Isosceles triangle height
I was doing a problem in which I was told the lengths of all sides of an isosceles triangle and asked to find the area.
I solved the problem by dividing the isosceles triangle into two equal triangles to find the height which I used in the area formula for the original triangle.
Looking at the answer, my method resulted in the correct value but, it seems I could have used the legs of the isosceles triangle (both 8) as the base and height and skipped finding the height. Why does that shortcut work?
The question is below:
geometry triangle
asked 51 mins ago
Crt
1155
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Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I was doing a problem in which I was told the lengths of all sides of an isosceles triangle and asked to find the area.
I solved the problem by dividing the isosceles triangle into two equal triangles to find the height which I used in the area formula for the original triangle.
Looking at the answer, my method resulted in the correct value but, it seems I could have used the legs of the isosceles triangle (both 8) as the base and height and skipped finding the height. Why does that shortcut work?
The question is below:
geometry triangle
asked 51 mins ago
Crt
1155
New contributor
Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
– Deepak
44 mins ago
Might be a coincidence of the mathematics but it's hard to say without knowing your work explicitly. Or as Deepak said, could be an isosceles right triangle. Could you edit your original post to include the original problem and how you worked it using your shortcut?
– Eevee Trainer
43 mins ago
Also, note that with the triangle lengths available, regardless of it being any specific type of triangle, you didn't need to determine the height. Instead, you could have directly solved the problem using these values in Heron's formula. I just saw you added the question itself. In your case, you could have also used the Pythagorean Theorem to determine the other side lengths, then getting the area using those $2$ values, with it being simpler in this case then Heron's formula.
– John Omielan
40 mins ago
@Deepak you were right. It makes sense once you recognize that the triangle is a right angle and it's drawn out. I can see the height is one of the legs. Thanks
– Crt
35 mins ago
add a comment |
I was doing a problem in which I was told the lengths of all sides of an isosceles triangle and asked to find the area.
I solved the problem by dividing the isosceles triangle into two equal triangles to find the height which I used in the area formula for the original triangle.
Looking at the answer, my method resulted in the correct value but, it seems I could have used the legs of the isosceles triangle (both 8) as the base and height and skipped finding the height. Why does that shortcut work?
The question is below:
geometry triangle
asked 51 mins ago
Crt
1155
New contributor
Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I was doing a problem in which I was told the lengths of all sides of an isosceles triangle and asked to find the area.
I solved the problem by dividing the isosceles triangle into two equal triangles to find the height which I used in the area formula for the original triangle.
Looking at the answer, my method resulted in the correct value but, it seems I could have used the legs of the isosceles triangle (both 8) as the base and height and skipped finding the height. Why does that shortcut work?
The question is below:
geometry triangle
geometry triangle
asked 51 mins ago
Crt
1155
New contributor
Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 51 mins ago
Crt
1155
New contributor
Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 40 mins ago
asked 51 mins ago
Crt
1155
New contributor
Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 51 mins ago
Crt
1155
asked 51 mins ago
Crt
1155
1155
New contributor
Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
– Deepak
44 mins ago
Might be a coincidence of the mathematics but it's hard to say without knowing your work explicitly. Or as Deepak said, could be an isosceles right triangle. Could you edit your original post to include the original problem and how you worked it using your shortcut?
– Eevee Trainer
43 mins ago
Also, note that with the triangle lengths available, regardless of it being any specific type of triangle, you didn't need to determine the height. Instead, you could have directly solved the problem using these values in Heron's formula. I just saw you added the question itself. In your case, you could have also used the Pythagorean Theorem to determine the other side lengths, then getting the area using those $2$ values, with it being simpler in this case then Heron's formula.
– John Omielan
40 mins ago
@Deepak you were right. It makes sense once you recognize that the triangle is a right angle and it's drawn out. I can see the height is one of the legs. Thanks
– Crt
35 mins ago
add a comment |
2
That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
– Deepak
44 mins ago
Might be a coincidence of the mathematics but it's hard to say without knowing your work explicitly. Or as Deepak said, could be an isosceles right triangle. Could you edit your original post to include the original problem and how you worked it using your shortcut?
– Eevee Trainer
43 mins ago
Also, note that with the triangle lengths available, regardless of it being any specific type of triangle, you didn't need to determine the height. Instead, you could have directly solved the problem using these values in Heron's formula. I just saw you added the question itself. In your case, you could have also used the Pythagorean Theorem to determine the other side lengths, then getting the area using those $2$ values, with it being simpler in this case then Heron's formula.
