PPLZ help this question is soo confusin? Triangles

sarcastic

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Farmer Sylvester was skyping with his twin sister Sylvia who is also a farmer in Canada. “I’ve just put a perimeter fence around a new triangular field” said Sylvester. “Life is full of coincidences” said Sylvia, “so have I.”
Sylvia told Sylvester the three different angles in her field. “Another coincidence” said Sylvester, “each of my three angles is the same as each of yours”
“Each of the three lengths of fencing I used was an whole number of decametres” said Sylvester. “Same for me” said Sylvia.
Sylvia gave the length of one side of fencing of her field. “I have that length as well” said Sylvester.
Sylvia gave the length of another side of fencing of her field. “I have that length too” said Sylvester.
Sylvester told his sister the length of fence needed for the third side.
“Well, we just have more space and fencing here,” said Sylvia. “My triangular field is bigger than yours.”
“I think that my field is as small as it could be with whole number side lengths” said Sylvester.
“So I think I know how much fencing you used around your field” said Sylvia (she is really good at mathematics). Sylvester logged off! What length of fencing did Sylvia think that he had used?


i think it might have something to do with sine, coz and tan!!:rolleyes::confused:
 
Denis. My daughter has the same problem and the post by sarcastic is correct. It seems to me to be a problem of scale. Two triangles, different sizes. They both have the same angles and two of the sides (but not necessarily the same sides) have the same length. The smaller triangle has sides that are integers and are decametres.

So as I see it what is the smallest possible triangle with whole number side lengths that can be scaled up to produce a triangle where the angles are the same and two of the sides have the same length ??

For example (I'm not suggesting these are correct numbers and I've no idea if the angles would be the same here) if the smaller triangle had sides of 20, 30 and 40 metres and the larger had sides of 30, 40 and 50 metres, they both have two sides which are the same length (30 and 40). So, smallest possible triangle that can be scaled up, keeping the angles the same, to a point where two of the sides are the same length as any two of the larger triangle.
 
Hello, sarcastic!

Farmer Sylvester was skyping with his twin sister Sylvia who is also a farmer in Canada.
“I’ve just put a perimeter fence around a new triangular field” said Sylvester.
. . “Life is full of coincidences” said Sylvia, “so have I.”
Sylvia told Sylvester the three different angles in her field.
. . “Another coincidence” said Sylvester, “each of my three angles is the same as each of yours”
“Each of the three lengths of fencing I used was an whole number of decametres” said Sylvester.
. . “Same for me” said Sylvia.
Sylvia gave the length of one side of fencing of her field.
. . “I have that length as well” said Sylvester.
Sylvia gave the length of another side of fencing of her field.
. . “I have that length too” said Sylvester.
Sylvester told his sister the length of fence needed for the third side.
“Well, we just have more space and fencing here,” said Sylvia. “My triangular field is bigger than yours.”

“I think that my field is as small as it could be with whole number side lengths” said Sylvester.
“So I think I know how much fencing you used around your field” said Sylvia. (She's really good at math!)
What length of fencing did Sylvia think that he had used?

The two triangles are similar with two pairs of equal sides.
They look like this . . .

Code:
                              *
                             *  *
          *               b *     *  c
       a *  *  b           *        *
        *     *           *           *
       *        *        *              *
      *  *  *  *  *     *  *  *  *  *  *  *
            c                    d

From similar triangles, we have: .\(\displaystyle \dfrac{a}{b} \:=\:\dfrac{b}{c} \:=\:\dfrac{c}{d}\)

. . Hence: .\(\displaystyle b^2 \,=\,ac\,\text{ and }\,c^2\,=\,bd\)


We seek a set of four numbers: .\(\displaystyle a,b,c,d\) such that:
. . \(\displaystyle b\) is the geometric mean of \(\displaystyle a\text{ and }c.\)
. . \(\displaystyle c\) is the geometric mean of \(\displaystyle b\text{ and }d.\)

I believe the smallest set is: .\(\displaystyle \{8,\,12,\,18,\,27\}\)


The smaller triangle has sides: .\(\displaystyle 8,\,12,\,18\)
The larger triangle has sides: .\(\displaystyle 12,\,18,\,27\)
. . (Both triangles have sides in the ratio \(\displaystyle 4:6:9.\))
 
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