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How to get the area of the trapezoid?
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up vote
1
down vote
favorite
I have the following basic geometry exercise, but I only know 3 sides. And I must not solve it with trigonometry.
Well, with trigonometry it is trivial. But with the formula of the trapeze area, I can not get the height.
So, How can I get the area? Thanks in advance.
geometry
asked Jul 22 at 5:36
Mattiu
766316
 |Â
show 5 more comments
up vote
1
down vote
favorite
I have the following basic geometry exercise, but I only know 3 sides. And I must not solve it with trigonometry.
Well, with trigonometry it is trivial. But with the formula of the trapeze area, I can not get the height.
So, How can I get the area? Thanks in advance.
geometry
asked Jul 22 at 5:36
Mattiu
766316
It isn’t possible to solve this without the use of trigonometry. You must use trigo to solve for the height
– QuIcKmAtHs
Jul 22 at 5:43
A trapezoid has a pair of parallel lines. Maybe you can try to cut it into parts.
– poyea
Jul 22 at 5:44
poyea, can you help me in that? i already tried it..
– Mattiu
Jul 22 at 5:47
The height of the trapezium is $BCsin 30^circ=6$. If you abhor trigonometry you can derive this by drawing a suitable equilateral triangle.
– Lord Shark the Unknown
Jul 22 at 5:50
3
"It isn’t possible to solve this without the use of trigonometry. " You don't need trig for a 30-60-90 triangle.
– fleablood
Jul 22 at 5:52
 |Â
show 5 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have the following basic geometry exercise, but I only know 3 sides. And I must not solve it with trigonometry.
Well, with trigonometry it is trivial. But with the formula of the trapeze area, I can not get the height.
So, How can I get the area? Thanks in advance.
geometry
asked Jul 22 at 5:36
Mattiu
766316
I have the following basic geometry exercise, but I only know 3 sides. And I must not solve it with trigonometry.
Well, with trigonometry it is trivial. But with the formula of the trapeze area, I can not get the height.
So, How can I get the area? Thanks in advance.
geometry
asked Jul 22 at 5:36
Mattiu
766316
asked Jul 22 at 5:36
Mattiu
766316
asked Jul 22 at 5:36
Mattiu
766316
asked Jul 22 at 5:36
Mattiu
766316
766316
It isn’t possible to solve this without the use of trigonometry. You must use trigo to solve for the height
– QuIcKmAtHs
Jul 22 at 5:43
A trapezoid has a pair of parallel lines. Maybe you can try to cut it into parts.
– poyea
Jul 22 at 5:44
poyea, can you help me in that? i already tried it..
– Mattiu
Jul 22 at 5:47
The height of the trapezium is $BCsin 30^circ=6$. If you abhor trigonometry you can derive this by drawing a suitable equilateral triangle.
– Lord Shark the Unknown
Jul 22 at 5:50
3
"It isn’t possible to solve this without the use of trigonometry. " You don't need trig for a 30-60-90 triangle.
– fleablood
Jul 22 at 5:52
 |Â
show 5 more comments
It isn’t possible to solve this without the use of trigonometry. You must use trigo to solve for the height
– QuIcKmAtHs
Jul 22 at 5:43
A trapezoid has a pair of parallel lines. Maybe you can try to cut it into parts.
– poyea
Jul 22 at 5:44
poyea, can you help me in that? i already tried it..
– Mattiu
Jul 22 at 5:47
The height of the trapezium is $BCsin 30^circ=6$. If you abhor trigonometry you can derive this by drawing a suitable equilateral triangle.
– Lord Shark the Unknown
Jul 22 at 5:50
3
"It isn’t possible to solve this without the use of trigonometry. " You don't need trig for a 30-60-90 triangle.
– fleablood
Jul 22 at 5:52
It isn’t possible to solve this without the use of trigonometry. You must use trigo to solve for the height
– QuIcKmAtHs
Jul 22 at 5:43
It isn’t possible to solve this without the use of trigonometry. You must use trigo to solve for the height
– QuIcKmAtHs
Jul 22 at 5:43
A trapezoid has a pair of parallel lines. Maybe you can try to cut it into parts.
