Shirley is drawing triangles that have the same area. The base of each triangle varies inversely with the height. What are the possible base and height of a second triangle if the first triangle's base is 14

Question

and its height is 6
?

Answer 1

To find the possible base and height of a second triangle with the same area as the first triangle, we start by calculating the area of the first triangle using the formula for the area of a triangle:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

For the first triangle, the base is \( b_1 = 14 \) and the height is \( h_1 = 6 \). Therefore, the area of the first triangle is:

\[ \text{Area}_1 = \frac{1}{2} \times 14 \times 6 = 42 \]

Now that we know the area of the first triangle, we can set up the area for the second triangle, which must also equal 42. Let's denote the base of the second triangle as \( b_2 \) and the height as \( h_2 \). Thus, we have:

\[ \frac{1}{2} \times b_2 \times h_2 = 42 \]

Multiplying both sides by 2 gives:

\[ b_2 \times h_2 = 84 \]

From the problem statement, we know that the base of each triangle varies inversely with the height. This means:

\[ b \times h = k \]

for some constant \( k \). In our case, the constant \( k \) can be represented by the area product from the first triangle:

\[ k = b_1 \times h_1 = 14 \times 6 = 84 \]

Thus, for the second triangle:

\[ b_2 \times h_2 = 84 \]

Since \( b_2 \) varies inversely with \( h_2 \), we can express this as:

\[ b_2 = \frac{84}{h_2} \]

Now we can use this equation to find possible values for \( b_2 \) and \( h_2 \). By substituting \( b_2 \) into the equation \( b_2 \times h_2 = 84 \):

\[ \frac{84}{h_2} \times h_2 = 84 \]

This will always hold true as long as both \( h_2 \) and \( b_2 \) remain positive non-zero values.

For example, if we choose a value for \( h_2 \), we can calculate \( b_2 \):

1. If \( h_2 = 1 \), then \( b_2 = \frac{84}{1} = 84 \).
2. If \( h_2 = 2 \), then \( b_2 = \frac{84}{2} = 42 \).
3. If \( h_2 = 3 \), then \( b_2 = \frac{84}{3} = 28 \).
4. If \( h_2 = 4 \), then \( b_2 = \frac{84}{4} = 21 \).
5. If \( h_2 = 6 \), then \( b_2 = \frac{84}{6} = 14 \).

This pattern continues for any positive height. Thus, the pairs \((b_2, h_2)\) that keep the area the same while conforming to the inverse relationship are:

• \( (84, 1) \)
• \( (42, 2) \)
• \( (28, 3) \)
• \( (21, 4) \)
• \( (14, 6) \) (which is the same as the first triangle)
• \( (12, 7) \)
• and so on...

In summary, the possible pairs of base and height \((b_2, h_2)\) for the second triangle must satisfy \( b_2 \times h_2 = 84\) and can be derived from choosing a positive height \( h_2 \).