Find Rectangle Sides Given Perimeter & Area

[Government$]

Homework Statement

Find the lengths of the sides of rectangle if perimeter is 7,4 m and area is 3m2.

Homework Equations

S= 2a+2b and P=ab

The Attempt at a Solution

So i tried getting a form P --> a=P/b and then subsstituting in perimeter formula a with P/b ---> S=2P/b+2b but i am stuck here.

Any help would be appreciated. Thank you

 

[Mark44]

Homework Statement

Find the lengths of the sides of rectangle if perimeter is 7,4 m and area is 3m2.

Homework Equations

S= 2a+2b and P=ab

The Attempt at a Solution

So i tried getting a form P --> a=P/b and then subsstituting in perimeter formula a with P/b ---> S=2P/b+2b but i am stuck here.

Any help would be appreciated. Thank you

Start by substituting the given information about the perimeter and area into your two formulas.

 

[Government$]

Ok, but still i don't get anywhere:
7,4=a2+b2 and 3m2=ab

7,4=2a+2b---> 7,4=2(a+b)---> a+b=3.7

3=ab---> a=3/b

 

[Mark44]

Ok, but still i don't get anywhere:
7,4=a2+b2 and 3m2=ab

The first equation should be written as 2a + 2b = 7.4
The second equation should not have the units in it. ab = 3

Solve either equation for a or b, and then substitute the variable you solved for into the other equation.

 

[Government$]

2a+2b=7.4 --> 2a=7.4-2b--> a=7.4/2-2b/2

3=ab---> a=3/b so

3/b=7.4/2-2b/2 Is that what you meant?

 

[Mark44]

2a+2b=7.4 --> 2a=7.4-2b--> a=7.4/2-2b/2

Or more simply, a = 3.7 - b

Now substitute 3.7 - b for a in the equation ab = 3.

3=ab---> a=3/b so

3/b=7.4/2-2b/2 Is that what you meant?

 

[Government$]

So a = 3.7 - b then (3.7-b)b=3 --> 3.7b-b^2=3 --> b-b^2=3-3.7 - b-b^2= -0.7 ??

Is that the way or i am wrong and sry if i am boring.

P.S. I have looked at the end of book for result and a=1.2m and b=2.5 but there is no process.

 

[Mark44]

So a = 3.7 - b then (3.7-b)b=3 --> 3.7b-b^2=3 --> b-b^2=3-3.7 - b-b^2= -0.7 ??

(3.7-b)b=3 Yes
3.7b-b^2=3 Yes
b-b^2=3-3.7 - b-b^2= -0.7 No

If you subtract 3.7 from 3.7b, you don't get b. 3.7 and 3.7b are not like terms, so can't be added or subtracted to produce something simpler.

The equation 3.7b-b^2=3 is a quadratic equation. Do you know how to solve this type of equation?

Is that the way or i am wrong and sry if i am boring.

P.S. I have looked at the end of book for result and a=1.2m and b=2.5 but there is no process.

 

[Government$]

(3.7-b)b=3 Yes
3.7b-b^2=3 Yes
b-b^2=3-3.7 - b-b^2= -0.7 No

If you subtract 3.7 from 3.7b, you don't get b. 3.7 and 3.7b are not like terms, so can't be added or subtracted to produce something simpler.

The equation 3.7b-b^2=3 is a quadratic equation. Do you know how to solve this type of equation?

Thank you very much. I know to solve quadratic equation i just haven't recognized it. Thing is i am doing this for next year during summer break, and title of section is area of geometric body in plain(i don't know if you would call it like that in English language i am translating this from my own language) and i haven't know that they are going to mix it up from last year.

Anyway thanks!

 

[Mark44]

You mentioned in another post that you will be studying calculus in September. Since you had so many difficulties with this problem, it's good that you are reviewing this material now.

To be honest, though, it is not a good sign that you had so much trouble getting this relatively simple problem started, and you really should not be thinking that 3.7b - 3.7 = b.

 

[Government$]

You mentioned in another post that you will be studying calculus in September. Since you had so many difficulties with this problem, it's good that you are reviewing this material now.

To be honest, though, it is not a good sign that you had so much trouble getting this relatively simple problem started, and you really should not be thinking that 3.7b - 3.7 = b.

Yes i did, but truth is we still have to do other stuff before calculus like analytical geometry and other stuff. So i don't know if i am starting calculus this year or next i'll be starting third grade in september. And yes i really need to review pre-cal material. Thanks again.