How to define the height of some triangles

[chijioke]

I know that the area of a triangle $$=\frac{1}{2}~base×height$$ And that all triangles has area. I also know that to calculate the area of triangle where the height is not given and all the three sides are given, the area can also be calculated.
My focus here is how do I define the height of some triangles to enable me calculate the area. Now here is a problem. I am required to find the area of these triangles
I have done #2. Here is my work on it. B= base, h= height.

When I tried solving #3, I was stucked because I found it difficult defining and obtaining the height. Nevertheless, I managed to do something. I will show what I managed to do on post #2

 

[chijioke]

Is the above correct? If it is correct, how do I then obtain the height h?
For this one below:
This is what I did
Is it correct If, how do I then obtain the height by way of calculation?

 

[Dr.Peterson]

Your answer for #2 is correct; you could have done less work if you realized that it is isosceles, so the two halves are identical.

Problem #3 is invalid; the information is inconsistent, not satisfying the Law of Sines. Also, the h you show is meaningless. The height of a triangle is the perpendicular distance from one vertex to whatever side you are calling the base.

The other problem is nonsense. I don't know what they might mean by the "length" of a triangle. But you might suppose that they mean "height", Can you solve it then?

 

[khansaheb]

I know that the area of a triangle $$=\frac{1}{2}~base×height$$ And that all triangles has area. I also know that to calculate the area of triangle where the height is not given and all the three sides are given, the area can also be calculated.
My focus here is how do I define the height of some triangles to enable me calculate the area. Now here is a problem. I am required to find the area of these trianglesView attachment 35635View attachment 35636
I have done #2. Here is my work on it. B= base, h= height.
View attachment 35648
When I tried solving #3, I was stucked because I found it difficult defining and obtaining the height. Nevertheless, I managed to do something. I will show what I managed to do on post #2

Do you know the

Laws of sines of a triangle and

Laws of cosines of a triangle ?

In most of these types of problems, where you are specifically interested in calculating the are enclosed, I try to calculate the lengths of 3 sides (using laws of sines & cosines) and calculate the area.

 

[chijioke]

The height of a triangle is the perpendicular distance from one vertex to whatever side you are calling the base.

So could you help draw it indicating the height how it is supposed to appear let me just take a look at it?

 

[stapel]

So could you help draw it indicating the height how it is supposed to appear let me just take a look at it?

There are probably *three* heights of the triangle. *You* decide which vertex is to be your apex. The side opposite that vertex (extended, if necessary) is then your base.

Take your ruler, put it perpendicular (roughly) to the base, and slide it from side to side until the ruler also crosses the vertex you chose. That's your height line.

If you get stuck, please reply showing your work in following the above steps. Thank you!

Eliz.

 

[Dr.Peterson]

So could you help draw it indicating the height how it is supposed to appear let me just take a look at it?

Here are two possible heights:



Here, h₁ is the height corresponding to the 7 cm base, and h₂ is the height corresponding to the 14 cm base.

But, again, this triangle is impossible.

 

[chijioke]

There are probably *three* heights of the triangle. *You* decide which vertex is to be your apex. The side opposite that vertex (extended, if necessary) is then your base.

Take your ruler, put it perpendicular (roughly) to the base, and slide it from side to side until the ruler also crosses the vertex you chose. That's your height line.

If you get stuck, please reply showing your work in following the above steps. Thank you!

Eliz.

The area of triangle is $$25.97~cm^2$$

Here are two possible heights:

View attachment 35658

Here, h₁ is the height corresponding to the 7 cm base, and h₂ is the height corresponding to the 14 cm base.

But, again, this triangle is impossible.

Thanks. I only needed the height in order to calculate the area of triangle of triangle 3.

 

[Dr.Peterson]

Yes, that's the correct area if such a triangle existed. And you'll get the same area if you use the other known base and the corresponding height.

Unfortunately, if you draw a triangle with the base of 7 and the angles given, here is what you get:

​

Side a is wrong!

And if you keep the lengths of a and c, and the angle at B (which is the one angle that was not given), you get this triangle, which has the area you calculated:

​

But it doesn't have the two given angles!

Do you see what I mean about the triangle not existing? You can either get all the angles and one side, or the two given sides and the angle they didn't give. And the Law of Sines told me that long ago: the problem is invalid!

 

[chijioke]

Thank you.

 

[chijioke]

The other problem is nonsense. I don't know what they might mean by the "length" of a triangle. But you might suppose that they mean "height", Can you solve it then?

If they mean that the length is height then area of the triangle is $$\frac{1}{2}×14*6=42~cm^2$$

 

[Dr.Peterson]

If they mean that the length is height then area of the triangle is $$\frac{1}{2}×14*6=42~cm^2$$

Yes. But I have no idea what they actually meant.