A trapezoid and a parallelogram. What must the value of x be for the trapezoid to have the same area as the parallelogram?
Trapezoid: b:x B:x+2 h:3
Parallelogram: b:x+4 h:2
+129850 Trapezoid: b:x B:x+2 h:3
Area of the Trapezoid = (3/2)[x + x + 2] = (3/2)[2x + 2] = (3/2)*2[x + 1] = 3[x + 1]
Parallelogram: b:x+4 h:2
Area of the Parallelogram = [x + 4]*2
So....setting these equal, we have......
3*[x + 1] = [x + 4] *2
3x + 3 = 2x + 8 subtract 2x, 3 from both sides
x = 5
The area is 18 sq units for each figure.....
