Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 5.2
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Exercise 105 Page 297

Practice makes perfect
a We want to find the measure of the triangle's third side. We can determine it by adding the two sides labeled x and y in the diagram below.
The sides we labeled, x and y, are both legs of right triangles. Since we know the hypotenuse and second leg in each of these, we can calculate x and y using the Pythagorean Theorem.
a^2+b^2=c^2
3^2+x^2=6^2
Solve for x
9+x^2=36
x^2=27
x=± sqrt(27)

x > 0

x= sqrt(27)
Next, we will calculate the value of y.
a^2+b^2=c^2
3^2+y^2=9^2
Solve for y
9+y^2=81
y^2=72
y=± sqrt(72)

y > 0

y= sqrt(72)
When we know the value of x and y, we can determine the unknown side of the original triangle.
b We also want to find the triangle's area. Since the dashed line is drawn from a vertex and is perpendicular to the side whose length we just determined, we know that these make up the triangle's height and base, respectively. We have enough information to calculate the area.
A=1/2bh
A=1/2( sqrt(27)+sqrt(72))( 3)
A=3/2(sqrt(27)+sqrt(72))
A=1.5(sqrt(27)+sqrt(72))
The exact value of the area is 1.5(sqrt(27)+sqrt(72))cm^2. We also need to find the approximate value.
A=1.5(sqrt(27)+sqrt(72))
A=20.5221506954...
A≈20.52
The area is approximately 20.52 cm^2
<listcircle icon="c">Since we know all the sides of the triangle, we have enough information to calculate the perimeter.</listc.ircle> P&=6+9+sqrt(27)+sqrt(72)cm & ⇓ P&=15+sqrt(27)+sqrt(72) cm We also need to find the approximate value of the perimeter.
P=15+sqrt(27)+sqrt(72)
P=28.6814337969...
P≈28.68