Big Ideas Math Geometry, 2014
BI
Big Ideas Math Geometry, 2014 View details
2. Bisectors of Triangles
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Exercise 48 Page 318

Drawing an isosceles right triangle makes things easy for us.

Example Solution:

Practice makes perfect

Let's think about how to make the triangle before we attempt to find the side lengths.

Making the triangle

Let's make it easy and draw an isosceles right triangle. Such a triangle has a right angle and two 45^(∘) base angles. We will start by drawing it's legs, one vertical and one horizontal.

How do we draw the third side? Well, since this is an isoceles triangle, an angle bisector will divide the hypotenuse into two equal parts. Therefore, if we measure a 45^(∘) angle with a protractor, the angle bisector's second point of intersection with the center circle, is where the hypotenuse of the triangle will run.

Now we can draw the angle bisector to the right angle, marking the second point of intersection with the center circle.


Next, we will draw the third side. Using a protractor, we can make sure it forms a right angle with the circle's radius.

Finally, we connect the leg's with the hypotenuse.

Let's isolate our triangle and label the positions of the players.

Finding the side lengths

Let's find the side lengths of the triangle. We will need to draw the angle bisectors and the distances between the center of the circle and the edges of the triangle.

Now that we have all of the necessary parts shown and labeled, we can find the lengths of the sides. Note that each of the distances from D to the side is 15 ft as that is the radius of the circle. A square of side length 15 ft is formed by CEDG. Let the remaining lengths of AC and BC be x ft.

By the HL Congruence Theorem, △ ADE ≅ △ ADF and △ BDG ≅ △ BDF. This means that AE=BG=AF=BF=x. To find the distance of CD, we can use the Pythagorean Theorem.
a^2+b^2=c^2
15^2+ 15^2=CD^2
Solve for CD
225+225=CD^2
225* 2=CD^2
CD^2=225* 2
CD=± sqrt(225* 2)
CD=± sqrt(225)sqrt(2)

c > 0

CD= sqrt(225)sqrt(2)
CD= 15sqrt(2)
Let's add all of the side lengths to our diagram.
Now we have enough information to solve for x by using the Pythagorean Theorem on the side lengths of △ AFC.
a^2+b^2=c^2
(15+15sqrt(2))^2+x^2=(x+15)^2
Solve for x
225+450sqrt(2)+450+x^2=x^2+30x+225
450sqrt(2)+675+x^2=x^2+30x+225
450sqrt(2)+675=30x+225
450sqrt(2)+450=30x
450sqrt(2)+450/30=x
x=450sqrt(2)+450/30
x=30(15sqrt(2)+15)/30
x=15sqrt(2)+15

Now that we know the value of x, we can label the side lengths of the triangle.

Let's clean up the diagram a bit and round the lengths to one decimal.