SOLUTION: The height of a door is 1.3 feet longer than its​ width, and its front area is 1579.6 square feet. Find the width and height of the door.

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Question 1129232: The height of a door is 1.3 feet longer than its​ width, and its front area is 1579.6 square feet. Find the width and height of the door.
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

The height of a door is h and the width w
if h is+1.3 feet longer than its width, we have
h=w%2B1.3

and, if its front area is 1579.6 square feet, we have

h%2Aw=1579.6 ...substitute h

%28w%2B1.3%29%2Aw=1579.6

w%5E2%2B1.3w=1579.6

w%5E2%2B1.3w-1579.6=0...... use quadratic formula


w+=+%28-1.3+%2B-+sqrt%28+1.3%5E2-4%2A1%2A%28-1579.6%29+%29%29%2F%282%2A1%29+

w+=+%28-1.3+%2B-+sqrt%28+1.69%2B6318.4%29+%29%29%2F2+

w+=+%28-1.3+%2B-+sqrt%28+6320.09%29+%29%2F2+

w+=+%28-1.3+%2B-+79.49899370432308%29%2F2+

we will need only positive solution

w+=+%28-1.3+%2B+79.49899370432308%29%2F2+

w+=+%2878.19899370432308%29%29%2F2+
w+=39.09949685216154->the width of the door

now find h
h=w%2B1.3
h=40.39949685216154->the height of the door

these above are exact solutions

if we round to one decimal place we will get:h=40.4 and w+=39.1
the area will be 40.4%2A39.1=1579.64 which is basicaly same; so, you can round it


Answer by ikleyn(52756) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let "u" be the unknown value exactly half way between the length L and the width W.


Then  u = L - 1.3%2F2 = L - 0.65 = W + 1.3%2F2 = W + 0.65,   or

      L = u + 0.65;  W = u - 0.65.


Then the area = 1579.6 = LW = (u+0.65)*(u-0.65) = u^2 - 0.65^2,  which implies


    u^2 = 1579.6 + 0.65^2 = 1580.0225.


======================>  u = sqrt%281580.0225%29 = 39.75 (approximately, rounded to two decimal places after the decimal point).


Then L = 39.75 + 0.65 = 40.4 ft;  W = 39.75 - 0.65 = 39.1 ft.


Check.  40.4 * 39.1 = 1579.6 square feet.    ! Correct !

Answer.   The dimensions are  40.4 ft  (length)  and  39.1 ft  (width).

Solved.

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There are several different ways to solve this problem / (such problems).

See the lesson
    - Three methods to find the dimensions of a rectangle when its area and the difference of two dimensions are given
in this site.

Of these three methods I presented here the most unexpected one.

It is the shortest solution method,  by the way,  and requires a minimum of calculations.

No quadratic equation should be solved. Only the square root should be calculated once.