Mixing
Given medians of a right triangle, find the length of one side of the triangle
$begingroup$
I wonder whether there is an easy way to solve the problem. The method I used is to find the length of AC and CB, using right triangle ACF and ECB, respectively. Then find the length of AB using right triangle ABC. It's a straightforward problem, but my answer is different from the answer sheet.
11
Problem: Medians are drawn from point A and B in a right triangle to divide BC and AC in half, respectively. The lengths of the medians are 6 and 2√ 11, respectively. What is the length of AB?
geometry
$endgroup$
add a comment |
$begingroup$
I wonder whether there is an easy way to solve the problem. The method I used is to find the length of AC and CB, using right triangle ACF and ECB, respectively. Then find the length of AB using right triangle ABC. It's a straightforward problem, but my answer is different from the answer sheet.
11
Problem: Medians are drawn from point A and B in a right triangle to divide BC and AC in half, respectively. The lengths of the medians are 6 and 2√ 11, respectively. What is the length of AB?
geometry
$endgroup$
$begingroup$
Hint: medians intersect themselves in ratio 2:1
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:50
$begingroup$
what is yours and what is answer sheet?
$endgroup$
– chenbai
Apr 6 '16 at 2:59
$begingroup$
Well you can turn this into an algebra question. Let x =AC and y = BC. $(x/2)^2 + y^2 =4*11$ while $x^2 + (y/2)^2 = 36$. Solve for x and y. AB= $sqrt {x^2+y^2} $.
$endgroup$
– fleablood
Apr 6 '16 at 22:25
add a comment |
$begingroup$
I wonder whether there is an easy way to solve the problem. The method I used is to find the length of AC and CB, using right triangle ACF and ECB, respectively. Then find the length of AB using right triangle ABC. It's a straightforward problem, but my answer is different from the answer sheet.
11
Problem: Medians are drawn from point A and B in a right triangle to divide BC and AC in half, respectively. The lengths of the medians are 6 and 2√ 11, respectively. What is the length of AB?
geometry
$endgroup$
I wonder whether there is an easy way to solve the problem. The method I used is to find the length of AC and CB, using right triangle ACF and ECB, respectively. Then find the length of AB using right triangle ABC. It's a straightforward problem, but my answer is different from the answer sheet.
11
Problem: Medians are drawn from point A and B in a right triangle to divide BC and AC in half, respectively. The lengths of the medians are 6 and 2√ 11, respectively. What is the length of AB?
geometry
geometry
asked Apr 6 '16 at 2:47
user321645
$begingroup$
Hint: medians intersect themselves in ratio 2:1
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:50
$begingroup$
what is yours and what is answer sheet?
$endgroup$
– chenbai
Apr 6 '16 at 2:59
$begingroup$
Well you can turn this into an algebra question. Let x =AC and y = BC. $(x/2)^2 + y^2 =4*11$ while $x^2 + (y/2)^2 = 36$. Solve for x and y. AB= $sqrt {x^2+y^2} $.
$endgroup$
– fleablood
Apr 6 '16 at 22:25
add a comment |
$begingroup$
Hint: medians intersect themselves in ratio 2:1
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:50
$begingroup$
what is yours and what is answer sheet?
$endgroup$
– chenbai
Apr 6 '16 at 2:59
$begingroup$
Well you can turn this into an algebra question. Let x =AC and y = BC. $(x/2)^2 + y^2 =4*11$ while $x^2 + (y/2)^2 = 36$. Solve for x and y. AB= $sqrt {x^2+y^2} $.
$endgroup$
– fleablood
Apr 6 '16 at 22:25
$begingroup$
Hint: medians intersect themselves in ratio 2:1
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:50
$begingroup$
Hint: medians intersect themselves in ratio 2:1
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:50
$begingroup$
what is yours and what is answer sheet?
$endgroup$
– chenbai
Apr 6 '16 at 2:59
$begingroup$
what is yours and what is answer sheet?
$endgroup$
– chenbai
Apr 6 '16 at 2:59
$begingroup$
Well you can turn this into an algebra question. Let x =AC and y = BC. $(x/2)^2 + y^2 =4*11$ while $x^2 + (y/2)^2 = 36$. Solve for x and y. AB= $sqrt {x^2+y^2} $.
$endgroup$
– fleablood
Apr 6 '16 at 22:25
$begingroup$
Well you can turn this into an algebra question. Let x =AC and y = BC. $(x/2)^2 + y^2 =4*11$ while $x^2 + (y/2)^2 = 36$. Solve for x and y. AB= $sqrt {x^2+y^2} $.
