Mixing
Finding the third side of a triangle given the area
I know the area and the lengths of two sides (a and b) of a non-right triangle. I also know P1 (vertex between a and c) and P2 (vertex between a and b).
I already know this much:
Perimeter = $ frac{(a+b+c)}{2} $
Area = $ A=sqrt{p(p-a)(p-b)(p-c)} $
How do I simplify the above two equations to solve for c? (Obviously, this is just algebra, but it's long enough that it is prone to error; I've tried this by hand now and gotten a different set of solutions each time.)
triangle
asked Jun 27 '16 at 17:40
Seth
11
add a comment |
I know the area and the lengths of two sides (a and b) of a non-right triangle. I also know P1 (vertex between a and c) and P2 (vertex between a and b).
I already know this much:
Perimeter = $ frac{(a+b+c)}{2} $
Area = $ A=sqrt{p(p-a)(p-b)(p-c)} $
How do I simplify the above two equations to solve for c? (Obviously, this is just algebra, but it's long enough that it is prone to error; I've tried this by hand now and gotten a different set of solutions each time.)
triangle
asked Jun 27 '16 at 17:40
Seth
11
The angle $theta$ between $a$ and $b$ is not fixed, hence neither it is the area $$Delta = frac{1}{2}absin(theta).$$
– Jack D'Aurizio
Jun 27 '16 at 17:42
Given two sides and area, most of the time when there is a triangle with these properties, there are exactly two.
– André Nicolas
Jun 27 '16 at 17:43
It is not very clear what is given and what is not. We have to find $c$ given $a,b,p$, or $a,b,Delta$, or something else?
– Jack D'Aurizio
Jun 27 '16 at 17:49
add a comment |
I know the area and the lengths of two sides (a and b) of a non-right triangle. I also know P1 (vertex between a and c) and P2 (vertex between a and b).
I already know this much:
Perimeter = $ frac{(a+b+c)}{2} $
Area = $ A=sqrt{p(p-a)(p-b)(p-c)} $
How do I simplify the above two equations to solve for c? (Obviously, this is just algebra, but it's long enough that it is prone to error; I've tried this by hand now and gotten a different set of solutions each time.)
triangle
asked Jun 27 '16 at 17:40
Seth
11
I know the area and the lengths of two sides (a and b) of a non-right triangle. I also know P1 (vertex between a and c) and P2 (vertex between a and b).
I already know this much:
Perimeter = $ frac{(a+b+c)}{2} $
Area = $ A=sqrt{p(p-a)(p-b)(p-c)} $
How do I simplify the above two equations to solve for c? (Obviously, this is just algebra, but it's long enough that it is prone to error; I've tried this by hand now and gotten a different set of solutions each time.)
triangle
triangle
asked Jun 27 '16 at 17:40
Seth
11
asked Jun 27 '16 at 17:40
Seth
11
asked Jun 27 '16 at 17:40
Seth
11
asked Jun 27 '16 at 17:40
Seth
11
asked Jun 27 '16 at 17:40
Seth
11
11
The angle $theta$ between $a$ and $b$ is not fixed, hence neither it is the area $$Delta = frac{1}{2}absin(theta).$$
– Jack D'Aurizio
Jun 27 '16 at 17:42
Given two sides and area, most of the time when there is a triangle with these properties, there are exactly two.
– André Nicolas
Jun 27 '16 at 17:43
It is not very clear what is given and what is not. We have to find $c$ given $a,b,p$, or $a,b,Delta$, or something else?
– Jack D'Aurizio
Jun 27 '16 at 17:49
add a comment |
The angle $theta$ between $a$ and $b$ is not fixed, hence neither it is the area $$Delta = frac{1}{2}absin(theta).$$
– Jack D'Aurizio
Jun 27 '16 at 17:42
Given two sides and area, most of the time when there is a triangle with these properties, there are exactly two.
– André Nicolas
Jun 27 '16 at 17:43
It is not very clear what is given and what is not. We have to find $c$ given $a,b,p$, or $a,b,Delta$, or something else?
– Jack D'Aurizio
Jun 27 '16 at 17:49
The angle $theta$ between $a$ and $b$ is not fixed, hence neither it is the area $$Delta = frac{1}{2}absin(theta).$$
– Jack D'Aurizio
Jun 27 '16 at 17:42
The angle $theta$ between $a$ and $b$ is not fixed, hence neither it is the area $$Delta = frac{1}{2}absin(theta).$$
– Jack D'Aurizio
Jun 27 '16 at 17:42
Given two sides and area, most of the time when there is a triangle with these properties, there are exactly two.
