Question

the equation $a = \frac{1}{2}(b_1 + b_2)h$ can be used to determine the area, $a$, of a trapezoid with height, $h$, and base lengths, $b_1$ and $b_2$. which are equivalent equations? choose two correct answers.
$\frac{a}{2h} - b_2 = b_1$ $\frac{2a}{h} - b_2 = b_1$ $\frac{a}{2(b_1 + b_2)} = h$
$\frac{2a - b_2}{h} = b_1$ $\frac{2a}{b_1 + b_2} = h$

Explanation:

We start with the original formula for the area of a trapezoid: \( a = \frac{1}{2}(b_1 + b_2)h \). We will manipulate this formula to find the equivalent equations.

Step 1: Solve for \( b_1 \)

First, multiply both sides of the equation by 2 to eliminate the fraction:
\( 2a = (b_1 + b_2)h \)
Then, divide both sides by \( h \):
\( \frac{2a}{h} = b_1 + b_2 \)
Subtract \( b_2 \) from both sides to solve for \( b_1 \):
\( \frac{2a}{h} - b_2 = b_1 \)
We can also rewrite \( \frac{2a}{h} - b_2 \) as \( \frac{2a - b_2h}{h} \)? Wait, no. Wait, \( \frac{2a}{h} - b_2 = \frac{2a - b_2h}{h} \)? Wait, no, let's check again. Wait, \( \frac{2a}{h} - b_2 = \frac{2a - b_2h}{h} \)? Wait, no, that's not correct. Wait, \( \frac{2a}{h} - b_2 = \frac{2a - b_2h}{h} \)? Wait, no, let's do it step by step. \( \frac{2a}{h} - b_2 = \frac{2a}{h} - \frac{b_2h}{h} = \frac{2a - b_2h}{h} \). Wait, but in the options, we have \( \frac{2a - b_2}{h} = b_1 \)? Wait, no, the option is \( \frac{2a - b_2}{h} = b_1 \)? Wait, no, looking at the options, one of them is \( \frac{2a - b_2}{h} = b_1 \)? Wait, no, the fourth option is \( \frac{2a - b_2}{h} = b_1 \)? Wait, no, the fourth option is \( \frac{2a - b_2}{h} = b_1 \)? Wait, no, let's check the original options again. Wait, the fourth option is \( \frac{2a - b_2}{h} = b_1 \)? Wait, no, the fourth option is \( \frac{2a - b_2}{h} = b_1 \)? Wait, no, looking at the image, the fourth option is \( \frac{2a - b_2}{h} = b_1 \)? Wait, no, maybe I misread. Wait, the original formula: \( a = \frac{1}{2}(b_1 + b_2)h \). Let's try solving for \( h \).

Step 2: Solve for \( h \)

Starting from \( a = \frac{1}{2}(b_1 + b_2)h \), multiply both sides by 2:
\( 2a = (b_1 + b_2)h \)
Then, divide both sides by \( (b_1 + b_2) \):
\( \frac{2a}{b_1 + b_2} = h \)

Now let's check each option:

1. \( \frac{a}{2h} - b_2 = b_1 \): Let's test with the original formula. If we start with \( a = \frac{1}{2}(b_1 + b_2)h \), solving for \( b_1 \) would not give this. So this is incorrect.

2. \( \frac{2a}{h} - b_2 = b_1 \): This is what we derived in Step 1, so this is correct.

3. \( \frac{a}{2(b_1 + b_2)} = h \): Let's check. Starting from \( a = \frac{1}{2}(b_1 + b_2)h \), if we solve for \( h \), we get \( h = \frac{2a}{b_1 + b_2} \), not \( \frac{a}{2(b_1 + b_2)} \). So this is incorrect.

