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If the diagonal and the area of a rectangle are 25 m and 168 m2, what is the length of the rectangle?
A. 17 m
B. 31 m
C. 12 m
D. 24 m

[vijaynarayanan]

answer is 24.

If you substitute the choices in for the Pythagoras Theorem... you will see that only,

24^2 + 7^2 = 25^2

and also 24x7=168

[jeremy8]

Is there any way to do algebra on this, using lenght + width as L and W, so that:

LxW=168

L^2+W^2=25^2

?

I tried, but came up with a ridiculous quadratic. I'm not very comfortable with exponents in equations.

[selango]

L*W=168

L^2+W^2=D^2

(L+W)^2-2LW=25^2

(L+W)^2-2*168=25^2

(L+W)^2=961

L+W=31 and L*W=168

With this you can get l=24 and B =7

[jeremy8]

L*W=168

L^2+W^2=D^2

(L+W)^2-2LW=25^2

(L+W)^2-2*168=25^2

(L+W)^2=961

L+W=31 and L*W=168

With this you can get l=24 and B =7

Thanks for your help.

It took me a while to understand how you got from L^2+W^2 to (L+W)^2-2LW.
Is this a mandatory step to finding L+W? Is factoring like this the only way to isolate and solve for L+W?

Is there any way of using the first equation in the form W=168/L and replacing W by this result in the other equation?

Something like: L^2+(168/L)^2=25^2

[selango]

Jeremy,

It can be solved by substituting L or W value.But it takes a bit longer time.

LW=168

L^2+W^2=625

Sub W=168/L

L^2+(168/L)^2=625

L^2+28224/L^2=625

L^4+28224=625L^2

L^4-625L^2+28224=0

As u can see this equation takes longer time to solve.

(L^2-576)*(L^2-49)=0

L^2=576,L=24 or L^2=49,L=7

L=24 or L=7

From the options only 24 is available.So L=24.

Pl let me know if this clarify ur doubt.

In algebra,u can solve the equation in many ways.The trick is you should know which one take less time to solve.

[jeremy8]

Pl let me know if this clarify ur doubt.

In algebra,u can solve the equation in many ways.The trick is you should know which one take less time to solve.

Anand,

Thanks again for your help. This shows me I'm not anywhere near ready to take the exam yet.
I really need to get a lot more comfortable with manipulating algebraic expressions.
I wasn't even sure it was possible to solve by substitution, although like you said, it's clearly not the fastest solution...
Back to my books....

[selango]

Jeremy,

There are lot of posts and materials available for algebra in BTG.Go through it and get familar with the topics.

Dont lose ur confidence.In this forum you can learn many things:)

go thru the below link,

https://www.platinumgmat.com/gmat_study_ ... _equations

[jeremy8]

Jeremy,

There are lot of posts and materials available for algebra in BTG.Go through it and get familar with the topics.

Dont lose ur confidence.In this forum you can learn many things:)

go thru the below link,

https://www.platinumgmat.com/gmat_study_ ... _equations

Thanks Anand,

I'm definitely glad I found this forum; I think it might really save me.
I'll check the link and try to really master this factorization process.