Finding the area of an orthic triangle (DEF) when given vertices of triangle ABC.
Given the triangle ABC whose vertices are endpoints of the altitudes from $A$,$B$ and $C$ is called the orthic triangle. The triangle $ABC$ has vertices at $A=(2,4)$; $B=(8,5)$ and $C=(3,9)$. The altitude from $B$ to $AC$, meets AC at point $D=(2.42,6.12)$. Find the area if the orthic triangle.
To attempt this problem I decided to use the formula
$$area = frac{abc|cosAcosBcosC|}{2R}$$ where $R$ is the circumradius of the triangle $ABC$. I calculated the length of each sides by using the length equation and got $a=sqrt41$, $b=sqrt26$, and $c=sqrt37$.
Next I used the cosine rule to get the angle at vertix $A$. Then used the sine rule to get angle at $B$ and subtracted these 2 angles from 180 to get the third angle. $A=69.23$, $B=48.12$ and $C=62.65$.
To get the circumradius of the triangle $ABC$ I used $$R=frac{abc}{4(Area)}$$ I found the area using heron's formula to be 14.5, putting all these values into the original equation I got the area of the orthic triangle to be 3.154. Is this correct? Is there an easier method? Thanks
geometry
asked Dec 5 '18 at 16:45
Matlab rookieMatlab rookie
357
add a comment |
Given the triangle ABC whose vertices are endpoints of the altitudes from $A$,$B$ and $C$ is called the orthic triangle. The triangle $ABC$ has vertices at $A=(2,4)$; $B=(8,5)$ and $C=(3,9)$. The altitude from $B$ to $AC$, meets AC at point $D=(2.42,6.12)$. Find the area if the orthic triangle.
To attempt this problem I decided to use the formula
$$area = frac{abc|cosAcosBcosC|}{2R}$$ where $R$ is the circumradius of the triangle $ABC$. I calculated the length of each sides by using the length equation and got $a=sqrt41$, $b=sqrt26$, and $c=sqrt37$.
Next I used the cosine rule to get the angle at vertix $A$. Then used the sine rule to get angle at $B$ and subtracted these 2 angles from 180 to get the third angle. $A=69.23$, $B=48.12$ and $C=62.65$.
To get the circumradius of the triangle $ABC$ I used $$R=frac{abc}{4(Area)}$$ I found the area using heron's formula to be 14.5, putting all these values into the original equation I got the area of the orthic triangle to be 3.154. Is this correct? Is there an easier method? Thanks
geometry
asked Dec 5 '18 at 16:45
Matlab rookieMatlab rookie
357
add a comment |
Given the triangle ABC whose vertices are endpoints of the altitudes from $A$,$B$ and $C$ is called the orthic triangle. The triangle $ABC$ has vertices at $A=(2,4)$; $B=(8,5)$ and $C=(3,9)$. The altitude from $B$ to $AC$, meets AC at point $D=(2.42,6.12)$. Find the area if the orthic triangle.
To attempt this problem I decided to use the formula
$$area = frac{abc|cosAcosBcosC|}{2R}$$ where $R$ is the circumradius of the triangle $ABC$. I calculated the length of each sides by using the length equation and got $a=sqrt41$, $b=sqrt26$, and $c=sqrt37$.
Next I used the cosine rule to get the angle at vertix $A$. Then used the sine rule to get angle at $B$ and subtracted these 2 angles from 180 to get the third angle. $A=69.23$, $B=48.12$ and $C=62.65$.
To get the circumradius of the triangle $ABC$ I used $$R=frac{abc}{4(Area)}$$ I found the area using heron's formula to be 14.5, putting all these values into the original equation I got the area of the orthic triangle to be 3.154. Is this correct? Is there an easier method? Thanks
geometry
asked Dec 5 '18 at 16:45
Matlab rookieMatlab rookie
357
Given the triangle ABC whose vertices are endpoints of the altitudes from $A$,$B$ and $C$ is called the orthic triangle. The triangle $ABC$ has vertices at $A=(2,4)$; $B=(8,5)$ and $C=(3,9)$. The altitude from $B$ to $AC$, meets AC at point $D=(2.42,6.12)$. Find the area if the orthic triangle.
