Jordan K. answered • 09/07/15
Tutor
4.9
(79)
Nationally Certified Math Teacher (grades 6 through 12)
Hi Bobi,
Let's begin by assigning letters to represent our three unknowns:
L = (length of rectangle in cm)
W = (width of rectangle in cm)
D = (diagonal of rectangle in cm)
Now let's represent width (W) of rectangle in terms of its length (L), according the relationship given in the problem:
W = 2L - 6 (width is 6 less than twice the length)
Now let's recall our area formula for a rectangle:
Area = length x width
Let's plug in 23 as our given area and our width (W) expression in terms of length (L) and solve for the length (L):
(L)(2L - 6) = 23
2L2 - 6L = 23
L2 - 3L = 23/2
L2 - 3L + (-3/2)2 = 23/2 + (-3/2)2
L2 - 3L + 9/4 = 23/2 + 9/4
L2 - 3L + 9/4 = 46/4 + 9/4
(L - 3/2)2 = 55/4
L - 3/2 = +/- √(55/4)
L = 3/2 +/- √(55/2)
L = 1.5 +/- 7.4/2
L = 1.5 +/- 3.7
L = 1.5 - 3.7
L = -2.2 (reject negative measurement)
L = 1.5 + 3.7
L = 5.2 (length of rectangle in cm)
We can now calculate width (W) of rectangle by plugging in our answer for length (L) in width (W) expression in terms of length (L):
W = 2L - 6
W = 2(5.2) - 6
W = 10.4 - 6
W = 4.4 (width of rectangle in cm)
Finally, we can calculate the diagonal (D) of the rectangle by applying the Pythagorean theorem (using our values for L and W), since the diagonal of a rectangle divides it into two congruent right triangles:
D2 = L2 + W2
D2 = (5.2)2 + (4.4)2
D2 = 27.04 + 19.36
D2 = 46.4
D = √46.4
D = 6.8 (diagonal of rectangle in cm)
The key to solving this problem was to express one unknown in terms of another unknown as per the relationship given in the problem. This allowed us to to solve for the first unknown and then use that value to calculate the second unknown and, finally, using both values to calculate the third unknown.
Thanks for submitting this problem and glad to help.
God bless, Jordan.
