How do you find the area of a triangle?
A triangle is formed by the points A(-1,1) B(4,3) and C (0,13).
Find the area of triangle ABC.
i know that fromula but which is the BASE and the HEIGHT??? HOW CAN i FIND THE BASE AND THE HEIGHT????????
- 1 decade agoFavorite Answer
Tiziana, there isn't a single formula for finding the area of triangles and in this case, you cannot simply use the formula 1/2*base*height. You would need complicated formulas for this 1. Since it would be pretty difficult to explain you the more complicated but faster formulas, I would use a less complicated but a fairly long formula. Please take some time to read it carefully and everything would go into your head. Here it goes..
First, since A,B and C are points on a co-ordinate axis and you do not know the sides of the triangle, you will first have to find them using the distance formula. If don't know what that is, then read on..
Acc to distance formula,
if A and B r points on the co-ordinate axis, such that
A=(x1,y1) and B=(x2,y2) then
AB^2 = (x2-x1)^2 + (y2-y1)^2
therefore, AB = [ (x2-x1)^2 + (y2-y1)^2]^1/2 (^1/2 means its sq root)
So now by applying this formula in the points which you gave which are A=(-1,1) , B=(4,3) , C=(0,13)
AB=[(-1-4)^2 + (1-3)^2]^1/2
BC=[(4-0)^2 + (3-13)^2]^1/2
BC=[16 + 100]^1/2
AC=[(-1-0)^2 + (1-13)^2]^1/2
Now, when you have all the sides, to find the area of the triangle we use the Heron's Formula which states-
Area of the triangle=[s(s-a)(s-b)(s-c)]^1/2
where a,b,c are sides of the triangle and s=(a+b+c)/2
In this case,
a=√29, b=√116, c=√145
s=(√29 + √116 + √145)/2
=(√29 + √4*29 + √5*29)/2
=(√29 + 2*√29 + √5*√29)/2
So now by put these values in the Heron's Formula and find the answer. I could have found you the answer but this post is getting just too long. You just need a calc for this part.
Hope this helps ;)
- 1 decade ago
What I would do is draw it out on a piece of paper. Then, draw a rectangle around the sides of the triangle, containing the vertices of the triangle. So the lines of the rectangle would be: x= -1, x = 4, y = 13, and y = 1.
Next, find the area of the rectangle (length times width). So the length would be 12 and the width would be 5, so the area would be 60!
Then, all you have to do is find the areas of the right triangles on the inside of the rectangle, and subtract that from the area of the rectangle. For example, you would have a right triangle whose vertices were (-1,1), (-1,13), and (0,13). Finding the area of that triangle would be easy, because you already know the height and the base. So just do that for all 3 right triangle and subtract that from the area of the rectangle!
Sorry, that might have been super confusing!
- 1 decade ago
You wouldn't be able to work out the area by using Area= (1/2)*h*b since there is no way of knowing the base and height.
As the point ABC are given as co-ordinates in a plane, you could calculate the area by taking the points A,B,C as position vectors from the origin and using:
Area = (1/2)*a*b*Sin(C)
where a = length of vector AB
b = length of vector b
and C = the angle between a and b.
Are you familiar with vectors?Source(s): Maths Student
- Peter HLv 71 decade ago
You don't need to know about vectors or determinants to solve this. You need Pythagoras and a useful formula. First, sketch the positions of A, B and C on paper, and we go on to find the lengths of the sides. AC is the hypotenuse of a right-angled triangle, of which the 3rd apex is at (0,1). From your sketch you can see what the lengths of the other 2 sides of this right-angled triangle are, and you can calculate AC. Repeat the exercise for AB and BC.
Then there is the useful formula
Area of Triangle when the length of the three sides is known =
1/4 * square-root [ P * (P-2a) * (P-2b) * (P-2c)]
where P is total perimeter (a=b+c)
and a, b and c are the lengths of the three sides.
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- 1 decade ago
A= Base times Height divided by 2.
- 1 decade ago
Are you familiar with determinants?
The area will be 1/2 of the value of this determinant
| -1 1 1 |
| 4 3 1|
| 0 13 1|
=29 units square
I could not figure out a way to draw the determinant in this space.
- Anonymous1 decade ago
Well i'm just gonna put this boldly, cause lets face it, maths is all about getting to the point. Its just simply:
a = length of vector AB
b = length of vector BC
and c = angle between AB.