Questions: Area of Parallelograms

Question 1

A parallelogram has a base of 8 cm, a slant side of 6 cm, and a perpendicular height of 5 cm. What is its area?

• A. 48 cm² (base × slant side)
• B. 40 cm² (base × perpendicular height)
• C. 24 cm² (½ × base × slant side)
• D. 20 cm² (½ × base × perpendicular height)

Answer: (B)

Area = base × perpendicular height = 8 × 5 = 40 cm². The slant side (6 cm) is not the height. Option A is the classic error: reaching for the slant side because it looks like a 'side length.' Options C and D apply the triangle formula (dividing by 2), which is wrong for a parallelogram. The perpendicular height is the straight-up distance from base to top — always shorter than the slant side.

Question 2

Why does the area formula for a parallelogram use the perpendicular height rather than the length of the slant side?

• A. Because the slant side is always longer, so using it would overcount the area of the angled corners
• B. Because cutting a right triangle from one end and sliding it to the other converts the parallelogram into a rectangle — and the rectangle's height is the perpendicular measurement, not the slant
• C. Because the slant side formula only applies when the parallelogram happens to be a rhombus
• D. Because mathematicians defined the formula arbitrarily and the perpendicular height is easier to measure in practice

Answer: (B)

The cut-and-slide argument is the geometric proof: slice a right triangle off one end of a parallelogram along a vertical line, slide it to the other end, and you have a rectangle with the same base and perpendicular height. Since rearranging pieces preserves area, the parallelogram's area equals the rectangle's area = base × perpendicular height. This also explains why using the slant side gives too large an answer — the slant side is longer than the perpendicular height and does not correspond to any dimension of the equivalent rectangle.

Question 3

The formula for the area of a parallelogram is different from the formula for the area of a rectangle.

• T. True
• F. False

Answer: False

Both use A = base × height. The formulas are identical because a parallelogram can be rearranged into a rectangle with the same base and perpendicular height. The cut-and-slide transformation proves this geometrically. This is not a coincidence — it is why the same formula works for both shapes, and it is the conceptual foundation for deriving the triangle and trapezoid formulas from the same starting point.

Question 4

The triangle area formula A = ½bh is derived from the parallelogram formula because two identical triangles can be joined to form a parallelogram with the same base and height.

• T. True
• F. False

Answer: True

If you duplicate a triangle and flip it 180°, the two pieces fit together along their shared edge to form a parallelogram. Therefore the triangle is exactly half of a parallelogram with the same base and perpendicular height, giving A = ½ × (bh) = ½bh. Understanding this derivation means you can reconstruct the triangle formula from the parallelogram formula rather than memorizing it as an independent fact.

Question 5

Explain why using the slant side of a parallelogram as the height always gives an area that is too large, and describe the geometric reasoning that shows why A = bh is correct.

Model answer: The slant side is always longer than the perpendicular height (the hypotenuse of a right triangle is always longer than either leg). Since the formula multiplies the base by this measurement, using the slant side produces a larger product than the true area. The correct formula comes from the cut-and-slide argument: cutting a right triangle from one end of the parallelogram and sliding it to the other end produces a rectangle with the same base and perpendicular height. Because rearranging preserves area, and the rectangle's area is base × perpendicular height, the parallelogram's area must be the same.

The geometric derivation is more than a memory aid — it reveals that area formulas are not arbitrary. The parallelogram formula works because of a conservation principle (rearranging doesn't change area) applied to a specific transformation (cut-and-slide). Students who understand this argument can reconstruct it and apply the same logic to derive related formulas for triangles and trapezoids.