Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
1. Section 11.1
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Exercise 51 Page 689

Practice makes perfect
a Let's find the value of x for the given figure.
Given diagram of pentagon
Recall the Polygon Angle-Sum Theorem.

Polygon Angle-Sum Theorem

The sum of the measures of the interior angles of a convex n-gon is (n-2)*180^(∘).

In this case, expressions are given for the measures of the interior angles. We can write an equation where the sum of these expressions is equal to (n-2)180^(∘). 90^(∘)+4x+(8x-10^(∘)) +90^(∘)+(6x+4^(∘)) = (n-2)180^(∘) Our polygon has 5 sides, so we can substitute 5 for n and solve our equation to find x.
90^(∘)+4x+(8x-10^(∘)) +90^(∘)+(6x+4^(∘)) = (n-2)180^(∘)
90^(∘)+4x+(8x-10^(∘)) +90^(∘)+(6x+4^(∘)) = ( 5-2)180^(∘)
Solve for x
18x + 174^(∘) = (3)180^(∘)
18x + 174^(∘) = 540^(∘)
18x = 366^(∘)
x = 20.333333...^(∘)
x ≈ 20.3^(∘)
Therefore, the value of x is approximately 20.3^(∘).
b To find the value of x, we will use the Pythagorean Theorem.

a^2+b^2=c^2 In the formula, a and b are the legs and c is the hypotenuse of a right triangle. We are given a triangle with a=x, b=x+17, and c = 25.

Diagram of a right triangle
Let's substitute these values into the formula.
a^2+b^2=c^2
x^2+ (x+17)^2= 25^2
Solve for x
x^2+x^2+2(x)(17) + 17^2 = 25^2
x^2+x^2+34x + 289 = 625
2x^2+34x + 289 = 625
2x^2+34x-336 =0
2(x^2+17x -168) =0
x^2+17x -168=0
Now, we need to solve the quadratic equation above. We will use the Quadratic Formula to do so. ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a We first need to identify the values of a, b, and c. x^2+17x -168=0 ⇕ 1x^2+ 17x+( - 168)=0 We see that a= 1, b= 17, and c= - 168. Let's substitute these values into the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
x=- 17±sqrt(17^2-4( 1)( - 168))/2( 1)
Solve for x and Simplify
x=-17±sqrt(289-4(1)(- 168))/2(1)
x=-17±sqrt(289-4(- 168))/2
x=-17±sqrt(289+672)/2
x=-17±sqrt(961)/2
x=-17± 31/2
The solutions for this equation are x= -17± 312. Let's separate them into the positive and negative cases.
x=-17± 31/2
x_1=-17+31/2 x_2=-17-31/2
x_1=14/2 x_2=-48/2
x_1=7 x_2=-24

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=7 and x_2=-24. Since a negative side length does not make sense, we only need to consider positive solutions. Therefore, x = 7 is our solution.

c Consider the given diagram.
Given diagram with an isosceles triangle
The diagram shows an isosceles triangle with one exterior angle of known angle. By the Base Angles Theorem we know that in isosceles triangle, angles opposite the congruent side have the same measure. Let's add this measure to the diagram.
We have two x+4 degrees angles

Next, notice that the 134^(∘) and the last missing angle form a linear pair of angles. This means the sum of their measures equals 180^(∘). So, to find the missing angle, we can subtract 134^(∘) from 180^(∘). 180^(∘) - 134^(∘) = 46^(∘) Let's add this angle to our diagram.

Missing angle has measure of 46 degrees
Recall that the angles in a triangle sum to 180^(∘). We can use this fact to write an equation for x. Let's do it! ( x+4^(∘))+( x+4^(∘)) + 46^(∘) = 180^(∘) Finally, let's solve this equation for x.
(x+4^(∘))+(x+4^(∘)) + 46^(∘) = 180^(∘)
2x + 54^(∘) = 180^(∘)
2x = 126^(∘)
x = 63^(∘)
d We want to find the value of x from the following diagram.
Diagram of a parallelogram
The diagram shows a parallelogram. In a parallelogram, the measures of adjacent angles sum to 180^(∘). With this, we can write the following equation. (x+9^(∘))+(3x+15^(∘)) = 180^(∘) Let's solve it for x!
(x+9^(∘))+(3x+15^(∘)) = 180^(∘)
4x + 24^(∘) = 180^(∘)
4x = 156^(∘)
x = 39^(∘)