I need to find the length and width of a rectangle. The area is 24in squared. The diagonal measure exactly?

Question

square root 73. How do I set this problem up?

a^2 + b^2 = 73 is what I have so far.

Answer 1

Let L be the length, and W be the width:

This is what you know:
L*W = 24
L^2 + W^2 = 73

That's two equations in 2 unknowns, solve it using substitution:
L = 24/W
substitute this into the second equation:
(24/W)^2 + W^2 = 73
multiply both sides by W^2:
24^2 + W^4 = 73*W^2
rearrange:
W^4 -73W^2 +24^2 = 0
this looks funny but is a quadratic in W^2.
you can solve it for the postive solution and then take the
square root to get W:

You can use the quadratic formula, or try to factor:
W^2 = 73/2 + sqrt(73^ - 4*24^2)/2
W^2 = 73/2 + sqrt(3025)/2
W^2 = 73/2 + 55/2
W^2 = 64
so
W = 8.
L = 24/W = 3
Check:

8^2 + 3^2 = 64 + 9 = 73

so, you're good!

Answer 2

A^2 + B^2 = 73 is a good start.

You also can determine that A*B =24 in squared.

Solve for either A or B and use substitution into your first equation.

Answer 3

Let sides be x and y
xy = 24
y = 24 / x
x² + y² = 73
x² + 576/x² = 73
x^(4) + 576 = 73 x²
x^(4) - 73 x² + 576 = 0
Let x² = c
c² - 73 c + 576 = 0
c = [73 ± √(73² - 2304) ] / 2
c = [73 ± 55 ] / 2
c = 64 , c = 9
x = 8 , x = 3
Sides are 8 ins and 3 ins

Answer 4

Sounds like you are making the problem more difficult than it really is. Since the diagonal is exact then you have 2 right triangles each with an area of 12". The area of a triangle is 1/2bh. By this reasoning you have 2 triangles with a base of 6" and height of 4". Therefore the rectangle is 6" long and 4" wide. Hope this helped.

Answer 5

a^2 + b^2 = 73
Also ab =24
b=24/a
Substituting
a^2 + (24/a)^2 =73
a^4 -73 a^2 +24^2 =0

a^4 -73 a^2 +576=0
(a^2 -9)(a^2-64)=0
a=3 or a =8
we can ignore negative roots


So a =3, b=8 or vice versa

Answer 6

a*b=24
ae2 + be2 = 73

o wait, 8X3=24, and 64+9=73 ;)

Answer 7

since you know area is 24in sq, then a*b = 24
solve simultaneously.

Answer 8

weird problem