A line segment with a length of 24 makes a 90-degree angle with one of the legs of an isosceles trapezoid....
Given that $ABCD$ is an isosceles trapezoid and that $|EB|=24$, $|EC|=26$, and m(EBC)=$90^o$. Find $A(ABCD)= ?$
From the pythagorean theorem, I can find that $|BC|=|AD|=10$. Then, I can find the area of the triangle $EBC$. But after this point, I can't progress any further. How do I find the area of this trapezoid?
geometry euclidean-geometry quadrilateral
edited Dec 27 '18 at 18:56
greedoid
38k114794
asked Dec 27 '18 at 18:30
Eldar Rahimli
767
add a comment |
Given that $ABCD$ is an isosceles trapezoid and that $|EB|=24$, $|EC|=26$, and m(EBC)=$90^o$. Find $A(ABCD)= ?$
From the pythagorean theorem, I can find that $|BC|=|AD|=10$. Then, I can find the area of the triangle $EBC$. But after this point, I can't progress any further. How do I find the area of this trapezoid?
geometry euclidean-geometry quadrilateral
edited Dec 27 '18 at 18:56
greedoid
38k114794
asked Dec 27 '18 at 18:30
Eldar Rahimli
767
2
ABCD does not look very isosceles in the image ...
– Henning Makholm
Dec 27 '18 at 19:03
Oh, Sorry, I didn't notice when I drew it
– Eldar Rahimli
Dec 27 '18 at 19:04
add a comment |
Given that $ABCD$ is an isosceles trapezoid and that $|EB|=24$, $|EC|=26$, and m(EBC)=$90^o$. Find $A(ABCD)= ?$
From the pythagorean theorem, I can find that $|BC|=|AD|=10$. Then, I can find the area of the triangle $EBC$. But after this point, I can't progress any further. How do I find the area of this trapezoid?
geometry euclidean-geometry quadrilateral
edited Dec 27 '18 at 18:56
greedoid
38k114794
asked Dec 27 '18 at 18:30
Eldar Rahimli
767
Given that $ABCD$ is an isosceles trapezoid and that $|EB|=24$, $|EC|=26$, and m(EBC)=$90^o$. Find $A(ABCD)= ?$
From the pythagorean theorem, I can find that $|BC|=|AD|=10$. Then, I can find the area of the triangle $EBC$. But after this point, I can't progress any further. How do I find the area of this trapezoid?
geometry euclidean-geometry quadrilateral
geometry euclidean-geometry quadrilateral
edited Dec 27 '18 at 18:56
greedoid
38k114794
asked Dec 27 '18 at 18:30
Eldar Rahimli
767
edited Dec 27 '18 at 18:56
greedoid
38k114794
asked Dec 27 '18 at 18:30
Eldar Rahimli
767
edited Dec 27 '18 at 18:56
greedoid
38k114794
edited Dec 27 '18 at 18:56
greedoid
38k114794
edited Dec 27 '18 at 18:56
greedoid
38k114794
38k114794
asked Dec 27 '18 at 18:30
Eldar Rahimli
767
asked Dec 27 '18 at 18:30
Eldar Rahimli
767
asked Dec 27 '18 at 18:30
Eldar Rahimli
767
767
2
ABCD does not look very isosceles in the image ...
– Henning Makholm
Dec 27 '18 at 19:03
Oh, Sorry, I didn't notice when I drew it
– Eldar Rahimli
Dec 27 '18 at 19:04
add a comment |
2
ABCD does not look very isosceles in the image ...
– Henning Makholm
Dec 27 '18 at 19:03
Oh, Sorry, I didn't notice when I drew it
– Eldar Rahimli
Dec 27 '18 at 19:04
2
2
ABCD does not look very isosceles in the image ...
– Henning Makholm
Dec 27 '18 at 19:03
ABCD does not look very isosceles in the image ...
