Mixing
Finding out the value of $angle DQC$ in a trapezium $ABQD$ where $angle DCB$ = 30$^circ$
$begingroup$
In this below diagram, $angle ABC=60^circ, angle DCB=30^circ $, $AD$ is parallel to $BC$ and $AP$ is perpendicular to $BC$. Both the area and perimeter of $ABCD$ and $APQD$ are equal . What is the value of /angle $DQC$?
At first, I started with the condition of area of the both the trapezium (mentioned in the question) being equal. As $AD$ is parallel to $BC$, So we can show that,
$frac{1}{2}(AD+BC)$ ×$AP$ = $frac{1}{2}(AD+PQ)$ ×$AP$
$AD$+$BC$ = $AD$+$PQ$
$BC$ = $PQ$ .................. (1)
And now, we can show their perimeter are equal with the below equation:
$AB+BC+CD+AD$ = $AP+PQ+DQ+AD$
$AB+BC+CD$ = $AP+PQ+DQ$
$AB+PQ+CD$ = $AP+PQ+DQ$............(from 1)
$CD+AB$ = $AP+DQ$.............(2)
I drew a vertical line $DE$ which is parallel to $AP$. $DE$ is equal to $AP$ as $AD$ and $BC$ are parallel to each other. Here /angle $APB$ and /angle $DEQ$ are respectively right-angle. So, we get two right-angled triangle $APB$ and $DEQ$.
By trigonometry and from the traingle $APB$, we can write that:-
$frac{AB}{AP}$ = $mathrm{cosec} 60^circ$
$AB$ = $frac{2AP}{sqrt 3}$..........(3)
Similarly, from $DEC$, we can write that:-
$frac{DC}{DE}$ = $mathrm{cosec} 30^circ$
$DC$ = 2$DE$
$CD$ = 2$DE$..........(4)
Let us denote the $angle DQE = $$theta$
So, now from the equation (2), we can get:-
$2AP$ + $frac{2AP}{sqrt 3}$ = $AP+ DQ$.........(from 3 and 4)
$frac{2sqrt 3AP + 2AP - sqrt 3AP}{sqrt 3} $ = $DQ$
$AP × frac{2+ sqrt 3}{sqrt 3} $ = $DQ$
$frac{2+ sqrt 3}{sqrt 3} $ = $frac{DQ}{AP} $
$frac{DQ}{DE} $ = $frac{2+ sqrt 3}{sqrt 3} $......($AP = DE$ according to the diagram)
$frac{DE}{DQ} $ = $frac{sqrt 3}{2 +sqrt 3} $
$sin theta$ = $frac{sqrt 3}{2} + 1 $.................(5)
We know that if function of x is described as f(x) = $sin^text{-1} $x, than function of x will be real and valid if and only if its domain is [-1,1]. But there is a little bit problem in my calculation.
(5) is invalid because the real value of $sin theta$ is above 1 which is impossible in this case. For this reason, the value of $theta$ is unreal. So, there is an extensive mistake either in my calculation or in the question. I have been unable to find my error. So, I need some help to identify the mistake to solve the problem properly.
EDIT: My fault is making equation (5) from the past.
geometry proof-verification area alternative-proof angle
asked Jan 24 at 6:59
Anirban NiloyAnirban Niloy
8331218
$endgroup$
add a comment |
$begingroup$
In this below diagram, $angle ABC=60^circ, angle DCB=30^circ $, $AD$ is parallel to $BC$ and $AP$ is perpendicular to $BC$. Both the area and perimeter of $ABCD$ and $APQD$ are equal . What is the value of /angle $DQC$?
At first, I started with the condition of area of the both the trapezium (mentioned in the question) being equal. As $AD$ is parallel to $BC$, So we can show that,
$frac{1}{2}(AD+BC)$ ×$AP$ = $frac{1}{2}(AD+PQ)$ ×$AP$
$AD$+$BC$ = $AD$+$PQ$
$BC$ = $PQ$ .................. (1)
And now, we can show their perimeter are equal with the below equation:
$AB+BC+CD+AD$ = $AP+PQ+DQ+AD$
$AB+BC+CD$ = $AP+PQ+DQ$
$AB+PQ+CD$ = $AP+PQ+DQ$............(from 1)
$CD+AB$ = $AP+DQ$.............(2)
I drew a vertical line $DE$ which is parallel to $AP$. $DE$ is equal to $AP$ as $AD$ and $BC$ are parallel to each other. Here /angle $APB$ and /angle $DEQ$ are respectively right-angle. So, we get two right-angled triangle $APB$ and $DEQ$.
By trigonometry and from the traingle $APB$, we can write that:-
$frac{AB}{AP}$ = $mathrm{cosec} 60^circ$
$AB$ = $frac{2AP}{sqrt 3}$..........(3)
Similarly, from $DEC$, we can write that:-
$frac{DC}{DE}$ = $mathrm{cosec} 30^circ$
$DC$ = 2$DE$
$CD$ = 2$DE$..........(4)
Let us denote the $angle DQE = $$theta$
So, now from the equation (2), we can get:-
$2AP$ + $frac{2AP}{sqrt 3}$ = $AP+ DQ$.........(from 3 and 4)
$frac{2sqrt 3AP + 2AP - sqrt 3AP}{sqrt 3} $ = $DQ$
$AP × frac{2+ sqrt 3}{sqrt 3} $ = $DQ$
$frac{2+ sqrt 3}{sqrt 3} $ = $frac{DQ}{AP} $
$frac{DQ}{DE} $ = $frac{2+ sqrt 3}{sqrt 3} $......($AP = DE$ according to the diagram)
$frac{DE}{DQ} $ = $frac{sqrt 3}{2 +sqrt 3} $
$sin theta$ = $frac{sqrt 3}{2} + 1 $.................(5)
We know that if function of x is described as f(x) = $sin^text{-1} $x, than function of x will be real and valid if and only if its domain is [-1,1]. But there is a little bit problem in my calculation.
