For example, you might have a triangle with a base measuring 5 cm long, and a height measuring 3 cm long.

For example, if the base of your triangle is 5 cm and the height is 3 cm, you would calculate:Area=12(bh){\displaystyle {\text{Area}}={\frac {1}{2}}(bh)}Area=12(5)(3){\displaystyle {\text{Area}}={\frac {1}{2}}(5)(3)}Area=12(15){\displaystyle {\text{Area}}={\frac {1}{2}}(15)}Area=7. 5{\displaystyle {\text{Area}}=7. 5} So, the area of a triangle with a base of 5 cm and a height of 3 cm is 7. 5 square centimeters.

You can also use this formula if you know one side length, plus the length of the hypotenuse. The hypotenuse is the longest side of a right triangle and is opposite the right angle. Remember that you can find a missing side length of a right triangle using the Pythagorean Theorem (a2+b2=c2{\displaystyle a^{2}+b^{2}=c^{2}}). For example, if the hypotenuse of a triangle is side c, the height and base would be the other two sides (a and b). If you know that the hypotenuse is 5 cm, and the base is 4 cm, use the Pythagorean theorem to find the height:a2+b2=c2{\displaystyle a^{2}+b^{2}=c^{2}}a2+42=52{\displaystyle a^{2}+4^{2}=5^{2}}a2+16=25{\displaystyle a^{2}+16=25}a2+16−16=25−16{\displaystyle a^{2}+16-16=25-16}a2=9{\displaystyle a^{2}=9}a=3{\displaystyle a=3}Now, you can plug the two perpendicular sides (a and b) into the area formula, substituting for the base and height:Area=12(bh){\displaystyle {\text{Area}}={\frac {1}{2}}(bh)}Area=12(4)(3){\displaystyle {\text{Area}}={\frac {1}{2}}(4)(3)}Area=12(12){\displaystyle {\text{Area}}={\frac {1}{2}}(12)}Area=6{\displaystyle {\text{Area}}=6}

For example, if a triangle has three sides that are 5 cm, 4 cm, and 3 cm long, the semiperimeter is shown by:s=12(3+4+5){\displaystyle s={\frac {1}{2}}(3+4+5)}s=12(12)=6{\displaystyle s={\frac {1}{2}}(12)=6}

For example:Area=s(s−a)(s−b)(s−c){\displaystyle {\text{Area}}={\sqrt {s(s-a)(s-b)(s-c)}}}Area=6(6−3)(6−4)(6−5){\displaystyle {\text{Area}}={\sqrt {6(6-3)(6-4)(6-5)}}}

For example:Area=6(6−3)(6−4)(6−5){\displaystyle {\text{Area}}={\sqrt {6(6-3)(6-4)(6-5)}}}Area=6(3)(2)(1){\displaystyle {\text{Area}}={\sqrt {6(3)(2)(1)}}}Area=6(6){\displaystyle {\text{Area}}={\sqrt {6(6)}}}

For example:Area=6(6){\displaystyle {\text{Area}}={\sqrt {6(6)}}}Area=36{\displaystyle {\text{Area}}={\sqrt {36}}}Area=6{\displaystyle {\text{Area}}=6}So, the area of the triangle is 6 square centimeters.

For example, you might have a triangle with three sides that are 6 cm long.

For example if the equilateral triangle has sides that are 6 cm long, you would calculate:Area=(s2)34{\displaystyle {\text{Area}}=(s^{2}){\frac {\sqrt {3}}{4}}}Area=(62)34{\displaystyle {\text{Area}}=(6^{2}){\frac {\sqrt {3}}{4}}}Area=(36)34{\displaystyle {\text{Area}}=(36){\frac {\sqrt {3}}{4}}}

For example:Area=(36)34{\displaystyle {\text{Area}}=(36){\frac {\sqrt {3}}{4}}}Area=62. 3524{\displaystyle {\text{Area}}={\frac {62. 352}{4}}}

For example:Area=62. 3524{\displaystyle {\text{Area}}={\frac {62. 352}{4}}}Area=15. 588{\displaystyle {\text{Area}}=15. 588}So, the area of an equilateral triangle with sides 6 cm long is about 15. 59 square centimeters.

For example, you might have a triangle with two adjacent sides measuring 150 cm and 231 cm in length. The angle between them is 123 degrees.

For example:Area=bc2sin⁡A{\displaystyle {\text{Area}}={\frac {bc}{2}}\sin A}Area=(150)(231)2sin⁡A{\displaystyle {\text{Area}}={\frac {(150)(231)}{2}}\sin A}Area=(34,650)2sin⁡A{\displaystyle {\text{Area}}={\frac {(34,650)}{2}}\sin A}Area=17,325sin⁡A{\displaystyle {\text{Area}}=17,325\sin A}

For example, the sine of a 123-degree angle is . 83867, so the formula will look like this:Area=17,325sin⁡A{\displaystyle {\text{Area}}=17,325\sin A}Area=17,325(. 83867){\displaystyle {\text{Area}}=17,325(. 83867)}

For example:Area=17,325(. 83867){\displaystyle {\text{Area}}=17,325(. 83867)}Area=14,529. 96{\displaystyle {\text{Area}}=14,529. 96}. So, the area of the triangle is about 14,530 square centimeters.