– poyea
Jul 22 at 5:44
A trapezoid has a pair of parallel lines. Maybe you can try to cut it into parts.
– poyea
Jul 22 at 5:44
poyea, can you help me in that? i already tried it..
– Mattiu
Jul 22 at 5:47
poyea, can you help me in that? i already tried it..
– Mattiu
Jul 22 at 5:47
The height of the trapezium is $BCsin 30^circ=6$. If you abhor trigonometry you can derive this by drawing a suitable equilateral triangle.
– Lord Shark the Unknown
Jul 22 at 5:50
The height of the trapezium is $BCsin 30^circ=6$. If you abhor trigonometry you can derive this by drawing a suitable equilateral triangle.
– Lord Shark the Unknown
Jul 22 at 5:50
3
3
"It isn’t possible to solve this without the use of trigonometry. " You don't need trig for a 30-60-90 triangle.
– fleablood
Jul 22 at 5:52
"It isn’t possible to solve this without the use of trigonometry. " You don't need trig for a 30-60-90 triangle.
– fleablood
Jul 22 at 5:52
 |Â
show 5 more comments
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
That the angle is 30° is the giveaway. If you drop the perpendicular of $C$ onto $AB$ so the foot is $E$, $triangle CEB$ is half of an equilateral triangle. It follows that the height $CE$ is half of $CB$, or 6, and the whole area follows.
answered Jul 22 at 5:48
Parcly Taxel
33.6k136588
Well, how you know that the triangle CEB is a half of an equilateral triangle? is a special kind of triangle or something? And why CE is half of CB, what is that property? thanks
– Mattiu
Jul 22 at 6:08
@Mattiu If I place two copies of $CEB$ together I get a triangle whose angles are all 60°. Thus equilateral.
– Parcly Taxel
Jul 22 at 6:09
1
This is clearly the intended solution. I might want to pick a bone with the original problem poser, arguing that knowledge about the $30-60-90$ triangle is really (primitive) trigonometry.
– Ethan Bolker
Jul 22 at 13:46
add a comment |Â
up vote
0
down vote
The area of a trapezoid is given by $$A=hcdotfraca+c2$$ where $h$ is the height, $a$ is the lower base, $c$ is the upper base. In your case this it's $a=|AB|,c=|CD|$. We have those lengths given, so all that remains is the height $h$. Now consider a line $l$ going through point $C$ perpendicular to $AB$, denote the intersection point of $l$ and $AB$ for example $X$ then $Delta XBC$ is a triangle with right angle at $X$. The height of the trapezoid is then $|XC|=h$. By elementary trigonometry $$sin30°=frac12$$ also $$sinalpha=fractextoppositetexthypotenuse$$
which in this case is
$$sin30°=fracXCBC=frach12=frac12$$
from which you get that $h=6$ and so your area is then
$$A=hcdotfraca+c2=6cdotfrac30+82=114$$
answered Jul 22 at 13:42
Michal Dvořák
48912
Thank you, but clearly I have requested a solution without trigonometry
– Mattiu
Jul 22 at 16:53
Well, what do you mean by solution with trigonometry?
– Michal Dvořák
Jul 22 at 22:22
add a comment |Â
up vote
0
down vote
So start with drawing a perpendicular line on AB which connects point C, making that a new point on line AB called point "E". Do the same thing on the left side by making another point on line AB called the point "F". Now you have a rectangle proving DC : FE and DF : CE. Both sides measuring 8. If you subtract AB - FE and divide the answer with 2, you now have the the measure for AF : BE which is 11. As CB = 12 you can now use Pythagorean Theorem to get measures of DF and CE.BC^2 - BE^2 = CE^2, the square root √(BC^2 - BE^2) = CE. √(12^2 - 11^2) = 4.795831523.
Area of Trapezoid = ((30+8)÷2)4.795831523)
= 91.12079894
answered Jul 22 at 13:46
Alay Mistry
11
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 13:52
Alan, you can not be sure that the segments are equal, that is, they measure 11. So this answer is wrong
– Mattiu
Jul 22 at 16:52
I really don't know why but if I put these measures from you diagram in a Google based Area of Trapezoid calculator I get the same answer that I got.
