# qustion

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17 years ago
#1
i have a triangel one side is x, one is x+6 and one is x+7 what are the lengths of the sides using
simeltaneous equaionos and can u show the working

thanks
0
17 years ago
#2
Joseph Brewer <[email protected]> wrote in uk.education.maths:
[q1]>i have a triangel one side is x, one is x+6 and one is x+7 what are the lengths of the sides using[/q1]
[q1]>simeltaneous equaionos and can u show the working[/q1]

The lengths of the sides are x, x+6, and x+7. There is not enough information to say more. They
might be 3, 9, 10 or 103, 109, 110, for instance.

--
Stan Brown, Oak Road Systems, Cortland County, New York, USA http://oakroadsystems.com/ "What in
heaven's name brought you to Casablanca?" "My health. I came to Casablanca for the waters." "The
waters? What waters? We're in the desert." "I was misinformed."
0
17 years ago
#3
Stan Brown <[email protected]> wrote in message news:[email protected]...
[q1]> Joseph Brewer <[email protected]> wrote in uk.education.maths:[/q1]
[q2]> >i have a triangel one side is x, one is x+6 and one is x+7 what are the lengths of the sides[/q2]
[q2]> >using simeltaneous equaionos and can u show the[/q2]
working
[q1]>[/q1]
[q1]> The lengths of the sides are x, x+6, and x+7. There is not enough information to say more. They[/q1]
[q1]> might be 3, 9, 10 or 103, 109, 110, for instance.[/q1]
[q1]>[/q1]

The one thing you can say is x>1

--
Martin

(remove barrier to reply)
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17 years ago
#4
CAnnot slove.........

not enough information..... all we can say is
x > 0 thats the answer...

omega

"Joseph Brewer" <[email protected]> wrote in message
news:X%[email protected]...
[q1]> i have a triangel one side is x, one is x+6 and one is x+7 what are the lengths of the sides using[/q1]
[q1]> simeltaneous equaionos and can u show the[/q1]
working
[q1]>[/q1]
[q1]> thanks[/q1]
0
17 years ago
#5
[q1]> not enough information..... all we can say is[/q1]
[q1]> x > 0 thats the answer...[/q1]

we can say more, x>1 (as we can't have a 1,7,8 triangle)

however i'm going to assume there is a piece of missing information, that it is a right angled
triangle.....

we get

x^2+(x+6)^2 = (x+7)^2 => x^2-2x-13 = 0 => x=1+-sqrt(14) x+(x+6) > x+7 (can't be equal) => x > 1 as
i said....
=> x = 1+sqrt(14) gives us a right triangle

all the other triangles are acute, with them tending to being equilateral as x-> infinity

[q2]> > i have a triangel one side is x, one is x+6 and one is x+7 what are the lengths of the sides[/q2]
[q2]> > using simeltaneous equaionos and can u show the[/q2]
[q1]> working[/q1]
[q2]> >[/q2]
[q2]> > thanks[/q2]
0
17 years ago
#6
In article <[email protected] optusnet.com.au>, Omega <[email protected]> writes
[q1]>CAnnot slove.........[/q1]
[q1]>[/q1]
[q1]>not enough information..... all we can say is[/q1]
[q1]>x > 0 thats the answer...[/q1]

It's not the whole answer. For instance, there is no triangle with sides of lengths 0.5,
6.5 and 7.5.

The only information we have is the triangle inequality, which tells us that x + (x+6) > x+7, which
gives x>1.

--
Simon Nickerson Phoebe: "Plus, there's the added mystery of, you know, who gets who." Ross: "Who
gets *whom*. <pause> I don't know why I do that."
0
17 years ago
#7
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