(5) is invalid because the real value of $sin theta$ is above 1 which is impossible in this case. For this reason, the value of $theta$ is unreal. So, there is an extensive mistake either in my calculation or in the question. I have been unable to find my error. So, I need some help to identify the mistake to solve the problem properly.
EDIT: My fault is making equation (5) from the past.
geometry proof-verification area alternative-proof angle
asked Jan 24 at 6:59
Anirban NiloyAnirban Niloy
8331218
$endgroup$
1
$begingroup$
There is no solution. You cannot have both perimeters and areas the same.
$endgroup$
– Andrei
Jan 24 at 7:42
$begingroup$
That's why I got stucked. I often get this kind of faulty question but find no satisfied answer. The source of the problem is also enigmatic. Thank you for telling me that point. I frankly didn't know that point. Should I delete this post? Otherwise anyone will be confused.
$endgroup$
– Anirban Niloy
Jan 24 at 7:48
1
$begingroup$
I did not know it either. But it's easy to prove. I will post that as an answer.
$endgroup$
– Andrei
Jan 24 at 7:51
$begingroup$
Okay. I will check that out later. I'll be eagerly waiting.
$endgroup$
– Anirban Niloy
Jan 24 at 7:53
add a comment |
$begingroup$
In this below diagram, $angle ABC=60^circ, angle DCB=30^circ $, $AD$ is parallel to $BC$ and $AP$ is perpendicular to $BC$. Both the area and perimeter of $ABCD$ and $APQD$ are equal . What is the value of /angle $DQC$?
At first, I started with the condition of area of the both the trapezium (mentioned in the question) being equal. As $AD$ is parallel to $BC$, So we can show that,
$frac{1}{2}(AD+BC)$ ×$AP$ = $frac{1}{2}(AD+PQ)$ ×$AP$
$AD$+$BC$ = $AD$+$PQ$
$BC$ = $PQ$ .................. (1)
And now, we can show their perimeter are equal with the below equation:
$AB+BC+CD+AD$ = $AP+PQ+DQ+AD$
$AB+BC+CD$ = $AP+PQ+DQ$
$AB+PQ+CD$ = $AP+PQ+DQ$............(from 1)
$CD+AB$ = $AP+DQ$.............(2)
I drew a vertical line $DE$ which is parallel to $AP$. $DE$ is equal to $AP$ as $AD$ and $BC$ are parallel to each other. Here /angle $APB$ and /angle $DEQ$ are respectively right-angle. So, we get two right-angled triangle $APB$ and $DEQ$.
By trigonometry and from the traingle $APB$, we can write that:-
$frac{AB}{AP}$ = $mathrm{cosec} 60^circ$
$AB$ = $frac{2AP}{sqrt 3}$..........(3)
Similarly, from $DEC$, we can write that:-
$frac{DC}{DE}$ = $mathrm{cosec} 30^circ$
$DC$ = 2$DE$
$CD$ = 2$DE$..........(4)
Let us denote the $angle DQE = $$theta$
So, now from the equation (2), we can get:-
$2AP$ + $frac{2AP}{sqrt 3}$ = $AP+ DQ$.........(from 3 and 4)
$frac{2sqrt 3AP + 2AP - sqrt 3AP}{sqrt 3} $ = $DQ$
$AP × frac{2+ sqrt 3}{sqrt 3} $ = $DQ$
$frac{2+ sqrt 3}{sqrt 3} $ = $frac{DQ}{AP} $
$frac{DQ}{DE} $ = $frac{2+ sqrt 3}{sqrt 3} $......($AP = DE$ according to the diagram)
$frac{DE}{DQ} $ = $frac{sqrt 3}{2 +sqrt 3} $
$sin theta$ = $frac{sqrt 3}{2} + 1 $.................(5)
We know that if function of x is described as f(x) = $sin^text{-1} $x, than function of x will be real and valid if and only if its domain is [-1,1]. But there is a little bit problem in my calculation.
(5) is invalid because the real value of $sin theta$ is above 1 which is impossible in this case. For this reason, the value of $theta$ is unreal. So, there is an extensive mistake either in my calculation or in the question. I have been unable to find my error. So, I need some help to identify the mistake to solve the problem properly.
EDIT: My fault is making equation (5) from the past.
geometry proof-verification area alternative-proof angle
asked Jan 24 at 6:59
Anirban NiloyAnirban Niloy
8331218
$endgroup$
In this below diagram, $angle ABC=60^circ, angle DCB=30^circ $, $AD$ is parallel to $BC$ and $AP$ is perpendicular to $BC$. Both the area and perimeter of $ABCD$ and $APQD$ are equal . What is the value of /angle $DQC$?
At first, I started with the condition of area of the both the trapezium (mentioned in the question) being equal. As $AD$ is parallel to $BC$, So we can show that,
$frac{1}{2}(AD+BC)$ ×$AP$ = $frac{1}{2}(AD+PQ)$ ×$AP$
$AD$+$BC$ = $AD$+$PQ$
$BC$ = $PQ$ .................. (1)
And now, we can show their perimeter are equal with the below equation:
$AB+BC+CD+AD$ = $AP+PQ+DQ+AD$
$AB+BC+CD$ = $AP+PQ+DQ$
$AB+PQ+CD$ = $AP+PQ+DQ$............(from 1)
$CD+AB$ = $AP+DQ$.............(2)
I drew a vertical line $DE$ which is parallel to $AP$. $DE$ is equal to $AP$ as $AD$ and $BC$ are parallel to each other. Here /angle $APB$ and /angle $DEQ$ are respectively right-angle. So, we get two right-angled triangle $APB$ and $DEQ$.
By trigonometry and from the traingle $APB$, we can write that:-
$frac{AB}{AP}$ = $mathrm{cosec} 60^circ$
$AB$ = $frac{2AP}{sqrt 3}$..........(3)
Similarly, from $DEC$, we can write that:-
$frac{DC}{DE}$ = $mathrm{cosec} 30^circ$
$DC$ = 2$DE$
$CD$ = 2$DE$..........(4)
Let us denote the $angle DQE = $$theta$
So, now from the equation (2), we can get:-
$2AP$ + $frac{2AP}{sqrt 3}$ = $AP+ DQ$.........(from 3 and 4)
$frac{2sqrt 3AP + 2AP - sqrt 3AP}{sqrt 3} $ = $DQ$
$AP × frac{2+ sqrt 3}{sqrt 3} $ = $DQ$
$frac{2+ sqrt 3}{sqrt 3} $ = $frac{DQ}{AP} $
$frac{DQ}{DE} $ = $frac{2+ sqrt 3}{sqrt 3} $......($AP = DE$ according to the diagram)
$frac{DE}{DQ} $ = $frac{sqrt 3}{2 +sqrt 3} $
$sin theta$ = $frac{sqrt 3}{2} + 1 $.................(5)
We know that if function of x is described as f(x) = $sin^text{-1} $x, than function of x will be real and valid if and only if its domain is [-1,1]. But there is a little bit problem in my calculation.
(5) is invalid because the real value of $sin theta$ is above 1 which is impossible in this case. For this reason, the value of $theta$ is unreal. So, there is an extensive mistake either in my calculation or in the question. I have been unable to find my error. So, I need some help to identify the mistake to solve the problem properly.
EDIT: My fault is making equation (5) from the past.
geometry proof-verification area alternative-proof angle
geometry proof-verification area alternative-proof angle
asked Jan 24 at 6:59
Anirban NiloyAnirban Niloy
8331218
asked Jan 24 at 6:59
Anirban NiloyAnirban Niloy
8331218
edited Feb 13 at 5:01
Anirban Niloy
asked Jan 24 at 6:59
Anirban NiloyAnirban Niloy
8331218
asked Jan 24 at 6:59
Anirban NiloyAnirban Niloy
8331218
asked Jan 24 at 6:59
Anirban NiloyAnirban Niloy
8331218
8331218
1
$begingroup$
There is no solution. You cannot have both perimeters and areas the same.
$endgroup$
– Andrei
Jan 24 at 7:42
$begingroup$
That's why I got stucked. I often get this kind of faulty question but find no satisfied answer. The source of the problem is also enigmatic. Thank you for telling me that point. I frankly didn't know that point. Should I delete this post? Otherwise anyone will be confused.
$endgroup$
– Anirban Niloy
Jan 24 at 7:48
1
$begingroup$
I did not know it either. But it's easy to prove. I will post that as an answer.
$endgroup$
– Andrei
Jan 24 at 7:51
$begingroup$
Okay. I will check that out later. I'll be eagerly waiting.
$endgroup$
– Anirban Niloy
Jan 24 at 7:53
add a comment |
1
$begingroup$
There is no solution. You cannot have both perimeters and areas the same.
$endgroup$
– Andrei
Jan 24 at 7:42
$begingroup$
That's why I got stucked. I often get this kind of faulty question but find no satisfied answer. The source of the problem is also enigmatic. Thank you for telling me that point. I frankly didn't know that point. Should I delete this post? Otherwise anyone will be confused.
$endgroup$
– Anirban Niloy
Jan 24 at 7:48
1
$begingroup$
I did not know it either. But it's easy to prove. I will post that as an answer.
$endgroup$
– Andrei
Jan 24 at 7:51
$begingroup$
Okay. I will check that out later. I'll be eagerly waiting.
$endgroup$
– Anirban Niloy
Jan 24 at 7:53
1
1
$begingroup$
There is no solution. You cannot have both perimeters and areas the same.
$endgroup$
– Andrei
Jan 24 at 7:42
$begingroup$
There is no solution. You cannot have both perimeters and areas the same.
$endgroup$
– Andrei
Jan 24 at 7:42
$begingroup$
That's why I got stucked. I often get this kind of faulty question but find no satisfied answer. The source of the problem is also enigmatic. Thank you for telling me that point. I frankly didn't know that point. Should I delete this post? Otherwise anyone will be confused.