$endgroup$
– fleablood
Apr 6 '16 at 22:25
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
From
$AC^2+(BC/2)^2=AF^2$
$(AC/2)^2+BC^2=BE^2$
you find $AC$ and $BC$.
Then
$AB^2=AC^2+BC^2$
answered Apr 6 '16 at 2:55
kmitovkmitov
3,9902816
$endgroup$
$begingroup$
how do you find $AC$? you only know one of the sides out of the three in the first equation.
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:59
$begingroup$
Thanks. That is how I did, but the answer is different from the answer sheet. May be the answer key is wrong. Thanks again.
$endgroup$
– user321645
Apr 6 '16 at 2:59
$begingroup$
oh ok, you solve the system of equations?
$endgroup$
– Jorge Fernández
Apr 6 '16 at 3:03
add a comment |
$begingroup$
Let $x = CF$ and $y = CE$. Then:
begin{align*}
(2x)^2 + y^2 = (2sqrt{11})^2 &iff 4x^2 + y^2 = 44 \
x^2 + (2y)^2 = 6^2 &iff x^2 + 4y^2 = 36
end{align*}
Adding the two equations together yields:
$$
5x^2 + 5y^2 = 80 iff x^2 + y^2 = 16
$$
The desired hypotenuse $AB$ is given by:
$$
sqrt{(2x)^2 + (2y)^2} = sqrt{4x^2 + 4y^2} = 2sqrt{x^2 + y^2} = 2sqrt{16} = 8
$$
answered Apr 6 '16 at 3:10
AdrianoAdriano
36.3k33071
$endgroup$
add a comment |
$begingroup$
Let $BC = 2a quad AC = 2b quad AB = 2c$
Then $EB^2 = b^2 + 4a^2 quad AF^2 = a^2 + 4b^2 quad c^2 = 4a^2 + 4b^2$
$EB^2 + AF^2 = 5a^2 + 5b^2 = dfrac 54 c^2$
So $c^2 = dfrac 45(EB^2 + AF^2)$
$c = sqrt{dfrac 45(EB^2 + AF^2)}$
For your example,
begin{align}
c &= sqrt{dfrac 45((2sqrt{11})^2 + 6^2)} \
&= sqrt{dfrac 5(44 + 36)} \
&= sqrt{dfrac 45(80)} \
&= 8
end{align}
OR
From
$left{ begin{array}{c}
EB^2 = b^2 + 4a^2 \
AF^2 = a^2 + 4b^2
end{array} right}$
we find
$left{ begin{array}{c}
BC^2 = 4a^2 = 4dfrac{4,AF^2 - EB^2}{15} \
AC^2 = 4b^2 = 4dfrac{4,EB^2 - AF^2}{15} \
end{array} right}$
Since $AF^2 = 36$ and $EB^2 = 44$, we get
$left{ begin{array}{c}
BC^2 = 4dfrac{144 - 44}{15} = dfrac{80}{3} \
AC^2 = 4dfrac{176 - 36}{15} = dfrac{112}{3} \
end{array} right}$
Finally $AB=sqrt{AC^2 + BC^2} =sqrt{64} = 8$
answered Apr 6 '16 at 3:13
steven gregorysteven gregory
18k32258
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From
$AC^2+(BC/2)^2=AF^2$
$(AC/2)^2+BC^2=BE^2$
you find $AC$ and $BC$.
Then
$AB^2=AC^2+BC^2$
answered Apr 6 '16 at 2:55
kmitovkmitov
3,9902816
$endgroup$
$begingroup$
how do you find $AC$? you only know one of the sides out of the three in the first equation.
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:59
$begingroup$
Thanks. That is how I did, but the answer is different from the answer sheet. May be the answer key is wrong. Thanks again.
$endgroup$
– user321645
Apr 6 '16 at 2:59
$begingroup$
oh ok, you solve the system of equations?
$endgroup$
– Jorge Fernández
Apr 6 '16 at 3:03
add a comment |
$begingroup$
From
$AC^2+(BC/2)^2=AF^2$
$(AC/2)^2+BC^2=BE^2$
you find $AC$ and $BC$.
Then
$AB^2=AC^2+BC^2$
answered Apr 6 '16 at 2:55
kmitovkmitov
3,9902816
$endgroup$
$begingroup$
how do you find $AC$? you only know one of the sides out of the three in the first equation.
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:59
$begingroup$
Thanks. That is how I did, but the answer is different from the answer sheet. May be the answer key is wrong. Thanks again.
