Mixing
Get leg length depending on another leg knowing the perimeter of a rectangle triangle [closed]
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Knowing the perimeter of the triangle, how can i find the side (leg) b in function of the leg a and it's perimeter p.
geometry trigonometry
edited Jan 19 at 19:59
mrtaurho
5,59051440
asked Jan 19 at 19:58
DeFabregasDeFabregas
12
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closed as off-topic by mrtaurho, Martin R, Leucippus, Shailesh, max_zorn Jan 20 at 3:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Martin R, Leucippus, Shailesh, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
Knowing the perimeter of the triangle, how can i find the side (leg) b in function of the leg a and it's perimeter p.
geometry trigonometry
edited Jan 19 at 19:59
mrtaurho
5,59051440
asked Jan 19 at 19:58
DeFabregasDeFabregas
12
$endgroup$
closed as off-topic by mrtaurho, Martin R, Leucippus, Shailesh, max_zorn Jan 20 at 3:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Martin R, Leucippus, Shailesh, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
The hypotenuse has length $sqrt{a^2+b^2}$, by the Pythagorean identity. So we have that $a + b+sqrt{a^2+b} = 31$. We can now solve for $b$ in terms of $a$.
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– jordan_glen
Jan 19 at 20:09
add a comment |
$begingroup$
Knowing the perimeter of the triangle, how can i find the side (leg) b in function of the leg a and it's perimeter p.
geometry trigonometry
edited Jan 19 at 19:59
mrtaurho
5,59051440
asked Jan 19 at 19:58
DeFabregasDeFabregas
12
$endgroup$
Knowing the perimeter of the triangle, how can i find the side (leg) b in function of the leg a and it's perimeter p.
geometry trigonometry
geometry trigonometry
edited Jan 19 at 19:59
mrtaurho
5,59051440
asked Jan 19 at 19:58
DeFabregasDeFabregas
12
edited Jan 19 at 19:59
mrtaurho
5,59051440
asked Jan 19 at 19:58
DeFabregasDeFabregas
12
edited Jan 19 at 19:59
mrtaurho
5,59051440
edited Jan 19 at 19:59
mrtaurho
5,59051440
edited Jan 19 at 19:59
mrtaurho
5,59051440
5,59051440
asked Jan 19 at 19:58
DeFabregasDeFabregas
12
asked Jan 19 at 19:58
DeFabregasDeFabregas
12
asked Jan 19 at 19:58
DeFabregasDeFabregas
12
12
closed as off-topic by mrtaurho, Martin R, Leucippus, Shailesh, max_zorn Jan 20 at 3:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Martin R, Leucippus, Shailesh, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by mrtaurho, Martin R, Leucippus, Shailesh, max_zorn Jan 20 at 3:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Martin R, Leucippus, Shailesh, max_zorn
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
The hypotenuse has length $sqrt{a^2+b^2}$, by the Pythagorean identity. So we have that $a + b+sqrt{a^2+b} = 31$. We can now solve for $b$ in terms of $a$.
$endgroup$
– jordan_glen
Jan 19 at 20:09
add a comment |
$begingroup$
The hypotenuse has length $sqrt{a^2+b^2}$, by the Pythagorean identity. So we have that $a + b+sqrt{a^2+b} = 31$. We can now solve for $b$ in terms of $a$.
$endgroup$
– jordan_glen
Jan 19 at 20:09
$begingroup$
The hypotenuse has length $sqrt{a^2+b^2}$, by the Pythagorean identity. So we have that $a + b+sqrt{a^2+b} = 31$. We can now solve for $b$ in terms of $a$.
$endgroup$
– jordan_glen
Jan 19 at 20:09
$begingroup$
The hypotenuse has length $sqrt{a^2+b^2}$, by the Pythagorean identity. So we have that $a + b+sqrt{a^2+b} = 31$. We can now solve for $b$ in terms of $a$.
$endgroup$
– jordan_glen
Jan 19 at 20:09
add a comment |
2 Answers
2
active
oldest
votes
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Using the Pythagorean theorem, the hypotenuse, $$c,=sqrt{a^2+b^2}$$ $$a+b+c=p$$ Combining the two equations gives $$a+b+sqrt{a^2+b^2}=p$$ Subtracting by $a+b$ and squaring yields $$a^2+b^2=(p-a-b)^2=p^2+a^2+b^2-2ap-2bp+2ab$$ Subtracting by $a^2-b^2$ yields $$p^2-2ap-2bp+2ab=0$$ This can then be rearranged and factored as $$b(2a-2p)=2ap-p^2$$ Dividing by $2a-2p$ yields $$b = frac{2ap-p^2}{2a-2p}$$
answered Jan 19 at 20:17
LittleKnownMathematicianLittleKnownMathematician
507
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We have $$p=a+b+sqrt{a^2+b^2}$$ so $$p-sqrt{a^2+b^2}=a+b$$ after squaring we get
$$p^2-2ab=2psqrt{a^2+b^2}$$ and now square again.
