Problems 10 through 13 are cut off or illegible, so I'm not able to answer them.
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Problem 14
The diagram shows that GE = 4. Because of this, this means DG = 4 as well.
This is because points D, E and F are all the same distance away from the center point G. We can draw a circle centered at G to have the circle edge pass through point E. This same circle will also have D and F on the edge of the circle. This circle is known as an "incircle" because it's completely contained inside the triangle, such that it's the largest circle possible without spilling outside the triangle.
Answer: 4
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Problem 15
Triangles BGD and BGE are congruent due to the hypotenuse leg theorem.
The hypotenuse lengths overlap at BG (this is a shared length). The congruent legs are DG and GE which are each 4 units.
Since the triangles are congruent, this then means that BD = BE = 11
Answer: 11
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Problem 16
Apply the pythagorean theorem to find the length of the hypotenuse BG
[tex]a^2+b^2 = c^2\\\\(BD)^2+(DG)^2 = (BG)^2\\\\BG = \sqrt{(BD)^2+(DG)^2}\\\\BG = \sqrt{(11)^2+(4)^2}\\\\BG = \sqrt{137}\\\\[/tex]
Answer: [tex]\sqrt{137}[/tex]
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Problem 17
Through a similar idea to problem 15, we can find that EC = 20 because triangles GFC and GEC are congruent (also by the hypotenuse leg theorem).
Now apply the pythagorean theorem to find GC
[tex]a^2+b^2 = c^2\\\\(GE)^2+(EC)^2 = (GC)^2\\\\GC = \sqrt{(GE)^2+(EC)^2}\\\\GC = \sqrt{(4)^2+(20)^2}\\\\GC = \sqrt{416}\\\\GC = \sqrt{16*26}\\\\GC = \sqrt{16}*\sqrt{26}\\\\GC = 4\sqrt{26}\\\\[/tex]
Answer: [tex]4\sqrt{26}[/tex]