$endgroup$
– Anirban Niloy
Jan 24 at 7:48
$begingroup$
That's why I got stucked. I often get this kind of faulty question but find no satisfied answer. The source of the problem is also enigmatic. Thank you for telling me that point. I frankly didn't know that point. Should I delete this post? Otherwise anyone will be confused.
$endgroup$
– Anirban Niloy
Jan 24 at 7:48
1
1
$begingroup$
I did not know it either. But it's easy to prove. I will post that as an answer.
$endgroup$
– Andrei
Jan 24 at 7:51
$begingroup$
I did not know it either. But it's easy to prove. I will post that as an answer.
$endgroup$
– Andrei
Jan 24 at 7:51
$begingroup$
Okay. I will check that out later. I'll be eagerly waiting.
$endgroup$
– Anirban Niloy
Jan 24 at 7:53
$begingroup$
Okay. I will check that out later. I'll be eagerly waiting.
$endgroup$
– Anirban Niloy
Jan 24 at 7:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The problem has no solution. Let's assume that there is one. Then we notice that increasing or decreasing the length of $AD$ will not change the fact that areas/ perimeters are the same. Indeed, if we increase the length by $x$, the perimeter of both trapezoids will increase by $2x$, and the areas will both increase by $xcdot AE$. Then let's do $x=-|AD|$. Then in your picture $A=D$ and $P=E$. I will therefore drop any reference to $D$ and $E$ from now on.
$ABC$ is a right angle triangle, with the acute angles $30^circ$ and $60^circ$. Then $$AC=frac{sqrt 3}2 BC$$ and $$AB=frac 12 BC$$
In the right angle triangle $ABC$ I can write the area in two ways to get $$BCcdot AP=ABcdot AC=BC^2frac{sqrt 3}{4}$$
Therefore $AP=BC frac{sqrt 3}{4}$. You already showed that $PQ=BC$. So in the right angle triangle $APQ$ you can write $$tantheta=frac{AP}{PQ}=frac{sqrt 3}4$$
Now lets see that this does not verify equal perimeter requirement: $$AB+AC+BC=BCleft(frac 12+frac{sqrt 3}2+1right)$$
$$AP+PQ+QA=BCleft(frac{sqrt 3}{4}+1+sqrt{1^2+left(frac{sqrt 3}{4}right)^2}right)$$
You can see that there is something with $sqrt{19}$ in the second equation, that it's not there in the first.
Note: you could get to the same conclusion even if you explicitly carry around $ADne 0$.
answered Jan 24 at 8:13
AndreiAndrei
13.1k21230
$endgroup$
$begingroup$
I didn't understand the fact of triangle ABC being a right angled triangle. Would you please explain that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:41
$begingroup$
$APperp BC$. For the other one, if you have in a triangle an angle of $60^circ$ and an angle of $30^circ$, the last one has to be $90^circ$
$endgroup$
– Andrei
Jan 24 at 8:44
$begingroup$
But here we can see in that diagram that $angle DCE = 30^circ$. Steven gregory solved the value of $angle DQC but how can we justify that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:46
1
$begingroup$
Make $A=D$. In the picture in the other answer use $y=0$
$endgroup$
– Andrei
Jan 24 at 8:48
$begingroup$
Oops, sorry. You told me in your answer. It's a great honour that you gave your precious time to me. Tnx for your help and support.
$endgroup$
– Anirban Niloy
Jan 24 at 8:50
add a comment |
$begingroup$
begin{align}
operatorname{perimeter}(APQD) &= operatorname{perimeter}(ABCD) \
(3+sqrt 3)x + 2y + CQ+DQ &= (6+2sqrt 3)x + 2y \
CQ +DQ &= (3+sqrt 3)x
end{align}
begin{align}
operatorname{area}(APQD) &= operatorname{area}(ABCD) \
frac 12(3x+2y+CQ)(sqrt 3x) &= frac 12(4x+2y)((sqrt 3x)) \
CQ &= x \
hline
DQ &= (2+sqrt 3)x \
(DQ)^2 &= (7+4sqrt 3)x^2
end{align}
But, then, $(7+4sqrt 3)x^2=(DQ)^2 = (sqrt 3x)^2+(4x)^2=19x^2$
So, by contradiction, there is no solution.
answered Jan 24 at 8:39
steven gregorysteven gregory
18.3k32258
$endgroup$
1
$begingroup$
In triangle $DEQ$ you have $EQ=EC+CQ=4x$. Then $DQ^2=DE^2+EQ^2=3x^2+16x^2=19x^2$ but $xsqrt{19}ne(2+sqrt 3)x$
$endgroup$
– Andrei
Jan 24 at 8:52
1
$begingroup$
@stevengregory If you use the tangent function, $tanangle DQC=frac{sqrt 3 x}{4x}$ you get a different value for the angle, only about $23.4^circ$
$endgroup$
– Andrei
Jan 24 at 9:15
$begingroup$
Yeah, you're right. If we consider the condition as a right one (including area and perimeter are equal of both the field mentioned in the question), then $27.65^circ$ $approx$ $23.4^circ$. That makes no sense.
