Find the area of quadrilateral ABCD having vertices A(-1,5) , B(-2,6) , C(4,10) and D(3,5)
Did you make a diagram? If not, do it now!
Draw BE parallel to AD. It intersects DC at (3.2,6), 1/5 of the way from D to C.
Now you just have
a triangle BEC with base 5.2 and height 4
a trapezoid with bases 5.2 and 4, and height 1
Now just find their areas and add them up.
Can you tell how the DC(3.2,6) comes?
Thanks
https://www.wolframalpha.com/input/?i=plot+%7B%28-1%2C5%29%2C%28-2%2C6%29%2C%284%2C10%29%2C%283%2C5%29%2C%28-1%2C5%29%7D
Note that the scaling for x and y is not the same
Several ways to to this.
1. Join AC to get two triangles. Then find the area of each triangle.
using Heron's formula or some other way to find the area of triangle.
2. From each of the points, drop perpendiculars to the x-axis.
Call these points A', B', C', and D'
area ABCD = area BB'CC' - BB'AA' - AA'DD' - DD'C'C
= (1/2)(6+10)(6) - (1/2)(6+5)(1) - (1/2)(5+5)(4) - (1/2(5+10)(1)
= 48 - 11/2 - 20 - 15/2
= 28 - 13
= 15
3. You could find angle B in triangle ABC, and angle D in triangle ADC
and use
area ABC = (1/2)(AB)(BC) , do the same for the other triangle
An easy way is to use the following algorithm
list the points in a column going counter-clockwise, starting with any point but repeating that point at the end
-1 5
3 5
4 10
-2 6
-1 5
area = (1/2)(downproducts - upproducts)
= (1/2)(-5 + 30 + 24 -10 - (15 + 20 - 20 - 6))
= (1/2)( 39 - 9) = 15
To find the area of the quadrilateral ABCD, you can use the formula for the area of a quadrilateral given its coordinates.
The formula is:
Area = 1/2 * |x1*(y2 - y3) + x2*(y3 - y1) + x3*(y1 - y2) + x1*(y3 - y4) + x3*(y4 - y1) + x4*(y1 - y3)|
where (x1, y1), (x2, y2), (x3, y3), and (x4, y4) are the coordinates of the vertices of the quadrilateral in a specific order.
Now, let's substitute the coordinates of the vertices A(-1,5), B(-2,6), C(4,10), and D(3,5) into the formula:
Area = 1/2 * |-1*(6 - 10) + -2*(10 - 5) + 4*(5 - 6) + -1*(10 - 5) + 4*(5 - 10) + 3*(10 - 5)|
Simplifying further:
Area = 1/2 * |-4 + 10 - 4 - 5 - 20 + 15|
Area = 1/2 * |-8|
Area = 1/2 * 8
Area = 4 square units
Therefore, the area of the quadrilateral ABCD is 4 square units.