Loading video...
Finding the Area of a Triangle
Main formulas:
- Master formula for right triangles: SOHCAHTOA!
sinθ = opposite side hypotenusecosθ = adjacent side hypotenusetanθ = opposite side adjacent side - The Law of Cosines (in any triangle):
c2 = a2 + b2 − 2abcosC - Heron's Formula:
where s = [1/2](a+b+c) is the semiperimeter of the triangle.Area =
√
s(s− a)(s− b)(s− b),
Example 1:
Find the area of a triangle that has two sides of lengths 8 and 12 with an included angle of 45° .Example 2:
Find the area of a triangle with side lengths 10, 14, and 16.Example 3:
Without using Heron's formula, find the area of a triangle whose side lengths are 7, 9, and 14.Example 4:
Find the area of a triangle that has two sides of lengths 7 and 13 with an included angle of 32° .Example 5:
A triangle has two sides of length 5 and 6 with an included angle of 60° . Find the area of the triangle.Find the area of a triangle that has two sides of length 6 and 10 with an included angle of 54°. (Do not use Heron's Formula to solve)
- Start by drawing a picture of the triangle with the sides and angles labeled
- Area of a Triangle = [1/2]bh
- Use SOHCAHTOA to find the height of the triangle
- sinθ = [Opposite/Hypotenuse] ⇒ sin54° = [h/6] ⇒ h = 6sin54° ⇒ h ≈ 4.85
- A = [1/2](10)(4.85)
A ≈ 24.3
Find the area of a triangle that has two sides of length 11 and 13 with an included angle of 63°. (Do not use Heron's Formula to solve)
- Start by drawing a picture of the triangle with the sides and angles labeled
- Area of a Triangle = [1/2]bh
- Use SOHCAHTOA to find the height of the triangle
- sinθ = [Opposite/Hypotenuse] ⇒ sin63° = [h/11] ⇒ h = 11sin63° ⇒ h ≈ 9.8
- A = [1/2](13)(9.8)
A ≈ 63.7
A triangle has two sides of length 5 and 7 with an included angle of 34°. Find the area of the triangle.
- Start by drawing a picture of the triangle with the sides and angles labeled
- Use the Law of Cosines to find the length of the missing side
- a2 = b2 + c2 − 2bc cosA ⇒ a2 = 72 + 52 − 2(7)(5)cos34° ⇒ a2 = 15.9672 ⇒ a = √{15.9672}
- a ≈ 4.0
- Now use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
- s = [1/2](4 + 7 + 5) ⇒ s = 8
- A = √{8(8 − 4)(8 − 7)(8 − 5)} ⇒ √{8(4)(1)(3)} √{96}
A ≈ 9.8
Find the area of a triangle with side lengths 11, 15, and 17
- Use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
- a = 11, b = 15, c = 17
- s = [1/2](11 + 15 + 17) ⇒ s = 21.5
- A = √{21.5(21.5 − 11)(21.5 − 15)(21.5 − 17)} ⇒ √{21.5(10.5)(6.5)(4.5)} = √{6603.19}
A ≈ 81.26
Find the area of a triangle with side lengths 7, 12, and 15
- Use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
- a = 7, b = 12, c = 15
- s = [1/2](7 + 12 + 15) ⇒ s = 17
- A = √{17(17 − 7)(17 − 12)(17 − 15)} ⇒ √{17(10)(5)(2)} = √{1700}
A ≈ 41.23
Find the area of a triangle with side lengths 10, 17, and 23
- Use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
- a = 10, b = 17, c = 23
- s = [1/2](10 + 17 + 23) ⇒ s = 25
- A = √{25(25 − 10)(25 − 17)(25 − 23)} ⇒ √{25(15)(8)(2)} = √{600}
A ≈ 77.46
Find the area of a triangle that has two sides of length 15 and 20 with an included angle of 50°.
- Start by drawing a picture of the triangle with the sides and angles labeled
- Area of a Triangle = [1/2]bh
- Use SOHCAHTOA to find the height of the triangle
- sinθ = [Opposite/Hypotenuse] ⇒ sin50° = [h/15] ⇒ h = 15sin50° ⇒ h ≈ 11.49
- A = [1/2](20)(11.49)
A ≈ 114.9
A triangle has two sides of length 3 and 7 with an included angle of 20°. Find the area of the triangle.
- Start by drawing a picture of the triangle with the sides and angles labeled
- Use the Law of Cosines to find the length of the missing side
- a2 = b2 + c2 − 2bc cosA ⇒ a2 = 72 + 32 − 2(7)(3)cos34° ⇒ a2 = 18.533 ⇒ a = √{18.533}
- a ≈ 4.3
- Now use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
- s = [1/2](4.3 + 3 + 7) ⇒ s = 7.15
- A = √{7.15(7.15 − 4.3)(7.15 − 3)(7.15 − 7)} ⇒ √{7.15(2.85)(4.15)(0.15)} = √{12.685}
A ≈ 3.56
Find the area of a triangle with side lengths 8, 12, and 16
- Use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
- a = 8, b = 12, c = 16
- s = [1/2](8 + 12 + 16) ⇒ s = 18
- A = √{18(18 − 8)(18 − 12)(18 − 16)} ⇒ √{18(10)(4)(2)} = √{1440}
A ≈ 37.95
Find the area of a triangle with side lengths 13, 15, and 18
- Use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
- a = 13, b = 15, c = 18
- s = [1/2](13 + 15 + 18) ⇒ s = 23
- A = √{23(23 − 13)(23 − 15)(23 − 18)} ⇒ √{17(10)(8)(5)} = √{9200}
A ≈ 95.92
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Finding the Area of a Triangle
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
- Intro
- Master Right Triangle Formula and Law of Cosines
- Heron's Formula
- Example 1: Area of Triangle with Two Sides and One Angle
- Example 2: Area of Triangle with Three Sides
- Example 3: Area of Triangle with Three Sides, No Heron's Formula
- Extra Example 1: Area of Triangle with Two Sides and One Angle
- Extra Example 2: Area of Triangle with Two Sides and One Angle






























Start Learning Now
Our free lessons will get you started (Flash® 10 required).
Sign up for Educator.comGet immediate access to our entire library.
Features Overview