– John Omielan
40 mins ago
@Deepak you were right. It makes sense once you recognize that the triangle is a right angle and it's drawn out. I can see the height is one of the legs. Thanks
– Crt
35 mins ago
2
2
That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
– Deepak
44 mins ago
That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
– Deepak
44 mins ago
Might be a coincidence of the mathematics but it's hard to say without knowing your work explicitly. Or as Deepak said, could be an isosceles right triangle. Could you edit your original post to include the original problem and how you worked it using your shortcut?
– Eevee Trainer
43 mins ago
Might be a coincidence of the mathematics but it's hard to say without knowing your work explicitly. Or as Deepak said, could be an isosceles right triangle. Could you edit your original post to include the original problem and how you worked it using your shortcut?
– Eevee Trainer
43 mins ago
Also, note that with the triangle lengths available, regardless of it being any specific type of triangle, you didn't need to determine the height. Instead, you could have directly solved the problem using these values in Heron's formula. I just saw you added the question itself. In your case, you could have also used the Pythagorean Theorem to determine the other side lengths, then getting the area using those $2$ values, with it being simpler in this case then Heron's formula.
– John Omielan
40 mins ago
Also, note that with the triangle lengths available, regardless of it being any specific type of triangle, you didn't need to determine the height. Instead, you could have directly solved the problem using these values in Heron's formula. I just saw you added the question itself. In your case, you could have also used the Pythagorean Theorem to determine the other side lengths, then getting the area using those $2$ values, with it being simpler in this case then Heron's formula.
– John Omielan
40 mins ago
@Deepak you were right. It makes sense once you recognize that the triangle is a right angle and it's drawn out. I can see the height is one of the legs. Thanks
– Crt
35 mins ago
@Deepak you were right. It makes sense once you recognize that the triangle is a right angle and it's drawn out. I can see the height is one of the legs. Thanks
– Crt
35 mins ago
add a comment |
3 Answers
3
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Pythagoras' theorem ($a^2 + b^2 = c^2$) is generally used to compute one "missing" side when two others are known in a triangle known to be a right triangle.
However, there is a converse of Pythagoras' theorem, whereby you can show a given triangle is right if the side lengths satisfy the relation.
In this case, $8^2 + 8^2 = (8sqrt 2)^2$, so the triangle is right-angled, and you can immediately find the area as $frac 12 (8)(8) = 32$.
(In this case, you don't need the converse. It's already given to be a right triangle. But in the case you're given the side lengths of $(8,8,8sqrt 2)$, you can apply the converse to simplify your work a bit).
answered 35 mins ago
Deepak
16.8k11436
add a comment |
If you know the lengths $L_1$, $L_2$ of two legs of a triangle and the angle between them $theta$, then you can calculate the area as
$$A = frac12L_1L_2sintheta$$
In your case $theta$ was 90 degrees, making the formula $A = frac12L_1L_2$.
answered 36 mins ago
Erik Parkinson
1164
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
After the edit to the OP, yeah, as pointed out by Deepak in the comments: it is because the triangle is not just any isosceles triangle, but an isosceles right triangle. That the question specifies this also may be indicative that your "shortcut" was the intended method (though kudos to you for finding an additional method either way!).
As is probably obvious whenever you draw right triangles, its area can be given by
$$A = frac12 ab$$
where $a,b$ are the legs of the triangle. In right triangles, the legs can be used as the height and the base. To actually further this discussion and extend to isosceles right triangles, suppose you have only the hypotenuse $h$. Since the triangle is isosceles and right, the legs are equal ($a=b$) and are given by $h/sqrt 2$. In that light we could make this even shorter by noting:
$$A = frac12 ab = frac12 frac{h}{sqrt 2} frac{h}{sqrt 2} = frac{h^2}{4}$$
so in a sense you don't even need to find the legs: in an isosceles right triangle, the hypotenuse uniquely determines the legs, and vice versa.
answered 33 mins ago
Eevee Trainer
4,6371634
add a comment |
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3 Answers
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3 Answers
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Pythagoras' theorem ($a^2 + b^2 = c^2$) is generally used to compute one "missing" side when two others are known in a triangle known to be a right triangle.
However, there is a converse of Pythagoras' theorem, whereby you can show a given triangle is right if the side lengths satisfy the relation.
In this case, $8^2 + 8^2 = (8sqrt 2)^2$, so the triangle is right-angled, and you can immediately find the area as $frac 12 (8)(8) = 32$.
(In this case, you don't need the converse. It's already given to be a right triangle. But in the case you're given the side lengths of $(8,8,8sqrt 2)$, you can apply the converse to simplify your work a bit).
answered 35 mins ago
Deepak
16.8k11436
add a comment |
Pythagoras' theorem ($a^2 + b^2 = c^2$) is generally used to compute one "missing" side when two others are known in a triangle known to be a right triangle.