– André Nicolas
Jun 27 '16 at 17:43
Given two sides and area, most of the time when there is a triangle with these properties, there are exactly two.
– André Nicolas
Jun 27 '16 at 17:43
It is not very clear what is given and what is not. We have to find $c$ given $a,b,p$, or $a,b,Delta$, or something else?
– Jack D'Aurizio
Jun 27 '16 at 17:49
It is not very clear what is given and what is not. We have to find $c$ given $a,b,p$, or $a,b,Delta$, or something else?
– Jack D'Aurizio
Jun 27 '16 at 17:49
add a comment |
2 Answers
2
active
oldest
votes
Suppose we know $$K = |triangle ABC|, quad a, b,$$ and we wish to determine $c$. Then
$$begin{align*} 16K^2 &= (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \
&= left((a+b)^2 - c^2 right)left(c^2 - (a-b)^2 right) \
&= -c^4 + left( (a-b)^2 + (a+b)^2 right) c^2 - (a+b)^2 (a-b)^2 \
&= -c^4 + 2(a^2 + b^2) c^2 - (a^2 - b^2)^2. end{align*}$$
This of course is a quadratic in $c^2$. We can complete the square to get
$$begin{align*} 0 &= (c^2 - (a^2+b^2))^2 + (a^2-b^2)^2 - (a^2+b^2)^2 + 16K^2 \
&= (c^2 - (a^2+b^2))^2 - 4a^2 b^2 + 16K^2, end{align*}$$
from which we obtain $$c^2 = a^2 + b^2 pm 2sqrt{a^2 b^2 - 4K^2}.$$ Note that we must always have $ab ge 2K$, as the area of the triangle is maximized when the angle between sides of fixed length $a$ and $b$ is right; thus the square root is always well-defined. Furthermore, the AM-GM inequality guarantees that when $a, b > 0$, $a^2 + b^2 ge 2 sqrt{a^2 b^2}$, hence $c^2$ admits exactly two positive solutions whenever the strict inequality $ab > 2K$ is satisfied, and one unique solution when $ab = 2K$ and the triangle is right.
answered Jun 27 '16 at 21:32
heropup
62.5k66099
add a comment |
$s;$ is the conventional way of representing the semiperimeter $frac{a+b+c}2$ of the triangle with sides $a,,b,,c$.
$sqrt{s(s-a)(s-b)(s-c)}=
frac{sqrt{-a^4-b^4-c^4+2b^2c^2+2c^2a^2+2a^2b^2}}4=Delta$
Solving for $c$,
$$c=sqrt{a^2+b^2pm 2sqrt{a^2b^2-4Delta^2}}$$
where area $=Delta$
edited Dec 27 '18 at 18:44
amWhy
192k28224439
answered Jun 27 '16 at 20:58
Senex Ægypti Parvi
2,2031816
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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Suppose we know $$K = |triangle ABC|, quad a, b,$$ and we wish to determine $c$. Then
$$begin{align*} 16K^2 &= (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \
&= left((a+b)^2 - c^2 right)left(c^2 - (a-b)^2 right) \
&= -c^4 + left( (a-b)^2 + (a+b)^2 right) c^2 - (a+b)^2 (a-b)^2 \
&= -c^4 + 2(a^2 + b^2) c^2 - (a^2 - b^2)^2. end{align*}$$
This of course is a quadratic in $c^2$. We can complete the square to get
$$begin{align*} 0 &= (c^2 - (a^2+b^2))^2 + (a^2-b^2)^2 - (a^2+b^2)^2 + 16K^2 \
&= (c^2 - (a^2+b^2))^2 - 4a^2 b^2 + 16K^2, end{align*}$$
from which we obtain $$c^2 = a^2 + b^2 pm 2sqrt{a^2 b^2 - 4K^2}.$$ Note that we must always have $ab ge 2K$, as the area of the triangle is maximized when the angle between sides of fixed length $a$ and $b$ is right; thus the square root is always well-defined. Furthermore, the AM-GM inequality guarantees that when $a, b > 0$, $a^2 + b^2 ge 2 sqrt{a^2 b^2}$, hence $c^2$ admits exactly two positive solutions whenever the strict inequality $ab > 2K$ is satisfied, and one unique solution when $ab = 2K$ and the triangle is right.