– Alay Mistry
Jul 23 at 1:18
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
That the angle is 30° is the giveaway. If you drop the perpendicular of $C$ onto $AB$ so the foot is $E$, $triangle CEB$ is half of an equilateral triangle. It follows that the height $CE$ is half of $CB$, or 6, and the whole area follows.
answered Jul 22 at 5:48
Parcly Taxel
33.6k136588
Well, how you know that the triangle CEB is a half of an equilateral triangle? is a special kind of triangle or something? And why CE is half of CB, what is that property? thanks
– Mattiu
Jul 22 at 6:08
@Mattiu If I place two copies of $CEB$ together I get a triangle whose angles are all 60°. Thus equilateral.
– Parcly Taxel
Jul 22 at 6:09
1
This is clearly the intended solution. I might want to pick a bone with the original problem poser, arguing that knowledge about the $30-60-90$ triangle is really (primitive) trigonometry.
– Ethan Bolker
Jul 22 at 13:46
add a comment |Â
up vote
3
down vote
accepted
That the angle is 30° is the giveaway. If you drop the perpendicular of $C$ onto $AB$ so the foot is $E$, $triangle CEB$ is half of an equilateral triangle. It follows that the height $CE$ is half of $CB$, or 6, and the whole area follows.
answered Jul 22 at 5:48
Parcly Taxel
33.6k136588
Well, how you know that the triangle CEB is a half of an equilateral triangle? is a special kind of triangle or something? And why CE is half of CB, what is that property? thanks
– Mattiu
Jul 22 at 6:08
@Mattiu If I place two copies of $CEB$ together I get a triangle whose angles are all 60°. Thus equilateral.
– Parcly Taxel
Jul 22 at 6:09
1
This is clearly the intended solution. I might want to pick a bone with the original problem poser, arguing that knowledge about the $30-60-90$ triangle is really (primitive) trigonometry.
– Ethan Bolker
Jul 22 at 13:46
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
That the angle is 30° is the giveaway. If you drop the perpendicular of $C$ onto $AB$ so the foot is $E$, $triangle CEB$ is half of an equilateral triangle. It follows that the height $CE$ is half of $CB$, or 6, and the whole area follows.
answered Jul 22 at 5:48
Parcly Taxel
33.6k136588
That the angle is 30° is the giveaway. If you drop the perpendicular of $C$ onto $AB$ so the foot is $E$, $triangle CEB$ is half of an equilateral triangle. It follows that the height $CE$ is half of $CB$, or 6, and the whole area follows.
answered Jul 22 at 5:48
Parcly Taxel
33.6k136588
answered Jul 22 at 5:48
Parcly Taxel
33.6k136588
answered Jul 22 at 5:48
Parcly Taxel
33.6k136588
answered Jul 22 at 5:48
Parcly Taxel
33.6k136588
33.6k136588
Well, how you know that the triangle CEB is a half of an equilateral triangle? is a special kind of triangle or something? And why CE is half of CB, what is that property? thanks
– Mattiu
Jul 22 at 6:08
@Mattiu If I place two copies of $CEB$ together I get a triangle whose angles are all 60°. Thus equilateral.
– Parcly Taxel
Jul 22 at 6:09
1
This is clearly the intended solution. I might want to pick a bone with the original problem poser, arguing that knowledge about the $30-60-90$ triangle is really (primitive) trigonometry.
– Ethan Bolker
Jul 22 at 13:46
add a comment |Â
Well, how you know that the triangle CEB is a half of an equilateral triangle? is a special kind of triangle or something? And why CE is half of CB, what is that property? thanks
– Mattiu
Jul 22 at 6:08
@Mattiu If I place two copies of $CEB$ together I get a triangle whose angles are all 60°. Thus equilateral.
– Parcly Taxel
Jul 22 at 6:09
1
This is clearly the intended solution. I might want to pick a bone with the original problem poser, arguing that knowledge about the $30-60-90$ triangle is really (primitive) trigonometry.
