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avatar+51 

Figure ABCD has vertices A(−3, 2), B(2, 2), C(2, −4), and D(−3, −2). What is the area of figure ABCD?

Sunrisesk8r  Aug 1, 2017
 #1
avatar+3189 
+1

Is it 25?????

tertre  Aug 1, 2017
 #2
avatar+90023 
+1

 

Look at the following diagram :

 

 

 

 

If you include  point E as (2, - 2)....we have a rectangle ABED  whose width  = AB = 5  and height  = BE  = 4

 

So...its total area is just  width * height  = 5 * 4  = 20 sq units

 

And triangle DEC has a base DE  = 5  and a height EC  = 2....so  its area  is [ b * h ] / 2  = [ 5 * 2 ] / 2  = 10 / 2  = 5 sq units

 

So....the toal area is just 20 + 5   = 25 sq units

 

 

cool cool cool

CPhill  Aug 1, 2017
 #3
avatar+2248 
0

Graphing is technically not necessary to solve the problem as you could be aware of the conditions of the different types of quadrilaterals, but doing so helps the solving process immensely.

TheXSquaredFactor  Aug 1, 2017

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