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Help on triangle problem

A rectangle has a diagonal of sqrt45. The length is twice the length of the width, find the length and width of this rectangle.

Unfortunately, exhaustion seems to have got to me and I cannot work out this question when I should normally be able to within a matter of minutes.

Any help is appreciated. :smile:
Reply 1
Avatar for Loyale
Loyale
15
Give the sides the value of x to the width and work from there
Illuminati? BTW I have no idea what you're on about :smile:
Reply 3
Avatar for joepiekos
joepiekos
OP
Original post by Loyale
Give the sides the value of x to the width and work from there


I've done that but I can't seem to work it out any further than, x^2 + 2x^2 = sqrt45^2
Reply 4
Avatar for Loyale
Loyale
15
Use Pythagoras once you've used algebra for the length and the width and create an equation out of this so,

Width is x
Length is 2x

X^2+2x^2=sqrt45
Reply 5
Avatar for Loyale
Loyale
15
Simply the sqrt45^2?
Reply 6
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Loyale
15
You get x^2+2x^2=45 if I'm not wrong and solve the qudratic
Reply 7
Avatar for joepiekos
joepiekos
OP
Original post by Loyale
You get x^2+2x^2=45 if I'm not wrong and solve the qudratic


Thanks, I did think that but i'll have another go and see what I get :smile:
Reply 8
Avatar for joepiekos
joepiekos
OP
I seem to get very long decimal answers, but the actual answer is width is 3cm but I don't know how to get there?
Original post by Loyale
Use Pythagoras once you've used algebra for the length and the width and create an equation out of this so,

Width is x
Length is 2x

X^2+2x^2=sqrt45


But if the length is 2x2x, then surely the square of the length is (2x)2=4x2(2x)^2=4x^2, not 2x22x^2.

The equation would then be x2+4x2=45\displaystyle x^2+4x^2=45.
Reply 10
Avatar for Loyale
Loyale
15
Original post by Dingooose
But if the length is 2x2x, then surely the square of the length is (2x)2=4x2(2x)^2=4x^2, not 2x22x^2.

The equation would then be x2+4x2=45\displaystyle x^2+4x^2=45.


Oh Yh my bad, just very tired too
Original post by joepiekos
I seem to get very long decimal answers, but the actual answer is width is 3cm but I don't know how to get there?


This is because the equation that Loyale derived is incorrect. Check my previous post.
Original post by Loyale
Oh Yh my bad, just very tired too


It's ok, everyone makes mistakes.
Reply 13
Avatar for joepiekos
joepiekos
OP
Original post by Dingooose
This is because the equation that Loyale derived is incorrect. Check my previous post.


Thank you so much for your help, I don't know why I didn't get this earlier XD
Reply 14
Avatar for joepiekos
joepiekos
OP
wait, if it's x^2 + 4x^2 = 45, I can't solve that like a normal quadratic can I?
Reply 15
Avatar for joepiekos
joepiekos
OP
Original post by joepiekos
wait, if it's x^2 + 4x^2 = 45, I can't solve that like a normal quadratic can I?


Oh wait I know what to do now, factorise to get X^2(1+4)=45 then X^2=45/5 which is x^2 =9 so x= 9?
Original post by joepiekos
Oh wait I know what to do now, factorise to get X^2(1+4)=45 then X^2=45/5 which is x^2 =9 so x= 9?


If x2=9x^2=9, then x=3x=3. Recall what xx represents and also keep in mind that you're being asked for the width AND length of the rectangle.
Reply 17
Avatar for joepiekos
joepiekos
OP
Yeah so the length is then 6 because it's just double the width (x)
Original post by joepiekos
Yeah so the length is then 6 because it's just double the width (x)


Indeed.
Reply 19
Avatar for joepiekos
joepiekos
OP
[QUOTE=Dingooose;59381661]Indeed.

I greatly appreciate your help :smile: thank you Very much

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