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3 Responses
Great question-I’ve tripped on that same pitfall before. The area of a trapezium depends on the two parallel bases and the perpendicular distance (the “height”) between them: A = (b1 + b2)/2 × h. So with bases 8 cm and 14 cm and perpendicular distance 6 cm, your A = (8 + 14)/2 × 6 = 66 cm² is spot on. The non‑parallel, slanted sides don’t enter the formula directly; you can shear the trapezium (tilt the slanted sides) while keeping the bases and the perpendicular gap the same, and the area doesn’t change. Even when the perpendicular from one base lands outside the other base (an obtuse trapezium), you still use that same perpendicular distance. The only “exception” I keep in mind is when the trapezium is isosceles-then using the slanted side as the height also works because the altitude lines up with it-but that’s a special case I try not to rely on to avoid mixing up which length is truly perpendicular.
A quick example: take bases 4 cm and 10 cm with a perpendicular distance of 5 cm. The area is (4 + 10)/2 × 5 = 35 cm². Now imagine two different trapeziums with those same bases and the same 5 cm perpendicular gap-one “right” trapezium where one non‑parallel side is exactly perpendicular, and another where both non‑parallel sides are quite slanted. Both shapes still have area 35 cm², even though the slanted sides are very different. If the foot of the perpendicular falls outside, some people treat that height as “negative,” but I usually just keep it positive and apply the same formula; the area comes out the same either way. For a clear visual proof and more examples, this short Khan Academy explainer is handy: https://www.khanacademy.org/math/geometry/hs-geo-foundations/hs-geo-foundations-area/v/area-of-a-trapezoid
The slant is just flair, not area: use A = (8+14)/2 * 6 = 66 cm², because only the perpendicular distance between the parallel bases matters (even if it lands outside). Also, with height 6 cm no slanted side can be as short as 5 cm, so those side lengths can’t all happen together.
The area depends only on the two parallel bases and the perpendicular distance between their lines (even if that drop lands outside), so A = (8+14)/2 × 6 = 66 cm² if the 6 cm is truly perpendicular. That said, a 5 cm slanted side can’t accommodate a 6 cm perpendicular height, so I suspect one of those measurements isn’t consistent.