Finding the area of an isosceles right triangle given its hypotenuse
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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:
If the hypotenuse of an isosceles right triangle is $8 sqrt 2$, what is the area of the triangle?
- (A) 18
- (B) 24
- (C) 32
- (D) 48
- (E) 64
geometry triangle
edited Jan 2 at 12:36
JoeTaxpayer
2,19121226
asked Jan 2 at 2:48
CrtCrt
1205
$endgroup$
add a comment |
$begingroup$
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:
If the hypotenuse of an isosceles right triangle is $8 sqrt 2$, what is the area of the triangle?
- (A) 18
- (B) 24
- (C) 32
- (D) 48
- (E) 64
geometry triangle
edited Jan 2 at 12:36
JoeTaxpayer
2,19121226
asked Jan 2 at 2:48
CrtCrt
1205
$endgroup$
2
$begingroup$
That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
$endgroup$
– Deepak
Jan 2 at 2:55
$begingroup$
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?
$endgroup$
– Eevee Trainer
Jan 2 at 2:56
$begingroup$
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.
$endgroup$
– John Omielan
Jan 2 at 2:59
$begingroup$
@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
$endgroup$
– Crt
Jan 2 at 3:03
add a comment |
$begingroup$
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:
If the hypotenuse of an isosceles right triangle is $8 sqrt 2$, what is the area of the triangle?
- (A) 18
- (B) 24
- (C) 32
- (D) 48
- (E) 64
geometry triangle
edited Jan 2 at 12:36
JoeTaxpayer
2,19121226
asked Jan 2 at 2:48
CrtCrt
1205
$endgroup$
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:
If the hypotenuse of an isosceles right triangle is $8 sqrt 2$, what is the area of the triangle?
- (A) 18
- (B) 24
- (C) 32
- (D) 48
- (E) 64
geometry triangle
geometry triangle
edited Jan 2 at 12:36
JoeTaxpayer
2,19121226
asked Jan 2 at 2:48
CrtCrt
1205
edited Jan 2 at 12:36
JoeTaxpayer
2,19121226
asked Jan 2 at 2:48
CrtCrt
1205
edited Jan 2 at 12:36
JoeTaxpayer
2,19121226
edited Jan 2 at 12:36
JoeTaxpayer
2,19121226
edited Jan 2 at 12:36
JoeTaxpayer
2,19121226
2,19121226
asked Jan 2 at 2:48
CrtCrt
1205
asked Jan 2 at 2:48
CrtCrt
1205
asked Jan 2 at 2:48
CrtCrt
1205
1205
2
$begingroup$
That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
$endgroup$
– Deepak
Jan 2 at 2:55
$begingroup$
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?
$endgroup$
– Eevee Trainer
Jan 2 at 2:56
$begingroup$
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.
$endgroup$
– John Omielan
Jan 2 at 2:59
$begingroup$
@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
$endgroup$
– Crt
Jan 2 at 3:03
add a comment |
2
$begingroup$
That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
$endgroup$
– Deepak
Jan 2 at 2:55
$begingroup$
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?
$endgroup$
– Eevee Trainer
Jan 2 at 2:56
$begingroup$
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.
$endgroup$
– John Omielan
Jan 2 at 2:59
$begingroup$
@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
$endgroup$
– Crt
Jan 2 at 3:03
2
2
$begingroup$
That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
$endgroup$
– Deepak
Jan 2 at 2:55
$begingroup$
That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
$endgroup$
– Deepak
Jan 2 at 2:55
$begingroup$
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?
$endgroup$
– Eevee Trainer
Jan 2 at 2:56
$begingroup$
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?
$endgroup$
– Eevee Trainer
Jan 2 at 2:56
$begingroup$
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.
$endgroup$
– John Omielan
Jan 2 at 2:59
$begingroup$
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.
$endgroup$
– John Omielan
Jan 2 at 2:59
$begingroup$
@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
$endgroup$
– Crt
Jan 2 at 3:03
$begingroup$
@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
$endgroup$
– Crt
Jan 2 at 3:03
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 Jan 2 at 3:03
DeepakDeepak
16.8k11436
$endgroup$
add a comment |
$begingroup$
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:
$$A = frac12 ab = frac12 frachsqrt 2 frachsqrt 2 = frach^24$$
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 Jan 2 at 3:06
Eevee TrainerEevee Trainer
5,4431936
$endgroup$
add a comment |
$begingroup$
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 Jan 2 at 3:03
Erik ParkinsonErik Parkinson
8549
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
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active
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active
oldest
votes
$begingroup$
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 Jan 2 at 3:03
DeepakDeepak
16.8k11436
$endgroup$
add a comment |
$begingroup$
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 Jan 2 at 3:03
DeepakDeepak
16.8k11436
$endgroup$
add a comment |
$begingroup$
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 Jan 2 at 3:03
DeepakDeepak
16.8k11436
$endgroup$
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 Jan 2 at 3:03
DeepakDeepak
16.8k11436
edited Jan 2 at 3:09
answered Jan 2 at 3:03
DeepakDeepak
16.8k11436
answered Jan 2 at 3:03
DeepakDeepak
16.8k11436
answered Jan 2 at 3:03
DeepakDeepak
16.8k11436
16.8k11436
add a comment |
add a comment |
$begingroup$
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:
$$A = frac12 ab = frac12 frachsqrt 2 frachsqrt 2 = frach^24$$
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 Jan 2 at 3:06
Eevee TrainerEevee Trainer
5,4431936
$endgroup$
add a comment |
$begingroup$
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:
$$A = frac12 ab = frac12 frachsqrt 2 frachsqrt 2 = frach^24$$
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 Jan 2 at 3:06
Eevee TrainerEevee Trainer
5,4431936
$endgroup$
add a comment |
$begingroup$
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:
$$A = frac12 ab = frac12 frachsqrt 2 frachsqrt 2 = frach^24$$
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 Jan 2 at 3:06
Eevee TrainerEevee Trainer
5,4431936
$endgroup$
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:
$$A = frac12 ab = frac12 frachsqrt 2 frachsqrt 2 = frach^24$$
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 Jan 2 at 3:06
Eevee TrainerEevee Trainer
5,4431936
answered Jan 2 at 3:06
Eevee TrainerEevee Trainer
5,4431936
answered Jan 2 at 3:06
Eevee TrainerEevee Trainer
5,4431936
answered Jan 2 at 3:06
Eevee TrainerEevee Trainer
5,4431936
5,4431936
add a comment |
add a comment |
$begingroup$
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 Jan 2 at 3:03
Erik ParkinsonErik Parkinson
8549
$endgroup$
add a comment |
$begingroup$
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 Jan 2 at 3:03
Erik ParkinsonErik Parkinson
8549
$endgroup$
add a comment |
$begingroup$
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 Jan 2 at 3:03
Erik ParkinsonErik Parkinson
8549
$endgroup$
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 Jan 2 at 3:03
Erik ParkinsonErik Parkinson
8549
answered Jan 2 at 3:03
Erik ParkinsonErik Parkinson
8549
answered Jan 2 at 3:03
Erik ParkinsonErik Parkinson
8549
answered Jan 2 at 3:03
Erik ParkinsonErik Parkinson
8549
8549
add a comment |
add a comment |
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2
$begingroup$
That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
$endgroup$
– Deepak
Jan 2 at 2:55
$begingroup$
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?
$endgroup$
– Eevee Trainer
Jan 2 at 2:56
$begingroup$
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.
$endgroup$
– John Omielan
Jan 2 at 2:59
$begingroup$
@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
$endgroup$
– Crt
Jan 2 at 3:03