One base is 3 inches and other base is 9 inches.
Solution:
Height of the trapezoid = 8 in
Let [tex]b_1 \ \text{and} \ b_2[/tex] be the bases of the trapezoid.
One base = 6 in + other base
[tex]b_1=6+b_2[/tex]
Area of the trapezoid = 48 in²
Using area of trapezoid formula:
[tex]$\frac{1}{2} (b_1+b_2)\times h= 48[/tex]
[tex]$\frac{1}{2} (6+b_2+b_2)\times 8= 48[/tex]
[tex]$(6+2b_2)\times 4= 48[/tex]
[tex]$24+8b_2= 48[/tex]
Subtract 24 from both sides, we get
[tex]8 b_{2}=24[/tex]
Divide by 8 on both sides, we get
[tex]b_2=3[/tex]
One base is 3 inches.
Other base = 6 + 3 = 9
Other base is 9 inches.
