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### Finding the Area of a Triangle

Main formulas:

• Master formula for right triangles: SOHCAHTOA!
 sinθ = opposite side hypotenuse cosθ = adjacent side hypotenuse tanθ = opposite side adjacent side
• The Law of Cosines (in any triangle):
 c2 = a2 + b2 − 2abcosC
• Heron's Formula:
 Area = √ s(s− a)(s− b)(s− b) ,
where s = [1/2](a+b+c) is the semiperimeter of the triangle.

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)

## 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 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}

## 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}

## 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}

## 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}

## 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 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}

## 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}

## 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.