– Henning Makholm
Dec 27 '18 at 19:03
Oh, Sorry, I didn't notice when I drew it
– Eldar Rahimli
Dec 27 '18 at 19:04
Oh, Sorry, I didn't notice when I drew it
– Eldar Rahimli
Dec 27 '18 at 19:04
add a comment |
3 Answers
3
active
oldest
votes
Option 1. Consider the isosceles trapezoid with $BE=BD$ as diagonal:
$hspace{3cm}$
We find:
$$BF=frac{2S_{Delta BCD}}{CD}=frac{240}{26}=frac{120}{13};\
CF=sqrt{BC^2-BF^2}=sqrt{100-frac{120^2}{13^2}}=frac{50}{13};\
S_{ABCD}=frac{AB+CD}{2}cdot BF=frac{(26-2cdot CF)+26}{2}cdot frac{120}{13}=\
=frac{34560}{169}approx color{red}{204.5}.\
$$
Option 2. Consider the point $E$ as a midpoint:
$hspace{3cm}$
We find:
$$EH=sqrt{BE^2+BH^2}=sqrt{24^2+5^2}=sqrt{601};\
BI=frac{2S_{Delta BEH}}{EH}=frac{2cdot 60}{sqrt{601}};\
S_{ABCD}=EHcdot BF=sqrt{601}cdot frac{4cdot 60}{sqrt{601}}=color{red}{240}.$$
Conclusion: The trapezoid is not unique.
answered Dec 27 '18 at 19:20
farruhota
19.3k2736
Interesting that the area in option 2 is exactly the same as when we take $A=E$. Perhaps that is what misled the problem setter?
– Henning Makholm
Dec 27 '18 at 19:28
Perhaps, the setter considered the two extreme cases and concluded. Though, the option 1 was the extreme.
– farruhota
Dec 27 '18 at 19:33
x @farruhota: But the extreme cases are $204.5$ and $240$. He would need to have considered one extreme and one in-the-middle case. (In your option 2, if we cut off triangle DEG and put it next to AE instead, we get a parallelogram that is obviously twice the area of BEC; the setter may have mistakenly thought that this generalizes).
– Henning Makholm
Dec 27 '18 at 19:37
1
@farruhota This was a problem from a high school geometry textbook. It was given in the properties section that when $E$ is the middle point the area of the triangle is half of the trapezoid's. Indeed, I believe that is where problem setter made a mistake by not mentioning that $AE=ED$. Thanks for the answer, by the way.
– Eldar Rahimli
Dec 28 '18 at 15:30
1
@TheGreatDuck, here it states "at least one pair of parallel sides" and shows special types of trapezoids that are parallelograms, rectangles, rhombi and squares.
– farruhota
Dec 29 '18 at 5:35
|
show 7 more comments
As a concrete example of how the figure is not determined:
One option is that $A$ and $E$ coincide, and $ABCD$ is a rectangle of area $24cdot 10=240$.
Another option is that $E$ and $D$ coincide, in which case the area of the trapezoid is $frac{10cdot 24}{26}(26-frac{10cdot 10}{26}) approx 204$.
answered Dec 27 '18 at 19:14
Henning Makholm
238k16303537
thanks for the answer. can you explain what do you mean by coincide,please? The answer is indeed 240
– Eldar Rahimli
Dec 27 '18 at 19:16
@EldarRahimli: "Coincide" means that $A$ and $E$ are the same point, so EC is simply the diagonal of the rectangle. It is true that $240$ is one possible answer, but based on the conditions you have disclosed it is not the only possible answer.
– Henning Makholm
Dec 27 '18 at 19:21
add a comment |
The triangle $BCE$ is uniqely determined, but other points $D$ and $A$ are not so this trapezoid does not have fixed area, it depend on $A$ (and then is $D$ determined also).
answered Dec 27 '18 at 18:38
greedoid
38k114794
If $|BCE|$ is fixed, then doesn't this fix A and D, too? Can you elaborate on your answer, please?
– Eldar Rahimli
Dec 27 '18 at 18:40
1
Note that the trapezoid is isosceles.
– Arthur
Dec 27 '18 at 18:43
Try to play in some aplet, say Geogebra. You can move $AD$ through $E$ and you will see you don't get a single trapezoid.
– greedoid
Dec 27 '18 at 18:43
How does that affect? @Arthur
– greedoid
Dec 27 '18 at 18:44
1
You can tilt the sides. Change the $BCD$ angle.