To attempt this problem I decided to use the formula
$$area = frac{abc|cosAcosBcosC|}{2R}$$ where $R$ is the circumradius of the triangle $ABC$. I calculated the length of each sides by using the length equation and got $a=sqrt41$, $b=sqrt26$, and $c=sqrt37$.
Next I used the cosine rule to get the angle at vertix $A$. Then used the sine rule to get angle at $B$ and subtracted these 2 angles from 180 to get the third angle. $A=69.23$, $B=48.12$ and $C=62.65$.
To get the circumradius of the triangle $ABC$ I used $$R=frac{abc}{4(Area)}$$ I found the area using heron's formula to be 14.5, putting all these values into the original equation I got the area of the orthic triangle to be 3.154. Is this correct? Is there an easier method? Thanks
geometry
geometry
asked Dec 5 '18 at 16:45
Matlab rookieMatlab rookie
357
asked Dec 5 '18 at 16:45
Matlab rookieMatlab rookie
357
asked Dec 5 '18 at 16:45
Matlab rookieMatlab rookie
357
asked Dec 5 '18 at 16:45
Matlab rookieMatlab rookie
357
asked Dec 5 '18 at 16:45
Matlab rookieMatlab rookie
357
357
add a comment |
add a comment |
2 Answers
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Find the feet of the altitudes $D$, $E$, $F$ (you already found $D$). The area of $DEF$ is $displaystyle frac12 leftVert overrightarrow{DE}timesoverrightarrow{DF}rightVert$.
To find $E$, we first find the equation of $(BC)$: $displaystyle y=dfrac{9-5}{3-8}(x-3)+9=-4/5x+11.4$. The equation of the altitude $(AE)$ is $displaystyle y=5/4 (x-2)+4=5/4 x+3/2$. To find the intersection $E$ of these two lines we solve $displaystyle 5/4 x+3/2=-4/5x +11.4$ we get $displaystyle Eleft(frac{198}{41},frac{309}{41}right)$.
After you find $D$, $E$ and $F$ you can use the formula I gave above. If you don't know what's a cross product, you can use the answers in this question.
answered Dec 5 '18 at 17:32
BPPBPP
2,149927
How do I find the feet of altitudes?
– Matlab rookie
Dec 5 '18 at 17:45
add a comment |
I suggest to use
$$ Delta_{orthic} = 2Deltaleft|cos Acos Bcos Cright|$$
and to compute $Delta$ from the shoelace formula and the remaining part from the cosine theorem, such that we do not need to extract any square root. By the shoelace formula
$$ 2Delta = |2cdot 5+8cdot 9+3cdot 4-4cdot 8-5cdot 3-9cdot 2|=29 $$
and by the cosine theorem
$$left|cos Acos Bcos Cright|=frac{(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^2+b^2+c^2)}{8a^2b^2 c^2} $$
where $a^2=41,b^2=26,c^2=37$ are given by the Pythagorean theorem. It follows that
$$Delta_{orthic} = 29cdotfrac{30cdot 52cdot 22 }{8cdot 41cdot 26cdot 37}=frac{4785}{1517}. $$
answered Dec 6 '18 at 0:24
Jack D'AurizioJack D'Aurizio
288k33280659
add a comment |
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2 Answers
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2 Answers
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Find the feet of the altitudes $D$, $E$, $F$ (you already found $D$). The area of $DEF$ is $displaystyle frac12 leftVert overrightarrow{DE}timesoverrightarrow{DF}rightVert$.