– Ethan Bolker
Jul 22 at 13:46
Well, how you know that the triangle CEB is a half of an equilateral triangle? is a special kind of triangle or something? And why CE is half of CB, what is that property? thanks
– Mattiu
Jul 22 at 6:08
Well, how you know that the triangle CEB is a half of an equilateral triangle? is a special kind of triangle or something? And why CE is half of CB, what is that property? thanks
– Mattiu
Jul 22 at 6:08
@Mattiu If I place two copies of $CEB$ together I get a triangle whose angles are all 60°. Thus equilateral.
– Parcly Taxel
Jul 22 at 6:09
@Mattiu If I place two copies of $CEB$ together I get a triangle whose angles are all 60°. Thus equilateral.
– Parcly Taxel
Jul 22 at 6:09
1
1
This is clearly the intended solution. I might want to pick a bone with the original problem poser, arguing that knowledge about the $30-60-90$ triangle is really (primitive) trigonometry.
– Ethan Bolker
Jul 22 at 13:46
This is clearly the intended solution. I might want to pick a bone with the original problem poser, arguing that knowledge about the $30-60-90$ triangle is really (primitive) trigonometry.
– Ethan Bolker
Jul 22 at 13:46
add a comment |Â
up vote
0
down vote
The area of a trapezoid is given by $$A=hcdotfraca+c2$$ where $h$ is the height, $a$ is the lower base, $c$ is the upper base. In your case this it's $a=|AB|,c=|CD|$. We have those lengths given, so all that remains is the height $h$. Now consider a line $l$ going through point $C$ perpendicular to $AB$, denote the intersection point of $l$ and $AB$ for example $X$ then $Delta XBC$ is a triangle with right angle at $X$. The height of the trapezoid is then $|XC|=h$. By elementary trigonometry $$sin30°=frac12$$ also $$sinalpha=fractextoppositetexthypotenuse$$
which in this case is
$$sin30°=fracXCBC=frach12=frac12$$
from which you get that $h=6$ and so your area is then
$$A=hcdotfraca+c2=6cdotfrac30+82=114$$
answered Jul 22 at 13:42
Michal Dvořák
48912
Thank you, but clearly I have requested a solution without trigonometry
– Mattiu
Jul 22 at 16:53
Well, what do you mean by solution with trigonometry?
– Michal Dvořák
Jul 22 at 22:22
add a comment |Â
up vote
0
down vote
The area of a trapezoid is given by $$A=hcdotfraca+c2$$ where $h$ is the height, $a$ is the lower base, $c$ is the upper base. In your case this it's $a=|AB|,c=|CD|$. We have those lengths given, so all that remains is the height $h$. Now consider a line $l$ going through point $C$ perpendicular to $AB$, denote the intersection point of $l$ and $AB$ for example $X$ then $Delta XBC$ is a triangle with right angle at $X$. The height of the trapezoid is then $|XC|=h$. By elementary trigonometry $$sin30°=frac12$$ also $$sinalpha=fractextoppositetexthypotenuse$$
which in this case is
$$sin30°=fracXCBC=frach12=frac12$$
from which you get that $h=6$ and so your area is then
$$A=hcdotfraca+c2=6cdotfrac30+82=114$$
answered Jul 22 at 13:42
Michal Dvořák
48912
Thank you, but clearly I have requested a solution without trigonometry
– Mattiu
Jul 22 at 16:53
Well, what do you mean by solution with trigonometry?