However, there is a converse of Pythagoras' theorem, whereby you can show a given triangle is right if the side lengths satisfy the relation.
In this case, $8^2 + 8^2 = (8sqrt 2)^2$, so the triangle is right-angled, and you can immediately find the area as $frac 12 (8)(8) = 32$.
(In this case, you don't need the converse. It's already given to be a right triangle. But in the case you're given the side lengths of $(8,8,8sqrt 2)$, you can apply the converse to simplify your work a bit).
answered 35 mins ago
Deepak
16.8k11436
add a comment |
Pythagoras' theorem ($a^2 + b^2 = c^2$) is generally used to compute one "missing" side when two others are known in a triangle known to be a right triangle.
However, there is a converse of Pythagoras' theorem, whereby you can show a given triangle is right if the side lengths satisfy the relation.
In this case, $8^2 + 8^2 = (8sqrt 2)^2$, so the triangle is right-angled, and you can immediately find the area as $frac 12 (8)(8) = 32$.
(In this case, you don't need the converse. It's already given to be a right triangle. But in the case you're given the side lengths of $(8,8,8sqrt 2)$, you can apply the converse to simplify your work a bit).
answered 35 mins ago
Deepak
16.8k11436
Pythagoras' theorem ($a^2 + b^2 = c^2$) is generally used to compute one "missing" side when two others are known in a triangle known to be a right triangle.
However, there is a converse of Pythagoras' theorem, whereby you can show a given triangle is right if the side lengths satisfy the relation.
In this case, $8^2 + 8^2 = (8sqrt 2)^2$, so the triangle is right-angled, and you can immediately find the area as $frac 12 (8)(8) = 32$.
(In this case, you don't need the converse. It's already given to be a right triangle. But in the case you're given the side lengths of $(8,8,8sqrt 2)$, you can apply the converse to simplify your work a bit).
answered 35 mins ago
Deepak
16.8k11436
edited 30 mins ago
answered 35 mins ago
Deepak
16.8k11436
answered 35 mins ago
Deepak
16.8k11436
answered 35 mins ago
Deepak
16.8k11436
16.8k11436
add a comment |
add a comment |
If you know the lengths $L_1$, $L_2$ of two legs of a triangle and the angle between them $theta$, then you can calculate the area as
$$A = frac12L_1L_2sintheta$$
In your case $theta$ was 90 degrees, making the formula $A = frac12L_1L_2$.
answered 36 mins ago
Erik Parkinson
1164
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
If you know the lengths $L_1$, $L_2$ of two legs of a triangle and the angle between them $theta$, then you can calculate the area as
$$A = frac12L_1L_2sintheta$$
In your case $theta$ was 90 degrees, making the formula $A = frac12L_1L_2$.
answered 36 mins ago
Erik Parkinson
1164
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
If you know the lengths $L_1$, $L_2$ of two legs of a triangle and the angle between them $theta$, then you can calculate the area as
$$A = frac12L_1L_2sintheta$$
In your case $theta$ was 90 degrees, making the formula $A = frac12L_1L_2$.
answered 36 mins ago
Erik Parkinson
1164
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If you know the lengths $L_1$, $L_2$ of two legs of a triangle and the angle between them $theta$, then you can calculate the area as
$$A = frac12L_1L_2sintheta$$
In your case $theta$ was 90 degrees, making the formula $A = frac12L_1L_2$.
answered 36 mins ago
Erik Parkinson
1164
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 36 mins ago
Erik Parkinson
1164
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 36 mins ago
Erik Parkinson
1164
answered 36 mins ago
Erik Parkinson
1164
1164
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
After the edit to the OP, yeah, as pointed out by Deepak in the comments: it is because the triangle is not just any isosceles triangle, but an isosceles right triangle. That the question specifies this also may be indicative that your "shortcut" was the intended method (though kudos to you for finding an additional method either way!).
As is probably obvious whenever you draw right triangles, its area can be given by
$$A = frac12 ab$$
where $a,b$ are the legs of the triangle. In right triangles, the legs can be used as the height and the base. To actually further this discussion and extend to isosceles right triangles, suppose you have only the hypotenuse $h$. Since the triangle is isosceles and right, the legs are equal ($a=b$) and are given by $h/sqrt 2$. In that light we could make this even shorter by noting:
$$A = frac12 ab = frac12 frac{h}{sqrt 2} frac{h}{sqrt 2} = frac{h^2}{4}$$
so in a sense you don't even need to find the legs: in an isosceles right triangle, the hypotenuse uniquely determines the legs, and vice versa.
answered 33 mins ago
Eevee Trainer
4,6371634
add a comment |
After the edit to the OP, yeah, as pointed out by Deepak in the comments: it is because the triangle is not just any isosceles triangle, but an isosceles right triangle. That the question specifies this also may be indicative that your "shortcut" was the intended method (though kudos to you for finding an additional method either way!).
