A Quadrilateral and A Triangle in a Trapizium
$begingroup$
In the above diagram, $ABCD$ is a Trapizium with $AD || BC$ and $BC perp AB$
$AB = 20, ; AD = 6,; BC = 30$
$M$ is a point on $DC$ such that $[ADMB] = [BMC]$, where $[x]$ denotes the area of $x$.
The length of $MC$ can be expressed as $dfrac{a}{b} sqrt{c}$.
Find the length of $MC$.
I can't find a way to start. Any hint will be helpful.
triangle area ratio
$endgroup$
add a comment |
$begingroup$
In the above diagram, $ABCD$ is a Trapizium with $AD || BC$ and $BC perp AB$
$AB = 20, ; AD = 6,; BC = 30$
$M$ is a point on $DC$ such that $[ADMB] = [BMC]$, where $[x]$ denotes the area of $x$.
The length of $MC$ can be expressed as $dfrac{a}{b} sqrt{c}$.
Find the length of $MC$.
I can't find a way to start. Any hint will be helpful.
triangle area ratio
$endgroup$
add a comment |
$begingroup$
In the above diagram, $ABCD$ is a Trapizium with $AD || BC$ and $BC perp AB$
$AB = 20, ; AD = 6,; BC = 30$
$M$ is a point on $DC$ such that $[ADMB] = [BMC]$, where $[x]$ denotes the area of $x$.
The length of $MC$ can be expressed as $dfrac{a}{b} sqrt{c}$.
Find the length of $MC$.
I can't find a way to start. Any hint will be helpful.
triangle area ratio
$endgroup$
In the above diagram, $ABCD$ is a Trapizium with $AD || BC$ and $BC perp AB$
$AB = 20, ; AD = 6,; BC = 30$
$M$ is a point on $DC$ such that $[ADMB] = [BMC]$, where $[x]$ denotes the area of $x$.
The length of $MC$ can be expressed as $dfrac{a}{b} sqrt{c}$.
Find the length of $MC$.
I can't find a way to start. Any hint will be helpful.
triangle area ratio
triangle area ratio
asked Feb 9 '16 at 15:39
Rezwan ArefinRezwan Arefin
1,683631
1,683631
add a comment |
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Hint: You can set up coordinates with the origin at A, D at (6,0), B at (0,20), C at (30,20). Then you can let t be the ratio DM/DC so that M becomes D+t(C-D). Note if you find t you can finish.
The other thing is to express each area in terms of t and equate the two expressions. There is a formula for area of a convex polygon once one lists the vertices in counterclockwise order. It is 1/2 times the "downs" minus the "ups". Here each down or up is a product of some coordinates. For example for vertices (1,3), (2,4), (0,2) the downs are 1*4,2*2,0*3 while the ups are 1*2,2*3,0*4. This works for more points, there are then just more "downs" and "ups"
Added note: The equation of equal areas will be linear in t with integer coefficients, so t will end up rational. Also side DC has length the squareroot of an integer. So getting an answer $(a/b)sqrt{c}$ will follow.
Some more details:
The formula for area mentioned above, if $(a_1,b_1),(a_2,b_2),...,(a_k,b_k)$ is a list of the vertices of a convex polygon going around the boundary in the counterclockwise direction, is $A=(1/2)[D-U]$ where
$$D=a_1b_2+a_2b_3+...+a_{n-1}b_n+a_nb_1,$$
and
$$U=a_1b_n+a_2b_1+...+a_n b_{n-1}.$$
Now consider a general version of your diagram, where $A=(0,0), B=(0,b), C=(c,b), D=(a,0).$ Then the point $M=D+t(C-D)$ has coordinates $(a+t(c-a),tb).$ When these coordinates are put in to find the areas of the quadrilateral part and the triangle, we find that twice their areas come out $b(ct+a)$ for the quadrilateral and $b(c-ct)$ for the triangle. Equating these (doubled) areas then gives $t=(c-a)/2c.$ [Interesting here is no $b$ occurs in this $t.$]
For the actual length of $MC$ we need to use $(1-t)$ since I set up the ratio as $DM/DC$ rather than what I should have done, namely $MC/DC.$ Then since $1-t=(c+a)/2c,$ it is this fraction which is to be multiplied by the length of $DC$ and that length is $sqrt{(c-a)^2+b^2}.$
Note the answer expected may assume that one removes the square part of the integer inside the radical, and combining it with the fraction for $1-t.$
$endgroup$
add a comment |
$begingroup$
Note:- In the following, quantities a ~ g can easily be found.