– Michal Dvořák
Jul 22 at 22:22
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The area of a trapezoid is given by $$A=hcdotfraca+c2$$ where $h$ is the height, $a$ is the lower base, $c$ is the upper base. In your case this it's $a=|AB|,c=|CD|$. We have those lengths given, so all that remains is the height $h$. Now consider a line $l$ going through point $C$ perpendicular to $AB$, denote the intersection point of $l$ and $AB$ for example $X$ then $Delta XBC$ is a triangle with right angle at $X$. The height of the trapezoid is then $|XC|=h$. By elementary trigonometry $$sin30°=frac12$$ also $$sinalpha=fractextoppositetexthypotenuse$$
which in this case is
$$sin30°=fracXCBC=frach12=frac12$$
from which you get that $h=6$ and so your area is then
$$A=hcdotfraca+c2=6cdotfrac30+82=114$$
answered Jul 22 at 13:42
Michal Dvořák
48912
The area of a trapezoid is given by $$A=hcdotfraca+c2$$ where $h$ is the height, $a$ is the lower base, $c$ is the upper base. In your case this it's $a=|AB|,c=|CD|$. We have those lengths given, so all that remains is the height $h$. Now consider a line $l$ going through point $C$ perpendicular to $AB$, denote the intersection point of $l$ and $AB$ for example $X$ then $Delta XBC$ is a triangle with right angle at $X$. The height of the trapezoid is then $|XC|=h$. By elementary trigonometry $$sin30°=frac12$$ also $$sinalpha=fractextoppositetexthypotenuse$$
which in this case is
$$sin30°=fracXCBC=frach12=frac12$$
from which you get that $h=6$ and so your area is then
$$A=hcdotfraca+c2=6cdotfrac30+82=114$$
answered Jul 22 at 13:42
Michal Dvořák
48912
answered Jul 22 at 13:42
Michal Dvořák
48912
answered Jul 22 at 13:42
Michal Dvořák
48912
answered Jul 22 at 13:42
Michal Dvořák
48912
48912
Thank you, but clearly I have requested a solution without trigonometry
– Mattiu
Jul 22 at 16:53
Well, what do you mean by solution with trigonometry?
– Michal Dvořák
Jul 22 at 22:22
add a comment |Â
Thank you, but clearly I have requested a solution without trigonometry
– Mattiu
Jul 22 at 16:53
Well, what do you mean by solution with trigonometry?
– Michal Dvořák
Jul 22 at 22:22
Thank you, but clearly I have requested a solution without trigonometry
– Mattiu
Jul 22 at 16:53
Thank you, but clearly I have requested a solution without trigonometry
– Mattiu
Jul 22 at 16:53
Well, what do you mean by solution with trigonometry?
– Michal Dvořák
Jul 22 at 22:22
Well, what do you mean by solution with trigonometry?
– Michal Dvořák
Jul 22 at 22:22
add a comment |Â
up vote
0
down vote
So start with drawing a perpendicular line on AB which connects point C, making that a new point on line AB called point "E". Do the same thing on the left side by making another point on line AB called the point "F". Now you have a rectangle proving DC : FE and DF : CE. Both sides measuring 8. If you subtract AB - FE and divide the answer with 2, you now have the the measure for AF : BE which is 11. As CB = 12 you can now use Pythagorean Theorem to get measures of DF and CE.BC^2 - BE^2 = CE^2, the square root √(BC^2 - BE^2) = CE. √(12^2 - 11^2) = 4.795831523.
Area of Trapezoid = ((30+8)÷2)4.795831523)
= 91.12079894
answered Jul 22 at 13:46
Alay Mistry
11
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 13:52
Alan, you can not be sure that the segments are equal, that is, they measure 11. So this answer is wrong
– Mattiu
Jul 22 at 16:52
I really don't know why but if I put these measures from you diagram in a Google based Area of Trapezoid calculator I get the same answer that I got.
– Alay Mistry
Jul 23 at 1:18
add a comment |Â
up vote
0
down vote
So start with drawing a perpendicular line on AB which connects point C, making that a new point on line AB called point "E". Do the same thing on the left side by making another point on line AB called the point "F". Now you have a rectangle proving DC : FE and DF : CE. Both sides measuring 8. If you subtract AB - FE and divide the answer with 2, you now have the the measure for AF : BE which is 11. As CB = 12 you can now use Pythagorean Theorem to get measures of DF and CE.BC^2 - BE^2 = CE^2, the square root √(BC^2 - BE^2) = CE. √(12^2 - 11^2) = 4.795831523.