As is probably obvious whenever you draw right triangles, its area can be given by
$$A = frac12 ab$$
where $a,b$ are the legs of the triangle. In right triangles, the legs can be used as the height and the base. To actually further this discussion and extend to isosceles right triangles, suppose you have only the hypotenuse $h$. Since the triangle is isosceles and right, the legs are equal ($a=b$) and are given by $h/sqrt 2$. In that light we could make this even shorter by noting:
$$A = frac12 ab = frac12 frac{h}{sqrt 2} frac{h}{sqrt 2} = frac{h^2}{4}$$
so in a sense you don't even need to find the legs: in an isosceles right triangle, the hypotenuse uniquely determines the legs, and vice versa.
answered 33 mins ago
Eevee Trainer
4,6371634
add a comment |
After the edit to the OP, yeah, as pointed out by Deepak in the comments: it is because the triangle is not just any isosceles triangle, but an isosceles right triangle. That the question specifies this also may be indicative that your "shortcut" was the intended method (though kudos to you for finding an additional method either way!).
As is probably obvious whenever you draw right triangles, its area can be given by
$$A = frac12 ab$$
where $a,b$ are the legs of the triangle. In right triangles, the legs can be used as the height and the base. To actually further this discussion and extend to isosceles right triangles, suppose you have only the hypotenuse $h$. Since the triangle is isosceles and right, the legs are equal ($a=b$) and are given by $h/sqrt 2$. In that light we could make this even shorter by noting:
$$A = frac12 ab = frac12 frac{h}{sqrt 2} frac{h}{sqrt 2} = frac{h^2}{4}$$
so in a sense you don't even need to find the legs: in an isosceles right triangle, the hypotenuse uniquely determines the legs, and vice versa.
answered 33 mins ago
Eevee Trainer
4,6371634
After the edit to the OP, yeah, as pointed out by Deepak in the comments: it is because the triangle is not just any isosceles triangle, but an isosceles right triangle. That the question specifies this also may be indicative that your "shortcut" was the intended method (though kudos to you for finding an additional method either way!).
As is probably obvious whenever you draw right triangles, its area can be given by
$$A = frac12 ab$$
where $a,b$ are the legs of the triangle. In right triangles, the legs can be used as the height and the base. To actually further this discussion and extend to isosceles right triangles, suppose you have only the hypotenuse $h$. Since the triangle is isosceles and right, the legs are equal ($a=b$) and are given by $h/sqrt 2$. In that light we could make this even shorter by noting:
$$A = frac12 ab = frac12 frac{h}{sqrt 2} frac{h}{sqrt 2} = frac{h^2}{4}$$
so in a sense you don't even need to find the legs: in an isosceles right triangle, the hypotenuse uniquely determines the legs, and vice versa.
answered 33 mins ago
Eevee Trainer
4,6371634
answered 33 mins ago
Eevee Trainer
4,6371634
answered 33 mins ago
Eevee Trainer
4,6371634
answered 33 mins ago
Eevee Trainer
4,6371634
4,6371634
add a comment |
add a comment |
Crt is a new contributor. Be nice, and check out our Code of Conduct.
Crt is a new contributor. Be nice, and check out our Code of Conduct.
Crt is a new contributor. Be nice, and check out our Code of Conduct.
Crt is a new contributor. Be nice, and check out our Code of Conduct.
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2
That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
– Deepak
44 mins ago
Might be a coincidence of the mathematics but it's hard to say without knowing your work explicitly. Or as Deepak said, could be an isosceles right triangle. Could you edit your original post to include the original problem and how you worked it using your shortcut?
– Eevee Trainer
43 mins ago
Also, note that with the triangle lengths available, regardless of it being any specific type of triangle, you didn't need to determine the height. Instead, you could have directly solved the problem using these values in Heron's formula. I just saw you added the question itself. In your case, you could have also used the Pythagorean Theorem to determine the other side lengths, then getting the area using those $2$ values, with it being simpler in this case then Heron's formula.
– John Omielan
40 mins ago
@Deepak you were right. It makes sense once you recognize that the triangle is a right angle and it's drawn out. I can see the height is one of the legs. Thanks
– Crt
35 mins ago