Problem 1)
Let p = distance from V to Y
By the angle bisector theorem, we can form the proportion
VT/VY = TK/KY
57/p = 129.2/68
57*68 = p*129.2
3876 = 129.2p
3876/129.2 = 129.2p/129.2
30 = p
p = 30
Use the value of p to find x
x = VY+YK
x = p+68
x = 30+68
x = 98
Answer: 98 mm
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Problem 2)
The triangles are similar, so we can form a proportion of the corresponding sides
MN/MA = MP/MB
67.2/(67.2-32) = 81.9/x
67.2/35.2 = 81.9/x
67.2*x = 35.2*81.9
67.2*x = 2882.88
67.2*x/67.2 = 2882.88/67.2
x = 42.9
Answer: 42.9 meters
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Problem 3)
Similar to problem 1, we use the same theorem
EG/GD = EH/HD
44.8/(x+4) = 56/35
44.8*35 = 56(x+4)
1568 = 56x+224
1568-224 = 56x+224-224
1344 = 56x
56x = 1344
56x/56 = 1344/56
x = 24
Answer: 24 millimeters
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Problem 4)
We'll use a similar idea used in problem 2
9/(9+72) = (3x-20)/((3x-20)+56)
9/81 = (3x-20)/(3x+36)
1/9 = (3x-20)/(3x+36)
1*(3x+36) = 9(3x-20)
3x+36 = 27x-180
3x+36+180 = 27x-180+180
3x+216 = 27x
3x+216-3x = 27x-3x
216 = 24x
24x = 216
24x/24 = 216/24
x = 9
Answer: 9 centimeters
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Problem 5)
Area of triangle
A = 0.5*b*h
Plug in A = 210 and h = 35; solve for b
A = 0.5*b*h
210 = 0.5*b*35
210 = 0.5*35*b
210 = 17.5*b
12 = b
b = 12
Now use the base and height to find the hypotenuse
a^2 + b^2 = c^2
h^2 + b^2 = c^2
35^2 + 12^2 = c^2
1369 = c^2
c^2 = 1369
sqrt(c^2) = sqrt(1369)
c = 37
Answer: 37 inches
