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# Help on triangle problem watch

1. 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.
2. Give the sides the value of x to the width and work from there
3. Illuminati? BTW I have no idea what you're on about
4. (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
5. 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
6. Simply the sqrt45^2?
7. You get x^2+2x^2=45 if I'm not wrong and solve the qudratic
8. (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
9. 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?
10. (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 , then surely the square of the length is , not .

The equation would then be .
11. (Original post by Dingooose)
But if the length is , then surely the square of the length is , not .

The equation would then be .
Oh Yh my bad, just very tired too
12. (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.
13. (Original post by Loyale)
Oh Yh my bad, just very tired too
It's ok, everyone makes mistakes.
14. (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
15. wait, if it's x^2 + 4x^2 = 45, I can't solve that like a normal quadratic can I?
16. (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?
17. (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 , then . Recall what represents and also keep in mind that you're being asked for the width AND length of the rectangle.
18. Yeah so the length is then 6 because it's just double the width (x)
19. (Original post by joepiekos)
Yeah so the length is then 6 because it's just double the width (x)
Indeed.
20. (Original post by Dingooose;[url="tel:59381661")
59381661[/url]]Indeed.
I greatly appreciate your help thank you Very much

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