 cm. (b) Find height h: 10^2 + h^2 = 26^2 -> 100 + h^2 = 676 -> h^2 = 576 -> h = 24. (recognize triple 5,12,13 x 2). (c) Area(Triangle ABC) = 1/2 (20cm x 24cm) = 240 cm^2. 16. This is a (5, 12, 13) right triangle. x = 13 blocks. Verification: 5^2 + 12^2 = 25 + 144 = 169. sqrt(169) = 13. 17 (a) Find x: (45-45-90 triangle) x = (5 * sqrt(2)) * sqrt(2) = 5 * 2 = 10. 17 (b) Find x and z: (30-60-90 triangle) z = 4 * 2 = 8 (hypotenuse is 2x short leg). x = 4 * sqrt(3) (long leg is short leg * sqrt(3)). 17 (c) Find y and z: (30-60-90 triangle) Long leg is 9 cm. Short leg y = 9 / sqrt(3) = (9 * sqrt(3)) / 3 = 3 * sqrt(3). Hypotenuse z = (3 * sqrt(3)) * 2 = 6 * sqrt(3). Page 12 17 (d) Find the area of a square whose diagonal is 8 cm. The lower half of the square is an isosceles right triangle whose legs have length x. x = 8 / sqrt(2). Area of square = (8 / sqrt(2)) * (8 / sqrt(2)) = 64 / 2 = 32 cm^2. Note: formula Area = 1/2 c*d also works (1/2 * 8 * 8 = 32). 17 (e) The diagonal of a rectangle is 12 cm and makes a 30 degree angle with one side. Find the area. Short leg = 12 / 2 = 6 cm. Longer leg = 6 * sqrt(3) cm. Area = 6 * 6 * sqrt(3) = 36 * sqrt(3) cm^2. 18. Triangle problem: (a) BD = sqrt(8) / sqrt(2) = sqrt(4) = 2 cm. (b) Triangle BDC is isosceles, so DC = BD = 2 cm. (c) AB = 2 * 2 = 4 cm. (d) AD = 2 * sqrt(3) cm. (e) AC = (2 + 2*sqrt(3)) cm = 2(1 + sqrt(3)) cm. 19. Equilateral triangle with side 2 in. Find exact area. Draw line thru B perpendicular to AC. Triangle BDC is a 30-60-90 triangle. DC = 2 / 2 = 1 in. h = 1 * sqrt(3) = sqrt(3) in. Area = 1/2 (2 in * sqrt(3) in) = sqrt(3) in^2.