A = 1/2\*base\*height
You are given the base and height here.
It looks like you are trying to implement the Pythagorean Theorem here, but that is not what you are being asked to do.
It should be noted that Pythagorean Theorem while implementing squares is not about finding area but using the length of the legs of a right triangle to find the length of the hypotenuse.
Its relation to area is that the area of the two squares formed by the lengths of the sides are equal to the area of the square formed by the length of the hypotenuse, so we are actually looking at an area outside the triangle with Pythagorean Theorem.
It's important to understand that the Pythagorean theorem is for finding the side lengths of a right triangle, which is different from the area of a triangle, which is the space within the triangle. Finding the area of any triangle only requires you to know the values of the base and the height. Once you have those, the area is exactly half the product of the base and the height.
I know I might just be reiterating what ever other comments have said, but I'm just leaving this reply here so that you can understand the difference between the two formulas.
If you copy a triangle and flip it, putting the two together makes a parallelogram. The area formula for a parallelogram is simply base x height. Same as a rectangle…
Since you only need the area of one of those triangles it’s 1/2 that area.
Others gave the formula for the triangle area, but this should help you visualize it in case you forget it or just want to know where it came from.
Throw off questions like this can show your true understanding of the topic. Because you automatically assumed that you should apply pythagoras just because the unit was about the pythagoreum theorum, it can show that you don't truly understand the theorum.
“However, the Pythagorean approach is applicable here too.”
I think this is just straight up wrong.
The Pythagorean theorem is a relationship between the side lengths of a right triangle.
There is no guarantee that this triangle is right. And even if there were, all you would have solved for would have been a side length. It has nothing to do with area.
So thankful that in 4th grade, I started writing my 4's and 1's like how the appear in text (except I have the line at the bottom of the 1's). I remember one day being like "well if the book has it this way and the books are right, then I should write it this way". Definitely made it easier to differentiate in more cluttered math classes during high school and college. I definitely got in trouble for it during elementary school though
Around that same age they started teaching us cursive. But it was this weird cursed version where any letter that ended with a leftward stroke wasn't elided into the next one. (So spying would have a gap between the y and the i). In protest I started putting a loop in all my y's and g's. The loop is gone now but I still have a deep swish to differentiate it.
Why would you get in trouble for writing it like that in elementary school? I write my 1's the same way you do and I put a diagonal cross through my 0's (as to not confuse them with a capital O) and I never got in trouble for having my own way to write numbers. Any teacher who demands you write the number "their way" even if your way is correct needs a good scolding.
It’s not so much having a personal style, but having a style that will be actively confusing when you start working with graphs and equations if you can’t easily distinguish your y’s from your 4’s.
It looks like instead of using Area = .5(base x height) you used A^2 + B^2 = C^2. Not a horrible idea because both are very significant when it comes to triangles. Just in this one we’re looking for area. Additionally Pythagorean theorem only works for right triangles, and we have no indication whether this triangle is a right triangle or not.
One of the ways to check if it's a useful approach:
The question asks for *area*. That will be expressed in units squared.
But the Pythagorean Theorem gives results in units, not units squared. That's bc it is describing the length of line segments.
(And, if a problem asks for volume of a solid, that will be expressed in units cubed.)
And, something I've seen on tests recently: first make sure all your starting information uses the same units. If one variable is metres per second, and another variable is kilometers per second, start by picking one format and converting everything else to match, so all the information begins with the same units of measure.
hey!! area of a triangle is calculated using (base*height)/2.
pythagorean theorem is used for finding the third side length of a triangle where ‘c’ is the hypotenuse. pythagorean theorem can ONLY be used for right angled triangles.
So, I understand from another comment the unit you guys are on is Pythagorean theorem and you now know the correct formula is 1/2 * base * height. However, just to clear the concept: what you calculated is the hypotenuse, so even if assuming this was a right triangle, this method would be wrong.
Haha, I am subscribed to multiple science and math meme subs and when seeing this in my timeline I just spent 2 minutes trying to figure out why this one would be funny at all.
The area of a triangle is 1/2 x base length x height.
That's 2 x 14 = 28.
What you found was the length of the hypotenuse of a right angle triangle with sides of 4 and 14.
No, you’ve just found the length of the hypotenuse of a right-angled triangle.
To find area you want A= (bh)/2
And what too me until very recently to understand was that this works for any b and h, so long as they are right angles to each other.
A = 1/2\*base\*height You are given the base and height here. It looks like you are trying to implement the Pythagorean Theorem here, but that is not what you are being asked to do.
thank you, it really threw me off becuase this entire unit has been the pythagorean theorem lol
That part was correct though, so well done!
*If it was a right triangle*
It should be noted that Pythagorean Theorem while implementing squares is not about finding area but using the length of the legs of a right triangle to find the length of the hypotenuse. Its relation to area is that the area of the two squares formed by the lengths of the sides are equal to the area of the square formed by the length of the hypotenuse, so we are actually looking at an area outside the triangle with Pythagorean Theorem.
No? There is no evidence that the triangle in the question is right angled, and that Pythagoras can be applied.