And we get $$a=frac{1}{2}frac{p(2b-p)}{b-p}$$
answered Jan 19 at 20:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using the Pythagorean theorem, the hypotenuse, $$c,=sqrt{a^2+b^2}$$ $$a+b+c=p$$ Combining the two equations gives $$a+b+sqrt{a^2+b^2}=p$$ Subtracting by $a+b$ and squaring yields $$a^2+b^2=(p-a-b)^2=p^2+a^2+b^2-2ap-2bp+2ab$$ Subtracting by $a^2-b^2$ yields $$p^2-2ap-2bp+2ab=0$$ This can then be rearranged and factored as $$b(2a-2p)=2ap-p^2$$ Dividing by $2a-2p$ yields $$b = frac{2ap-p^2}{2a-2p}$$
answered Jan 19 at 20:17
LittleKnownMathematicianLittleKnownMathematician
507
$endgroup$
add a comment |
$begingroup$
Using the Pythagorean theorem, the hypotenuse, $$c,=sqrt{a^2+b^2}$$ $$a+b+c=p$$ Combining the two equations gives $$a+b+sqrt{a^2+b^2}=p$$ Subtracting by $a+b$ and squaring yields $$a^2+b^2=(p-a-b)^2=p^2+a^2+b^2-2ap-2bp+2ab$$ Subtracting by $a^2-b^2$ yields $$p^2-2ap-2bp+2ab=0$$ This can then be rearranged and factored as $$b(2a-2p)=2ap-p^2$$ Dividing by $2a-2p$ yields $$b = frac{2ap-p^2}{2a-2p}$$
answered Jan 19 at 20:17
LittleKnownMathematicianLittleKnownMathematician
507
$endgroup$
add a comment |
$begingroup$
Using the Pythagorean theorem, the hypotenuse, $$c,=sqrt{a^2+b^2}$$ $$a+b+c=p$$ Combining the two equations gives $$a+b+sqrt{a^2+b^2}=p$$ Subtracting by $a+b$ and squaring yields $$a^2+b^2=(p-a-b)^2=p^2+a^2+b^2-2ap-2bp+2ab$$ Subtracting by $a^2-b^2$ yields $$p^2-2ap-2bp+2ab=0$$ This can then be rearranged and factored as $$b(2a-2p)=2ap-p^2$$ Dividing by $2a-2p$ yields $$b = frac{2ap-p^2}{2a-2p}$$
answered Jan 19 at 20:17
LittleKnownMathematicianLittleKnownMathematician
507
$endgroup$
Using the Pythagorean theorem, the hypotenuse, $$c,=sqrt{a^2+b^2}$$ $$a+b+c=p$$ Combining the two equations gives $$a+b+sqrt{a^2+b^2}=p$$ Subtracting by $a+b$ and squaring yields $$a^2+b^2=(p-a-b)^2=p^2+a^2+b^2-2ap-2bp+2ab$$ Subtracting by $a^2-b^2$ yields $$p^2-2ap-2bp+2ab=0$$ This can then be rearranged and factored as $$b(2a-2p)=2ap-p^2$$ Dividing by $2a-2p$ yields $$b = frac{2ap-p^2}{2a-2p}$$
answered Jan 19 at 20:17
LittleKnownMathematicianLittleKnownMathematician
507
answered Jan 19 at 20:17
LittleKnownMathematicianLittleKnownMathematician
507
answered Jan 19 at 20:17
LittleKnownMathematicianLittleKnownMathematician
507
answered Jan 19 at 20:17
LittleKnownMathematicianLittleKnownMathematician
507
507
add a comment |
add a comment |
$begingroup$
We have $$p=a+b+sqrt{a^2+b^2}$$ so $$p-sqrt{a^2+b^2}=a+b$$ after squaring we get
$$p^2-2ab=2psqrt{a^2+b^2}$$ and now square again.
And we get $$a=frac{1}{2}frac{p(2b-p)}{b-p}$$
answered Jan 19 at 20:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
add a comment |
$begingroup$
We have $$p=a+b+sqrt{a^2+b^2}$$ so $$p-sqrt{a^2+b^2}=a+b$$ after squaring we get
$$p^2-2ab=2psqrt{a^2+b^2}$$ and now square again.
And we get $$a=frac{1}{2}frac{p(2b-p)}{b-p}$$
answered Jan 19 at 20:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
add a comment |
$begingroup$
We have $$p=a+b+sqrt{a^2+b^2}$$ so $$p-sqrt{a^2+b^2}=a+b$$ after squaring we get
$$p^2-2ab=2psqrt{a^2+b^2}$$ and now square again.
And we get $$a=frac{1}{2}frac{p(2b-p)}{b-p}$$
answered Jan 19 at 20:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
We have $$p=a+b+sqrt{a^2+b^2}$$ so $$p-sqrt{a^2+b^2}=a+b$$ after squaring we get
$$p^2-2ab=2psqrt{a^2+b^2}$$ and now square again.
And we get $$a=frac{1}{2}frac{p(2b-p)}{b-p}$$
answered Jan 19 at 20:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
answered Jan 19 at 20:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
answered Jan 19 at 20:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
answered Jan 19 at 20:13
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
76.7k42866
add a comment |
add a comment |
$begingroup$
The hypotenuse has length $sqrt{a^2+b^2}$, by the Pythagorean identity. So we have that $a + b+sqrt{a^2+b} = 31$. We can now solve for $b$ in terms of $a$.
$endgroup$
– jordan_glen
Jan 19 at 20:09