[⊿ABD] = a
[Trap ABCD] = b
[⊿BDC] = c
CD = g
Since BM divides the whole figure into two equal halves (of area = d each), [⊿BDM] = e (= d - a).
If this is so, we can let [⊿BMC] = f. M should be at a place that divides [⊿BDC] in the ratio $e : f$. $MC = dfrac {f}{e + f} g$.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Hint: You can set up coordinates with the origin at A, D at (6,0), B at (0,20), C at (30,20). Then you can let t be the ratio DM/DC so that M becomes D+t(C-D). Note if you find t you can finish.
The other thing is to express each area in terms of t and equate the two expressions. There is a formula for area of a convex polygon once one lists the vertices in counterclockwise order. It is 1/2 times the "downs" minus the "ups". Here each down or up is a product of some coordinates. For example for vertices (1,3), (2,4), (0,2) the downs are 1*4,2*2,0*3 while the ups are 1*2,2*3,0*4. This works for more points, there are then just more "downs" and "ups"
Added note: The equation of equal areas will be linear in t with integer coefficients, so t will end up rational. Also side DC has length the squareroot of an integer. So getting an answer $(a/b)sqrt{c}$ will follow.
Some more details:
The formula for area mentioned above, if $(a_1,b_1),(a_2,b_2),...,(a_k,b_k)$ is a list of the vertices of a convex polygon going around the boundary in the counterclockwise direction, is $A=(1/2)[D-U]$ where
$$D=a_1b_2+a_2b_3+...+a_{n-1}b_n+a_nb_1,$$
and
$$U=a_1b_n+a_2b_1+...+a_n b_{n-1}.$$
Now consider a general version of your diagram, where $A=(0,0), B=(0,b), C=(c,b), D=(a,0).$ Then the point $M=D+t(C-D)$ has coordinates $(a+t(c-a),tb).$ When these coordinates are put in to find the areas of the quadrilateral part and the triangle, we find that twice their areas come out $b(ct+a)$ for the quadrilateral and $b(c-ct)$ for the triangle. Equating these (doubled) areas then gives $t=(c-a)/2c.$ [Interesting here is no $b$ occurs in this $t.$]
For the actual length of $MC$ we need to use $(1-t)$ since I set up the ratio as $DM/DC$ rather than what I should have done, namely $MC/DC.$ Then since $1-t=(c+a)/2c,$ it is this fraction which is to be multiplied by the length of $DC$ and that length is $sqrt{(c-a)^2+b^2}.$
Note the answer expected may assume that one removes the square part of the integer inside the radical, and combining it with the fraction for $1-t.$
$endgroup$
add a comment |
$begingroup$
Hint: You can set up coordinates with the origin at A, D at (6,0), B at (0,20), C at (30,20). Then you can let t be the ratio DM/DC so that M becomes D+t(C-D). Note if you find t you can finish.
The other thing is to express each area in terms of t and equate the two expressions. There is a formula for area of a convex polygon once one lists the vertices in counterclockwise order. It is 1/2 times the "downs" minus the "ups". Here each down or up is a product of some coordinates. For example for vertices (1,3), (2,4), (0,2) the downs are 1*4,2*2,0*3 while the ups are 1*2,2*3,0*4. This works for more points, there are then just more "downs" and "ups"
Added note: The equation of equal areas will be linear in t with integer coefficients, so t will end up rational. Also side DC has length the squareroot of an integer. So getting an answer $(a/b)sqrt{c}$ will follow.
