Math (Quadrilaterals and Polygons)?

Question

Hey guys, 
I have been stuck on these two questions for the longest time. I've tried everything. PLEASE HELP!!! 
No.1) 
Find the value of x and y. 

No. 2) 

Find the area of the figure 

Answer 1

1) From the figure , the trepezoid is isosceles trepezoid .

In an isosceles trepezoid when the sides that are n't parallel are equal in length and both angles coming from parallel sides are equal.

2x + 3y = 12 → (1)

30x + 11y  = 112 → (2)

Solving above two equations

To get two equations that contain opposite terms multiply the first equation by 15.

Write the equations in column form and add the equations to eliminate x - variable.

The resultant equation is - 34y  = - 68

y  = (-68)/(-34)

y  = 2

Substitute the value of  y  = 2 in either of the original equations and solve for x.

The first equation:  2x + 3y  = 12

2x + 3(2) = 12

2x  + 6 = 12

2x  = 12 - 6

2x  = 6

x  = 6/2

x  = 3

Therfore , the values of x  = 3 and y  = 3.

2) From the figure it is a irregular pentagon.

In this case the irregular pentagon can be split into one rihtangle triangle and trepezoid.

Now we need to find area of each of them and add them together.

• Area of right angle triangle = 1/2 * base * height

= (13.42 * 12)/2

= 13.42*6

= 80.52 in²

• Area of the trepezoid = (a + b)h/2

= [(11.75 + 18)6.12]/2

= 91.035 in²

Total area of irregular pentagon = Area of the trepezoid + Area of right angle triangle

= 80.52 in² + 91.035 in ²

= 171.555 in²

Therfore, area of figure is 171.555 in².