The area of a triangle is one of the most “useful forever” geometry skills. The main idea is simple: take a base and a perpendicular height, multiply them, then take half. The only tricky part is making sure the height you use is really perpendicular to the chosen base.
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For any triangle, the area is:
The height is not “any side.” It is the perpendicular distance from the base to the opposite vertex. If the triangle is tilted, you can still use the formula—you just have to drop an altitude (sometimes outside the triangle for obtuse triangles).
Problem: A triangle has base 12 and height 7. Find the area.
Final answer: 42 square units.
Problem: A triangle has base 10. An altitude to the base splits the base into segments 6 and 4. The side adjacent to the 6 segment has length 8. Find the area.
The altitude creates a right triangle with hypotenuse 8 and one leg 6. So the height h is:
Now use area = (1/2)·base·height:
That’s an exact answer. If you need a decimal, √7 ≈ 2.6458 so area ≈ 26.46 square units.
For triangle vertices A(x1, y1), B(x2, y2), C(x3, y3), the area can be computed using:
This is especially useful when the triangle is tilted and base/height are not obvious.
Problem: Find the area of the triangle with vertices A(1, 2), B(5, 2), C(3, 7).
Here, A and B share the same y-coordinate, so AB is a horizontal base of length 4, and the height from C to the line y = 2 is 5. That’s an easy base-height approach:
You can also confirm using the coordinate formula, but this “spot the horizontal/vertical side” trick is often faster.
Sometimes a problem gives three side lengths but no height. In that case, Heron’s formula is a direct method. Let sides be a, b, c and let s be the semiperimeter:
This looks heavy, but it’s mechanical. It’s also common in contest-style geometry or word problems.
Problem: Find the area of a triangle with side lengths 13, 14, and 15.
Step 1: Compute the semiperimeter:
Step 2: Plug into Heron’s formula:
Final answer: 84 square units. This is a nice “clean” triangle, but even when the square root doesn’t simplify, the steps are the same.
A common confusion happens with obtuse triangles (one angle greater than 90°). The perpendicular altitude to a chosen base may land outside the triangle. That’s still fine: the height is defined as the perpendicular distance to the line containing the base. In practice, you extend the base line and drop a perpendicular.
Quick rule: If the triangle “leans,” you might need to extend a side to draw the altitude. The area formula stays the same: (1/2)·b·h.
Sometimes you don’t have a height, but you do have two sides and the angle between them. In that case, you can use this area formula:
Why it works: if you drop an altitude, the height becomes b·sin(C) (or a·sin(C)), so it is still a base-height calculation—just written in a more direct way.
Problem: Two sides of a triangle are 9 and 12, and the included angle between them is 30°. Find the area.
Use area = (1/2)·a·b·sin(C):
Since sin(30°) = 1/2, this one stays clean. If the sine value is not a “special angle,” you can use a calculator for the sine and still get a correct numerical area.
Height must be perpendicular to the base. If you’re not sure, draw the altitude and mark the right angle.
Length is in units. Area is in square units. Always write something like cm².
The most common mistake is using b·h instead of (1/2)·b·h.
If you’re using Heron’s formula with side lengths, make sure the sides can form a triangle.
If you want an additional explanation of when Heron’s formula is useful (and how to compute it cleanly), many geometry references summarize it well; the short overview on Heron’s formulacan be a quick reminder of the steps.
For #4, first find the height with the Pythagorean theorem, then use (1/2)·b·h.
Describe the triangle or paste coordinates and get the best method + clean area.
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