SOLUTION: the diagonal of a rectangle is 15cm, and the perimeter is 38cm. what is the area?

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Question 1150827: the diagonal of a rectangle is 15cm, and the perimeter is 38cm. what is the area?
Found 4 solutions by Alan3354, MathLover1, ikleyn, MathTherapy:
Answer by Alan3354(69443) About Me  (Show Source):
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the diagonal of a rectangle is 15cm, and the perimeter is 38cm. what is the area?
--------
Side lengths = a & b
a^2 + b^2 = 15^2
a + b = 38/2 = 19
=================
(a+b)^2 = a^2 + 2ab + b^2 = 361
..........a^2 + b^2 = 225
------------------------------- Subtract
2ab = 136
a*b = 68
Sub for b
a*(19 - a) = 68
a^2 - 19a + 68 = 0
Solve for a
etc

Answer by MathLover1(20849) About Me  (Show Source):
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the diagonal of a rectangle is d=15cm,
recall:
d%5E2=a%5E2%2Bb%5E2 where a and b are sides of the rectangle
so, 15%5E2=a%5E2%2Bb%5E2
225=a%5E2%2Bb%5E2..........eq.1

and, if the perimeter is P=38, means 2%28a%2Bb%29=38
a%2Bb=38%2F2
a%2Bb=19.......solve for a
a=19-b..........eq.2
go to
225=a%5E2%2Bb%5E2..........eq.1...substitute a
225=%2819-b%29%5E2%2Bb%5E2.........solve for b
225=19%5E2-38b%2Bb%5E2%2Bb%5E2
225=361-38b%2B2b%5E2...swap the sides
2b%5E2-38b%2B361=225
2b%5E2-38b%2B361-225=0
2b%5E2-38b%2B136=0....both sides divide by 2
b%5E2-19b%2B68=0.......use quadratic formula
b+=+%28-%28-19%29+%2B-+sqrt%28+%28-19%29%5E2-4%2A1%2A68+%29%29%2F%282%2A1%29+
b+=+%2819+%2B-+sqrt%28+361-272+%29%29%2F2+
b+=+%2819+%2B-+sqrt%28+89+%29%29%2F2+.........since we are looking for side length, we need only positive root
b+=+%2819+%2Bsqrt%28+89+%29%29%2F2+-> exact solution
b+=+14.216991-> approximately

now find a

a=19-b..........eq.2
a=19-%2819+%2Bsqrt%28+89+%29%29%2F2
a=%2838-19+-sqrt%28+89+%29%29%2F2
a=%2819+-sqrt%28+89+%29%29%2F2-> exact solution
a=4.78301-> approximately

then, the area will be:
A=a%2Ab......using exact solutions we have
A=%28%2819+-sqrt%28+89+%29%29%2F2%29%2A%28%2819+%2Bsqrt%28+89+%29%29%2F2%29
A=68cm%5E2
or, using approximate solutions
A=a%2Ab
A=4.78301%2A14.216991
A=68.00001012291.....rounded it is
A=68cm%5E2

Answer by ikleyn(52781) About Me  (Show Source):
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.
the diagonal of a rectangle is 15cm, and the perimeter is 38cm. what is the area?
~~~~~~~~~~~~~~~~~~~~ 

Let "a" and "b" be the lengths of two adjacent sides of the rectangle.


Then  

    a + b = 19  =  38%2F2      (1) 

    a%5E2 + b%5E2 = 15%5E2 = 225   (2)


Square both sides of the equation (1);  keep equation (2) as is.


    a%5E2+%2B+2ab+%2B+b%5E2 = 19%5E2 = 361   (1')

    a%5E2  +  b%5E2 = 225         (2')


Subtract equation (2')  from equation (1')

     2ab = 361 - 225 = 136.


Now,  2ab  is two times the area of the rectangle;  hence, the area of the rectangle is  136%2F2 = 68 square centimeters.


ANSWER.  The area of the rectangle is 68 square centimeters.

Solved.

------------------

The lesson to learn: 

    In order to solve this problem, you do not need solve quadratic equation.


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
the diagonal of a rectangle is 15cm, and the perimeter is 38cm. what is the area?
I agree with @IKLEYN. Why spend a "day" solving a problem when it can take just 2 minutes?  And, almost no-one who 

I've ever dealt with, wants to use the quadratic equation or "Completing the Square" to solve a quadratic equation.



Let length and width be L and W, respectively

With perimeter being 38 cm, we get: 2(L + W) = 2(19)____L + W = 19 ---- eq (i)

With diagonal being 15 cm, we also get: matrix%281%2C3%2C+L%5E2+%2B+W%5E2%2C+%22=%22%2C+15%5E2%29 ----- eq (ii)

 ------ Squaring eq (i) ------ eq (iii)

matrix%281%2C3%2C+15%5E2%2C+%22=%22%2C+19%5E2+-+2LW%29 ------- Substituting matrix%281%2C3%2C+15%5E2%2C+for%2C+L%5E2+%2B+W%5E2%29

matrix%281%2C3%2C+15%5E2+-+19%5E2%2C+%22=%22%2C+-+2LW%29

We then get: