SOLUTION: The length of a rectangle is 5 cm more than 3 times its width. The area of the rectangle is 1428 square cm. Find the length and width of the rectangle.

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Question 1185996: The length of a rectangle is 5 cm more than 3 times its width. The area of the rectangle is 1428 square cm. Find the length and width of the rectangle.

Found 3 solutions by ikleyn, MathLover1, greenestamps:
Answer by ikleyn(52784) About Me  (Show Source):
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.
The length of a rectangle is 5 cm more than 3 times its width.
The area of the rectangle is 1428 square cm. Find the length and width of the rectangle.
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width = x

length = 3x + 5


Area equation


    x*(3x + 5) = 1428.


Simplify and solve


    3x^2 + 5x - 1428 = 0

    x%5B1%2C2%5D = %28-5+%2B-+sqrt%28%28-5%29%5E2+%2B+4%2A3%2A1428%29%29%2F%282%2A3%29 = %28-5+%2B-+131%29%2F6.


Use the positive root, only:  x = %28-5+%2B+131%29%2F6 = 21.


ANSWER.  The width is 21 cm;  The length is 3*21+5 = 68 cm.


CHECK.  21*68 = 1428 cm^2.    ! Correct !

Solved.



Answer by MathLover1(20850) About Me  (Show Source):
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The length of a rectangle is 5 cm more than 3 times its width:
L=3W%2B5.......eq.1
The area of the rectangle is 1428 square cm.
A=L%2AW
1428=L%2AW........substitute L from eq.1
1428=%283W%2B5%29%2AW
1428=3W%5E2%2B5W
3W%5E2%2B5W-1428=0....factor
3W%5E2-63W%2B68W-1428=0
%283W%5E2-63W+%29%2B%2868W-1428%29=0
3W%28W-21+%29%2B68%28W-21%29=0
%28W+-+21%29+%283W+%2B+68%29+=+0

=> %28W+-+21%29+++=+0 =>W+=+21cm
go to L=3W%2B5.......eq.1, substitute W
L=3%2A21%2B5
L=68cm



Answer by greenestamps(13200) About Me  (Show Source):
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The responses from the other two tutors both use formal algebraic methods to solve the problem, which is probably what the student was supposed to do.

However, solving the problem with formal algebra requires solving a quadratic equation using either difficult factoring or the quadratic formula.

So, if formal algebra is not required, the problem is solved much more quickly with a bit of estimation and mental arithmetic.

Note also that, even if a formal algebraic solution is required for the assignment, solving the problem with logical reasoning and mental arithmetic gives you valuable brain exercise.

It might go something like this.

For a very rough estimate, since the length is "a bit more than 3 times the width", let the width be x and the length be "a bit more than 3x", so that the area (length times width) is 1482.

1482 is a bit more than 3x^2
1482/3 = 496 is a bit more than x^2

The square root of 496 is a bit more than 22; so the width should be either 22 or perhaps 21.

Then mental arithmetic determines that 21 is a factor of 1428 while 22 is not, so the width is probably 21.

Dividing 1428 by 21 gives 68; and 68 is 5 more than 3 times 21. So we have the answer.

ANSWERS: width 21, length 68