– Andrei
Dec 27 '18 at 18:48
|
show 6 more comments
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Option 1. Consider the isosceles trapezoid with $BE=BD$ as diagonal:
$hspace{3cm}$
We find:
$$BF=frac{2S_{Delta BCD}}{CD}=frac{240}{26}=frac{120}{13};\
CF=sqrt{BC^2-BF^2}=sqrt{100-frac{120^2}{13^2}}=frac{50}{13};\
S_{ABCD}=frac{AB+CD}{2}cdot BF=frac{(26-2cdot CF)+26}{2}cdot frac{120}{13}=\
=frac{34560}{169}approx color{red}{204.5}.\
$$
Option 2. Consider the point $E$ as a midpoint:
$hspace{3cm}$
We find:
$$EH=sqrt{BE^2+BH^2}=sqrt{24^2+5^2}=sqrt{601};\
BI=frac{2S_{Delta BEH}}{EH}=frac{2cdot 60}{sqrt{601}};\
S_{ABCD}=EHcdot BF=sqrt{601}cdot frac{4cdot 60}{sqrt{601}}=color{red}{240}.$$
Conclusion: The trapezoid is not unique.
answered Dec 27 '18 at 19:20
farruhota
19.3k2736
Interesting that the area in option 2 is exactly the same as when we take $A=E$. Perhaps that is what misled the problem setter?
– Henning Makholm
Dec 27 '18 at 19:28
Perhaps, the setter considered the two extreme cases and concluded. Though, the option 1 was the extreme.
– farruhota
Dec 27 '18 at 19:33
x @farruhota: But the extreme cases are $204.5$ and $240$. He would need to have considered one extreme and one in-the-middle case. (In your option 2, if we cut off triangle DEG and put it next to AE instead, we get a parallelogram that is obviously twice the area of BEC; the setter may have mistakenly thought that this generalizes).
– Henning Makholm
Dec 27 '18 at 19:37
1
@farruhota This was a problem from a high school geometry textbook. It was given in the properties section that when $E$ is the middle point the area of the triangle is half of the trapezoid's. Indeed, I believe that is where problem setter made a mistake by not mentioning that $AE=ED$. Thanks for the answer, by the way.
– Eldar Rahimli
Dec 28 '18 at 15:30
1
@TheGreatDuck, here it states "at least one pair of parallel sides" and shows special types of trapezoids that are parallelograms, rectangles, rhombi and squares.
– farruhota
Dec 29 '18 at 5:35
|
show 7 more comments
Option 1. Consider the isosceles trapezoid with $BE=BD$ as diagonal:
$hspace{3cm}$
We find:
$$BF=frac{2S_{Delta BCD}}{CD}=frac{240}{26}=frac{120}{13};\
CF=sqrt{BC^2-BF^2}=sqrt{100-frac{120^2}{13^2}}=frac{50}{13};\
S_{ABCD}=frac{AB+CD}{2}cdot BF=frac{(26-2cdot CF)+26}{2}cdot frac{120}{13}=\
=frac{34560}{169}approx color{red}{204.5}.\
$$
Option 2. Consider the point $E$ as a midpoint:
$hspace{3cm}$
We find:
$$EH=sqrt{BE^2+BH^2}=sqrt{24^2+5^2}=sqrt{601};\
BI=frac{2S_{Delta BEH}}{EH}=frac{2cdot 60}{sqrt{601}};\
S_{ABCD}=EHcdot BF=sqrt{601}cdot frac{4cdot 60}{sqrt{601}}=color{red}{240}.$$
Conclusion: The trapezoid is not unique.
answered Dec 27 '18 at 19:20
farruhota
19.3k2736
Interesting that the area in option 2 is exactly the same as when we take $A=E$. Perhaps that is what misled the problem setter?
– Henning Makholm
Dec 27 '18 at 19:28
Perhaps, the setter considered the two extreme cases and concluded. Though, the option 1 was the extreme.
– farruhota
Dec 27 '18 at 19:33
x @farruhota: But the extreme cases are $204.5$ and $240$. He would need to have considered one extreme and one in-the-middle case. (In your option 2, if we cut off triangle DEG and put it next to AE instead, we get a parallelogram that is obviously twice the area of BEC; the setter may have mistakenly thought that this generalizes).
– Henning Makholm
Dec 27 '18 at 19:37
1
@farruhota This was a problem from a high school geometry textbook. It was given in the properties section that when $E$ is the middle point the area of the triangle is half of the trapezoid's. Indeed, I believe that is where problem setter made a mistake by not mentioning that $AE=ED$. Thanks for the answer, by the way.
– Eldar Rahimli
Dec 28 '18 at 15:30
1
@TheGreatDuck, here it states "at least one pair of parallel sides" and shows special types of trapezoids that are parallelograms, rectangles, rhombi and squares.
