Answer: 19 cm
Step-by-step explanation:
[tex]A_{trapezoid}=\frac{b_{1}+b_{2}}{2}*h[/tex]
99 = [tex]\frac{(x+4 + x+9}{2}*(x - 4)[/tex]
99 = [tex]\frac{(2x + 13}{2}*(x - 4)[/tex]
198 = (2x + 13)(x - 4)
198 = 2x² + 5x - 52
0 = 2x² + 5x - 250
0 = 2x²- 20x + 25x - 250
0 = 2x(x - 10) + 25( x - 10)
0 = (2x + 25)(x - 10)
0 = 2x + 25 or 0 = x - 10
[tex]-\frac{25}{2}[/tex] = x or x = 10
Since length cannot be negative, [tex]-\frac{25}{2}[/tex] can be disregarded
Larger base: x + 9 = 10 + 9 = 19
Answer:
Length of the larger base is 19 cm.
Step-by-step explanation:
Height of the trapezoid = (x - 4)
Bases of the trapezoid = (x + 4) and (x + 9)
Area of the trapezoid = 99 cm²
We know the formula,
Area of trapezoid = [tex]\frac{1}{2}(\text{Sum of bases})\times (\text{Distance between the bases}})[/tex]
99 = [tex]\frac{1}{2}[(x + 4)+(x + 9)](x - 4)[/tex]
99 = [tex]\frac{1}{2}[2x + 13](x - 4)[/tex]
99×2 = (2x + 13)(x - 4)
198 = 2x² - 8x + 13x - 52
2x² + 5x - 52 - 198 = 0
2x² + 5x - 250 = 0
2x² + 25x - 20x - 250 = 0
x(2x + 25) - 10(2x + 25) = 0
(2x + 25)(x - 10) = 0
(2x + 25) = 0
2x = -25
x = -[tex]\frac{25}{2}[/tex]
Since length of the base can not be negative.
Therefore, (x - 10) = 0 will be the solution
x = 10 cm
Length of the larger base = x + 9
= 10 + 9
= 19 cm