Some more details:
The formula for area mentioned above, if $(a_1,b_1),(a_2,b_2),...,(a_k,b_k)$ is a list of the vertices of a convex polygon going around the boundary in the counterclockwise direction, is $A=(1/2)[D-U]$ where
$$D=a_1b_2+a_2b_3+...+a_{n-1}b_n+a_nb_1,$$
and
$$U=a_1b_n+a_2b_1+...+a_n b_{n-1}.$$
Now consider a general version of your diagram, where $A=(0,0), B=(0,b), C=(c,b), D=(a,0).$ Then the point $M=D+t(C-D)$ has coordinates $(a+t(c-a),tb).$ When these coordinates are put in to find the areas of the quadrilateral part and the triangle, we find that twice their areas come out $b(ct+a)$ for the quadrilateral and $b(c-ct)$ for the triangle. Equating these (doubled) areas then gives $t=(c-a)/2c.$ [Interesting here is no $b$ occurs in this $t.$]
For the actual length of $MC$ we need to use $(1-t)$ since I set up the ratio as $DM/DC$ rather than what I should have done, namely $MC/DC.$ Then since $1-t=(c+a)/2c,$ it is this fraction which is to be multiplied by the length of $DC$ and that length is $sqrt{(c-a)^2+b^2}.$
Note the answer expected may assume that one removes the square part of the integer inside the radical, and combining it with the fraction for $1-t.$
$endgroup$
add a comment |
$begingroup$
Hint: You can set up coordinates with the origin at A, D at (6,0), B at (0,20), C at (30,20). Then you can let t be the ratio DM/DC so that M becomes D+t(C-D). Note if you find t you can finish.
The other thing is to express each area in terms of t and equate the two expressions. There is a formula for area of a convex polygon once one lists the vertices in counterclockwise order. It is 1/2 times the "downs" minus the "ups". Here each down or up is a product of some coordinates. For example for vertices (1,3), (2,4), (0,2) the downs are 1*4,2*2,0*3 while the ups are 1*2,2*3,0*4. This works for more points, there are then just more "downs" and "ups"
Added note: The equation of equal areas will be linear in t with integer coefficients, so t will end up rational. Also side DC has length the squareroot of an integer. So getting an answer $(a/b)sqrt{c}$ will follow.
Some more details:
The formula for area mentioned above, if $(a_1,b_1),(a_2,b_2),...,(a_k,b_k)$ is a list of the vertices of a convex polygon going around the boundary in the counterclockwise direction, is $A=(1/2)[D-U]$ where
$$D=a_1b_2+a_2b_3+...+a_{n-1}b_n+a_nb_1,$$
and
$$U=a_1b_n+a_2b_1+...+a_n b_{n-1}.$$
Now consider a general version of your diagram, where $A=(0,0), B=(0,b), C=(c,b), D=(a,0).$ Then the point $M=D+t(C-D)$ has coordinates $(a+t(c-a),tb).$ When these coordinates are put in to find the areas of the quadrilateral part and the triangle, we find that twice their areas come out $b(ct+a)$ for the quadrilateral and $b(c-ct)$ for the triangle. Equating these (doubled) areas then gives $t=(c-a)/2c.$ [Interesting here is no $b$ occurs in this $t.$]
For the actual length of $MC$ we need to use $(1-t)$ since I set up the ratio as $DM/DC$ rather than what I should have done, namely $MC/DC.$ Then since $1-t=(c+a)/2c,$ it is this fraction which is to be multiplied by the length of $DC$ and that length is $sqrt{(c-a)^2+b^2}.$
Note the answer expected may assume that one removes the square part of the integer inside the radical, and combining it with the fraction for $1-t.$
$endgroup$
Hint: You can set up coordinates with the origin at A, D at (6,0), B at (0,20), C at (30,20). Then you can let t be the ratio DM/DC so that M becomes D+t(C-D). Note if you find t you can finish.