Area of Trapezoid = ((30+8)÷2)4.795831523)
= 91.12079894
answered Jul 22 at 13:46
Alay Mistry
11
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 13:52
Alan, you can not be sure that the segments are equal, that is, they measure 11. So this answer is wrong
– Mattiu
Jul 22 at 16:52
I really don't know why but if I put these measures from you diagram in a Google based Area of Trapezoid calculator I get the same answer that I got.
– Alay Mistry
Jul 23 at 1:18
add a comment |Â
up vote
0
down vote
up vote
0
down vote
So start with drawing a perpendicular line on AB which connects point C, making that a new point on line AB called point "E". Do the same thing on the left side by making another point on line AB called the point "F". Now you have a rectangle proving DC : FE and DF : CE. Both sides measuring 8. If you subtract AB - FE and divide the answer with 2, you now have the the measure for AF : BE which is 11. As CB = 12 you can now use Pythagorean Theorem to get measures of DF and CE.BC^2 - BE^2 = CE^2, the square root √(BC^2 - BE^2) = CE. √(12^2 - 11^2) = 4.795831523.
Area of Trapezoid = ((30+8)÷2)4.795831523)
= 91.12079894
answered Jul 22 at 13:46
Alay Mistry
11
So start with drawing a perpendicular line on AB which connects point C, making that a new point on line AB called point "E". Do the same thing on the left side by making another point on line AB called the point "F". Now you have a rectangle proving DC : FE and DF : CE. Both sides measuring 8. If you subtract AB - FE and divide the answer with 2, you now have the the measure for AF : BE which is 11. As CB = 12 you can now use Pythagorean Theorem to get measures of DF and CE.BC^2 - BE^2 = CE^2, the square root √(BC^2 - BE^2) = CE. √(12^2 - 11^2) = 4.795831523.
Area of Trapezoid = ((30+8)÷2)4.795831523)
= 91.12079894
answered Jul 22 at 13:46
Alay Mistry
11
answered Jul 22 at 13:46
Alay Mistry
11
answered Jul 22 at 13:46
Alay Mistry
11
answered Jul 22 at 13:46
Alay Mistry
11
11
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 13:52
Alan, you can not be sure that the segments are equal, that is, they measure 11. So this answer is wrong
– Mattiu
Jul 22 at 16:52
I really don't know why but if I put these measures from you diagram in a Google based Area of Trapezoid calculator I get the same answer that I got.
– Alay Mistry
Jul 23 at 1:18
add a comment |Â
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 13:52
Alan, you can not be sure that the segments are equal, that is, they measure 11. So this answer is wrong
– Mattiu
Jul 22 at 16:52
I really don't know why but if I put these measures from you diagram in a Google based Area of Trapezoid calculator I get the same answer that I got.
– Alay Mistry
Jul 23 at 1:18
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 13:52
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 22 at 13:52
Alan, you can not be sure that the segments are equal, that is, they measure 11. So this answer is wrong
– Mattiu
Jul 22 at 16:52
Alan, you can not be sure that the segments are equal, that is, they measure 11. So this answer is wrong
– Mattiu
Jul 22 at 16:52
I really don't know why but if I put these measures from you diagram in a Google based Area of Trapezoid calculator I get the same answer that I got.
– Alay Mistry
Jul 23 at 1:18
I really don't know why but if I put these measures from you diagram in a Google based Area of Trapezoid calculator I get the same answer that I got.
– Alay Mistry
Jul 23 at 1:18
add a comment |Â
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It isn’t possible to solve this without the use of trigonometry. You must use trigo to solve for the height
– QuIcKmAtHs
Jul 22 at 5:43
A trapezoid has a pair of parallel lines. Maybe you can try to cut it into parts.
– poyea
Jul 22 at 5:44
poyea, can you help me in that? i already tried it..
– Mattiu
Jul 22 at 5:47
The height of the trapezium is $BCsin 30^circ=6$. If you abhor trigonometry you can derive this by drawing a suitable equilateral triangle.
– Lord Shark the Unknown
Jul 22 at 5:50
3
"It isn’t possible to solve this without the use of trigonometry. " You don't need trig for a 30-60-90 triangle.
– fleablood
Jul 22 at 5:52