SOLUTION: Find the area of a right triangle which has a perimeter of length 16 meters and a hypotenuse with a length of 7 meters. a) sqrt (56)m^2 b) 8m2 c) 10m2 d) 12m2 e) 16m2

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Question 251322: Find the area of a right triangle which has a perimeter of length 16 meters and a hypotenuse with a length of 7 meters.
a) sqrt (56)m^2 b) 8m2 c) 10m2 d) 12m2 e) 16m2
Found 2 solutions by solver91311, edjones:
Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

If the hypotenuse is 7 and the perimeter is 16, then the sum of the measures of the other two sides must be 16 - 7 = 9. Therefore, letting represent the measure of one leg and represent the measure of the other leg, we can say:

which is to say:

But since we know that the hypotenuse is 7, we can also say:

Substituting:

Simplifying (verification of this step is left as an exercise for the student):

This quadratic does not factor, so using the quadratic formula:

.

Since

We can say one leg of the triangle is:

And the other is:

Since the area of a triangle is given by:

And since the legs of a right triangle are perpendicular so either can be considered the base and the other the height, all you need to do now is multiply the two roots of the quadratic we just solved by each other and then divide by 2. Hint: The two roots are a conjugate pair -- think difference of two squares.

John

Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website!
c=7
16-7=9=a+b
b=9-a
a+(9-a)+7=16
a^2+b^2=c^2
a^2+(9-a)^2=c^2
a^2+81-18a+a^2=49
2a^2-18a+32=0
a^2-9a+16=0
a^2-9a =-16
a^2-9a+(81/4)=(81/4)-16 completing the square.
(x-(9/2))^2=17/4
x-(9/2)=+-sqrt(17)/2
x=(9+sqrt17)/2, x=(9-sqrt(17))/2
((9+sqrt17)/2) * ((9-sqrt(17))/2) * (1/2)
=8m^2
.
Ed