I’m stuck on finding the area of a triangle when the base isn’t horizontal or vertical. I know the area formula is 1/2 × base × height, but I keep tripping over what the “height” actually is when the base is slanted.
Concrete example: A(1,1), B(7,4), C(4,8). If I pick AB as the base, I can get its length (sqrt(45)). The slope of AB is 1/2, so I wrote the line as y = 1 + 0.5(x − 1). Then at x = 4, the line’s y-value is 2.5, so I tried using 8 − 2.5 = 5.5 as the height (vertical difference). I also tried a simpler shortcut using 8 − 4 = 4 as the height (difference in y between C and B), which I’m pretty sure is wrong, and I get different areas depending on which one I pick.
I know the height is supposed to be the perpendicular distance from C to line AB, not just a vertical or horizontal gap. I’m just not sure how to compute that perpendicular distance cleanly from two points on the base. Could someone show me the correct way to get the height from C to AB in this example, and point out where my reasoning goes off? I want to understand how to do this properly when the base is slanted.
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3 Responses
Great question! When the base is slanted, the height is just the vertical gap from C straight up to the line at the same x-value, so for AB (y = 1 + 0.5(x − 1)), at x = 4 you get y = 2.5, hence h = 8 − 2.5 = 5.5 and Area = 1/2 · √45 · 5.5.
Think of a tilted plank: you measure “height” by going straight up to it, not by chasing a perpendicular.
You’re right that the height must be perpendicular to AB; a clean way is to use the point-to-line (or cross-product) formula: h = |(xB−xA)(yA−yC) − (xA−xC)(yB−yA)| / sqrt((xB−xA)^2 + (yB−yA)^2), since vertical gaps like 8−2.5 or 8−4 aren’t perpendicular unless AB is horizontal.
Worked example: A(1,1), B(7,4), C(4,8) gives AB=(6,3), AC=(3,7) so h=|6·7−3·3|/sqrt(6^2+3^2)=33/(3√5)=11/√5 and area = ½·|AB|·h = ½·(3√5)·(11/√5) = 33/2.
Totally get the confusion-“height” only lines up nicely with vertical or horizontal when the base is horizontal or vertical. When the base is slanted, the height has to be the length of the segment that meets the base at a right angle. A clean way to get that is: write the line through your base in the form Ax + By + C = 0, then use the point-to-line distance formula distance = |Ax0 + By0 + C| / sqrt(A^2 + B^2). That distance is exactly the height. (Another nice shortcut: if you like vectors, the area is 1/2 times the magnitude of AB × AC, so height = |AB × AC| / |AB|.)
Let’s do your example. Line AB through A(1,1) and B(7,4) has slope 1/2, so y − 1 = (1/2)(x − 1). In standard form this is x − 2y + 1 = 0. Now the perpendicular distance from C(4,8) to AB is |4 − 2·8 + 1| / sqrt(1^2 + (−2)^2) = |4 − 16 + 1| / sqrt(5) = 11/√5. That’s the correct height (about 4.919). The base length |AB| is √((6)^2 + (3)^2) = √45 = 3√5. So the area is (1/2) × base × height = (1/2) × 3√5 × (11/√5) = 33/2 = 16.5. Your 5.5 came from a vertical drop, which isn’t perpendicular to the slanted base, so it overestimates the true height here; the 4 from 8 − 4 compares C to point B, which isn’t a drop to the line at all. The perpendicular distance 11/√5 is the one you want.