Word Problem

1. Jul 12, 2005

powp

Hello All

I have this word problem that I am having problems with

Here it is

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 in^2. What are dimensions of the rectangle??

Here is what I have figured out. Not sure if it of any use

Formula of Circle is

10^2 = y ^ 2 + x ^ 2

Area of rectangle

160 = LW

Thanks

2. Jul 12, 2005

saltydog

So, as I see it you have a circle of radius 10 and want to draw a rectangle in it with a total area of 160 right? Look at the plot below (drawn to scale) and where I have theta. Can you figure out what the length and height has to be using some trig expressions?

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3. Jul 12, 2005

powp

Don't think I have enough info to use trig expessions. All I know for that triangle is the hypotenuse is 10 in and is a right angle triangle.

How can I determine the remaining info?
if I use sin I need opposite side?

4. Jul 12, 2005

plucker_08

IT’S ALL GEOMETRY…

X^2 + Y^2 = 20^2 = 400 ----------------------DIAGONAL OF THE RECTANGLE

XY = 160
2XY = 320

X^2 + 2XY + Y^2 = 400 + 320 = 720
(X+Y) ^2= 720

X+Y = 12 √5

X^2 – 2XY + Y^2 = 400 – 320 = 80
(X-Y)^ 2 = 80

X-Y = 4 √5

2 EQS., 2 UNKNOWNS

2X = 4 √5 + 12 √5 = 16 √5

X = 8 √5
Y = 4√5

:surprised

5. Jul 12, 2005

MaxwellPhill

Nice explanation plucker.

6. Jul 12, 2005

powp

thanks why did you multiply xy = 160 by 2 ??

7. Jul 12, 2005

plucker_08

adding 2xy to x^2 + y^2 will yield to an expression x^2 + 2xy + y^2 which is the square of the sum of the sides of the rectangle...(x + y)

subtracting 2xy to x^2 + y^2 will yield to an expression x^2 - 2xy + y^2 which is the square of the difference of the sides of the rectangle...(x - y)

you will have 2 eqs and 2 unknows...

8. Jul 13, 2005

powp

how can you just add values to the equation?? How do you know when to do this?

9. Jul 13, 2005

saltydog

Hey Powp. Sorry for not get getting back with you last night but looks like these guys helped you ok. Yea, I didnt' see the obvious algegra:

$$x^2+y^2=400$$

$$xy=160$$

Solving for x or y in the second equation and substituting into the first leads to the same equation for both x and y:

$$r^4-400r^2+160^2=0$$

Now, using the quadratic formula, just figure out which of the 4 roots make sense.

I drew a pretty graph though.

Last edited: Jul 13, 2005