– farruhota
Dec 29 '18 at 5:35
|
show 7 more comments
Option 1. Consider the isosceles trapezoid with $BE=BD$ as diagonal:
$hspace{3cm}$
We find:
$$BF=frac{2S_{Delta BCD}}{CD}=frac{240}{26}=frac{120}{13};\
CF=sqrt{BC^2-BF^2}=sqrt{100-frac{120^2}{13^2}}=frac{50}{13};\
S_{ABCD}=frac{AB+CD}{2}cdot BF=frac{(26-2cdot CF)+26}{2}cdot frac{120}{13}=\
=frac{34560}{169}approx color{red}{204.5}.\
$$
Option 2. Consider the point $E$ as a midpoint:
$hspace{3cm}$
We find:
$$EH=sqrt{BE^2+BH^2}=sqrt{24^2+5^2}=sqrt{601};\
BI=frac{2S_{Delta BEH}}{EH}=frac{2cdot 60}{sqrt{601}};\
S_{ABCD}=EHcdot BF=sqrt{601}cdot frac{4cdot 60}{sqrt{601}}=color{red}{240}.$$
Conclusion: The trapezoid is not unique.
answered Dec 27 '18 at 19:20
farruhota
19.3k2736
Option 1. Consider the isosceles trapezoid with $BE=BD$ as diagonal:
$hspace{3cm}$
We find:
$$BF=frac{2S_{Delta BCD}}{CD}=frac{240}{26}=frac{120}{13};\
CF=sqrt{BC^2-BF^2}=sqrt{100-frac{120^2}{13^2}}=frac{50}{13};\
S_{ABCD}=frac{AB+CD}{2}cdot BF=frac{(26-2cdot CF)+26}{2}cdot frac{120}{13}=\
=frac{34560}{169}approx color{red}{204.5}.\
$$
Option 2. Consider the point $E$ as a midpoint:
$hspace{3cm}$
We find:
$$EH=sqrt{BE^2+BH^2}=sqrt{24^2+5^2}=sqrt{601};\
BI=frac{2S_{Delta BEH}}{EH}=frac{2cdot 60}{sqrt{601}};\
S_{ABCD}=EHcdot BF=sqrt{601}cdot frac{4cdot 60}{sqrt{601}}=color{red}{240}.$$
Conclusion: The trapezoid is not unique.
answered Dec 27 '18 at 19:20
farruhota
19.3k2736
answered Dec 27 '18 at 19:20
farruhota
19.3k2736
answered Dec 27 '18 at 19:20
farruhota
19.3k2736
answered Dec 27 '18 at 19:20
farruhota
19.3k2736
19.3k2736
Interesting that the area in option 2 is exactly the same as when we take $A=E$. Perhaps that is what misled the problem setter?
– Henning Makholm
Dec 27 '18 at 19:28
Perhaps, the setter considered the two extreme cases and concluded. Though, the option 1 was the extreme.
– farruhota
Dec 27 '18 at 19:33
x @farruhota: But the extreme cases are $204.5$ and $240$. He would need to have considered one extreme and one in-the-middle case. (In your option 2, if we cut off triangle DEG and put it next to AE instead, we get a parallelogram that is obviously twice the area of BEC; the setter may have mistakenly thought that this generalizes).
– Henning Makholm
Dec 27 '18 at 19:37
1
@farruhota This was a problem from a high school geometry textbook. It was given in the properties section that when $E$ is the middle point the area of the triangle is half of the trapezoid's. Indeed, I believe that is where problem setter made a mistake by not mentioning that $AE=ED$. Thanks for the answer, by the way.
– Eldar Rahimli
Dec 28 '18 at 15:30
1
@TheGreatDuck, here it states "at least one pair of parallel sides" and shows special types of trapezoids that are parallelograms, rectangles, rhombi and squares.
– farruhota
Dec 29 '18 at 5:35
|
show 7 more comments
Interesting that the area in option 2 is exactly the same as when we take $A=E$. Perhaps that is what misled the problem setter?
– Henning Makholm
Dec 27 '18 at 19:28
Perhaps, the setter considered the two extreme cases and concluded. Though, the option 1 was the extreme.
– farruhota
Dec 27 '18 at 19:33
x @farruhota: But the extreme cases are $204.5$ and $240$. He would need to have considered one extreme and one in-the-middle case. (In your option 2, if we cut off triangle DEG and put it next to AE instead, we get a parallelogram that is obviously twice the area of BEC; the setter may have mistakenly thought that this generalizes).
