find a length of a triangle having two known sides and a median line
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Can anyone help me solve this problem in a simple way? I used area of ACD+ABD=ABC to solve the problem, but it is quite complicated. Thank you very much!
Problem: Two sides of a triangle have lengths 25 and 20, and the median to the third side has length 19.5. Find the length of the third side.
geometry
$endgroup$
add a comment |
$begingroup$
Can anyone help me solve this problem in a simple way? I used area of ACD+ABD=ABC to solve the problem, but it is quite complicated. Thank you very much!
Problem: Two sides of a triangle have lengths 25 and 20, and the median to the third side has length 19.5. Find the length of the third side.
geometry
$endgroup$
$begingroup$
Stewart's Theorem works well here.
$endgroup$
– lulu
Mar 26 '16 at 0:21
1
$begingroup$
Exactly, Stewart is good but the formula $4(alpha)^2=2b^2+2c^2-a^2$ gives you $a^2$ directly (you have $alpha$, $b$ and $c$)
$endgroup$
– Piquito
Mar 26 '16 at 0:38
add a comment |
$begingroup$
Can anyone help me solve this problem in a simple way? I used area of ACD+ABD=ABC to solve the problem, but it is quite complicated. Thank you very much!
Problem: Two sides of a triangle have lengths 25 and 20, and the median to the third side has length 19.5. Find the length of the third side.
geometry
$endgroup$
Can anyone help me solve this problem in a simple way? I used area of ACD+ABD=ABC to solve the problem, but it is quite complicated. Thank you very much!
Problem: Two sides of a triangle have lengths 25 and 20, and the median to the third side has length 19.5. Find the length of the third side.
geometry
geometry
asked Mar 26 '16 at 0:15
user321645
$begingroup$
Stewart's Theorem works well here.
$endgroup$
– lulu
Mar 26 '16 at 0:21
1
$begingroup$
Exactly, Stewart is good but the formula $4(alpha)^2=2b^2+2c^2-a^2$ gives you $a^2$ directly (you have $alpha$, $b$ and $c$)
$endgroup$
– Piquito
Mar 26 '16 at 0:38
add a comment |
$begingroup$
Stewart's Theorem works well here.
$endgroup$
– lulu
Mar 26 '16 at 0:21
1
$begingroup$
Exactly, Stewart is good but the formula $4(alpha)^2=2b^2+2c^2-a^2$ gives you $a^2$ directly (you have $alpha$, $b$ and $c$)
$endgroup$
– Piquito
Mar 26 '16 at 0:38
$begingroup$
Stewart's Theorem works well here.
$endgroup$
– lulu
Mar 26 '16 at 0:21
$begingroup$
Stewart's Theorem works well here.
$endgroup$
– lulu
Mar 26 '16 at 0:21
1
1
$begingroup$
Exactly, Stewart is good but the formula $4(alpha)^2=2b^2+2c^2-a^2$ gives you $a^2$ directly (you have $alpha$, $b$ and $c$)
$endgroup$
– Piquito
Mar 26 '16 at 0:38
$begingroup$
Exactly, Stewart is good but the formula $4(alpha)^2=2b^2+2c^2-a^2$ gives you $a^2$ directly (you have $alpha$, $b$ and $c$)
$endgroup$
– Piquito
Mar 26 '16 at 0:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The length of a median to side $a$ is:
$$frac{1}{2} sqrt{2b^2 + 2c^2 - a^2}$$
Here $a=25$, $b=20$ and length of median = $19.5$.
So,
$$19.5=frac{1}{2} sqrt{2(20)^2 + 2(25)^2 - a^2}$$
Squaring,
$$1521 = 800 + 1250 - a^2$$
We get $a^2 = 529$ so $a=23$ as $a=-23$ is not possible.
answered Mar 26 '16 at 2:23
mayank budhwanimayank budhwani
1897
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The length of a median to side $a$ is:
$$frac{1}{2} sqrt{2b^2 + 2c^2 - a^2}$$
Here $a=25$, $b=20$ and length of median = $19.5$.
So,
$$19.5=frac{1}{2} sqrt{2(20)^2 + 2(25)^2 - a^2}$$
Squaring,
$$1521 = 800 + 1250 - a^2$$
We get $a^2 = 529$ so $a=23$ as $a=-23$ is not possible.
answered Mar 26 '16 at 2:23
mayank budhwanimayank budhwani
1897
$endgroup$
add a comment |
$begingroup$
The length of a median to side $a$ is:
$$frac{1}{2} sqrt{2b^2 + 2c^2 - a^2}$$
Here $a=25$, $b=20$ and length of median = $19.5$.
So,
$$19.5=frac{1}{2} sqrt{2(20)^2 + 2(25)^2 - a^2}$$
Squaring,
$$1521 = 800 + 1250 - a^2$$
We get $a^2 = 529$ so $a=23$ as $a=-23$ is not possible.
answered Mar 26 '16 at 2:23
mayank budhwanimayank budhwani
1897
$endgroup$
add a comment |
$begingroup$
The length of a median to side $a$ is:
$$frac{1}{2} sqrt{2b^2 + 2c^2 - a^2}$$
Here $a=25$, $b=20$ and length of median = $19.5$.
So,
$$19.5=frac{1}{2} sqrt{2(20)^2 + 2(25)^2 - a^2}$$
Squaring,
$$1521 = 800 + 1250 - a^2$$
We get $a^2 = 529$ so $a=23$ as $a=-23$ is not possible.
answered Mar 26 '16 at 2:23
mayank budhwanimayank budhwani
1897
$endgroup$
The length of a median to side $a$ is:
$$frac{1}{2} sqrt{2b^2 + 2c^2 - a^2}$$
Here $a=25$, $b=20$ and length of median = $19.5$.
So,
$$19.5=frac{1}{2} sqrt{2(20)^2 + 2(25)^2 - a^2}$$
Squaring,
$$1521 = 800 + 1250 - a^2$$
We get $a^2 = 529$ so $a=23$ as $a=-23$ is not possible.
answered Mar 26 '16 at 2:23
mayank budhwanimayank budhwani
1897
edited Mar 31 '16 at 6:32
answered Mar 26 '16 at 2:23
mayank budhwanimayank budhwani
1897
answered Mar 26 '16 at 2:23
mayank budhwanimayank budhwani
1897
answered Mar 26 '16 at 2:23
mayank budhwanimayank budhwani
1897
1897
add a comment |
add a comment |
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$begingroup$
Stewart's Theorem works well here.
$endgroup$
– lulu
Mar 26 '16 at 0:21
1
$begingroup$
Exactly, Stewart is good but the formula $4(alpha)^2=2b^2+2c^2-a^2$ gives you $a^2$ directly (you have $alpha$, $b$ and $c$)
$endgroup$
– Piquito
Mar 26 '16 at 0:38