Geometry Puzzle – Challenge 69

Challenge:

A length of a rectangle is increased by \(20\%\). What percent of its width should be decreased to keep the area of the rectangle the same?

A- \(15\%\)

B- \(16\frac{2}{3}\%\)

C- \(20\%\)

D- \(25\%\)

E- \(30\%\)

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The area of a rectangle is the product of its width and length. Let x be the area of the rectangle.
Area of a rectangle = Width × Length = x
Length → 1.2Length → Area of a rectangle = Width × 1.2Length = x
Let W1 be the width of the first rectangle and W2 be the width of the second rectangle.
W1 × 1.2L = W2 × L → W1 × 1.2 = W2 → W1 \(= \frac{1}{1.2}\)W2 →
W1 \(= \frac{5}{6 }\)W2
To keep the area of the rectangle the same \(\frac{1}{6 }\) or \(16\frac{2}{3}\) percent of the width should be decreased.

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Frequently Asked Questions

How do I set up the equation?

Original area = W times L. New area = W’ times 1.2L. Setting them equal: W’ times 1.2L = W times L. Divide both sides by 1.2L: W’ = W/1.2.

Why is W/1.2 the same as (5/6)W?

1.2 = 6/5, so dividing by 6/5 is the same as multiplying by 5/6. Therefore W/1.2 = (5/6) W.

What is the percent decrease?

Decrease = 1 – 5/6 = 1/6, which is approximately 0.1667, or 16.67% (also written 16 and 2/3 percent).

Why isn’t the answer 20%?

Common intuition says “increase by 20%, decrease by 20%” — but the compensations are NOT symmetric. A 20% increase requires only about a 16.67% decrease to compensate, because the new (larger) base shrinks proportionally less than 20%.

Can I check with a concrete example?

Yes. Original 10 by 10, area = 100. New length 12 (20% increase). To keep area 100, new width must be 100/12 = 8.33 (which is 10 – 1.67, a 16.67% decrease).

What if length increases by 50% instead?

New length = 1.5L. New width = W/1.5 = (2/3)W, a decrease of 1/3 ≈ 33.33%.

What if length doubles?

New length = 2L. New width = W/2 = 0.5W, a decrease of 50%.

Why is percent compensation not symmetric?

Because the BASE for the inverse change differs from the original base. A 20% increase from 10 gives 12. To get back to 10 from 12 requires losing 2 out of 12, which is about 16.67% — not 20% out of 12.

What is the general formula?

If length increases by p (as a decimal), the width must decrease by p/(1+p) to keep area constant. For p = 0.2: decrease = 0.2/1.2 = 1/6.

Where does this principle show up?

Pricing (a discount and re-markup are not equal), efficiency calculations (a 25% productivity loss requires more than 25% to recover), economics (inflation-adjusted values).

Related Lessons You May Like

• How to find the area of rectangles
• How to find the perimeter of rectangles
• How to find volume and surface area of cubes
• How to solve percent of change
• How to solve multi-step word problems

If your student enjoys puzzles like this, Geometry for Beginners works the same relationships inside a full curriculum. Pre-Algebra for Beginners covers the algebraic foundations gently.