– Henning Makholm
Dec 27 '18 at 19:37
1
@farruhota This was a problem from a high school geometry textbook. It was given in the properties section that when $E$ is the middle point the area of the triangle is half of the trapezoid's. Indeed, I believe that is where problem setter made a mistake by not mentioning that $AE=ED$. Thanks for the answer, by the way.
– Eldar Rahimli
Dec 28 '18 at 15:30
1
@TheGreatDuck, here it states "at least one pair of parallel sides" and shows special types of trapezoids that are parallelograms, rectangles, rhombi and squares.
– farruhota
Dec 29 '18 at 5:35
Interesting that the area in option 2 is exactly the same as when we take $A=E$. Perhaps that is what misled the problem setter?
– Henning Makholm
Dec 27 '18 at 19:28
Interesting that the area in option 2 is exactly the same as when we take $A=E$. Perhaps that is what misled the problem setter?
– Henning Makholm
Dec 27 '18 at 19:28
Perhaps, the setter considered the two extreme cases and concluded. Though, the option 1 was the extreme.
– farruhota
Dec 27 '18 at 19:33
Perhaps, the setter considered the two extreme cases and concluded. Though, the option 1 was the extreme.
– farruhota
Dec 27 '18 at 19:33
x @farruhota: But the extreme cases are $204.5$ and $240$. He would need to have considered one extreme and one in-the-middle case. (In your option 2, if we cut off triangle DEG and put it next to AE instead, we get a parallelogram that is obviously twice the area of BEC; the setter may have mistakenly thought that this generalizes).
– Henning Makholm
Dec 27 '18 at 19:37
x @farruhota: But the extreme cases are $204.5$ and $240$. He would need to have considered one extreme and one in-the-middle case. (In your option 2, if we cut off triangle DEG and put it next to AE instead, we get a parallelogram that is obviously twice the area of BEC; the setter may have mistakenly thought that this generalizes).
– Henning Makholm
Dec 27 '18 at 19:37
1
1
@farruhota This was a problem from a high school geometry textbook. It was given in the properties section that when $E$ is the middle point the area of the triangle is half of the trapezoid's. Indeed, I believe that is where problem setter made a mistake by not mentioning that $AE=ED$. Thanks for the answer, by the way.
– Eldar Rahimli
Dec 28 '18 at 15:30
@farruhota This was a problem from a high school geometry textbook. It was given in the properties section that when $E$ is the middle point the area of the triangle is half of the trapezoid's. Indeed, I believe that is where problem setter made a mistake by not mentioning that $AE=ED$. Thanks for the answer, by the way.
– Eldar Rahimli
Dec 28 '18 at 15:30
1
1
@TheGreatDuck, here it states "at least one pair of parallel sides" and shows special types of trapezoids that are parallelograms, rectangles, rhombi and squares.
– farruhota
Dec 29 '18 at 5:35
@TheGreatDuck, here it states "at least one pair of parallel sides" and shows special types of trapezoids that are parallelograms, rectangles, rhombi and squares.
– farruhota
Dec 29 '18 at 5:35
|
show 7 more comments
As a concrete example of how the figure is not determined:
One option is that $A$ and $E$ coincide, and $ABCD$ is a rectangle of area $24cdot 10=240$.
Another option is that $E$ and $D$ coincide, in which case the area of the trapezoid is $frac{10cdot 24}{26}(26-frac{10cdot 10}{26}) approx 204$.
answered Dec 27 '18 at 19:14
Henning Makholm
238k16303537
thanks for the answer. can you explain what do you mean by coincide,please? The answer is indeed 240
– Eldar Rahimli
Dec 27 '18 at 19:16
@EldarRahimli: "Coincide" means that $A$ and $E$ are the same point, so EC is simply the diagonal of the rectangle. It is true that $240$ is one possible answer, but based on the conditions you have disclosed it is not the only possible answer.
– Henning Makholm
Dec 27 '18 at 19:21
add a comment |
As a concrete example of how the figure is not determined:
One option is that $A$ and $E$ coincide, and $ABCD$ is a rectangle of area $24cdot 10=240$.
