# Find the area of the triangle with component measurements listed below: A=5 B=8 C=10 I did some work on this and I came up with a series of equations. I just got stuck and have no idea how to...

Find the area of the triangle with component measurements listed below:

A=5

B=8

C=10

I did some work on this and I came up with a series of equations. I just got stuck and have no idea how to proceed. I provided a pic of what I did so far.

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When given with the lengths of the three sides of a triangle, it is better to use Heron's formula in solving its area.

The Heron's formula is:

`A =sqrt(s(s-a)(s-b)(s-c))`

where

a, b, and c are the lengths of the sides of the triangle and

s is the semi-perimeter of the triangle.

To apply that, solve for the semi-perimeter of the triangle. To do so, add the three sides and take half of it.

`s=(a+b+c)/2 = (5+8+10)/2=23/2`

`s=11.5`

Then, plug-in the values of a, b, c and s to Heron's formula.

`A=sqrt(11.5(11.5-5)(11.5-8)(11.5-10))`

`A=sqrt(11.5(6.5)(3.5)(1.5))`

`A=sqrt(392.4375)`

`A=19.81`

**Hence, the area of the given triangle is 19.81 square units.**

If you are given a problem with lengths of sides of a triangle, you can simply use Heron's formula to solve for the area.

Heron's formula is as follows:

`A = sqrt((s)(s-a)(s-b)(s-c))`

where a, b, and c are lengths of a side of the triangle and

`s = (a+b+c)/2`

which is also known as the semiperimeter.

To solve this problem, we must first solve for s.

`s = (5+8+10)/2`

`s = 23/2 = 11.5`

Next, plug this into the original area formula

`A = sqrt((11.5)(11.5-5)(11.5-8)(11.5-10))`

`A = sqrt((11.5)(6.5)(3.5)(1.5))`

`A = sqrt((392.44))`

`A = 19.81`

Thus, the area equals 19.81

Heron's formula

,

is what would be best in this problem. However, you can continue with the process that you had started, just with a little tweeks. I'll show you how to do this, and the second link is a proof that shows how the following result is equivalent, even if not very desirable.

Set up the triangle as shown in the image.

Then set up according to Pythagoras:

a^2 = h^2 + x^2

b^2 = h^2 +(c-x)^2

Solve for h^2 and substitute:

h^2 = a^2 -x^2

b^2 = (a^2 -x^2) +(c-x)^2

Solve for x:

b^2 = (a^2 -x^2) + (c^2 -2xc +x^2)

x^2 cancels out

b^2 = a^2 + c^2 -2xc

2xc = a^2 + c^2 - b^2

x = (a^2 + c^2 - b^2)/2c

Then use x to find h:

h^2 = a^2 -x^2

h^2 = a^2 - ((a^2 + c^2 - b^2)/2c)^2

h = sqrt [ a^2 - ((a^2 + c^2 - b^2)/2c)^2]

Then use the basic equation for area of a triangle, A = base * height, base being c in this case and height above.

A = c * sqrt [ a^2 - ((a^2 + c^2 - b^2)/2c)^2]

The following link shows how this is equivalent by simplifying the expression immensely. And although this will get you the same answer of 19.81, its almost impossible to keep straight, or memorize, so using the form

is much more preferable, however, that doesn't make your process any less right, just complicated.

Heron's formula for finding area of triangle:

A = sqrrt(s×(s-a)×(s-b)×(s-c))

s = (a+b+c)/2 {s is semi perimeter of triangle}

Here a =5,b = 8,c = 10

s = (5+8+10)/2

= 11.5m

Area = sqrrt(11.5×(11.5-5)×(11.5-8)×(11.5-10))

=19.81m^2