SOLUTION: I need to find the area of a trapezoid but the trapezoid isn't a isosceles trapezoid. The bases of this trapezoid is 10 and 24 and it has one leg that is 13 and the second leg is

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Question 250457: I need to find the area of a trapezoid but the trapezoid isn't a isosceles trapezoid.
The bases of this trapezoid is 10 and 24 and it has one leg that is 13 and the second leg is 15
I have to find the exact area of this problem

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
Use of Heron's formula may help.

That gives you the area of a triangle given 3 sides of the triangle.

The formula is:

Area=sqrt(s(s-a)(s-b)(s-c))

s = sum of the sides of the triangle divided by 2.
a,b,c are the sides of the triangle.

your trapezoid looks like and is labeled like this:

                           A                           B
                           xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
                         x x                           xx
                      x    x                           x x
                   x       x                           x  x
                x          x                           x   x
             x             x                           x    x
          x                x                           x     x
       x                   x                           x      x
     xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
     C                     D                           E       F 

your trapezoid is ABFC
you drop an altitude from A to D and from B to E.
you have a rectangle at ABED
you have 2 triangles at ADC and BEF
CA is equal to 15
AB is equal to 10
BF is equal to 13
CF is equal to 24
DE is equal to 10 because it is the opposite side of AB in rectangle ABED.

the length of your top base is AB = 10
the length of your bottom base is CF = 24 with DE representing 10 of it.

if you remove the rectangle from the diagram, you are left with 2 triangles that connect together and form 1 triangle that has sides of 15, 13, and a base of 14.

your modified diagram looks like this:
                           A                          
                           x
                         x xx
                      x    x x
                   x       x  x
                x          x   x
             x             x    x
          x                x     x
       x                   x      x
     xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
     C                     D       F 

CA = 15
AF = 13
CF = 24 - 10 = 14

the sum of the perimeter is equal to 15 + 13 + 14 = 42

Heron's formula says the area of this triangle is equal to:

Area=sqrt(s(s-a)(s-b)(s-c))

where s = (a+b+c)/2 = 42/2 = 21

Formula becomes:

Area = sqrt (21 * (21-15) * (21-13) * (21-14))

This resolves to:

Area = 84

Now that we have the area of this triangle, we can find the height of the triangle which is also the height of the trapezoid.

The area the triangle is equal to 1/2 * b * h.

Since the area of the triangle is 84, this becomes:

84 = 1/2 * b * h

Since the base of the triangle is 14, this becomes:

84 = 1/2 * 14 * h

solve for h to get:

h = 84 * 2 / 14 = 12

now that we have h we can solve for the area of the trapezoid.

Area of the trapezoid = (1/2) * (b1+b2)*h.

This equals to (1/2) * (10+24) * 12 = (1/2) * (34) * 12

Area of the trapezoid = 204

Here's a link to Heron's formula.

http://mste.illinois.edu/dildine/heron/triarea.html

I don't know if there's another way to solve this. I tried to find a way to get the height but was unsuccessful. It just didn't appear there was enough information provided.





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