Right triangles and trigonometry #5 - Q&A

Question 1 If a line parallel to a side of a triangle intersects the other two sides, then it divides those sides _______________________. Answer 1 This is the Triangle Proportionality Theorem: If a line parallel to a side of a triangle intersects the other two sides, then it divides those sides proportionally. Question 2 A rectangle with dimensions 21.6 × 12 has a right triangle with a base 9.6 and a height of 7.2 cut out of the rectangle. What is the area of the shaded region? Answer 1 The area of the shaded region is the area of the rectangle minus the area of the right triangle: 𝐴 shaded = 𝐴 rectangle − 𝐴 triangle A shaded ​ =A rectangle ​ −A triangle ​ 𝐴 shaded = 𝑙 𝑤 − 1 2 𝑏 ℎ A shaded ​ =lw− 2 1 ​ bh 𝐴 shaded = ( 21.6 ) ( 12 ) − 1 2 ( 9.6 ) ( 7.2 ) A shaded ​ =(21.6)(12)− 2 1 ​ (9.6)(7.2) 𝐴 shaded = 259.2 − 34.56 A shaded ​ =259.2−34.56 𝐴 shaded = 224.64   square units A shaded ​ =224.64square units Question 3 The pentagon at the right is equilateral and equiangular. a. What two triangles must be congruent to prove 𝐻 𝐵 = 𝐻 𝐸 HB=HE? b. Write a proof to show 𝐻 𝐵 ≅ 𝐻 𝐸 HB≅HE. Answer 1 a. 𝐻 𝐵 HB and 𝐻 𝐸 HE are side lengths of △ 𝐻 𝐵 𝐶 △HBC and △ 𝐻 𝐷 𝐸 △HDE. Therefore, to prove 𝐻 𝐵 ≅ 𝐻 𝐸 HB≅HE, we must prove △ 𝐻 𝐵 𝐶 ≅ △ 𝐻 𝐷 𝐸 △HBC≅△HDE. b. Proof Outline: Since the pentagon is equilateral, we know 𝐸 𝐷 ≅ 𝐵 𝐶 ED≅BC. Vertical angles are congruent so we also know ∠ 𝐸 𝐻 𝐷 ≅ ∠ 𝐶 𝐵 𝐻 ∠EHD≅∠CBH. This is not enough to prove the triangles are congruent though. Since the pentagon is equiangular, we know ∠ 𝐵 𝐶 𝐷 ≅ ∠ 𝐸 𝐷 𝐶 ∠BCD≅∠EDC. We can then prove that △ 𝐵 𝐶 𝐷 ≅ △ 𝐸 𝐷 𝐶 △BCD≅△EDC by SAS. Using CPCTC, we then have ∠ 𝐻 𝐵 𝐶 ≅ ∠ 𝐻 𝐸 𝐷 ∠HBC≅∠HED. We now have two pairs of congruent corresponding angles and a pair of congruent non-included sides, so then △ 𝐻 𝐵 𝐶 ≅ △ 𝐻 𝐷 𝐸 △HBC≅△HDE by AAS. Proof: Statements Reasons 1. ABCDE is an equilateral and equiangular pentagon 1. Given equiangular pentagon 2. 