Geometry Puzzle – Challenge 63

Challenge:

A rectangle is made of 7 equal squares. If the area of the rectangle is 567 feet, what is the perimeter of the rectangle?

A- 120 feet

B- 132 feet

C- 144 feet

D- 169 feet

E- 180 feet

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The rectangle is made of 7 equal squares. Therefore, the ratio of the width and length of the rectangle is 1 to 7. Why?
Let X be the width of the rectangle. So,
\(X × 7X = 567 → 7X^2 = 567 → X^2 = 81 → X = 9\)
The width of the rectangle is 9 feet and the length is 63 feet.
(7 × 9 = 63)
The perimeter of the rectangle is:
2 × (9 + 63) = 144 feet

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Frequently Asked Questions

Why is the length 7 times the width?

Because the rectangle is made of 7 equal squares lined up end-to-end. Each square has the same side length, so 7 squares in a row give a rectangle whose length equals 7 of those sides — that is, 7 times the width.

How do I set up the area equation?

Let X be the side of each small square. The rectangle has width X and length 7X, so its area is X times 7X, which equals 7X squared. Set that equal to 567: 7X^2 = 567.

Walk through solving 7X^2 = 567.

Divide both sides by 7: X^2 = 81. Take the positive square root: X = 9 (the side length must be positive). So each small square is 9 ft by 9 ft.

How do I find the perimeter once I know the dimensions?

Use P = 2(length + width). With length = 63 ft and width = 9 ft: P = 2(63 + 9) = 2(72) = 144 ft.

Why does the puzzle specify equal squares?

Equal squares pin down the length-to-width ratio (here 7:1). Without that constraint, many rectangles could have area 567, each with a different perimeter. The equal-squares condition picks out exactly one rectangle.

What if the rectangle were made of 5 equal squares with area 405 sq ft?

Same method. Length = 5 times width, so 5X^2 = 405, X^2 = 81, X = 9. The rectangle is 9 by 45, and perimeter = 2(9 + 45) = 108 ft.

How does this puzzle exercise algebraic thinking?

It rewards translation. The hardest step is reading the words and writing the right equation; once 7X^2 = 567 is on the page, the rest is one division and one square root. That translation skill is the algebra muscle that pays off for years.

What about a sanity check on the answer?

Multiply: 9 by 63 = 567 (matches the given area). Add: 9 + 63 = 72, doubled = 144 (matches choice C). Both checks confirm the answer.

What grade level can solve this?

Strong middle schoolers (Grade 6-8) once they have area, perimeter, and basic equation solving with squaring/square root. The puzzle is also a nice anchor problem for early Algebra I.

Where would this kind of reasoning appear in real life?

Anywhere a known ratio constrains dimensions: garden planning, picture framing, fabric cutting, packaging design, room layouts. The same idea — fixing a ratio, then solving for one unknown from a known area or volume — shows up constantly.

Related Lessons You May Like

• How to find the area of rectangles
• How to find the perimeter of rectangles
• How to find complementary, supplementary, vertical, and adjacent angles
• How to solve angles and angle measure
• How to solve multi-step word problems

If your student enjoys puzzles like this, Geometry for Beginners dives into the same kinds of relationships inside a full curriculum. For the algebra you will lean on, Pre-Algebra for Beginners fills in the foundations gently.