$endgroup$
– user321645
Apr 6 '16 at 2:59
$begingroup$
oh ok, you solve the system of equations?
$endgroup$
– Jorge Fernández
Apr 6 '16 at 3:03
add a comment |
$begingroup$
From
$AC^2+(BC/2)^2=AF^2$
$(AC/2)^2+BC^2=BE^2$
you find $AC$ and $BC$.
Then
$AB^2=AC^2+BC^2$
answered Apr 6 '16 at 2:55
kmitovkmitov
3,9902816
$endgroup$
From
$AC^2+(BC/2)^2=AF^2$
$(AC/2)^2+BC^2=BE^2$
you find $AC$ and $BC$.
Then
$AB^2=AC^2+BC^2$
answered Apr 6 '16 at 2:55
kmitovkmitov
3,9902816
answered Apr 6 '16 at 2:55
kmitovkmitov
3,9902816
answered Apr 6 '16 at 2:55
kmitovkmitov
3,9902816
answered Apr 6 '16 at 2:55
kmitovkmitov
3,9902816
3,9902816
$begingroup$
how do you find $AC$? you only know one of the sides out of the three in the first equation.
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:59
$begingroup$
Thanks. That is how I did, but the answer is different from the answer sheet. May be the answer key is wrong. Thanks again.
$endgroup$
– user321645
Apr 6 '16 at 2:59
$begingroup$
oh ok, you solve the system of equations?
$endgroup$
– Jorge Fernández
Apr 6 '16 at 3:03
add a comment |
$begingroup$
how do you find $AC$? you only know one of the sides out of the three in the first equation.
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:59
$begingroup$
Thanks. That is how I did, but the answer is different from the answer sheet. May be the answer key is wrong. Thanks again.
$endgroup$
– user321645
Apr 6 '16 at 2:59
$begingroup$
oh ok, you solve the system of equations?
$endgroup$
– Jorge Fernández
Apr 6 '16 at 3:03
$begingroup$
how do you find $AC$? you only know one of the sides out of the three in the first equation.
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:59
$begingroup$
how do you find $AC$? you only know one of the sides out of the three in the first equation.
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:59
$begingroup$
Thanks. That is how I did, but the answer is different from the answer sheet. May be the answer key is wrong. Thanks again.
$endgroup$
– user321645
Apr 6 '16 at 2:59
$begingroup$
Thanks. That is how I did, but the answer is different from the answer sheet. May be the answer key is wrong. Thanks again.
$endgroup$
– user321645
Apr 6 '16 at 2:59
$begingroup$
oh ok, you solve the system of equations?
$endgroup$
– Jorge Fernández
Apr 6 '16 at 3:03
$begingroup$
oh ok, you solve the system of equations?
$endgroup$
– Jorge Fernández
Apr 6 '16 at 3:03
add a comment |
$begingroup$
Let $x = CF$ and $y = CE$. Then:
begin{align*}
(2x)^2 + y^2 = (2sqrt{11})^2 &iff 4x^2 + y^2 = 44 \
x^2 + (2y)^2 = 6^2 &iff x^2 + 4y^2 = 36
end{align*}
Adding the two equations together yields:
$$
5x^2 + 5y^2 = 80 iff x^2 + y^2 = 16
$$
The desired hypotenuse $AB$ is given by:
$$
sqrt{(2x)^2 + (2y)^2} = sqrt{4x^2 + 4y^2} = 2sqrt{x^2 + y^2} = 2sqrt{16} = 8
$$
answered Apr 6 '16 at 3:10
AdrianoAdriano
36.3k33071
$endgroup$
add a comment |
$begingroup$