Another option is that $E$ and $D$ coincide, in which case the area of the trapezoid is $frac{10cdot 24}{26}(26-frac{10cdot 10}{26}) approx 204$.
answered Dec 27 '18 at 19:14
Henning Makholm
238k16303537
thanks for the answer. can you explain what do you mean by coincide,please? The answer is indeed 240
– Eldar Rahimli
Dec 27 '18 at 19:16
@EldarRahimli: "Coincide" means that $A$ and $E$ are the same point, so EC is simply the diagonal of the rectangle. It is true that $240$ is one possible answer, but based on the conditions you have disclosed it is not the only possible answer.
– Henning Makholm
Dec 27 '18 at 19:21
add a comment |
As a concrete example of how the figure is not determined:
One option is that $A$ and $E$ coincide, and $ABCD$ is a rectangle of area $24cdot 10=240$.
Another option is that $E$ and $D$ coincide, in which case the area of the trapezoid is $frac{10cdot 24}{26}(26-frac{10cdot 10}{26}) approx 204$.
answered Dec 27 '18 at 19:14
Henning Makholm
238k16303537
As a concrete example of how the figure is not determined:
One option is that $A$ and $E$ coincide, and $ABCD$ is a rectangle of area $24cdot 10=240$.
Another option is that $E$ and $D$ coincide, in which case the area of the trapezoid is $frac{10cdot 24}{26}(26-frac{10cdot 10}{26}) approx 204$.
answered Dec 27 '18 at 19:14
Henning Makholm
238k16303537
edited Dec 27 '18 at 19:25
answered Dec 27 '18 at 19:14
Henning Makholm
238k16303537
answered Dec 27 '18 at 19:14
Henning Makholm
238k16303537
answered Dec 27 '18 at 19:14
Henning Makholm
238k16303537
238k16303537
thanks for the answer. can you explain what do you mean by coincide,please? The answer is indeed 240
– Eldar Rahimli
Dec 27 '18 at 19:16
@EldarRahimli: "Coincide" means that $A$ and $E$ are the same point, so EC is simply the diagonal of the rectangle. It is true that $240$ is one possible answer, but based on the conditions you have disclosed it is not the only possible answer.
– Henning Makholm
Dec 27 '18 at 19:21
add a comment |
thanks for the answer. can you explain what do you mean by coincide,please? The answer is indeed 240
– Eldar Rahimli
Dec 27 '18 at 19:16
@EldarRahimli: "Coincide" means that $A$ and $E$ are the same point, so EC is simply the diagonal of the rectangle. It is true that $240$ is one possible answer, but based on the conditions you have disclosed it is not the only possible answer.
– Henning Makholm
Dec 27 '18 at 19:21
thanks for the answer. can you explain what do you mean by coincide,please? The answer is indeed 240
– Eldar Rahimli
Dec 27 '18 at 19:16
thanks for the answer. can you explain what do you mean by coincide,please? The answer is indeed 240
– Eldar Rahimli
Dec 27 '18 at 19:16
@EldarRahimli: "Coincide" means that $A$ and $E$ are the same point, so EC is simply the diagonal of the rectangle. It is true that $240$ is one possible answer, but based on the conditions you have disclosed it is not the only possible answer.
– Henning Makholm
Dec 27 '18 at 19:21
@EldarRahimli: "Coincide" means that $A$ and $E$ are the same point, so EC is simply the diagonal of the rectangle. It is true that $240$ is one possible answer, but based on the conditions you have disclosed it is not the only possible answer.
– Henning Makholm
Dec 27 '18 at 19:21
add a comment |
The triangle $BCE$ is uniqely determined, but other points $D$ and $A$ are not so this trapezoid does not have fixed area, it depend on $A$ (and then is $D$ determined also).
answered Dec 27 '18 at 18:38
greedoid
38k114794
If $|BCE|$ is fixed, then doesn't this fix A and D, too? Can you elaborate on your answer, please?
– Eldar Rahimli
Dec 27 '18 at 18:40
1
Note that the trapezoid is isosceles.
– Arthur
Dec 27 '18 at 18:43
Try to play in some aplet, say Geogebra. You can move $AD$ through $E$ and you will see you don't get a single trapezoid.
– greedoid
Dec 27 '18 at 18:43
How does that affect? @Arthur
– greedoid
Dec 27 '18 at 18:44
1
You can tilt the sides. Change the $BCD$ angle.
