# Finding area of a non right angled triangle

• Eobardrush
In summary: Slinky.This will create a right triangle, with the height of the Slinky as the third side.Or is this the shortest method there isThe Slinky method is the shortest method, but it is not always possible. Sometimes the original triangle must be sliced into two pieces in order to create a right triangle.
Eobardrush
Homework Statement
Find the area of this triangle
Relevant Equations
0.5 x a x c x Sin B
I just simply used the formula to solve. Note the "x" represents multiplication in this case

0.5 x a x c Sin B

This is based on the conditions given in the textbook I am using which quotes "Use this formula to find the area of any triangle when you know 2 sides and an angle between them"

So I got my answer as follows: 0.5 x 3.5 x 2.5 x Sin 100 = 4.31 cm2

It is correct but what I am struggling is in understanding how I can do this without relying on the formula, like splitting this into a right angled triangle(by finding height for example)
I attached a diagram I drew about this triangle. As you can see I tried to split this into a right angled triangle

Can someone give me some insights on how I can try to find this without relying on the formula so that I would better understand how this works?

Note that I did see the proof of the formula but it was for the normal looking triangles(with acute angles) so I am not sure how to apply the proof to this type of triangle

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Last edited:
Suppose that ##a## is the length of the base. Then ##c \sin(B)## is simply the height perpendicular to the base. So the equation that you gave as the Relevant Equation really is ##area = 0.5 * base * height##.

Delta2
Eobardrush said:
Homework Statement:: Find the area of this triangle
Relevant Equations:: 0.5 x a x c x Sin B

Can someone give me some insights on how I can try to find this without relying on the formula so that I would better understand how this works?
Imagine the triangle as a bunch of thin horizontal slices, stacked up vertically. Obviously, the stack here is slanted to the left.

Now slide all of the slices to the right so that their left edges line up. Obviously this does not change the total area. And obviously the result is a right triangle.

FactChecker
Eobardrush said:
As you can see I tried to split this into a right angled triangle
I can't tell from your picture, but assume that the leftmost point in the triangle is A and the rightmost point is C.

The line segment that you labeled h is not an altitude in ABC.

Draw the line containing BC. Draw a line through A perpendicular to the line containing BC. Say it intersects that line at X. Then AX is the altitude to BC in the triangle ABC. From the picture you will see that ## \triangle ABC = \triangle AXC - \triangle AXB ##

FactChecker, Delta2 and Eobardrush
Prof B said:
I can't tell from your picture, but assume that the leftmost point in the triangle is A and the rightmost point is C.

The line segment that you labeled h is not an altitude in ABC.

Draw the line containing BC. Draw a line through A perpendicular to the line containing BC. Say it intersects that line at X. Then AX is the altitude to BC in the triangle ABC. From the picture you will see that ## \triangle ABC = \triangle AXC - \triangle AXB ##
Thank you! I was able to do it, although it was pretty long. Sorry if the workings look unclear. Was wondering since it was pretty long if there is a shorter way of doing this?(Excluding using the formula I mentioned)
I mean like I might have used some extra unnecessary steps here
Any steps that I should exclude in my workings here?

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Eobardrush said:
Was wondering since it was pretty long if there is a shorter way of doing this?
As @jbriggs444 said in post #3, you can see that the top of the original triangle can get slid to the right, making a right triangle, without changing the area. That gives you a simpler ## area = 1/2 * base * height## for a right triangle without worrying about the other triangle with ##b##.
Eobardrush said:
(Excluding using the formula I mentioned)
Or is this the shortest method there is
You will still need to use the ## area = 1/2 * base * height## equation for a right triangle, but that is so simply related to the ##base * height## formula for the rectangle that it is practically intuitive.

FactChecker said:
As @jbriggs444 said in post #3, you can see that the top of the original triangle can get slid to the right, making a right triangle, without changing the area. That gives you a simpler ## area = 1/2 * base * height## for a right triangle without worrying about the other triangle with ##b##.
I am really bad at visualizing things so I was not able to visualize what he said

Delta2
Eobardrush said:
I am really bad at visualizing things so I was not able to visualize what he said
Take a deck of cards with ##area = deckBase * deckHeight##. Now slide the top slightly to the left until the left edge is tilted like your triangle. Clearly, the stack of cards still has the same ##area = deckBase * deckHeight## as the original deck. Half of that is your triangle area.

jbriggs444
Eobardrush said:
I was able to do it, although it was pretty long. Sorry if the workings look unclear. Was wondering since it was pretty long if there is a shorter way of doing this?(Excluding using the formula I mentioned)
I'd suggest you recalibrate your sense of what a long problem is. :)

One way of reducing the number of calculations would be to not use numbers until the end.
\begin{align*}
\triangle AXC &= \frac 12 h (b+c) \\
\triangle AXB &= \frac 12 h b \\
\triangle ABC &= \triangle AXC - \triangle AXB = \frac 12 h (b+c) - \frac 12 h b = \frac 12 h c
\end{align*} Note that this eliminates the need to find the length ##b## because it cancels out in the end. Use trig to find ##h##, and you've essentially derived the formula you used in the beginning, but now you can see why it works.

Prof B and FactChecker
Eobardrush said:
Note that I did see the proof of the formula but it was for the normal looking triangles(with acute angles) so I am not sure how to apply the proof to this type of triangle
It is good that you are interested in this sort of thing. Note that in the case that X is between A and B it is not necessary to calculate |AX| or |BX|. That sort of observation is helpful when you try to work out other cases yourself.

## 1. How do you find the area of a non right angled triangle?

To find the area of a non right angled triangle, you can use the formula A = 1/2 * base * height. The base and height can be any two sides of the triangle, as long as they form a right angle. If you do not have the base and height, you can use Heron's formula, which takes into account all three sides of the triangle.

## 2. Can you use the same formula for all types of non right angled triangles?

Yes, the formula A = 1/2 * base * height can be used for any non right angled triangle, as long as you have the base and height. However, for triangles with sides of different lengths, Heron's formula may be more accurate.

## 3. What if I don't know the base and height of the triangle?

If you do not know the base and height of the triangle, you can use Heron's formula, which takes into account all three sides of the triangle. The formula is A = √(s(s-a)(s-b)(s-c)), where s is the semi-perimeter of the triangle and a, b, and c are the lengths of the three sides.

## 4. Can you use trigonometry to find the area of a non right angled triangle?

Yes, you can use trigonometry to find the area of a non right angled triangle if you have at least two sides and an angle measurement. The formula is A = 1/2 * a * b * sin(C), where a and b are the two sides and C is the angle between them.

## 5. Is there a specific unit for measuring the area of a non right angled triangle?

The area of a non right angled triangle is typically measured in square units, such as square centimeters or square meters. However, any unit of length can be used for the base and height as long as they are consistent.

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