Let $x = CF$ and $y = CE$. Then:
begin{align*}
(2x)^2 + y^2 = (2sqrt{11})^2 &iff 4x^2 + y^2 = 44 \
x^2 + (2y)^2 = 6^2 &iff x^2 + 4y^2 = 36
end{align*}
Adding the two equations together yields:
$$
5x^2 + 5y^2 = 80 iff x^2 + y^2 = 16
$$
The desired hypotenuse $AB$ is given by:
$$
sqrt{(2x)^2 + (2y)^2} = sqrt{4x^2 + 4y^2} = 2sqrt{x^2 + y^2} = 2sqrt{16} = 8
$$
answered Apr 6 '16 at 3:10
AdrianoAdriano
36.3k33071
$endgroup$
add a comment |
$begingroup$
Let $x = CF$ and $y = CE$. Then:
begin{align*}
(2x)^2 + y^2 = (2sqrt{11})^2 &iff 4x^2 + y^2 = 44 \
x^2 + (2y)^2 = 6^2 &iff x^2 + 4y^2 = 36
end{align*}
Adding the two equations together yields:
$$
5x^2 + 5y^2 = 80 iff x^2 + y^2 = 16
$$
The desired hypotenuse $AB$ is given by:
$$
sqrt{(2x)^2 + (2y)^2} = sqrt{4x^2 + 4y^2} = 2sqrt{x^2 + y^2} = 2sqrt{16} = 8
$$
answered Apr 6 '16 at 3:10
AdrianoAdriano
36.3k33071
$endgroup$
Let $x = CF$ and $y = CE$. Then:
begin{align*}
(2x)^2 + y^2 = (2sqrt{11})^2 &iff 4x^2 + y^2 = 44 \
x^2 + (2y)^2 = 6^2 &iff x^2 + 4y^2 = 36
end{align*}
Adding the two equations together yields:
$$
5x^2 + 5y^2 = 80 iff x^2 + y^2 = 16
$$
The desired hypotenuse $AB$ is given by:
$$
sqrt{(2x)^2 + (2y)^2} = sqrt{4x^2 + 4y^2} = 2sqrt{x^2 + y^2} = 2sqrt{16} = 8
$$
answered Apr 6 '16 at 3:10
AdrianoAdriano
36.3k33071
answered Apr 6 '16 at 3:10
AdrianoAdriano
36.3k33071
answered Apr 6 '16 at 3:10
AdrianoAdriano
36.3k33071
answered Apr 6 '16 at 3:10
AdrianoAdriano
36.3k33071
36.3k33071
add a comment |
add a comment |
$begingroup$
Let $BC = 2a quad AC = 2b quad AB = 2c$
Then $EB^2 = b^2 + 4a^2 quad AF^2 = a^2 + 4b^2 quad c^2 = 4a^2 + 4b^2$
$EB^2 + AF^2 = 5a^2 + 5b^2 = dfrac 54 c^2$
So $c^2 = dfrac 45(EB^2 + AF^2)$
$c = sqrt{dfrac 45(EB^2 + AF^2)}$
For your example,
begin{align}
c &= sqrt{dfrac 45((2sqrt{11})^2 + 6^2)} \
&= sqrt{dfrac 5(44 + 36)} \
&= sqrt{dfrac 45(80)} \
&= 8
end{align}
OR
From
$left{ begin{array}{c}
EB^2 = b^2 + 4a^2 \
AF^2 = a^2 + 4b^2
end{array} right}$
we find
$left{ begin{array}{c}
BC^2 = 4a^2 = 4dfrac{4,AF^2 - EB^2}{15} \
AC^2 = 4b^2 = 4dfrac{4,EB^2 - AF^2}{15} \
end{array} right}$
Since $AF^2 = 36$ and $EB^2 = 44$, we get
$left{ begin{array}{c}
BC^2 = 4dfrac{144 - 44}{15} = dfrac{80}{3} \
AC^2 = 4dfrac{176 - 36}{15} = dfrac{112}{3} \
end{array} right}$
Finally $AB=sqrt{AC^2 + BC^2} =sqrt{64} = 8$
answered Apr 6 '16 at 3:13
steven gregorysteven gregory
18k32258
$endgroup$
add a comment |
$begingroup$
Let $BC = 2a quad AC = 2b quad AB = 2c$
Then $EB^2 = b^2 + 4a^2 quad AF^2 = a^2 + 4b^2 quad c^2 = 4a^2 + 4b^2$
$EB^2 + AF^2 = 5a^2 + 5b^2 = dfrac 54 c^2$
So $c^2 = dfrac 45(EB^2 + AF^2)$
$c = sqrt{dfrac 45(EB^2 + AF^2)}$
For your example,
begin{align}
c &= sqrt{dfrac 45((2sqrt{11})^2 + 6^2)} \
&= sqrt{dfrac 5(44 + 36)} \
&= sqrt{dfrac 45(80)} \
&= 8
end{align}
OR
From
$left{ begin{array}{c}
EB^2 = b^2 + 4a^2 \
AF^2 = a^2 + 4b^2
end{array} right}$
we find
$left{ begin{array}{c}
