SOLUTION: The diagonal of a rectangle field is y^2+16cm and one side is 8y cm. For what value of y the perimeter of the rectangle will be 62 cm I. 3cm II.5cm III.7cm I II I AND II II AN

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: The diagonal of a rectangle field is y^2+16cm and one side is 8y cm. For what value of y the perimeter of the rectangle will be 62 cm I. 3cm II.5cm III.7cm I II I AND II II AN      Log On


   

Question 1133678: The diagonal of a rectangle field is y^2+16cm and one side is 8y cm. For what value of y the perimeter of the rectangle will be 62 cm
I. 3cm II.5cm III.7cm
I
II
I AND II
II AND III
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

The diagonal of a rectangle field is y%5E2%2B16cm and one side is 8y cm. For what value of y the perimeter of the rectangle will be 62+cm
To find the width, multiply the length that you have been given by+2, and subtract the result from the perimeter. You now have the total length for the remaining 2 sides. This number divided by 2 is the width.
2W=+62-2L
2W=+62-2%2A8y
W=+62%2F2-2%2A8y%2F2
W=+31-8y
diagonal cuts rectangle into two right triangles, so if
d=y%5E2%2B16, L=8y, and W=+31-8y we have
d%5E2=L%5E2%2BW%5E2+
d=sqrt%28L%5E2%2BW%5E2%29
y%5E2%2B16=sqrt%28%288y%29%5E2%2B%2831-8y%29%5E2%29
y%5E2%2B16=sqrt%2864y%5E2%2B64y%5E2+-+496+y+%2B+961%29
y%5E2%2B16=sqrt%28128y%5E2+-+496+y+%2B+961%29
%28y%5E2%2B16%29%5E2=%28sqrt%28128y%5E2+-+496+y+%2B+961%29%29%5E2
y%5E4+%2B+32+y%5E2+%2B+256=+128y%5E2+-+496+y+%2B+961+
128y%5E2-y%5E4++-+496+y-+32+y%5E2%2B+961+-256=0
-y%5E4+%2B+96+y%5E2+-+496+y+%2B+705+=+0
-%28y+-+5%29+%28y+-+3%29+%28y+%28y+%2B+8%29+-+47%29+=+0

if -%28y+-+5%29=0+=>-y%2B5=0=>5=y....one solution to use
if %28y+-+3%29=0=>y=3....another solution to use
%28y+%28y+%2B+8%29+-+47%29+=+0=>y%5E2+%2B+8y+-+47=+0=>y+=+-4+-+3+sqrt%287%29 , y+=+3+sqrt%287%29+-+4.....disregard these solutions
go with:
=>y=5
L=8%2A5
L=40cm
and
W=+31-8%2A5
W=+31-40
W=+-9cm=> disregard negative solution

go with:
=>y=3
L=8%2A3
L=24cm
and
W=+31-8%2A3
W=+31-24
W=+7cm
check the perimeter:
2%2A24cm%2B2%2A7cm=62cm
48cm%2B14cm=62cm
62cm=62cm
diagonal:
d=y%5E2%2B16
d=3%5E2%2B16
d=25
d=sqrt%28L%5E2%2BW%5E2%29
25=sqrt%2824%5E2%2B7%5E2%29
25=sqrt%28625%29
25=25

so, your answer is: I. 3cm

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Solving the problem formally using algebra gets quite messy, as you can see in the solution from the other tutor. And it involves finding the roots of a 4th degree polynomial, which involves some trial and error with synthetic division.

The way the problem is presented, with answer choices, it makes FAR more sense simply to see which answers work. y = 3, 5, and 7 all give right triangles that are Pythagorean triples; but only y = 3 gives a perimeter of 62 for the rectangle.