Page 174 - ArithBook5thEd ~ BCC

Solution. In the formulae, the legs are denoted a and b.Suppose a is the unkown leg. Putting c =169
and b = 119 in the first formula

√
! !
2
2
2
2
a = c − b = 169 − 119 = 14400 = 120.
The other leg is 120 feet.
Example 218. Find the length of the hypotenuse of a right triangle whose legs have lengths 5 and 12
feet.

Solution. We put a =5 and b = 12 in the third formula.

√
. √
!
2
2
2
2
c = a + b = (5) +(12) = 25 + 144 = 169 = 13 feet.
Example 219. The semi-perimeter of a triangle with side lengths a, b and c is given by the formula
1
s = (a + b + c).
2
Find the semi-perimeter of a triangle with side lengths a =6 ft 8 in, b =9 ft 4 in, and c =11 ft 10 in.

Solution. We convert all measurements to inches, using the fact that 1 ft = 12 in. Thus

a =80 in, b =112 in, and c =142 in.

The semi-perimeter is

1 1
s = (a + b + c)= (80 + 112 + 142)
2 2
=167 in
=13 ft 11 in.

Example 220. The child’s dosage for a medicine is given by the formula
t
C = · A,
t +12

where C is the child’s dosage, A is the adult dosage, and t is the child’s age in years. Find the dosage
for a four-year old child if the adult dosage is 48 mg (milligrams).

Solution. Substituting t =4 and A =48,

4 1 48
C = · 48 = · =12.
4+ 12 4 1
The child’s dosage is 12 mg.

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