Question

\(\triangle XQY\) and \(\triangle ZYR\) are equilateral triangles, and the ratio of ZR to PR is 1 to 4.
 

Column A	Column B
The perimeter of \(\triangle XQY\)	The perimeter of parallelogram PXYZ

 

Hint:

In complex geometry problems, the first step is always to translate all the given information (ratios, shape properties) into algebraic expressions with a common variable. This turns the geometry problem into a simpler algebra problem.

Answer 1

Step 1: Understanding the Concept:
This geometry problem requires us to find the perimeters of two different shapes, a triangle and a parallelogram, using information about their properties and relationships. The key is to express all lengths in terms of a single variable.
Step 2: Key Formula or Approach:
- Perimeter of an equilateral triangle with side \(s\): \(P = 3s\).
- Perimeter of a parallelogram with adjacent sides \(a\) and \(b\): \(P = 2(a+b)\).
- Properties of a parallelogram: Opposite sides are equal in length.
Step 3: Detailed Explanation:
Let's define the side lengths based on the given ratio.
The ratio of ZR to PR is 1 to 4. Let \(ZR = a\). Then \(PR = 4a\).
The diagram shows that the points P, Z, and R are collinear, so the length of the segment PZ is \(PR + ZR = 4a + a = 5a\).
Now, let's use the properties of the given shapes.
We are given that \(\triangle ZYR\) is equilateral. This means all its sides are equal to ZR.
\[ ZY = YR = ZR = a \] We are given that PXYZ is a parallelogram. In a parallelogram, opposite sides are equal.
\[ XY = PZ = 5a \] \[ PX = ZY = a \] We are given that \(\triangle XQY\) is equilateral. This means all its sides are equal to XY.
\[ XQ = QY = XY = 5a \] Now we can calculate the perimeters for both columns.
For Column A: The perimeter of \(\triangle XQY\)
The side length is \(XY = 5a\).
\[ \text{Perimeter} = 3 \times (\text{side length}) = 3 \times (5a) = 15a \] For Column B: The perimeter of parallelogram PXYZ
The adjacent sides are PZ and ZY. Their lengths are \(PZ = 5a\) and \(ZY = a\).
\[ \text{Perimeter} = 2 \times (PZ + ZY) = 2 \times (5a + a) = 2 \times (6a) = 12a \] Comparison:
We are comparing 15a (Column A) and 12a (Column B). Since \(a\) represents a geometric length, it must be positive (\(a>0\)). Therefore, \(15a>12a\).
Step 4: Final Answer:
The perimeter of the triangle (15a) is greater than the perimeter of the parallelogram (12a). The quantity in Column A is greater.