Help: 3 Geometry questions

1.     Draw VR....this creates triangle VMR  and rectangle VRDE

Let VR be the base of triangle VMR  = 6 units   and the height drawn from M  to the base  = 3 units

So...the area of this triangle is  (1/2)base* height  = (1/2) 6 *3  =  (1/2) 18  =

9 units^2   (1)

And rectangle VRDE has area VR * VE   =  6 * 5  =  30 units^2  (2)

So.....adding (1) and (2)  gives  the total area  of  39 units^2 

2.  Connect  ( 2, 1)  and (3, -3)....this creates a rectangle  with vertices (2,1) , (3, -3), (-5, -5) 

and (-6, -1)

The distance from  (2,1) to (3, -3)  forms one side of this rectangle  = √17 units

The distance  from (-6, -1) to (2,1)  forms the other side of this rectangle  = 2√17 units

So....the area of this rectangle is  √17 * 2√17  =  2 * 17  =  34 units^2   (1)

Connect (2, 1)  and (3, 1)......this forms a triangle at the bottom right with vertices  (2,1), (3,1) 

and (3, -3)

The base of this triangle  is  (2,1) to (3,1)  = 1 unit

And the height of this triangle is (3, 1)  to (3, -3)  =  4 units

So...the area of this triangle is  (1/2) (1) (4)  =  2 units^2   (2)

Connect (2,3)  and (3,3)....this creates a rectangle with vertices  (2,3) (3,3), (2,1) and (3,1)

The height of this rectangle is  (3,1) to (3, 3)  = 2 units

And the widthof this rectangle is from (2,1) to (3,1)  = 1 unit

So...the area of the rectangle is  2 * 1  = 2 units^2  (3)

Finally....we have a triangle with the vertices (2,3), (3,3)  and (2,7)

The base of this triangle is from (2,3) to (3,3)  = 1 unit

And the height of this triangle is from   (2,3) to (2,7) =  4 units

So....the area of this triangle  is  (1/2) (1) (4)  =  2 units^2   (4)

So  the total area =   (1)  + (2)  + (3) + (4)  =  [34 + 2 + 2 + 2]  =

40 units^2

3)  We have a triangle at the top left with a base of 3 units  and a height of 4 units

So its area  is  (1/2) (3) (4)  =  6 inches^2

And we have a reatangle with dimensions 8 x 7   =  56 inches^2

So  the total area is   6 + 56   =    62 inches^2