McGraw Hill Glencoe Geometry, 2012
MH
McGraw Hill Glencoe Geometry, 2012 View details
1. Areas of Parallelograms and Triangles
Continue to next subchapter

Exercise 35 Page 785

Practice makes perfect
a We are given that the Heron's Formula relates the lengths of a triangle to the area of the triangle.
A=sqrt(s(s-a)(s-b)(s-c))In this formula a, b, and c are side lengths and s is the semiperimeter — half of the perimeter of this triangle. In our exercise we are asked to use this formula to find the area of a triangle with side lengths 7, 10, and 4. First let's evaluate the semiperimeter s.
s=1/2(a+b+c)
s=1/2( 7+ 10+ 4)
Simplify right-hand side
s=1/2(21)
s=21/2
s=10.5
The semiperimeter is 10.5. Now we will substitute the side lengths and semiperimeter into Heron's formula to evaluate the area.
A=sqrt(s(s-a)(s-b)(s-c))
A=sqrt(10.5(10.5- 7)(10.5- 10)(10.5- 4))
A=sqrt(10.5(3.5)(0.5)(6.5))
A=sqrt(119.4375)
A=10.9287...
A≈ 10.9
The area of this triangle is approximately 10.9 square units.
b In this part we are asked to show that the areas found for a 5- 12- 13 right triangle are the same using Heron's formula and the standard area formula. First we will evaluate the semiperimeter of this triangle.
s=1/2(a+b+c)
s=1/2( 5+ 12+ 13)
Simplify right-hand side
s=1/2(30)
s=30/2
s=15
The semiperimeter is 15. Now we will substitute side lengths and the semiperimeter into the Heron's formula and check if the result is the same as using the standard formula. Notice that in a right triangle one leg is a base and the other is a height.
sqrt(s(s-a)(s-b)(s-c))? =1/2bh
sqrt(15(15- 5)(15- 12)(15- 13))? =1/2( 5)( 12)
sqrt(15(10)(3)(2))? =1/2(5)(12)
sqrt(900)? =30
30=30
Since we ended with a true statement, we can see that both methods give us the same result.