4. \( \frac{2a - b_2}{h} = b_1 \): Wait, no, let's re-express \( \frac{2a}{h} - b_2 \). \( \frac{2a}{h} - b_2 = \frac{2a - b_2h}{h} \), but the option is \( \frac{2a - b_2}{h} = b_1 \), which is different. Wait, maybe there's a typo, or maybe I misread the option. Wait, looking at the image again, the fourth option is \( \frac{2a - b_2}{h} = b_1 \)? Wait, no, maybe it's \( \frac{2a - b_2 h}{h} = b_1 \)? But the option shows \( \frac{2a - b_2}{h} = b_1 \). Wait, maybe the option is \( \frac{2a - b_2 h}{h} = b_1 \), but written as \( \frac{2a - b_2}{h} = b_1 \) with a missing \( h \)? Alternatively, maybe the fourth option is correct as \( \frac{2a - b_2}{h} = b_1 \) is a simplification? Wait, no, let's test with numbers. Let's take \( a = 6 \), \( b_1 = 2 \), \( b_2 = 4 \), \( h = 2 \). Then \( a = \frac{1}{2}(2 + 4)*2 = 6 \), which works. Now let's test option 2: \( \frac{2*6}{2} - 4 = 6 - 4 = 2 = b_1 \), correct. Option 4: \( \frac{2*6 - 4}{2} = \frac{12 - 4}{2} = 4 \), which is not \( b_1 = 2 \). Wait, that's not correct. Wait, maybe the fourth option is \( \frac{2a - b_2 h}{h} = b_1 \). Let's test with the numbers: \( \frac{12 - 4*2}{2} = \frac{12 - 8}{2} = 2 = b_1 \), which is correct. So maybe the option has a typo, and it's \( \frac{2a - b_2 h}{h} = b_1 \), but written as \( \frac{2a - b_2}{h} = b_1 \) (missing the \( h \) in \( b_2 h \))? Alternatively, maybe the fourth option is correct as \( \frac{2a - b_2}{h} = b_1 \) is a mistake, but let's check again. Wait, when we derived \( \frac{2a}{h} - b_2 = b_1 \), that's equivalent to \( \frac{2a - b_2 h}{h} = b_1 \), but the option is \( \frac{2a - b_2}{h} = b_1 \). Maybe the option has a typo, and the \( b_2 \) should be \( b_2 h \)? But assuming the option is \( \frac{2a - b_2 h}{h} = b_1 \), which is the same as \( \frac{2a}{h} - b_2 = b_1 \), so maybe the fourth option is a misprint, but let's check the fifth option.

5. \( \frac{2a}{b_1 + b_2} = h \): This is what we derived when solving for \( h \), so this is correct.

Wait, so the correct options are the second one (\( \frac{2a}{h} - b_2 = b_1 \)) and the fifth one (\( \frac{2a}{b_1 + b_2} = h \)). Also, let's check the fourth option again. If we take \( \frac{2a - b_2}{h} = b_1 \), let's plug in the numbers: \( a = 6 \), \( b_1 = 2 \), \( b_2 = 4 \), \( h = 2 \). Then \( \frac{2*6 - 4}{2} = \frac{12 - 4}{2} = 4 \), which is not equal to \( b_1 = 2 \). So that's incorrect. Wait, maybe the fourth option is \( \frac{2a - b_2 h}{h} = b_1 \), which would be \( \frac{12 - 8}{2} = 2 \), which is correct. So maybe the option has a typo, and the \( b_2 \) should be \( b_2 h \). But given the options, the second one (\( \frac{2a}{h} - b_2 = b_1 \)) and the fifth one (\( \frac{2a}{b_1 + b_2} = h \)) are correct. Also, let's check the fourth option again. Wait, maybe I made a mistake in the algebra. Let's re-express \( \frac{2a}{h} - b_2 \). \( \frac{2a}{h} - b_2 = \frac{2a - b_2 h}{h} \), so if the fourth option is \( \frac{2a - b_2 h}{h} = b_1 \), that's correct. But the option shows \( \frac{2a - b_2}{h} = b_1 \), which is missing the \( h \) in \( b_2 h \). Maybe that's a typo, and the intended option is \( \frac{2a - b_2 h}{h} = b_1 \), which is equivalent to \( \frac{2a}{h} - b_2 = b_1 \). So perhaps the fourth option is also correct. Wait, let's check with the numbers again. If \( a = 6 \), \( b_1 = 2 \), \( b_2 = 4 \), \( h = 2 \), then \( \frac{2*6 - 4*2}{2} = \frac{12 - 8}{2} = 2 = b_1 \), which is correct. So if the fourth option is \( \frac{2a - b_2 h}{h} = b_1 \), but written as \( \frac{2a - b_2}{h} = b_1 \) (missing the \( h \) in \( b_2 h \)), then it's a typo, but the intended answer is correct. Alternatively, maybe the fourth option is \( \frac{2a - b_2}{h} = b_1 \) is incorrect, but the second and fifth are correct. Wait, the fifth option is \( \frac{2a}{b_1 + b_2} = h \), which is correct. The second option is \( \frac{2a}{h} - b_2 = b_1 \), which is correct. The fourth option: if we take \( \frac{2a - b_2}{h} = b_1 \), with our numbers, that gives 4, which is not \( b_1 \), so that's incorrect. So the correct options are the second one (\( \frac{2a}{h} - b_2 = b_1 \)) and the fifth one (\( \frac{2a}{b_1 + b_2} = h \)).

Answer:

The two correct equivalent equations are:

• \( \boldsymbol{\frac{2a}{h} - b_2 = b_1} \)
• \( \boldsymbol{\frac{2a}{b_1 + b_2} = h} \)