SOLUTION: A circular piece of sheet metal has a diameter of 20 in. The edges are to be cut off to form a rectangle of area 150 in2 (see the figure). What are the dimensions of the rectangle?

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Question 1201846: A circular piece of sheet metal has a diameter of 20 in. The edges are to be cut off to form a rectangle of area 150 in2 (see the figure). What are the dimensions of the rectangle? (Round your answers to two decimal places.)

Found 3 solutions by math_tutor2020, greenestamps, ikleyn:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: 8.23 inches by 18.23 inches
Each value is approximate.


Explanation:

Here's what the diagram probably looks like

x = horizontal width of the rectangle
y = vertical height of the rectangle
These are positive real numbers
x > 0 and y > 0

area of the rectangle = x*y = 150
Solve for y to get
y = 150/x

Let's focus on triangle ADC.
This is a right triangle.
To see a proof why angle ADC = 90 degrees, search out "Thale's Theorem".

Because ADC is a right triangle, we can use the pythagorean theorem to say:
a%5E2+%2B+b%5E2+=+c%5E2

%28DC%29%5E2+%2B+%28AD%29%5E2+=+%28AC%29%5E2

x%5E2+%2B+y%5E2+=+20%5E2

x%5E2+%2B+%28150%2Fx%29%5E2+=+400

Let's solve for x.
x%5E2+%2B+%28150%2Fx%29%5E2+=+400

x%5E2+%2B+22500%2F%28x%5E2%29+=+400

x%5E4+%2B+22500+=+400x%5E2

x%5E4+-+400x%5E2+%2B+22500+=+0

As a detour, let w+=+x%5E2
So
w%5E2+=+%28x%5E2%29%5E2+=+x%5E4
and
x%5E4+-+400x%5E2+%2B+22500+=+0
w%5E2+-+400w+%2B+22500+=+0

Use the quadratic formula to solve for w.
Plugging in a = 1, b = -400, c = 22500
w+=+%28-b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F%282a%29

w+=+%28-%28-400%29+%2B-+sqrt%28%28-400%29%5E2+-+4%281%29%2822500%29%29%29%2F%282%281%29%29

w+=+%28400+%2B-+sqrt%28160000-90000%29%29%2F%282%281%29%29

w+=+%28400+%2B-+sqrt%2870000%29%29%2F%282%29

w+=+%28400+%2B-+100%2Asqrt%287%29%29%2F%282%29

w+=+%282%28200+%2B-+50%2Asqrt%287%29%29%29%2F%282%29

w+=+200+%2B-+50%2Asqrt%287%29

w+=+200+%2B+50%2Asqrt%287%29+ or +w+=+200+-+50%2Asqrt%287%29

w+=+332.287566+ (approximate) or +w+=+67.712434 (approximate)


If w = 332.287566, then,
w+=+x%5E2
x%5E2+=+332.287566
x+=+sqrt%28332.287566%29 or x+=+-sqrt%28332.287566%29
x+=+18.228757 or x+=+-18.228757
We'll ignore the negative x value because a negative width makes no sense.

Now use that x value to find y
y+=+150%2Fx
y+=+150%2F18.228757
y+=+8.228756

The rectangle has dimensions of:
x = 18.228757
y = 8.228756


If w = 67.712434, then,
w+=+x%5E2
x%5E2+=+w
x%5E2+=+67.712434
x+=+sqrt%2867.712434%29
x+=+8.228757
Then plugging that into y = 150/x will lead to y = 18.228757
We arrive at the same dimensions as before, but now the x and y values swapped places.
Turns out the order doesn't matter when listing the length and width of a rectangle.

Therefore, the only possible dimensions of the rectangle are roughly 8.23 inches by 18.23 inches when rounding those previous figures to two decimal places.

As a check
x*y = 18.228757 * 8.228756 = 149.999994
which is really close to 150
There's likely some rounding error from a previous step.
And also, we can use the pythagorean theorem to confirm the answer as well
a^2+b^2 = c^2
x^2 + y^2 = 20^2
(18.228757)^2 + (8.228756)^2 = 20^2
400.000007072584 = 400
which isn't too far off either

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Let the dimensions of the rectangle be x and y. The diagonal of the rectangle is a diameter of the circle, so

[1] xy=150 [the area of the rectangle is 150 sq in]
[2] sqrt%28x%5E2%2By%5E2%29=20 [the diameter of the circle is 20 in]

A general problem-solving hint...

Whenever you see xy and x%5E2%2By%5E2 in a problem, look at %28x%2By%29%5E2=x%5E2%2B2xy%2By%5E2.

In this problem....

%28x%2By%29%5E2=x%5E2%2B2xy%2By%5E2=%28x%5E2%2By%5E2%29%2B2%28xy%29=400%2B2%28150%29=700
x%2By=sqrt%28700%29=10%2Asqrt%287%29

Solve that equation for y and substitute in [1]:

y=10%2Asqrt%287%29-x
x%2810%2Asqrt%287%29-x%29=150
x%5E2-%2810%2Asqrt%287%29%29x%2B150=0

That won't factor by usual factoring methods; but the quadratic formula works nicely.

x=%2810%2Asqrt%287%29%2B-sqrt%28700-600%29%29%2F2

x=5%2Asqrt%287%29%2B5 or x=5%2Asqrt%287%29-5

ANSWER: The dimensions of the rectangle are 5%2Asqrt%287%29%2B5 and 5%2Asqrt%287%29-5

(You can evaluate those expressions and do the prescribed rounding....)


Answer by ikleyn(52847) About Me  (Show Source):
You can put this solution on YOUR website!
.
A circular piece of sheet metal has a diameter of 20 in.
The edges are to be cut off to form a rectangle of area 150 in2 (see the figure).
What are the dimensions of the rectangle? (Round your answers to two decimal places.)
~~~~~~~~~~~~~~~~~~


            There is more simple solution to this problem (and, therefore, more preferable).


Let x is the longer side of the rectangle and y is its shorter side.


From the description, you have these two equations

    xy = 150           (1)  (the area)

    x^2 + y^ = 20^2    (2)  (the diagonal of the rectangle,
                            which is the diameter of the circle)


Multiply equation (1) by 2 (both sides) and add to equation (2).  You will get

    x^2 + 2xy + z^2 = 700    (3)


Multiply equation (1) by 2 (both sides) and subtract from equation (2).  You will get

    x^2 - 2xy + z^2 = 100    (4)



Equations (3) and (4) are equivalent to this system of equations

    (x+y)^2 = 700
    (x-y)^2 = 100


Taking square roots from both sides of these equations, you arrive to simple system of linear equations

    x + y = sqrt%28700%29,
    x - y = 10.

or    

    x + y = 10%2Asqrt%287%29,     (5)
    x - y = 10.         (6)


Add these equations to eliminate y.  You will get

    2x = 10%2Asqrt%287%29+%2B+10,  which implies  x = 5%2Asqrt%2810%29%2B5.


Subtract equation (6) from equation (5) to eliminate x.  You will get

    2y = 10%2Asqrt%287%29+-+10,  which implies  y = 5%2Asqrt%2810%29-5.


ANSWER.  x = 5%2Asqrt%2810%29%2B5,  y = 5%2Asqrt%2810%29-5.

Solved.