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3 Responses
Yes-area = 1/2 × base × height always works, no matter how the triangle is tilted or whether the perpendicular drops inside or outside. The key is that “base” can be any side, but the “height” must be the perpendicular distance from the opposite vertex to the line containing that base (extend the base line if needed for an obtuse triangle). Don’t pair a base with a slanted side unless they’re perpendicular; only the right-angle drop counts as the height. A quick way to see why: reflect the triangle across its chosen base to make a parallelogram-its area is base × height, so the triangle’s area is half that, and this doesn’t care where the foot of the altitude lands. If you prefer a formula that avoids hunting for the altitude, use 1/2 × a × b × sin(C), where a and b are two sides and C is the included angle. Think of the base as a road and the height as the shortest crosswalk to the point across it-the crosswalk must be at right angles to the road. As I check the logic: pick any side as the road, draw the shortest perpendicular to it, and 1/2 × (that side) × (that perpendicular) gives the same area every time.
Yes-1/2 × base × height always works, no matter how the triangle is tilted. The key is that “height” means the perpendicular distance from the opposite vertex to the line containing your chosen base; if the foot of that perpendicular lands outside the side, just imagine extending the base as a straight line and measure the right-angled distance to it. My routine is: pick any side as the base, draw the right angle from the opposite vertex to that base line, take that length as h, and compute A = 1/2 b h. The base and height values change if you pick a different side, but their product b·h is the same because they describe the same area. Simple example: let B = (0, 0), C = (6, 0), and A = (−2, 3). The perpendicular from A to the x-axis hits at (−2, 0), which is outside segment BC, yet the perpendicular distance is 3, so the area is 1/2 × 6 × 3 = 9. As an alternative, you can use A = 1/2 ab sin C with sides a, b and the angle C between them; in fact, you can pair any two sides with any angle and the sine takes care of the tilt. A quick self-check I like: if the “height” you picked doesn’t meet the base at 90 degrees, it’s not the right one; and if you switch to a longer base, the corresponding height gets smaller, but the area should stay consistent.
Yep-1/2 × base × height always works: you can choose any side as the base, but the matching “height” must be the perpendicular distance from the opposite vertex to the line containing that side (extend the side if it’s obtuse), not a slanted edge. Example: if the base is 10 and the perpendicular drop is 4 (even if it lands outside), area = 1/2 × 10 × 4 = 20; nice refresher here: https://www.khanacademy.org/math/geometry/hs-geo-foundations/hs-geo-area-triangle/v/area-of-triangles