Finding the Ratio of Triangle to Parallelogram Heights

This problem involves comparing the dimensions of a triangle and a parallelogram that have the same area but related bases. We need to find the ratio of their corresponding heights.

Understanding the Area Formulas

The area of a triangle and a parallelogram are calculated using their base and height. Let's denote the base of the triangle as \(b_t\) and its height as \(h_t\). Let the base of the parallelogram be \(b_p\) and its height be \(h_p\).

• Area of Triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} b_t h_t\)
• Area of Parallelogram = \(\text{base} \times \text{height} = b_p h_p\)

Setting up the Relationship Based on the Problem Statement

The problem gives us two key pieces of information:

1. The area of the triangle is the same as the area of the parallelogram.
2. The base of the triangle is five-sixth of the base of the parallelogram.

Let's write these relationships mathematically:

• Area of Triangle = Area of Parallelogram
• \(\frac{1}{2} b_t h_t = b_p h_p\) (Equation 1)
• \(b_t = \frac{5}{6} b_p\) (Equation 2)

Solving for the Ratio of Heights

Now we can substitute Equation 2 into Equation 1 to relate the heights:

Substitute \(b_t = \frac{5}{6} b_p\) into \(\frac{1}{2} b_t h_t = b_p h_p\):

\(\frac{1}{2} \left( \frac{5}{6} b_p \right) h_t = b_p h_p\)

Simplify the left side:

\(\frac{5}{12} b_p h_t = b_p h_p\)

We are looking for the ratio of the height of the triangle to the parallelogram, which is \(\frac{h_t}{h_p}\). To find this ratio, we can divide both sides of the equation by \(b_p h_p\) (assuming \(b_p \neq 0\) and \(h_p \neq 0\), which must be true for valid geometric figures with area):

\(\frac{\frac{5}{12} b_p h_t}{b_p h_p} = \frac{b_p h_p}{b_p h_p}\)

\(\frac{5}{12} \frac{h_t}{h_p} = 1\)

Now, isolate the ratio \(\frac{h_t}{h_p}\) by multiplying both sides by \(\frac{12}{5}\):

\(\frac{h_t}{h_p} = 1 \times \frac{12}{5}\)

\(\frac{h_t}{h_p} = \frac{12}{5}\)

So, the ratio of the corresponding heights of the triangle to the parallelogram is \(12 : 5\).

Summary of Calculation Steps

The ratio of the height of the triangle to the height of the parallelogram is \(12:5\).

Revision Table: Triangle and Parallelogram Ratios

Given Conditions:

• Area\(_{\text{Triangle}}\) = Area\(_{\text{Parallelogram}}\)
• \(b_t = \frac{5}{6} b_p\)

Resulting Ratio of Heights:

• \(h_t : h_p = 12 : 5\)

Additional Information: Area and Dimensions

When two geometric figures have the same area, their dimensions (like base and height) are inversely related. If the base of one figure is smaller compared to the other (and the shapes are similar or related like triangle/parallelogram), its corresponding height must be larger to maintain the same area, and vice versa.

In this case, the triangle's base (\(b_t\)) is five-sixth (\(\frac{5}{6}\)) of the parallelogram's base (\(b_p\)). Since \(\frac{5}{6} < 1\), the triangle's base is smaller than the parallelogram's base. To compensate for this and have the same area, the triangle's height must be larger than the parallelogram's height. Our calculated ratio \(12:5\) (\(\frac{12}{5} > 1\)) confirms this, showing the triangle's height is more than twice the parallelogram's height.