Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
1. Section 7.1
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Exercise 64 Page 395

Practice makes perfect
a When drawing the diagram, we will assume that the swing can go up to 5 feet away from its horizontal starting position without hitting either the bush or the shed. The maximum length of the chain we will label x in our diagram.
b To find the length of x we can use either of the right triangles from the diagram in Part A. Notice that both of them are 30-60-90 triangles, which means the hypotenuse, x, is twice the length of the shorter leg. Also, the longer leg is sqrt(3) times the shorter leg.
We can write an equation relating the long and short leg.
Long leg=sqrt(3)(Short leg)
5=sqrt(3)(Short leg)
Solve for "Short leg"
5/sqrt(3)=Short leg
Short leg=5/sqrt(3)
Short leg=2.88675...
Short leg=2.89
If the short leg is 2.89 feet, the hypotenuse — which describes the chain's length — will be twice that length. Max Length of Chain: 2.89(2)≈ 5.77 The maximum length of the chain is 5.77 feet.