$endgroup$
– Anirban Niloy
Jan 24 at 10:20
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The problem has no solution. Let's assume that there is one. Then we notice that increasing or decreasing the length of $AD$ will not change the fact that areas/ perimeters are the same. Indeed, if we increase the length by $x$, the perimeter of both trapezoids will increase by $2x$, and the areas will both increase by $xcdot AE$. Then let's do $x=-|AD|$. Then in your picture $A=D$ and $P=E$. I will therefore drop any reference to $D$ and $E$ from now on.
$ABC$ is a right angle triangle, with the acute angles $30^circ$ and $60^circ$. Then $$AC=frac{sqrt 3}2 BC$$ and $$AB=frac 12 BC$$
In the right angle triangle $ABC$ I can write the area in two ways to get $$BCcdot AP=ABcdot AC=BC^2frac{sqrt 3}{4}$$
Therefore $AP=BC frac{sqrt 3}{4}$. You already showed that $PQ=BC$. So in the right angle triangle $APQ$ you can write $$tantheta=frac{AP}{PQ}=frac{sqrt 3}4$$
Now lets see that this does not verify equal perimeter requirement: $$AB+AC+BC=BCleft(frac 12+frac{sqrt 3}2+1right)$$
$$AP+PQ+QA=BCleft(frac{sqrt 3}{4}+1+sqrt{1^2+left(frac{sqrt 3}{4}right)^2}right)$$
You can see that there is something with $sqrt{19}$ in the second equation, that it's not there in the first.
Note: you could get to the same conclusion even if you explicitly carry around $ADne 0$.
answered Jan 24 at 8:13
AndreiAndrei
13.1k21230
$endgroup$
$begingroup$
I didn't understand the fact of triangle ABC being a right angled triangle. Would you please explain that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:41
$begingroup$
$APperp BC$. For the other one, if you have in a triangle an angle of $60^circ$ and an angle of $30^circ$, the last one has to be $90^circ$
$endgroup$
– Andrei
Jan 24 at 8:44
$begingroup$
But here we can see in that diagram that $angle DCE = 30^circ$. Steven gregory solved the value of $angle DQC but how can we justify that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:46
1
$begingroup$
Make $A=D$. In the picture in the other answer use $y=0$
$endgroup$
– Andrei
Jan 24 at 8:48
$begingroup$
Oops, sorry. You told me in your answer. It's a great honour that you gave your precious time to me. Tnx for your help and support.
$endgroup$
– Anirban Niloy
Jan 24 at 8:50
add a comment |
$begingroup$
The problem has no solution. Let's assume that there is one. Then we notice that increasing or decreasing the length of $AD$ will not change the fact that areas/ perimeters are the same. Indeed, if we increase the length by $x$, the perimeter of both trapezoids will increase by $2x$, and the areas will both increase by $xcdot AE$. Then let's do $x=-|AD|$. Then in your picture $A=D$ and $P=E$. I will therefore drop any reference to $D$ and $E$ from now on.
$ABC$ is a right angle triangle, with the acute angles $30^circ$ and $60^circ$. Then $$AC=frac{sqrt 3}2 BC$$ and $$AB=frac 12 BC$$
In the right angle triangle $ABC$ I can write the area in two ways to get $$BCcdot AP=ABcdot AC=BC^2frac{sqrt 3}{4}$$
Therefore $AP=BC frac{sqrt 3}{4}$. You already showed that $PQ=BC$. So in the right angle triangle $APQ$ you can write $$tantheta=frac{AP}{PQ}=frac{sqrt 3}4$$
Now lets see that this does not verify equal perimeter requirement: $$AB+AC+BC=BCleft(frac 12+frac{sqrt 3}2+1right)$$
$$AP+PQ+QA=BCleft(frac{sqrt 3}{4}+1+sqrt{1^2+left(frac{sqrt 3}{4}right)^2}right)$$
You can see that there is something with $sqrt{19}$ in the second equation, that it's not there in the first.
Note: you could get to the same conclusion even if you explicitly carry around $ADne 0$.
answered Jan 24 at 8:13
AndreiAndrei
13.1k21230
$endgroup$
$begingroup$
I didn't understand the fact of triangle ABC being a right angled triangle. Would you please explain that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:41
$begingroup$
$APperp BC$. For the other one, if you have in a triangle an angle of $60^circ$ and an angle of $30^circ$, the last one has to be $90^circ$
$endgroup$
– Andrei
Jan 24 at 8:44
$begingroup$
But here we can see in that diagram that $angle DCE = 30^circ$. Steven gregory solved the value of $angle DQC but how can we justify that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:46
1
$begingroup$
Make $A=D$. In the picture in the other answer use $y=0$
$endgroup$
– Andrei
Jan 24 at 8:48
$begingroup$
Oops, sorry. You told me in your answer. It's a great honour that you gave your precious time to me. Tnx for your help and support.
$endgroup$
– Anirban Niloy
Jan 24 at 8:50
add a comment |
$begingroup$
The problem has no solution. Let's assume that there is one. Then we notice that increasing or decreasing the length of $AD$ will not change the fact that areas/ perimeters are the same. Indeed, if we increase the length by $x$, the perimeter of both trapezoids will increase by $2x$, and the areas will both increase by $xcdot AE$. Then let's do $x=-|AD|$. Then in your picture $A=D$ and $P=E$. I will therefore drop any reference to $D$ and $E$ from now on.