To find $E$, we first find the equation of $(BC)$: $displaystyle y=dfrac{9-5}{3-8}(x-3)+9=-4/5x+11.4$. The equation of the altitude $(AE)$ is $displaystyle y=5/4 (x-2)+4=5/4 x+3/2$. To find the intersection $E$ of these two lines we solve $displaystyle 5/4 x+3/2=-4/5x +11.4$ we get $displaystyle Eleft(frac{198}{41},frac{309}{41}right)$.
After you find $D$, $E$ and $F$ you can use the formula I gave above. If you don't know what's a cross product, you can use the answers in this question.
answered Dec 5 '18 at 17:32
BPPBPP
2,149927
How do I find the feet of altitudes?
– Matlab rookie
Dec 5 '18 at 17:45
add a comment |
Find the feet of the altitudes $D$, $E$, $F$ (you already found $D$). The area of $DEF$ is $displaystyle frac12 leftVert overrightarrow{DE}timesoverrightarrow{DF}rightVert$.
To find $E$, we first find the equation of $(BC)$: $displaystyle y=dfrac{9-5}{3-8}(x-3)+9=-4/5x+11.4$. The equation of the altitude $(AE)$ is $displaystyle y=5/4 (x-2)+4=5/4 x+3/2$. To find the intersection $E$ of these two lines we solve $displaystyle 5/4 x+3/2=-4/5x +11.4$ we get $displaystyle Eleft(frac{198}{41},frac{309}{41}right)$.
After you find $D$, $E$ and $F$ you can use the formula I gave above. If you don't know what's a cross product, you can use the answers in this question.
answered Dec 5 '18 at 17:32
BPPBPP
2,149927
How do I find the feet of altitudes?
– Matlab rookie
Dec 5 '18 at 17:45
add a comment |
Find the feet of the altitudes $D$, $E$, $F$ (you already found $D$). The area of $DEF$ is $displaystyle frac12 leftVert overrightarrow{DE}timesoverrightarrow{DF}rightVert$.
To find $E$, we first find the equation of $(BC)$: $displaystyle y=dfrac{9-5}{3-8}(x-3)+9=-4/5x+11.4$. The equation of the altitude $(AE)$ is $displaystyle y=5/4 (x-2)+4=5/4 x+3/2$. To find the intersection $E$ of these two lines we solve $displaystyle 5/4 x+3/2=-4/5x +11.4$ we get $displaystyle Eleft(frac{198}{41},frac{309}{41}right)$.
After you find $D$, $E$ and $F$ you can use the formula I gave above. If you don't know what's a cross product, you can use the answers in this question.
answered Dec 5 '18 at 17:32
BPPBPP
2,149927
Find the feet of the altitudes $D$, $E$, $F$ (you already found $D$). The area of $DEF$ is $displaystyle frac12 leftVert overrightarrow{DE}timesoverrightarrow{DF}rightVert$.
To find $E$, we first find the equation of $(BC)$: $displaystyle y=dfrac{9-5}{3-8}(x-3)+9=-4/5x+11.4$. The equation of the altitude $(AE)$ is $displaystyle y=5/4 (x-2)+4=5/4 x+3/2$. To find the intersection $E$ of these two lines we solve $displaystyle 5/4 x+3/2=-4/5x +11.4$ we get $displaystyle Eleft(frac{198}{41},frac{309}{41}right)$.
After you find $D$, $E$ and $F$ you can use the formula I gave above. If you don't know what's a cross product, you can use the answers in this question.
answered Dec 5 '18 at 17:32
BPPBPP
2,149927
edited Dec 5 '18 at 17:55
answered Dec 5 '18 at 17:32
BPPBPP
2,149927
answered Dec 5 '18 at 17:32
BPPBPP
2,149927
answered Dec 5 '18 at 17:32
BPPBPP
2,149927
2,149927
How do I find the feet of altitudes?
– Matlab rookie
Dec 5 '18 at 17:45
add a comment |
How do I find the feet of altitudes?
– Matlab rookie
Dec 5 '18 at 17:45
How do I find the feet of altitudes?
– Matlab rookie
Dec 5 '18 at 17:45
How do I find the feet of altitudes?