BC^2 = 4a^2 = 4dfrac{4,AF^2 - EB^2}{15} \
AC^2 = 4b^2 = 4dfrac{4,EB^2 - AF^2}{15} \
end{array} right}$
Since $AF^2 = 36$ and $EB^2 = 44$, we get
$left{ begin{array}{c}
BC^2 = 4dfrac{144 - 44}{15} = dfrac{80}{3} \
AC^2 = 4dfrac{176 - 36}{15} = dfrac{112}{3} \
end{array} right}$
Finally $AB=sqrt{AC^2 + BC^2} =sqrt{64} = 8$
answered Apr 6 '16 at 3:13
steven gregorysteven gregory
18k32258
$endgroup$
add a comment |
$begingroup$
Let $BC = 2a quad AC = 2b quad AB = 2c$
Then $EB^2 = b^2 + 4a^2 quad AF^2 = a^2 + 4b^2 quad c^2 = 4a^2 + 4b^2$
$EB^2 + AF^2 = 5a^2 + 5b^2 = dfrac 54 c^2$
So $c^2 = dfrac 45(EB^2 + AF^2)$
$c = sqrt{dfrac 45(EB^2 + AF^2)}$
For your example,
begin{align}
c &= sqrt{dfrac 45((2sqrt{11})^2 + 6^2)} \
&= sqrt{dfrac 5(44 + 36)} \
&= sqrt{dfrac 45(80)} \
&= 8
end{align}
OR
From
$left{ begin{array}{c}
EB^2 = b^2 + 4a^2 \
AF^2 = a^2 + 4b^2
end{array} right}$
we find
$left{ begin{array}{c}
BC^2 = 4a^2 = 4dfrac{4,AF^2 - EB^2}{15} \
AC^2 = 4b^2 = 4dfrac{4,EB^2 - AF^2}{15} \
end{array} right}$
Since $AF^2 = 36$ and $EB^2 = 44$, we get
$left{ begin{array}{c}
BC^2 = 4dfrac{144 - 44}{15} = dfrac{80}{3} \
AC^2 = 4dfrac{176 - 36}{15} = dfrac{112}{3} \
end{array} right}$
Finally $AB=sqrt{AC^2 + BC^2} =sqrt{64} = 8$
answered Apr 6 '16 at 3:13
steven gregorysteven gregory
18k32258
$endgroup$
Let $BC = 2a quad AC = 2b quad AB = 2c$
Then $EB^2 = b^2 + 4a^2 quad AF^2 = a^2 + 4b^2 quad c^2 = 4a^2 + 4b^2$
$EB^2 + AF^2 = 5a^2 + 5b^2 = dfrac 54 c^2$
So $c^2 = dfrac 45(EB^2 + AF^2)$
$c = sqrt{dfrac 45(EB^2 + AF^2)}$
For your example,
begin{align}
c &= sqrt{dfrac 45((2sqrt{11})^2 + 6^2)} \
&= sqrt{dfrac 5(44 + 36)} \
&= sqrt{dfrac 45(80)} \
&= 8
end{align}
OR
From
$left{ begin{array}{c}
EB^2 = b^2 + 4a^2 \
AF^2 = a^2 + 4b^2
end{array} right}$
we find
$left{ begin{array}{c}
BC^2 = 4a^2 = 4dfrac{4,AF^2 - EB^2}{15} \
AC^2 = 4b^2 = 4dfrac{4,EB^2 - AF^2}{15} \
end{array} right}$
Since $AF^2 = 36$ and $EB^2 = 44$, we get
$left{ begin{array}{c}
BC^2 = 4dfrac{144 - 44}{15} = dfrac{80}{3} \
AC^2 = 4dfrac{176 - 36}{15} = dfrac{112}{3} \
end{array} right}$
Finally $AB=sqrt{AC^2 + BC^2} =sqrt{64} = 8$
answered Apr 6 '16 at 3:13
steven gregorysteven gregory
18k32258
edited Apr 7 '16 at 4:33
answered Apr 6 '16 at 3:13
steven gregorysteven gregory
18k32258
answered Apr 6 '16 at 3:13
steven gregorysteven gregory
18k32258
answered Apr 6 '16 at 3:13
steven gregorysteven gregory
18k32258
18k32258
add a comment |
add a comment |
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$begingroup$
Hint: medians intersect themselves in ratio 2:1
$endgroup$
– Jorge Fernández
Apr 6 '16 at 2:50
$begingroup$
what is yours and what is answer sheet?
$endgroup$
– chenbai
Apr 6 '16 at 2:59
$begingroup$
Well you can turn this into an algebra question. Let x =AC and y = BC. $(x/2)^2 + y^2 =4*11$ while $x^2 + (y/2)^2 = 36$. Solve for x and y. AB= $sqrt {x^2+y^2} $.
$endgroup$
– fleablood
Apr 6 '16 at 22:25