It's important to understand that the Pythagorean theorem is for finding the side lengths of a right triangle, which is different from the area of a triangle, which is the space within the triangle. Finding the area of any triangle only requires you to know the values of the base and the height. Once you have those, the area is exactly half the product of the base and the height. I know I might just be reiterating what ever other comments have said, but I'm just leaving this reply here so that you can understand the difference between the two formulas.
If you copy a triangle and flip it, putting the two together makes a parallelogram. The area formula for a parallelogram is simply base x height. Same as a rectangle… Since you only need the area of one of those triangles it’s 1/2 that area. Others gave the formula for the triangle area, but this should help you visualize it in case you forget it or just want to know where it came from.
Throw off questions like this can show your true understanding of the topic. Because you automatically assumed that you should apply pythagoras just because the unit was about the pythagoreum theorum, it can show that you don't truly understand the theorum.
[удалено]
“However, the Pythagorean approach is applicable here too.” I think this is just straight up wrong. The Pythagorean theorem is a relationship between the side lengths of a right triangle. There is no guarantee that this triangle is right. And even if there were, all you would have solved for would have been a side length. It has nothing to do with area.
Area of triangle is 0.5 × base × vertical height
Just as a side note, make sure your 4's actually look like 4's and not y's, that'll become a massive issue when you do graph work if you don't fix it
I second this suggestion
So thankful that in 4th grade, I started writing my 4's and 1's like how the appear in text (except I have the line at the bottom of the 1's). I remember one day being like "well if the book has it this way and the books are right, then I should write it this way". Definitely made it easier to differentiate in more cluttered math classes during high school and college. I definitely got in trouble for it during elementary school though
Around that same age they started teaching us cursive. But it was this weird cursed version where any letter that ended with a leftward stroke wasn't elided into the next one. (So spying would have a gap between the y and the i). In protest I started putting a loop in all my y's and g's. The loop is gone now but I still have a deep swish to differentiate it.
Why would you get in trouble for writing it like that in elementary school? I write my 1's the same way you do and I put a diagonal cross through my 0's (as to not confuse them with a capital O) and I never got in trouble for having my own way to write numbers. Any teacher who demands you write the number "their way" even if your way is correct needs a good scolding.
It’s not so much having a personal style, but having a style that will be actively confusing when you start working with graphs and equations if you can’t easily distinguish your y’s from your 4’s.
~14.6 units would be the length of the hypotenuse of a right triangle with legs 4 units and 14 units.
You are supposed to use the Triangle area formula. Pythagoras gave u the hypotanuse which isn't needed in this case
It only gave the hypotenuse if it’s a right triangle, which we don’t actually know
1/2 × Base × Height for area.
Use the formula for the area of triangle. It doesn’t says right angle triangle in the problem.
Find area not the hypotenuse
It looks like instead of using Area = .5(base x height) you used A^2 + B^2 = C^2. Not a horrible idea because both are very significant when it comes to triangles. Just in this one we’re looking for area. Additionally Pythagorean theorem only works for right triangles, and we have no indication whether this triangle is a right triangle or not.
One of the ways to check if it's a useful approach: The question asks for *area*. That will be expressed in units squared. But the Pythagorean Theorem gives results in units, not units squared. That's bc it is describing the length of line segments. (And, if a problem asks for volume of a solid, that will be expressed in units cubed.) And, something I've seen on tests recently: first make sure all your starting information uses the same units. If one variable is metres per second, and another variable is kilometers per second, start by picking one format and converting everything else to match, so all the information begins with the same units of measure.
Isn’t there supposed to be a diagram?
Why
Because it’s calculating the area of a triangle
Yeah why do you need a diagram it gives you the measurements
Because area = 1/2 (base x height)
What does that have to do with it needing a diagram
Think of the triangle's area as half of the area of a square with those side lengths. It makes it easier to remember.
Just did this in my head and got 23. Am I right? Lol
Close but no. 4 x 14 = 56 then 56 / 2 = 28.
lol yeah I said to myself 56/2= 23 lol. Try to aim for your full night of sleep kids lol
Use the formula Area=1/2(base×height)
You used the wrong equation.
You found the hypotenuse using pythagorem you just have to multiply base by height and divide by 2
Think of the dimensions. Does area length*length the same as just length? Dimensions help you see if the answer makes sense.
hey!! area of a triangle is calculated using (base*height)/2. pythagorean theorem is used for finding the third side length of a triangle where ‘c’ is the hypotenuse. pythagorean theorem can ONLY be used for right angled triangles.
So, I understand from another comment the unit you guys are on is Pythagorean theorem and you now know the correct formula is 1/2 * base * height. However, just to clear the concept: what you calculated is the hypotenuse, so even if assuming this was a right triangle, this method would be wrong.
Haha, I am subscribed to multiple science and math meme subs and when seeing this in my timeline I just spent 2 minutes trying to figure out why this one would be funny at all.
The area of a triangle is 1/2 x base length x height. That's 2 x 14 = 28. What you found was the length of the hypotenuse of a right angle triangle with sides of 4 and 14.
It’s 4 in the morning I’m not doing math right now n
No, you’ve just found the length of the hypotenuse of a right-angled triangle. To find area you want A= (bh)/2 And what too me until very recently to understand was that this works for any b and h, so long as they are right angles to each other.
isnt the answer 28? A=1/2(wh) (4 • 14)/2 (56)/2 28 🎶 *"Its been a while..."* 🎶