– Andrei
Dec 27 '18 at 18:48
|
show 6 more comments
The triangle $BCE$ is uniqely determined, but other points $D$ and $A$ are not so this trapezoid does not have fixed area, it depend on $A$ (and then is $D$ determined also).
answered Dec 27 '18 at 18:38
greedoid
38k114794
If $|BCE|$ is fixed, then doesn't this fix A and D, too? Can you elaborate on your answer, please?
– Eldar Rahimli
Dec 27 '18 at 18:40
1
Note that the trapezoid is isosceles.
– Arthur
Dec 27 '18 at 18:43
Try to play in some aplet, say Geogebra. You can move $AD$ through $E$ and you will see you don't get a single trapezoid.
– greedoid
Dec 27 '18 at 18:43
How does that affect? @Arthur
– greedoid
Dec 27 '18 at 18:44
1
You can tilt the sides. Change the $BCD$ angle.
– Andrei
Dec 27 '18 at 18:48
|
show 6 more comments
The triangle $BCE$ is uniqely determined, but other points $D$ and $A$ are not so this trapezoid does not have fixed area, it depend on $A$ (and then is $D$ determined also).
answered Dec 27 '18 at 18:38
greedoid
38k114794
The triangle $BCE$ is uniqely determined, but other points $D$ and $A$ are not so this trapezoid does not have fixed area, it depend on $A$ (and then is $D$ determined also).
answered Dec 27 '18 at 18:38
greedoid
38k114794
answered Dec 27 '18 at 18:38
greedoid
38k114794
answered Dec 27 '18 at 18:38
greedoid
38k114794
answered Dec 27 '18 at 18:38
greedoid
38k114794
38k114794
If $|BCE|$ is fixed, then doesn't this fix A and D, too? Can you elaborate on your answer, please?
– Eldar Rahimli
Dec 27 '18 at 18:40
1
Note that the trapezoid is isosceles.
– Arthur
Dec 27 '18 at 18:43
Try to play in some aplet, say Geogebra. You can move $AD$ through $E$ and you will see you don't get a single trapezoid.
– greedoid
Dec 27 '18 at 18:43
How does that affect? @Arthur
– greedoid
Dec 27 '18 at 18:44
1
You can tilt the sides. Change the $BCD$ angle.
– Andrei
Dec 27 '18 at 18:48
|
show 6 more comments
If $|BCE|$ is fixed, then doesn't this fix A and D, too? Can you elaborate on your answer, please?
– Eldar Rahimli
Dec 27 '18 at 18:40
1
Note that the trapezoid is isosceles.
– Arthur
Dec 27 '18 at 18:43
Try to play in some aplet, say Geogebra. You can move $AD$ through $E$ and you will see you don't get a single trapezoid.
– greedoid
Dec 27 '18 at 18:43
How does that affect? @Arthur
– greedoid
Dec 27 '18 at 18:44
1
You can tilt the sides. Change the $BCD$ angle.
– Andrei
Dec 27 '18 at 18:48
If $|BCE|$ is fixed, then doesn't this fix A and D, too? Can you elaborate on your answer, please?
– Eldar Rahimli
Dec 27 '18 at 18:40
If $|BCE|$ is fixed, then doesn't this fix A and D, too? Can you elaborate on your answer, please?
– Eldar Rahimli
Dec 27 '18 at 18:40
1
1
Note that the trapezoid is isosceles.
– Arthur
Dec 27 '18 at 18:43
Note that the trapezoid is isosceles.
– Arthur
Dec 27 '18 at 18:43
Try to play in some aplet, say Geogebra. You can move $AD$ through $E$ and you will see you don't get a single trapezoid.
– greedoid
Dec 27 '18 at 18:43
Try to play in some aplet, say Geogebra. You can move $AD$ through $E$ and you will see you don't get a single trapezoid.
– greedoid
Dec 27 '18 at 18:43
How does that affect? @Arthur
– greedoid
Dec 27 '18 at 18:44
How does that affect? @Arthur
– greedoid
Dec 27 '18 at 18:44
1
1
You can tilt the sides. Change the $BCD$ angle.
– Andrei
Dec 27 '18 at 18:48
You can tilt the sides. Change the $BCD$ angle.
– Andrei
Dec 27 '18 at 18:48
|
show 6 more comments
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2
ABCD does not look very isosceles in the image ...
– Henning Makholm
Dec 27 '18 at 19:03
Oh, Sorry, I didn't notice when I drew it
– Eldar Rahimli
Dec 27 '18 at 19:04