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# Help

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The vertices of a square (-5,-4),(-2,-8),(1,-4),(-2,0). What is th area of the square

Guest May 24, 2017
#1
+7324
+1

This actually isn't a square since the sides are not perpendicular....It is a rhombus!

Here's what it looks like:

It is made up of two congruent triangles that split the rhombus in half.

(I'm looking at the top and bottom halves, but there are also left and right halves.)

area of rhombus = 2 * area of one of these triangles

= 2 * (1/2) * base of triangle * height of triangle

= base of triangle * height of triangle

=       [ 1 - -5 ]       *  [ (1/2) * (0 - -8) ]

= [ 6 ] * [ 4 ]

= 24   sq. units

hectictar  May 24, 2017
#2
+90052
+2

Here's one other method based on hectictar's pic......

The intersection of two grid lines represents a "point" in the following theorem

So.......the area of this figure  =

(Number of interior points in the figure) + (number of boundary points ) / 2  - 1    =

( 23)   + (4 ) / 2   - 1   =

23 +  2  - 1   =

24  sq units

[ This is known as Pick's  Theorem ]

CPhill  May 24, 2017
#3
+2248
+1

$$24units^2$$

Here's another way to find the area of this figure. The area for area for a rhombus is the following:

Let A= area of rhombus
Let d1= length of a diagonal of rhombus

Let d2= length of the second diagonal of rhombus

$$A=\frac{1}{2}d_1d_2$$

How will we find the lengths of the diagonals? Use the distance formula! That formula is:

$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Let's find the length of d1 first by substituting and then simplifying.

$$d_1=\sqrt{(-2-(-2))^2+(-8-0)^2}$$

$$d_1=\sqrt{0^2+(-8)^2}$$

$$d_1=\sqrt{0+64}$$

$$d_1=8$$

Let's find d2 by substituting and then simplifying:

$$d_2=\sqrt{(-5-1)^2+(-4-(-4))^2}$$

$$d_2=\sqrt{(-6)^2+0^2}$$

$$d_2=\sqrt{36+0}$$

$$d_2=6$$

Now, plug d1 and d2 back into the area formula for a rhombus:

$$\frac{1}{2}*8*6=4*6=24units^2$$

Isn't it beautiful how three unique and different methods arrive one to the same final answer? It fascinates me...

TheXSquaredFactor  May 24, 2017