$ABC$ is a right angle triangle, with the acute angles $30^circ$ and $60^circ$. Then $$AC=frac{sqrt 3}2 BC$$ and $$AB=frac 12 BC$$
In the right angle triangle $ABC$ I can write the area in two ways to get $$BCcdot AP=ABcdot AC=BC^2frac{sqrt 3}{4}$$
Therefore $AP=BC frac{sqrt 3}{4}$. You already showed that $PQ=BC$. So in the right angle triangle $APQ$ you can write $$tantheta=frac{AP}{PQ}=frac{sqrt 3}4$$
Now lets see that this does not verify equal perimeter requirement: $$AB+AC+BC=BCleft(frac 12+frac{sqrt 3}2+1right)$$
$$AP+PQ+QA=BCleft(frac{sqrt 3}{4}+1+sqrt{1^2+left(frac{sqrt 3}{4}right)^2}right)$$
You can see that there is something with $sqrt{19}$ in the second equation, that it's not there in the first.
Note: you could get to the same conclusion even if you explicitly carry around $ADne 0$.
answered Jan 24 at 8:13
AndreiAndrei
13.1k21230
$endgroup$
The problem has no solution. Let's assume that there is one. Then we notice that increasing or decreasing the length of $AD$ will not change the fact that areas/ perimeters are the same. Indeed, if we increase the length by $x$, the perimeter of both trapezoids will increase by $2x$, and the areas will both increase by $xcdot AE$. Then let's do $x=-|AD|$. Then in your picture $A=D$ and $P=E$. I will therefore drop any reference to $D$ and $E$ from now on.
$ABC$ is a right angle triangle, with the acute angles $30^circ$ and $60^circ$. Then $$AC=frac{sqrt 3}2 BC$$ and $$AB=frac 12 BC$$
In the right angle triangle $ABC$ I can write the area in two ways to get $$BCcdot AP=ABcdot AC=BC^2frac{sqrt 3}{4}$$
Therefore $AP=BC frac{sqrt 3}{4}$. You already showed that $PQ=BC$. So in the right angle triangle $APQ$ you can write $$tantheta=frac{AP}{PQ}=frac{sqrt 3}4$$
Now lets see that this does not verify equal perimeter requirement: $$AB+AC+BC=BCleft(frac 12+frac{sqrt 3}2+1right)$$
$$AP+PQ+QA=BCleft(frac{sqrt 3}{4}+1+sqrt{1^2+left(frac{sqrt 3}{4}right)^2}right)$$
You can see that there is something with $sqrt{19}$ in the second equation, that it's not there in the first.
Note: you could get to the same conclusion even if you explicitly carry around $ADne 0$.
answered Jan 24 at 8:13
AndreiAndrei
13.1k21230
answered Jan 24 at 8:13
AndreiAndrei
13.1k21230
answered Jan 24 at 8:13
AndreiAndrei
13.1k21230
answered Jan 24 at 8:13
AndreiAndrei
13.1k21230
13.1k21230
$begingroup$
I didn't understand the fact of triangle ABC being a right angled triangle. Would you please explain that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:41
$begingroup$
$APperp BC$. For the other one, if you have in a triangle an angle of $60^circ$ and an angle of $30^circ$, the last one has to be $90^circ$
$endgroup$
– Andrei
Jan 24 at 8:44
$begingroup$
But here we can see in that diagram that $angle DCE = 30^circ$. Steven gregory solved the value of $angle DQC but how can we justify that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:46
1
$begingroup$
Make $A=D$. In the picture in the other answer use $y=0$
$endgroup$
– Andrei
Jan 24 at 8:48
$begingroup$
Oops, sorry. You told me in your answer. It's a great honour that you gave your precious time to me. Tnx for your help and support.
$endgroup$
– Anirban Niloy
Jan 24 at 8:50
add a comment |
$begingroup$
I didn't understand the fact of triangle ABC being a right angled triangle. Would you please explain that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:41
$begingroup$
$APperp BC$. For the other one, if you have in a triangle an angle of $60^circ$ and an angle of $30^circ$, the last one has to be $90^circ$
$endgroup$
– Andrei
Jan 24 at 8:44
$begingroup$
But here we can see in that diagram that $angle DCE = 30^circ$. Steven gregory solved the value of $angle DQC but how can we justify that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:46
1
$begingroup$
Make $A=D$. In the picture in the other answer use $y=0$
$endgroup$
– Andrei
Jan 24 at 8:48
$begingroup$
Oops, sorry. You told me in your answer. It's a great honour that you gave your precious time to me. Tnx for your help and support.
$endgroup$
– Anirban Niloy
Jan 24 at 8:50
$begingroup$
I didn't understand the fact of triangle ABC being a right angled triangle. Would you please explain that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:41
$begingroup$
I didn't understand the fact of triangle ABC being a right angled triangle. Would you please explain that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:41
$begingroup$
$APperp BC$. For the other one, if you have in a triangle an angle of $60^circ$ and an angle of $30^circ$, the last one has to be $90^circ$
$endgroup$
– Andrei
Jan 24 at 8:44
$begingroup$
$APperp BC$. For the other one, if you have in a triangle an angle of $60^circ$ and an angle of $30^circ$, the last one has to be $90^circ$
$endgroup$
– Andrei
Jan 24 at 8:44
$begingroup$
But here we can see in that diagram that $angle DCE = 30^circ$. Steven gregory solved the value of $angle DQC but how can we justify that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:46
$begingroup$
But here we can see in that diagram that $angle DCE = 30^circ$. Steven gregory solved the value of $angle DQC but how can we justify that?