answered Jun 27 '16 at 21:32
heropup
62.5k66099
add a comment |
Suppose we know $$K = |triangle ABC|, quad a, b,$$ and we wish to determine $c$. Then
$$begin{align*} 16K^2 &= (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \
&= left((a+b)^2 - c^2 right)left(c^2 - (a-b)^2 right) \
&= -c^4 + left( (a-b)^2 + (a+b)^2 right) c^2 - (a+b)^2 (a-b)^2 \
&= -c^4 + 2(a^2 + b^2) c^2 - (a^2 - b^2)^2. end{align*}$$
This of course is a quadratic in $c^2$. We can complete the square to get
$$begin{align*} 0 &= (c^2 - (a^2+b^2))^2 + (a^2-b^2)^2 - (a^2+b^2)^2 + 16K^2 \
&= (c^2 - (a^2+b^2))^2 - 4a^2 b^2 + 16K^2, end{align*}$$
from which we obtain $$c^2 = a^2 + b^2 pm 2sqrt{a^2 b^2 - 4K^2}.$$ Note that we must always have $ab ge 2K$, as the area of the triangle is maximized when the angle between sides of fixed length $a$ and $b$ is right; thus the square root is always well-defined. Furthermore, the AM-GM inequality guarantees that when $a, b > 0$, $a^2 + b^2 ge 2 sqrt{a^2 b^2}$, hence $c^2$ admits exactly two positive solutions whenever the strict inequality $ab > 2K$ is satisfied, and one unique solution when $ab = 2K$ and the triangle is right.
answered Jun 27 '16 at 21:32
heropup
62.5k66099
add a comment |
Suppose we know $$K = |triangle ABC|, quad a, b,$$ and we wish to determine $c$. Then
$$begin{align*} 16K^2 &= (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \
&= left((a+b)^2 - c^2 right)left(c^2 - (a-b)^2 right) \
&= -c^4 + left( (a-b)^2 + (a+b)^2 right) c^2 - (a+b)^2 (a-b)^2 \
&= -c^4 + 2(a^2 + b^2) c^2 - (a^2 - b^2)^2. end{align*}$$
This of course is a quadratic in $c^2$. We can complete the square to get
$$begin{align*} 0 &= (c^2 - (a^2+b^2))^2 + (a^2-b^2)^2 - (a^2+b^2)^2 + 16K^2 \
&= (c^2 - (a^2+b^2))^2 - 4a^2 b^2 + 16K^2, end{align*}$$
from which we obtain $$c^2 = a^2 + b^2 pm 2sqrt{a^2 b^2 - 4K^2}.$$ Note that we must always have $ab ge 2K$, as the area of the triangle is maximized when the angle between sides of fixed length $a$ and $b$ is right; thus the square root is always well-defined. Furthermore, the AM-GM inequality guarantees that when $a, b > 0$, $a^2 + b^2 ge 2 sqrt{a^2 b^2}$, hence $c^2$ admits exactly two positive solutions whenever the strict inequality $ab > 2K$ is satisfied, and one unique solution when $ab = 2K$ and the triangle is right.
answered Jun 27 '16 at 21:32
heropup
62.5k66099
Suppose we know $$K = |triangle ABC|, quad a, b,$$ and we wish to determine $c$. Then
$$begin{align*} 16K^2 &= (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \
&= left((a+b)^2 - c^2 right)left(c^2 - (a-b)^2 right) \
&= -c^4 + left( (a-b)^2 + (a+b)^2 right) c^2 - (a+b)^2 (a-b)^2 \
&= -c^4 + 2(a^2 + b^2) c^2 - (a^2 - b^2)^2. end{align*}$$
This of course is a quadratic in $c^2$. We can complete the square to get
$$begin{align*} 0 &= (c^2 - (a^2+b^2))^2 + (a^2-b^2)^2 - (a^2+b^2)^2 + 16K^2 \
&= (c^2 - (a^2+b^2))^2 - 4a^2 b^2 + 16K^2, end{align*}$$
from which we obtain $$c^2 = a^2 + b^2 pm 2sqrt{a^2 b^2 - 4K^2}.$$ Note that we must always have $ab ge 2K$, as the area of the triangle is maximized when the angle between sides of fixed length $a$ and $b$ is right; thus the square root is always well-defined. Furthermore, the AM-GM inequality guarantees that when $a, b > 0$, $a^2 + b^2 ge 2 sqrt{a^2 b^2}$, hence $c^2$ admits exactly two positive solutions whenever the strict inequality $ab > 2K$ is satisfied, and one unique solution when $ab = 2K$ and the triangle is right.
