Respuesta :
The complete question is as follows.
The equation a = [tex]\frac{1}{2}(b_1 + b_2 )h[/tex] can be used to determine the area , a, of a trapezoid with height , h, and base lengths, [tex]b_1[/tex] and [tex]b_2[/tex]. Which are equivalent equations?
(a) [tex]\frac{2a}{h} - b_2 = b_1[/tex]
(b) [tex]\frac{a}{2h} - b_2 = b_1[/tex]
(c) [tex]\frac{2a - b_2}{h}[/tex] = [tex]b_1[/tex]
(d) [tex]\frac{2a}{b_1 + b_2} = h[/tex]
(e) [tex]\frac{a}{2(b_1 + b_2)}[/tex] = h
Answer: (a) [tex]\frac{2a}{h} - b_2 = b_1[/tex]; (d) [tex]\frac{2a}{b_1 + b_2} = h[/tex];
Step-by-step explanation: To determine [tex]b_1[/tex]:
a = [tex]\frac{1}{2}(b_1 + b_2 )h[/tex]
2a = ([tex]b_1 + b_2[/tex])h
[tex]\frac{2a}{h} = b_1 + b_2[/tex]
[tex]\frac{2a}{h} - b_2 = b_1[/tex]
To determine h:
a = [tex]\frac{1}{2}(b_1 + b_2 )h[/tex]
2a = [tex](b_1 + b_2)h[/tex]
[tex]\frac{2a}{(b_1 + b_2)}[/tex] = h
To determine [tex]b_2[/tex]
a = [tex]\frac{1}{2}(b_1 + b_2 )h[/tex]
2a = [tex](b_1 + b_2)h[/tex]
[tex]\frac{2a}{h} = (b_1 + b_2)[/tex]
[tex]\frac{2a}{h} - b_1 = b_2[/tex]
Checking the alternatives, you have that [tex]\frac{2a}{h} - b_2 = b_1[/tex] and [tex]\frac{2a}{(b_1 + b_2)}[/tex] = h, so alternatives A and D are correct.