$endgroup$
– Anirban Niloy
Jan 24 at 8:46
1
1
$begingroup$
Make $A=D$. In the picture in the other answer use $y=0$
$endgroup$
– Andrei
Jan 24 at 8:48
$begingroup$
Make $A=D$. In the picture in the other answer use $y=0$
$endgroup$
– Andrei
Jan 24 at 8:48
$begingroup$
Oops, sorry. You told me in your answer. It's a great honour that you gave your precious time to me. Tnx for your help and support.
$endgroup$
– Anirban Niloy
Jan 24 at 8:50
$begingroup$
Oops, sorry. You told me in your answer. It's a great honour that you gave your precious time to me. Tnx for your help and support.
$endgroup$
– Anirban Niloy
Jan 24 at 8:50
add a comment |
$begingroup$
begin{align}
operatorname{perimeter}(APQD) &= operatorname{perimeter}(ABCD) \
(3+sqrt 3)x + 2y + CQ+DQ &= (6+2sqrt 3)x + 2y \
CQ +DQ &= (3+sqrt 3)x
end{align}
begin{align}
operatorname{area}(APQD) &= operatorname{area}(ABCD) \
frac 12(3x+2y+CQ)(sqrt 3x) &= frac 12(4x+2y)((sqrt 3x)) \
CQ &= x \
hline
DQ &= (2+sqrt 3)x \
(DQ)^2 &= (7+4sqrt 3)x^2
end{align}
But, then, $(7+4sqrt 3)x^2=(DQ)^2 = (sqrt 3x)^2+(4x)^2=19x^2$
So, by contradiction, there is no solution.
answered Jan 24 at 8:39
steven gregorysteven gregory
18.3k32258
$endgroup$
1
$begingroup$
In triangle $DEQ$ you have $EQ=EC+CQ=4x$. Then $DQ^2=DE^2+EQ^2=3x^2+16x^2=19x^2$ but $xsqrt{19}ne(2+sqrt 3)x$
$endgroup$
– Andrei
Jan 24 at 8:52
1
$begingroup$
@stevengregory If you use the tangent function, $tanangle DQC=frac{sqrt 3 x}{4x}$ you get a different value for the angle, only about $23.4^circ$
$endgroup$
– Andrei
Jan 24 at 9:15
$begingroup$
Yeah, you're right. If we consider the condition as a right one (including area and perimeter are equal of both the field mentioned in the question), then $27.65^circ$ $approx$ $23.4^circ$. That makes no sense.
$endgroup$
– Anirban Niloy
Jan 24 at 10:20
add a comment |
$begingroup$
begin{align}
operatorname{perimeter}(APQD) &= operatorname{perimeter}(ABCD) \
(3+sqrt 3)x + 2y + CQ+DQ &= (6+2sqrt 3)x + 2y \
CQ +DQ &= (3+sqrt 3)x
end{align}
begin{align}
operatorname{area}(APQD) &= operatorname{area}(ABCD) \
frac 12(3x+2y+CQ)(sqrt 3x) &= frac 12(4x+2y)((sqrt 3x)) \
CQ &= x \
hline
DQ &= (2+sqrt 3)x \
(DQ)^2 &= (7+4sqrt 3)x^2
end{align}
But, then, $(7+4sqrt 3)x^2=(DQ)^2 = (sqrt 3x)^2+(4x)^2=19x^2$
So, by contradiction, there is no solution.
answered Jan 24 at 8:39
steven gregorysteven gregory
18.3k32258
$endgroup$
1
$begingroup$
In triangle $DEQ$ you have $EQ=EC+CQ=4x$. Then $DQ^2=DE^2+EQ^2=3x^2+16x^2=19x^2$ but $xsqrt{19}ne(2+sqrt 3)x$
$endgroup$
– Andrei
Jan 24 at 8:52
1
$begingroup$
@stevengregory If you use the tangent function, $tanangle DQC=frac{sqrt 3 x}{4x}$ you get a different value for the angle, only about $23.4^circ$
$endgroup$
– Andrei
Jan 24 at 9:15
$begingroup$
Yeah, you're right. If we consider the condition as a right one (including area and perimeter are equal of both the field mentioned in the question), then $27.65^circ$ $approx$ $23.4^circ$. That makes no sense.
$endgroup$
– Anirban Niloy
Jan 24 at 10:20
add a comment |
$begingroup$
begin{align}
operatorname{perimeter}(APQD) &= operatorname{perimeter}(ABCD) \
(3+sqrt 3)x + 2y + CQ+DQ &= (6+2sqrt 3)x + 2y \
CQ +DQ &= (3+sqrt 3)x
end{align}
begin{align}
operatorname{area}(APQD) &= operatorname{area}(ABCD) \
frac 12(3x+2y+CQ)(sqrt 3x) &= frac 12(4x+2y)((sqrt 3x)) \
CQ &= x \
hline
DQ &= (2+sqrt 3)x \
(DQ)^2 &= (7+4sqrt 3)x^2
end{align}
But, then, $(7+4sqrt 3)x^2=(DQ)^2 = (sqrt 3x)^2+(4x)^2=19x^2$
So, by contradiction, there is no solution.