answered Jun 27 '16 at 21:32
heropup
62.5k66099
answered Jun 27 '16 at 21:32
heropup
62.5k66099
answered Jun 27 '16 at 21:32
heropup
62.5k66099
answered Jun 27 '16 at 21:32
heropup
62.5k66099
62.5k66099
add a comment |
add a comment |
$s;$ is the conventional way of representing the semiperimeter $frac{a+b+c}2$ of the triangle with sides $a,,b,,c$.
$sqrt{s(s-a)(s-b)(s-c)}=
frac{sqrt{-a^4-b^4-c^4+2b^2c^2+2c^2a^2+2a^2b^2}}4=Delta$
Solving for $c$,
$$c=sqrt{a^2+b^2pm 2sqrt{a^2b^2-4Delta^2}}$$
where area $=Delta$
edited Dec 27 '18 at 18:44
amWhy
192k28224439
answered Jun 27 '16 at 20:58
Senex Ægypti Parvi
2,2031816
add a comment |
$s;$ is the conventional way of representing the semiperimeter $frac{a+b+c}2$ of the triangle with sides $a,,b,,c$.
$sqrt{s(s-a)(s-b)(s-c)}=
frac{sqrt{-a^4-b^4-c^4+2b^2c^2+2c^2a^2+2a^2b^2}}4=Delta$
Solving for $c$,
$$c=sqrt{a^2+b^2pm 2sqrt{a^2b^2-4Delta^2}}$$
where area $=Delta$
edited Dec 27 '18 at 18:44
amWhy
192k28224439
answered Jun 27 '16 at 20:58
Senex Ægypti Parvi
2,2031816
add a comment |
$s;$ is the conventional way of representing the semiperimeter $frac{a+b+c}2$ of the triangle with sides $a,,b,,c$.
$sqrt{s(s-a)(s-b)(s-c)}=
frac{sqrt{-a^4-b^4-c^4+2b^2c^2+2c^2a^2+2a^2b^2}}4=Delta$
Solving for $c$,
$$c=sqrt{a^2+b^2pm 2sqrt{a^2b^2-4Delta^2}}$$
where area $=Delta$
edited Dec 27 '18 at 18:44
amWhy
192k28224439
answered Jun 27 '16 at 20:58
Senex Ægypti Parvi
2,2031816
$s;$ is the conventional way of representing the semiperimeter $frac{a+b+c}2$ of the triangle with sides $a,,b,,c$.
$sqrt{s(s-a)(s-b)(s-c)}=
frac{sqrt{-a^4-b^4-c^4+2b^2c^2+2c^2a^2+2a^2b^2}}4=Delta$
Solving for $c$,
$$c=sqrt{a^2+b^2pm 2sqrt{a^2b^2-4Delta^2}}$$
where area $=Delta$
edited Dec 27 '18 at 18:44
amWhy
192k28224439
answered Jun 27 '16 at 20:58
Senex Ægypti Parvi
2,2031816
edited Dec 27 '18 at 18:44
amWhy
192k28224439
edited Dec 27 '18 at 18:44
amWhy
192k28224439
edited Dec 27 '18 at 18:44
amWhy
192k28224439
192k28224439
answered Jun 27 '16 at 20:58
Senex Ægypti Parvi
2,2031816
answered Jun 27 '16 at 20:58
Senex Ægypti Parvi
2,2031816
answered Jun 27 '16 at 20:58
Senex Ægypti Parvi
2,2031816
2,2031816
add a comment |
add a comment |
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The angle $theta$ between $a$ and $b$ is not fixed, hence neither it is the area $$Delta = frac{1}{2}absin(theta).$$
– Jack D'Aurizio
Jun 27 '16 at 17:42
Given two sides and area, most of the time when there is a triangle with these properties, there are exactly two.
– André Nicolas
Jun 27 '16 at 17:43
It is not very clear what is given and what is not. We have to find $c$ given $a,b,p$, or $a,b,Delta$, or something else?
– Jack D'Aurizio
Jun 27 '16 at 17:49