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Maths

How do you find the area of a rectangle with a given diagonal, width and length but the width and length is in algebra so x and 2x?
Write the diagonal in terms of x, then solve for x. Finally, calculate the area.
Original post by RogerOxon
Write the diagonal in terms of x, then solve for x. Finally, calculate the area.


My diagonal is 25 so how do I put that in terms of x? I've never done diagonal before so that's why am confused.
Reply 3
Avatar for Notnek
Notnek
21
Original post by Daydreamer3
My diagonal is 25 so how do I put that in terms of x? I've never done diagonal before so that's why am confused.

Try and form an equation using Pythagoras' Theorem. Please post your thoughts/working if you're unsure.
Original post by notnek
Try and form an equation using Pythagoras' Theorem. Please post your thoughts/working if you're unsure.


Ohhh Okay so
2x^2+x^2=25?
Reply 5
Avatar for Notnek
Notnek
21
Original post by Daydreamer3
Ohhh Okay so
2x^2+x^2=25?

Nearly but (2x)2=4x2(2x)^2 = 4x^2 not 2x22x^2.
Original post by notnek
Nearly but (2x)2=4x2(2x)^2 = 4x^2 not 2x22x^2.


Oh right I didnt know you had to put it into brackets....
So does x = 2.236067977?
Reply 7
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Notnek
21
Original post by Daydreamer3
Oh right I didnt know you had to put it into brackets....
So does x = 2.236067977?

You haven't said if the diagonal length is 5 but if it is then that is correct. You can write is as 5\sqrt{5} though - no need to use your calculator.

Also this looks like it wouldn't be a calculator question in the GCSE exam.

Can you find the area?
Original post by notnek
You haven't said if the diagonal length is 5 but if it is then that is correct. You can write is as 5\sqrt{5} though - no need to use your calculator.

Also this looks like it wouldn't be a calculator question in the GCSE exam.

Can you find the area?


The diagonal is 25....
I did
5x^2=25
x^2=5
x= square root of 5
Is that wrong?
Reply 9
Avatar for Notnek
Notnek
21
Original post by Daydreamer3
The diagonal is 25....
I did
5x^2=25
x^2=5
x= square root of 5
Is that wrong?

If the diagonal is 25 then you've made a mistake because you have to square the hypotenuse when using Pythagoras. So your equation should be:

(2x)2+x2=252(2x)^2 + x^2 = 25^2
Original post by notnek
If the diagonal is 25 then you've made a mistake because you have to square the hypotenuse when using Pythagoras. So your equation should be:

(2x)2+x2=252(2x)^2 + x^2 = 25^2


Oh right so x=11.18033989 or square root 125
Reply 11
Avatar for Notnek
Notnek
21
Original post by Daydreamer3
Oh right so x=11.18033989 or square root 125

That's correct. Can you now find the area?

Please post future maths questions in the maths help forum :smile:
Original post by notnek
That's correct. Can you now find the area?

Please post future maths questions in the maths help forum :smile:


I got 250
Thank you for the help! :smile:
Original post by Daydreamer3
I got 250
Thank you for the help! :smile:

Me too. Here's my working - note that a calculator is not required:

(2x)2+x2=252(2x)^2+x^2=25^2

5x2=2525x^2=25^2

x2=(52)25=545=53x^2=\frac{({5^2})^2}{5}=\frac{5^4}{5}=5^3

A=2x2=2.53=250A=2x^2=2.5^3=250

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