Which combination of shapes will balance the scale?

Steps to Solve

1. Analyze the first balance scale

The first scale shows that 2 squares and 2 triangles are equal to 2 half-circles and 1 triangle. We can write this as an equation, using $S$ for square, $T$ for triangle, and $H$ for half-circle:

$2S + 2T = 2H + 1T$

2. Simplify the equation

Subtract 1 triangle ($1T$) from both sides of the equation:

$2S + 2T - 1T = 2H + 1T - 1T$ $2S + T = 2H$

3. Solve for a half-circle

Divide both sides of the equation by 2:

$\frac{2S + T}{2} = \frac{2H}{2}$ $S + \frac{1}{2}T = H$

4. Determine what balances one half-circle

So, one half-circle is equivalent to one square and one-half of a triangle.

$H = S + \frac{1}{2}T$

Since we can't have half a triangle, let's multiply both sides of the equation by 2.

$2H = 2S + T$

5. Find the combination of shapes that balance the second scale

The second scale has one half-circle on the left side. From step 4, we know $H = S + \frac{1}{2}T$. Since we can't have half a triangle, let's go back to equation from step 2

$2S + T = 2H$ $H = S + \frac{1}{2}T$

Let's multiply both sides of equation by 2 to get rid of the fraction. That will give use 2 half circles, which we don't want, since we only have 1 half-circle.

Looking at the answer options, the only option that contains 1 square and 1 triangle is option 4. However, it has 1 square and 2 triangles. Therefore, let's examine the first equation again, $2S + 2T = 2H + T$. If we divide by 2, then we get $S + T = H + \frac{1}{2}T$. If we subtract the fraction of a triangle again, then we get $S + \frac{1}{2}T = H$, which we already know.

Let's go back to $2S + T = 2H$. If we divide by 3/2, we get $\frac{2S + T}{3/2} = \frac{2H}{3/2}$. That becomes $\frac{4S + 2T}{3} = \frac{4H}{3}$. However, this gets messy, so we will skip it.

Instead, let's think about what equals one half-circle

We know that $2S + T = 2H$

Divide by 2 on both sides: $S + \frac{1}{2}T = H$

Looking at the choices, the only answer that includes a square is numbers 1, 2, and 4. Number 2 includes a half circle, which is not correct, because it is the answer we are trying to solve for. Number 4 is one square and two triangles, which is not equivalent to one half-circle. The only answer that includes shapes that are equivalent to 1 half-circle is 4.

Therefore, the answer is 4.