Pearson Geometry Common Core, 2011
PG
Pearson Geometry Common Core, 2011 View details
1. The Pythagorean Theorem and Its Converse
Continue to next subchapter

Exercise 45 Page 497

Practice makes perfect
a Suppose that the side lengths of a triangle are a, b, and c, with c being the longest. If we compare a^2+b^2 to c^2, we can find the type of triangle these sides form.
Condition Type of Triangle
a^2+b^2 < c^2 Obtuse triangle
a^2+b^2 = c^2 Right triangle
a^2+b^2 > c^2 Acute triangle
We are given two sides lengths, a=6 and b=9. Let's start by finding a number c such that 6^2+9^2=c^2.
6^2+9^2=c^2
Solve for c
36+81=c^2
117=c^2
sqrt(117)=c
c=sqrt(117)
The side lengths 6, 9, and sqrt(117)≈ 10.8 form a right triangle. Therefore, for 6, 9, and the integer number j to be the sides of an acute triangle, j must be greater than 6 and 9, but less than sqrt(117)≈ 10.8. 6 < 9 < j < 10.8 A number that satisfies these conditions is j=10. Let's verify that 6^2+9^2 is greater than 10^2.
6^2+9^2 ? > 10^2
36+81? > 100
117>100 ✓
Please note that we could have also assumed that 9 was the longest side and solved for a, so our answer is only one possible correct answer.
b In Part A we found that the given two side lengths, 6 and 9, and the number sqrt(117)≈ 10.8 form a right triangle. Therefore, for 6, 9, and the integer k to represent the side lengths of an obtuse triangle, k must be greater than sqrt(117)≈ 10.8.
6 < 9 < 10.8 < k A number that satisfies this condition is k=11. Let's verify that 6^2+9^2 is less than 11^2.
6^2+9^2 ? < 11^2
36+81? < 121
117<121 ✓
Once again, our answer is only one possible solution.