McGraw Hill Glencoe Geometry, 2012
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McGraw Hill Glencoe Geometry, 2012 View details
2. The Pythagorean Theorem and Its Converse
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Exercise 39 Page 553

Perimeter: 36 units
Area: 60 square units

Practice makes perfect

We want to find the area and the perimeter of the given triangle.

Height

In order to find the area, we first need to examine the given triangle in order to find its height.

Notice that it is an isosceles triangle, so its height splits it into two congruent right triangles. Therefore, the base of each triangle is half the base of the bigger triangle.

To find the height of the isosceles triangle, we will apply the Pythagorean Theorem for one of the right triangles, whose sides are a, b=5, and c=13.

Let's substitute these values into the formula.
a^2+b^2=c^2
a^2+ 5^2= 13^2
â–Ľ
Solve for a
a^2+25=169
a^2=144
a=12
Since a negative side length does not make sense, we only need to consider positive solutions. Therefore, the height is 12.

Area

Here we will use the formula for calculating the area of a triangle. A=1/2bh Now we substitute the value of the base, b= 10, and the value of the height, h= 12, into the formula to calculate A.
A=1/2bh
A=1/2( 10)( 12)
A=1/2 ( 120 )
A=1 * 120/2
A=120/2
A=60
The area of the triangle is 60 square units.

Perimeter

The perimeter of a figure is the sum of the side lengths. For the given triangle, they are 13, 10, and 13. P= 13 + 10 + 13 ⇒ P = 36 The triangle's perimeter is 36 units.