Trigonometry (11th Edition) Clone

Published by Pearson

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 324: 72

Answer

(a), (b) , (c), (d) Since these triangles have integer sides and areas, these triangles are Heron triangles.

Work Step by Step

(a) We can find the semi-perimeter $S$: $S = \frac{11+13+20}{2} = 22$ We can find the area: $A = \sqrt{S~(S-a)(S-b)(S-c)}$ $A = \sqrt{(22)~(22-11)(22-13)(22-20)}$ $A = \sqrt{(22)~(11)(9)(2)}$ $A = \sqrt{4356}$ $A = 66$ Since this triangle has integer sides and area, this triangle is a Heron triangle. (b) We can find the semi-perimeter $S$: $S = \frac{13+14+15}{2} = 21$ We can find the area: $A = \sqrt{S~(S-a)(S-b)(S-c)}$ $A = \sqrt{(21)~(21-13)(21-14)(21-15)}$ $A = \sqrt{(21)~(8)(7)(6)}$ $A = \sqrt{7056}$ $A = 84$ Since this triangle has integer sides and area, this triangle is a Heron triangle. (c) We can find the semi-perimeter $S$: $S = \frac{7+15+20}{2} = 21$ We can find the area: $A = \sqrt{S~(S-a)(S-b)(S-c)}$ $A = \sqrt{(21)~(21-7)(21-15)(21-20)}$ $A = \sqrt{(21)~(14)(6)(1)}$ $A = \sqrt{1764}$ $A = 42$ Since this triangle has integer sides and area, this triangle is a Heron triangle. (d) We can find the semi-perimeter $S$: $S = \frac{9+10+17}{2} = 18$ We can find the area: $A = \sqrt{S~(S-a)(S-b)(S-c)}$ $A = \sqrt{(18)~(18-9)(18-10)(18-17)}$ $A = \sqrt{(18)~(9)(8)(1)}$ $A = \sqrt{1296}$ $A = 36$ Since this triangle has integer sides and area, this triangle is a Heron triangle.

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