𝐸 𝐷 ≅ 𝐵 𝐶 ED≅BC 2. Definition of equilateral 3. ∠ 𝐸 𝐻 𝐷 ≅ ∠ 𝐶 𝐵 𝐻 ∠EHD≅∠CBH 3. Vertical Angles Theorem 4. △ 𝐵 𝐶 𝐷 ≅ △ 𝐸 𝐷 𝐶 △BCD≅△EDC 4. Definition of equiangular 5. ∠ 𝐻 𝐵 𝐶 ≅ ∠ 𝐻 𝐸 𝐷 ∠HBC≅∠HED 5. CPCTC 6. △ 𝐻 𝐵 𝐶 ≅ △ 𝐻 𝐷 𝐸 △HBC≅△HDE 6. AAS Question 4 What is the length of the hypotenuse of a right triangle with legs of length 7 and 8? Answer 1 By the Pythagorean Theorem, the sum of the squares of the leg lengths 𝑎 a and 𝑏 b is equal to the square of the hypotenuse 𝑐 c in a right triangle: 𝑎 2 + 𝑏 2 = 𝑐 2 a 2 +b 2 =c 2 7 2 + 8 2 = 𝑐 2 7 2 +8 2 =c 2 49 + 64 = 𝑐 2 49+64=c 2 113 = 𝑐 2 113=c 2 𝑐 = 113 ≈ 10.6 c= 113 ​ ≈10.6 Question 5 Expanding isosceles triangle: The legs of an isosceles right triangle increase in length at a rate of 2 m/s. a. At what rate is the area of the triangle changing when the legs are 2 m long? b. At what rate is the area of the triangle changing when the hypotenuse is 1 m long? c. At what rate is the length of the hypotenuse changing? Answer 1 Consider 𝐴 𝐵 = 𝑥 AB=x as the length of a leg of an isosceles right triangle and 𝐴 𝐶 = 𝑦 AC=y as the hypotenuse. Given that: 𝑑 𝑥 𝑑 𝑡 = 2   m/s dt dx ​ =2m/s a) We know that the area of a right triangle is: 𝐴 = 1 2 × base × height A= 2 1 ​ ×base×height 𝐴 = 1 2 ( 𝑥 ) ( 𝑥 ) A= 2 1 ​ (x)(x) 𝐴 = 1 2 𝑥 2 A= 2 1 ​ x 2 Differentiate 𝐴 A with respect to 𝑡 t: 𝑑 𝐴 𝑑 𝑡 = 𝑑 𝑑 𝑡 ( 1 2 𝑥 2 ) dt dA ​ = dt d ​ ( 2 1 ​ x 2 ) 𝑑 𝐴 𝑑 𝑡 = 1 2 ( 2 𝑥 ) 𝑑 𝑥 𝑑 𝑡 dt dA ​ = 2 1 ​ (2x) dt dx ​ 𝑑 𝐴 𝑑 𝑡 = 𝑥 𝑑 𝑥 𝑑 𝑡 dt dA ​ =x dt dx ​ 𝑑 𝐴 𝑑 𝑡 = ( 2 ) ( 2 ) dt dA ​ =(2)(2) 𝑑 𝐴 𝑑 𝑡 = 4   m 2 / s dt dA ​ =4m 2 /s Hence, the area of the triangle is changing at 4   m 2 / s 4m 2 /s. b) In an isosceles right triangle 𝐴 𝐵 𝐶 ABC: 𝐴 𝐶 2 = 𝐴 𝐵 2 + 𝐵 𝐶 2 AC 2 =AB 2 +BC 2 𝑦 2 = 2 𝑥 2 y 2 =2x 2 Given 𝑦 = 1   m y=1m, solve: 𝑥 2 = 1 2 x 2 = 2 1 ​ 𝑥 = 1 2 x= 2 ​ 1 ​ Substitute 𝑥 = 1 2 x= 2 ​ 1 ​ and 𝑑 𝑥 𝑑 𝑡 = 2 dt dx ​ =2: 𝑑 𝐴 𝑑 𝑡 = 𝑥 𝑑 𝑥 𝑑 𝑡 dt dA ​ =x dt dx ​ 𝑑 𝐴 𝑑 𝑡 = ( 1 2 ) ( 2 ) dt dA ​ =( 2 ​ 1 ​ )(2) 𝑑 𝐴 𝑑 𝑡 = 2   m 2 / s dt dA ​ = 2 ​ m 2 /s c) In the isosceles right triangle 𝐴 𝐵 𝐶 ABC: 𝑦 2 = 2 𝑥 2 y 2 =2x 2 𝑦 = 2 𝑥 y= 2 ​ x Differentiate 𝑦 y with respect to 𝑡 t: 𝑑 𝑦 𝑑 𝑡 = 𝑑 𝑑 𝑡 ( 2 𝑥 ) dt dy ​ = dt d ​ ( 2 ​ x) 𝑑 𝑦 𝑑 𝑡 = 2 𝑑 𝑥 𝑑 𝑡 dt dy ​ = 2 ​ dt dx ​ 𝑑 𝑦 𝑑 𝑡 = 2 ( 2 ) dt dy ​ = 2 ​ (2) 𝑑 𝑦 𝑑 𝑡 = 2 2   m/s dt dy ​ =2 2 ​ m/s Hence, the hypotenuse is changing at a rate of 2 2   m/s 2 2 ​ m/s.