The other thing is to express each area in terms of t and equate the two expressions. There is a formula for area of a convex polygon once one lists the vertices in counterclockwise order. It is 1/2 times the "downs" minus the "ups". Here each down or up is a product of some coordinates. For example for vertices (1,3), (2,4), (0,2) the downs are 1*4,2*2,0*3 while the ups are 1*2,2*3,0*4. This works for more points, there are then just more "downs" and "ups"
Added note: The equation of equal areas will be linear in t with integer coefficients, so t will end up rational. Also side DC has length the squareroot of an integer. So getting an answer $(a/b)sqrt{c}$ will follow.
Some more details:
The formula for area mentioned above, if $(a_1,b_1),(a_2,b_2),...,(a_k,b_k)$ is a list of the vertices of a convex polygon going around the boundary in the counterclockwise direction, is $A=(1/2)[D-U]$ where
$$D=a_1b_2+a_2b_3+...+a_{n-1}b_n+a_nb_1,$$
and
$$U=a_1b_n+a_2b_1+...+a_n b_{n-1}.$$
Now consider a general version of your diagram, where $A=(0,0), B=(0,b), C=(c,b), D=(a,0).$ Then the point $M=D+t(C-D)$ has coordinates $(a+t(c-a),tb).$ When these coordinates are put in to find the areas of the quadrilateral part and the triangle, we find that twice their areas come out $b(ct+a)$ for the quadrilateral and $b(c-ct)$ for the triangle. Equating these (doubled) areas then gives $t=(c-a)/2c.$ [Interesting here is no $b$ occurs in this $t.$]
For the actual length of $MC$ we need to use $(1-t)$ since I set up the ratio as $DM/DC$ rather than what I should have done, namely $MC/DC.$ Then since $1-t=(c+a)/2c,$ it is this fraction which is to be multiplied by the length of $DC$ and that length is $sqrt{(c-a)^2+b^2}.$
Note the answer expected may assume that one removes the square part of the integer inside the radical, and combining it with the fraction for $1-t.$
edited Feb 10 '16 at 12:39
answered Feb 9 '16 at 16:00
coffeemathcoffeemath
27.2k22342
27.2k22342
add a comment |
add a comment |
$begingroup$
Note:- In the following, quantities a ~ g can easily be found.
[⊿ABD] = a
[Trap ABCD] = b
[⊿BDC] = c
CD = g
Since BM divides the whole figure into two equal halves (of area = d each), [⊿BDM] = e (= d - a).
If this is so, we can let [⊿BMC] = f. M should be at a place that divides [⊿BDC] in the ratio $e : f$. $MC = dfrac {f}{e + f} g$.
$endgroup$
add a comment |
$begingroup$
Note:- In the following, quantities a ~ g can easily be found.
[⊿ABD] = a
[Trap ABCD] = b
[⊿BDC] = c
CD = g
Since BM divides the whole figure into two equal halves (of area = d each), [⊿BDM] = e (= d - a).
If this is so, we can let [⊿BMC] = f. M should be at a place that divides [⊿BDC] in the ratio $e : f$. $MC = dfrac {f}{e + f} g$.
$endgroup$
add a comment |
$begingroup$
Note:- In the following, quantities a ~ g can easily be found.
[⊿ABD] = a
[Trap ABCD] = b
[⊿BDC] = c
CD = g
Since BM divides the whole figure into two equal halves (of area = d each), [⊿BDM] = e (= d - a).
If this is so, we can let [⊿BMC] = f. M should be at a place that divides [⊿BDC] in the ratio $e : f$. $MC = dfrac {f}{e + f} g$.
$endgroup$
Note:- In the following, quantities a ~ g can easily be found.
[⊿ABD] = a
[Trap ABCD] = b
[⊿BDC] = c
CD = g
Since BM divides the whole figure into two equal halves (of area = d each), [⊿BDM] = e (= d - a).
If this is so, we can let [⊿BMC] = f. M should be at a place that divides [⊿BDC] in the ratio $e : f$. $MC = dfrac {f}{e + f} g$.
edited Jan 6 at 9:00
answered Feb 10 '16 at 12:59
MickMick
11.9k21641
11.9k21641
add a comment |
add a comment |
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