– Matlab rookie
Dec 5 '18 at 17:45
add a comment |
I suggest to use
$$ Delta_{orthic} = 2Deltaleft|cos Acos Bcos Cright|$$
and to compute $Delta$ from the shoelace formula and the remaining part from the cosine theorem, such that we do not need to extract any square root. By the shoelace formula
$$ 2Delta = |2cdot 5+8cdot 9+3cdot 4-4cdot 8-5cdot 3-9cdot 2|=29 $$
and by the cosine theorem
$$left|cos Acos Bcos Cright|=frac{(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^2+b^2+c^2)}{8a^2b^2 c^2} $$
where $a^2=41,b^2=26,c^2=37$ are given by the Pythagorean theorem. It follows that
$$Delta_{orthic} = 29cdotfrac{30cdot 52cdot 22 }{8cdot 41cdot 26cdot 37}=frac{4785}{1517}. $$
answered Dec 6 '18 at 0:24
Jack D'AurizioJack D'Aurizio
288k33280659
add a comment |
I suggest to use
$$ Delta_{orthic} = 2Deltaleft|cos Acos Bcos Cright|$$
and to compute $Delta$ from the shoelace formula and the remaining part from the cosine theorem, such that we do not need to extract any square root. By the shoelace formula
$$ 2Delta = |2cdot 5+8cdot 9+3cdot 4-4cdot 8-5cdot 3-9cdot 2|=29 $$
and by the cosine theorem
$$left|cos Acos Bcos Cright|=frac{(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^2+b^2+c^2)}{8a^2b^2 c^2} $$
where $a^2=41,b^2=26,c^2=37$ are given by the Pythagorean theorem. It follows that
$$Delta_{orthic} = 29cdotfrac{30cdot 52cdot 22 }{8cdot 41cdot 26cdot 37}=frac{4785}{1517}. $$
answered Dec 6 '18 at 0:24
Jack D'AurizioJack D'Aurizio
288k33280659
add a comment |
I suggest to use
$$ Delta_{orthic} = 2Deltaleft|cos Acos Bcos Cright|$$
and to compute $Delta$ from the shoelace formula and the remaining part from the cosine theorem, such that we do not need to extract any square root. By the shoelace formula
$$ 2Delta = |2cdot 5+8cdot 9+3cdot 4-4cdot 8-5cdot 3-9cdot 2|=29 $$
and by the cosine theorem
$$left|cos Acos Bcos Cright|=frac{(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^2+b^2+c^2)}{8a^2b^2 c^2} $$
where $a^2=41,b^2=26,c^2=37$ are given by the Pythagorean theorem. It follows that
$$Delta_{orthic} = 29cdotfrac{30cdot 52cdot 22 }{8cdot 41cdot 26cdot 37}=frac{4785}{1517}. $$
answered Dec 6 '18 at 0:24
Jack D'AurizioJack D'Aurizio
288k33280659
I suggest to use
$$ Delta_{orthic} = 2Deltaleft|cos Acos Bcos Cright|$$
and to compute $Delta$ from the shoelace formula and the remaining part from the cosine theorem, such that we do not need to extract any square root. By the shoelace formula
$$ 2Delta = |2cdot 5+8cdot 9+3cdot 4-4cdot 8-5cdot 3-9cdot 2|=29 $$
and by the cosine theorem
$$left|cos Acos Bcos Cright|=frac{(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^2+b^2+c^2)}{8a^2b^2 c^2} $$
where $a^2=41,b^2=26,c^2=37$ are given by the Pythagorean theorem. It follows that
$$Delta_{orthic} = 29cdotfrac{30cdot 52cdot 22 }{8cdot 41cdot 26cdot 37}=frac{4785}{1517}. $$
answered Dec 6 '18 at 0:24
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 6 '18 at 0:24
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 6 '18 at 0:24
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 6 '18 at 0:24
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
add a comment |
add a comment |
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