Laura walks every evening on the edges of a sports field near her house. The field is in the shape of a rectangle 300 feet (ft) long and 200 ft wide, so 1 lap on the edges of the field is 1,000 ft. She enters through a gate at point G, located exactly halfway along the length of the field.
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One evening on her walk, Laura walks across the field from point W back to the gate at point G. What is the distance she walks, in feet, from point W to point G?
Correct answer: A
Laura's walk forms a right triangle, so the distance from W to G is the hypotenuse. Picture the rectangle: length 300 ft, width 200 ft. Point G sits halfway along the length, so G is 150 ft from the nearest corner along the long side. Point W is the opposite corner on the far long side. That gives a right triangle with legs of 200 ft (the width) and 150 ft (half the length). Use the Pythagorean theorem: a² + b² = c². Substitute: 200² + 150² = c². Compute each square: 200² = 40,000 and 150² = 22,500. Add them: 40,000 + 22,500 = 62,500. Now take the square root: √62,500 = 250. So the distance from W to G is 250 ft. Check the work by noticing this is a 3-4-5 triangle scaled by 50: 3 × 50 = 150, 4 × 50 = 200, 5 × 50 = 250. The numbers match, so the answer is confirmed. free adult literacy and education resources).
This is an answer-only item, so there are no choices to review. Still, it helps to look at the common wrong answers students produce, because each one points to a specific mistake. Knowing these errors helps you catch yourself on test day.
A common wrong answer is 350. A student gets this by adding 200 + 150 instead of using squares. Adding the legs only works if the path is a straight line along the edges, not a diagonal across the field. The diagonal is always shorter than the sum of the two legs, so 350 is too big.
Another common wrong answer is 500. This comes from adding the full length and width, 300 + 200, and ignoring that G is halfway along the length. The student reads "halfway" but does not use it, so the triangle leg becomes 300 instead of 150. Always mark the halfway point on your sketch.
A third wrong answer is about 360. A student squares 200 and 150, gets 62,500, but then takes the square root incorrectly or stops at 40,000 + 22,500 without finishing. Some students also confuse squaring with doubling: 2 × 200 = 400 and 2 × 150 = 300, which leads to wrong totals. Square means multiply a number by itself.
A fourth wrong answer is 150. This happens when a student uses only half the length and forgets the width leg of the triangle. The distance from W to G crosses both the width and part of the length, so both legs must be included.
A fifth wrong answer is 200. Here the student uses only the width and ignores the 150 ft leg. Again, the diagonal spans two directions, not one.
A sixth wrong answer is 300. This is just the full length of the field, used when a student skips the midpoint clue and the width entirely.
A seventh wrong answer is 100. This comes from subtracting 200 − 150 = 50 and then doubling, or from misreading the gate position as one-quarter of the way along the length.
To avoid all of these, sketch the rectangle, label the gate at the midpoint, draw the diagonal, and label the legs before computing. Then apply a² + b² = c² and take the square root last.
Conclusion
This item asks for the straight-line distance from corner W to gate G on a 300 ft by 200 ft rectangular field, with G at the midpoint of the length. The correct answer is 250 ft. Draw the rectangle, mark G 150 ft from the corner, and treat the path as the hypotenuse of a right triangle with legs 150 and 200. Use a² + b² = c², add the squares to get 62,500, and take the square root. Recognizing the scaled 3-4-5 pattern gives a fast check. This skill matters for real tasks like measuring diagonals, laying out fences, and reading maps.
Why the others are wrong
A common mistake is to add the two legs of the right triangle instead of using the Pythagorean theorem. For example, adding 200 and 150 gives 350, but that is the total distance along the edges, not the straight-line distance. The diagonal must be shorter than the sum of the legs.
Another error is to ignore the midpoint clue and use the full length of 300 feet as one leg. This leads to adding 300 and 200 to get 500, or trying to square 300 and 200. The gate is halfway along the length, so the leg is 150 feet, not 300.
Some students also forget to take the square root after adding the squares. They stop at 62,500 and think that is the distance. Others might use only one leg, like 150 or 200, forgetting that the diagonal spans both the width and part of the length.
Concept tested
Use the Pythagorean theorem to find the diagonal distance across a rectangle when the gate is at the midpoint of the length.
GED Tip
Sketch the rectangle, mark the gate at the midpoint, and label the legs (200 ft and 150 ft). Then apply a² + b² = c²: 200² + 150² = 62,500, so c = √62,500 = 250 ft. Check with the 3-4-5 pattern scaled by 50.
