perimeter and area help!

C

cpluva

New member
Joined
Oct 23, 2007
Messages
6
I have two questions.

The perimeter of a rectangle is 72 cm. The base is 3 times the height. What are the dimensions of the rectangle?

The perimeter of a rectangle is 80x. The base is 7 times the height. In terms of x, what are the dimensions of the rectangle?

If I know the dimensions, then I could figure out the area. Thanks if you can help!
 
The perimeter of a rectangle is 72 cm. The base is 3 times the height. What are the dimensions of the rectangle?
Click to expand...

That's the idea. To find the dimensions with the given info. Let x=base and y=height.

The perimeter is 72 cm. So, we have 2x+2y=72

We are told the base is 3 times the height. So, we have x=3y.

Now, sub the second into the first and solve for y. Then you can easily tell what x is. See?. Do the same for the other problem.
 
cpluva said:
I have two questions.

The perimeter of a rectangle is 72 cm. The base is 3 times the height. What are the dimensions of the rectangle?

The perimeter of a rectangle is 80x. The base is 7 times the height. In terms of x, what are the dimensions of the rectangle?

If I know the dimensions, then I could figure out the area. Thanks if you can help!
Click to expand...

Please show us your work and exactly where you are stuck - so that we know where to begin to help you.

#1

Base of rectangle = B

Height of rectangle = H

The base is 3 times the height.

B = 3 * H..................................................................(1)

The perimeter of a rectangle is 72 cm

How would you express the "perimeter of the rectangle" - in terms of its base and height? Equate that to 72. Then use (1) in there to eliminate one of the variables by substitution. Then solve for the remaining variable.

Then use (1) again to solve for the remaining variable.

Finally check your answer against the given two conditions.

for #2

start with naming variables and writing equations from the given conditions. Then show your work and tell us exactly where you are stuck (that is if you are).
 
Here's what I've done, step-by-step:
2b+2h=72, 2(3h)+2b=72
6h+2h=72 (the 6h is squared, idk how to fix that)
6h+2h-72=0
After that is where I'm stuck, but I just found 9 and 27 just by guessing numbers.

Second problem:
2b+2h=80x
2(7h)+2h=80x
14h+2h=80x
14h+2h-80x=0
 
cpluva said:
Here's what I've done, step-by-step:
2b+2h=72, 2(3h)+2b=72
6h+2h=72 (the 6h is squared, idk how to fix that)
6h+2h-72=0

8h - 72 = 0

8h = 72

h = 72/8 = 9

Now continue....
After that is where I'm stuck, but I just found 9 and 27 just by guessing numbers.

Second problem:
2b+2h=80x
2(7h)+2h=80x
14h+2h=80x
14h+2h-80x=0

16h = 80x

h = 5x

Now continue.....
Click to expand...
 
You must log in or register to reply here.
Top