Math Problem Statement

ABCD is a rectangle. A parallelogram is AECF is constructed within the rectangle such that point E is on AB and point F is on CD. If the area of the parallelogram is three-fifths the area of the rectangle and the length of CB is 25% less than the length of EB, then by what percentage is the perimeter of the rectangle ABCD greater than the perimeter of the parallelogram AECF?

Solution

Let's analyze and solve the problem step by step.

Step 1: Define Variables

Let:

  • The length of AB=lAB = l (the longer side of the rectangle).
  • The length of AD=wAD = w (the shorter side of the rectangle).

Then, the area of the rectangle ABCDABCD is given by: Area of ABCD=lw\text{Area of } ABCD = l \cdot w

Step 2: Given Information about the Parallelogram's Area

We know the area of parallelogram AECFAECF is three-fifths the area of rectangle ABCDABCD: Area of AECF=35(lw)\text{Area of } AECF = \frac{3}{5} \cdot (l \cdot w)

Step 3: Relationship between EBEB and CBCB

It is given that CBCB is 25% less than EBEB. So, if EB=xEB = x, then: CB=x0.25x=0.75xCB = x - 0.25x = 0.75x Since CB=wCB = w (the height of the rectangle), we can say: w=0.75xw = 0.75x or equivalently, x=w0.75=43wx = \frac{w}{0.75} = \frac{4}{3}w

Step 4: Perimeters of Rectangle and Parallelogram

The perimeter of the rectangle ABCDABCD is: Perimeter of ABCD=2(l+w)\text{Perimeter of } ABCD = 2(l + w)

To find the perimeter of the parallelogram AECFAECF, note that opposite sides of a parallelogram are equal. The two pairs of opposite sides are AEAE and CFCF, and AFAF and ECEC. Since AECFAECF is inscribed in ABCDABCD, the lengths of these sides align with parts of ABCDABCD:

  • AE=AB=lAE = AB = l (since EE lies on ABAB).
  • AF=AD=wAF = AD = w (since FF lies on CDCD).

Thus, the perimeter of parallelogram AECFAECF is: Perimeter of AECF=2(l+w)\text{Perimeter of } AECF = 2(l + w)

Step 5: Percentage Difference in Perimeters

Since both the rectangle and the parallelogram share the same perimeter, the percentage by which the perimeter of ABCDABCD is greater than the perimeter of AECFAECF is: (2(l+w)2(l+w))2(l+w)×100=0%\frac{(2(l + w) - 2(l + w))}{2(l + w)} \times 100 = 0\%

Conclusion

The perimeter of the rectangle ABCDABCD is 0% greater than the perimeter of the parallelogram AECFAECF.

Would you like a deeper explanation of any step?

Here are five related questions:

  1. How does changing the proportion of AEAE to ABAB affect the area of the parallelogram?
  2. How would the solution change if the area of the parallelogram were different from three-fifths of the rectangle's area?
  3. What is the perimeter relationship if the parallelogram were tilted within the rectangle?
  4. How would the answer change if CBCB were a different percentage less than EBEB?
  5. Can this approach be used for other quadrilaterals within rectangles?

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Math Problem Analysis

Mathematical Concepts

Geometry
Area of Rectangle
Area of Parallelogram
Perimeter of Rectangle
Perimeter of Parallelogram
Proportions
Algebra

Formulas

Area of Rectangle = l * w
Area of Parallelogram = (3/5) * (l * w)
Perimeter of Rectangle = 2 * (l + w)
Perimeter of Parallelogram = 2 * (l + w)

Theorems

Area and Perimeter of Parallelogram
Proportional Relationships

Suitable Grade Level

Grades 9-11