answered Jan 24 at 8:39
steven gregorysteven gregory
18.3k32258
$endgroup$
begin{align}
operatorname{perimeter}(APQD) &= operatorname{perimeter}(ABCD) \
(3+sqrt 3)x + 2y + CQ+DQ &= (6+2sqrt 3)x + 2y \
CQ +DQ &= (3+sqrt 3)x
end{align}
begin{align}
operatorname{area}(APQD) &= operatorname{area}(ABCD) \
frac 12(3x+2y+CQ)(sqrt 3x) &= frac 12(4x+2y)((sqrt 3x)) \
CQ &= x \
hline
DQ &= (2+sqrt 3)x \
(DQ)^2 &= (7+4sqrt 3)x^2
end{align}
But, then, $(7+4sqrt 3)x^2=(DQ)^2 = (sqrt 3x)^2+(4x)^2=19x^2$
So, by contradiction, there is no solution.
answered Jan 24 at 8:39
steven gregorysteven gregory
18.3k32258
edited Jan 24 at 11:01
answered Jan 24 at 8:39
steven gregorysteven gregory
18.3k32258
answered Jan 24 at 8:39
steven gregorysteven gregory
18.3k32258
answered Jan 24 at 8:39
steven gregorysteven gregory
18.3k32258
18.3k32258
1
$begingroup$
In triangle $DEQ$ you have $EQ=EC+CQ=4x$. Then $DQ^2=DE^2+EQ^2=3x^2+16x^2=19x^2$ but $xsqrt{19}ne(2+sqrt 3)x$
$endgroup$
– Andrei
Jan 24 at 8:52
1
$begingroup$
@stevengregory If you use the tangent function, $tanangle DQC=frac{sqrt 3 x}{4x}$ you get a different value for the angle, only about $23.4^circ$
$endgroup$
– Andrei
Jan 24 at 9:15
$begingroup$
Yeah, you're right. If we consider the condition as a right one (including area and perimeter are equal of both the field mentioned in the question), then $27.65^circ$ $approx$ $23.4^circ$. That makes no sense.
$endgroup$
– Anirban Niloy
Jan 24 at 10:20
add a comment |
1
$begingroup$
In triangle $DEQ$ you have $EQ=EC+CQ=4x$. Then $DQ^2=DE^2+EQ^2=3x^2+16x^2=19x^2$ but $xsqrt{19}ne(2+sqrt 3)x$
$endgroup$
– Andrei
Jan 24 at 8:52
1
$begingroup$
@stevengregory If you use the tangent function, $tanangle DQC=frac{sqrt 3 x}{4x}$ you get a different value for the angle, only about $23.4^circ$
$endgroup$
– Andrei
Jan 24 at 9:15
$begingroup$
Yeah, you're right. If we consider the condition as a right one (including area and perimeter are equal of both the field mentioned in the question), then $27.65^circ$ $approx$ $23.4^circ$. That makes no sense.
$endgroup$
– Anirban Niloy
Jan 24 at 10:20
1
1
$begingroup$
In triangle $DEQ$ you have $EQ=EC+CQ=4x$. Then $DQ^2=DE^2+EQ^2=3x^2+16x^2=19x^2$ but $xsqrt{19}ne(2+sqrt 3)x$
$endgroup$
– Andrei
Jan 24 at 8:52
$begingroup$
In triangle $DEQ$ you have $EQ=EC+CQ=4x$. Then $DQ^2=DE^2+EQ^2=3x^2+16x^2=19x^2$ but $xsqrt{19}ne(2+sqrt 3)x$
$endgroup$
– Andrei
Jan 24 at 8:52
1
1
$begingroup$
@stevengregory If you use the tangent function, $tanangle DQC=frac{sqrt 3 x}{4x}$ you get a different value for the angle, only about $23.4^circ$
$endgroup$
– Andrei
Jan 24 at 9:15
$begingroup$
@stevengregory If you use the tangent function, $tanangle DQC=frac{sqrt 3 x}{4x}$ you get a different value for the angle, only about $23.4^circ$
$endgroup$
– Andrei
Jan 24 at 9:15
$begingroup$
Yeah, you're right. If we consider the condition as a right one (including area and perimeter are equal of both the field mentioned in the question), then $27.65^circ$ $approx$ $23.4^circ$. That makes no sense.
$endgroup$
– Anirban Niloy
Jan 24 at 10:20
$begingroup$
Yeah, you're right. If we consider the condition as a right one (including area and perimeter are equal of both the field mentioned in the question), then $27.65^circ$ $approx$ $23.4^circ$. That makes no sense.
$endgroup$
– Anirban Niloy
Jan 24 at 10:20
add a comment |
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1
$begingroup$
There is no solution. You cannot have both perimeters and areas the same.
$endgroup$
– Andrei
Jan 24 at 7:42
$begingroup$
That's why I got stucked. I often get this kind of faulty question but find no satisfied answer. The source of the problem is also enigmatic. Thank you for telling me that point. I frankly didn't know that point. Should I delete this post? Otherwise anyone will be confused.
$endgroup$
– Anirban Niloy
Jan 24 at 7:48
1
$begingroup$
I did not know it either. But it's easy to prove. I will post that as an answer.
$endgroup$
– Andrei
Jan 24 at 7:51
$begingroup$
Okay. I will check that out later. I'll be eagerly waiting.
$endgroup$
– Anirban Niloy
Jan 24 at 7:53