SOLUTION: If the diagonal of a rectangle has length 7 feet and the rectangle’s area is 16 square feet, what is the perimeter of the rectangle in feet? a 10 feet b 15 feet c 18 feet d 2

Algebra ->  Rectangles -> SOLUTION: If the diagonal of a rectangle has length 7 feet and the rectangle’s area is 16 square feet, what is the perimeter of the rectangle in feet? a 10 feet b 15 feet c 18 feet d 2      Log On


   

Question 252288: If the diagonal of a rectangle has length 7 feet and the rectangle’s area is 16 square feet, what is the perimeter of the rectangle in feet?
a 10 feet b 15 feet c 18 feet d 23 feet e 26 feet
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
FOR THE AREA OF THE RECTANGLE:
XY=16 OR X=16/Y
FOR THE TRIANGLE:
X^2+Y^2=7^2
(16/Y)^2+Y^2=49
(256/Y^2)+Y^2=49
(256+Y^4)/Y^2=49
256+Y^4=49Y^2
Y^4-49Y^2+256=0
Y%5E2+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
Y^2=(49+-SQRT[-49^2-4*1*256])/2*1
Y^2=(49+-SQRT[2,401-1.024])/2
Y^2=(49+-SQRT1,377)/2
Y^2=(49+-37.108)/2
Y^2=(49+37.108)/2
Y^2=86.108/2
Y^2=43.05
Y=SQRT43.05
Y=6.56 ANS. THEN X=2.44
Y^2=(49-37.108)/2
Y^2=11.892/2
Y^2=5.95
Y=SQRT5.95
Y=2.44 ANS. THEN X=6.56
PROOFS:
6.56*2.44=16
16~16
6.56^2+2.44^2=49
43.03+5.954=49
49~49
tHUS THE PERIMETERIS:
2*6.56+2*2.44=13.12+4.88=18 FT. IS THE PERIMETER.
ANSWER C)