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• Dec 13th 2007, 09:56 AM
zazeem
Hi, i really need help figuring two problems out.

problem 1

the perimeter of a rectangular picture is 34cm. the diagonal is 13cm. find the dimensions of this picture.

how on earth do i figure this out?

problem 2

Find the distance (correct to three decimal places) from the point (-5,-2) to the line 3x+y=8. Hint: the distance from a point to a line is the perpendicular distance.

• Dec 13th 2007, 10:06 AM
topsquark
Quote:

Originally Posted by zazeem
Hi, i really need help figuring two problems out.

problem 1

the perimeter of a rectangular picture is 34cm. the diagonal is 13cm. find the dimensions of this picture.

how on earth do i figure this out?

Call the width of the rectangle x and the height y. Then we know that
$2x + 2y = 34$ (from the perimeter)
and
$x^2 + y^2 = 13^2 = 169$ (from the Pythagorean theorem)

Solve the top equation for y:
$y = 17 - x$

Now insert this value of y into the bottom equation:
$x^2 + (17 - x)^2 = 169$

Can you finish this from here?

-Dan
• Dec 13th 2007, 10:13 AM
zazeem
hmm i see how you got this now but i still cant seem to complete it, is there any way you could write it out for me so i can see then go back and i can do it out or try a different one? i would greatly appreciate it. thanks for helping
• Dec 13th 2007, 10:19 AM
topsquark
Quote:

Originally Posted by zazeem
hmm i see how you got this now but i still cant seem to complete it, is there any way you could write it out for me so i can see then go back and i can do it out or try a different one? i would greatly appreciate it. thanks for helping

Quote:

Originally Posted by topsquark
Call the width of the rectangle x and the height y. Then we know that
$2x + 2y = 34$ (from the perimeter)
and
$x^2 + y^2 = 13^2 = 169$ (from the Pythagorean theorem)

Solve the top equation for y:
$y = 17 - x$

Now insert this value of y into the bottom equation:
$x^2 + (17 - x)^2 = 169$

Can you finish this from here?

-Dan

To continue:
$x^2 + x^2 - 34x + 289 = 169$

$2x^2 - 34x + 120 = 0$

$x^2 - 17x + 60 = 0$

$(x - 12)(x - 5) = 0$

So x = 5 or x = 12.

Can you finish from here?

-Dan
• Dec 13th 2007, 10:36 AM
zazeem
it is already done isn't it? the dimensions would be x= 5 or x= 12? this problem is one of the hardest yet sorry for being such a noob lol
• Dec 13th 2007, 10:47 AM
Henderson
Quote:

Originally Posted by zazeem
problem 2

Find the distance (correct to three decimal places) from the point (-5,-2) to the line 3x+y=8.

The hint suggests that you'll need a perpendicular line through the point. Perpendicular to 3x + y is x - 3y. So:
$x - 3y = C$
$(-5) - 3(-2) = C$ (substituting in your point)
$1 = C$ (simplifying)
$x - 3y = 1$

This line goes from your point to your original line at a right angle. So, if you can find the intersection of these two lines, the distance would be the distance from your original point to this intersection.

Say the word if you need more past this.
• Dec 13th 2007, 10:48 AM
topsquark
Quote:

Originally Posted by zazeem
it is already done isn't it? the dimensions would be x= 5 or x= 12? this problem is one of the hardest yet sorry for being such a noob lol

It's virtually done. I've left you the (rather trivial) problem of finding y. (Hey, I had to leave you something to do! (Nod) )

-Dan
• Dec 13th 2007, 10:57 AM
zazeem
Quote:

Originally Posted by topsquark
It's virtually done. I've left you the (rather trivial) problem of finding y. (Hey, I had to leave you something to do! (Nod) )

-Dan

so wait how do i find y? can you finish it for me and then after i can try a similar problem set from my homework for understanding, it would be easiest that way. Thanks so much for the help!
• Dec 13th 2007, 10:58 AM
zazeem
Quote:

Originally Posted by Henderson
The hint suggests that you'll need a perpendicular line through the point. Perpendicular to 3x + y is x - 3y. So:
$x - 3y = C$
$(-5) - 3(-2) = C$ (substituting in your point)
$1 = C$ (simplifying)
$x - 3y = 1$

This line goes from your point to your original line at a right angle. So, if you can find the intersection of these two lines, the distance would be the distance from your original point to this intersection.

Say the word if you need more past this.

yes can you please complete the problem so i can do similar problems and use this as a reference example, i would appreciate it greatly. thanks!
• Dec 13th 2007, 11:05 AM
topsquark
Quote:

Originally Posted by zazeem
so wait how do i find y? can you finish it for me and then after i can try a similar problem set from my homework for understanding, it would be easiest that way. Thanks so much for the help!

Look back through my original post. You will see three equations you can use to calculate a value for y given an x value.

-Dan
• Dec 13th 2007, 11:12 AM
zazeem
should i get the same y value with either one of the x terms from before?
• Dec 13th 2007, 11:20 AM
zazeem
i got y = 12 or y = 5 after plugging the two x terms into y = 17 - x. its the same as the x?
• Dec 13th 2007, 11:22 AM
topsquark
Quote:

Originally Posted by zazeem
i got y = 12 or y = 5 after plugging the two x terms into y = 17 - x. its the same as the x?

Well, either the length is 5 cm and the height is 12 cm or the length is 12 cm and the height is 5 cm.

-Dan
• Dec 13th 2007, 11:27 AM
zazeem
o ok well since it is a rectangle y being the height must be the smaller number = 5, and x the width = 12

is this correct?
• Dec 13th 2007, 01:18 PM
topsquark
Quote:

Originally Posted by zazeem
o ok well since it is a rectangle y being the height must be the smaller number = 5, and x the width = 12

is this correct?

That just depends on the orientation of the rectangle. Either orientation will do to solve this problem, so both answers are correct. (If you want to call them separate answers given that the dimensions are the same.)

-Dan
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