{"id":2212,"date":"2025-08-11T16:53:34","date_gmt":"2025-08-11T16:53:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2212"},"modified":"2025-08-13T16:56:40","modified_gmt":"2025-08-13T16:56:40","slug":"non-right-triangles-with-law-of-cosines-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/non-right-triangles-with-law-of-cosines-learn-it-3\/","title":{"raw":"Non-right Triangles with Law of Cosines: Learn It 3","rendered":"Non-right Triangles with Law of Cosines: Learn It 3"},"content":{"raw":"

Using Heron's Formula to Find the Area of a Triangle<\/h2>\r\nWe already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use Heron\u2019s formula<\/strong> instead of finding the height. Heron of Alexandria<\/strong> was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known.\r\n\r\n
\r\n

Heron's formula<\/h3>\r\nHeron\u2019s formula finds the area of oblique triangles in which sides [latex]a,b[\/latex], and [latex]c[\/latex] are known.\r\n

[latex]\\text{Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}[\/latex]<\/p>\r\nwhere [latex]s=\\frac{\\left(a+b+c\\right)}{2}[\/latex] is one half of the perimeter of the triangle, sometimes called the semi-perimeter.\r\n\r\n<\/section>

Find the area of the triangle using Heron\u2019s formula.\"A\r\n\r\n[reveal-answer q=\"256710\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"256710\"]\r\n\r\nFirst, we calculate [latex]s[\/latex].\r\n

[latex]\\begin{align} s&=\\frac{\\left(a+b+c\\right)}{2} \\\\ s&=\\frac{\\left(10+15+7\\right)}{2}=16 \\end{align}[\/latex]<\/p>\r\nThen we apply the formula.\r\n

[latex]\\begin{align} \\text{Area}&=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)} \\\\ \\text{Area}&=\\sqrt{16\\left(16 - 10\\right)\\left(16 - 15\\right)\\left(16 - 7\\right)} \\\\ \\text{Area}&\\approx 29.4 \\end{align}[\/latex]<\/p>\r\nThe area is approximately 29.4 square units.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Use Heron\u2019s formula to find the area of a triangle with sides of lengths [latex]a=29.7\\text{ft},b=42.3\\text{ft}[\/latex], and [latex]c=38.4\\text{ft}[\/latex].[reveal-answer q=\"159200\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"159200\"]Area = 552 square feet[\/hidden-answer]\r\n\r\n<\/section>
[ohm_question hide_question_numbers=1]149312[\/ohm_question]<\/section>
A Chicago city developer wants to construct a building consisting of artist\u2019s lofts on a triangular lot bordered by Rush Street, Wabash Avenue, and Pearson Street. The frontage along Rush Street is approximately 62.4 meters, along Wabash Avenue it is approximately 43.5 meters, and along Pearson Street it is approximately 34.1 meters. How many square meters are available to the developer?\"A\r\n\r\n[reveal-answer q=\"63725\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"63725\"]\r\n\r\nFind the measurement for [latex]s[\/latex], which is one-half of the perimeter.\r\n

[latex]\\begin{align}s&=\\frac{\\left(62.4+43.5+34.1\\right)}{2} \\\\ s&=70\\text{m} \\end{align}[\/latex]<\/p>\r\nApply Heron\u2019s formula.\r\n

[latex]\\begin{align}\\text{Area}&=\\sqrt{70\\left(70 - 62.4\\right)\\left(70 - 43.5\\right)\\left(70 - 34.1\\right)} \\\\ \\text{Area}&=\\sqrt{506,118.2} \\\\ \\text{Area}&\\approx 711.4 \\end{align}[\/latex]<\/p>\r\nThe developer has about 711.4 square meters.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

\r\n
\r\n\r\nFind the area of a triangle given [latex]a=4.38\\text{ft},b=3.79\\text{ft,}[\/latex] and [latex]c=5.22\\text{ft}\\text{.}[\/latex]\r\n\r\n[reveal-answer q=\"942985\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"942985\"]\r\n\r\nabout 8.15 square feet\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
[ohm_question hide_question_numbers=1]8473[\/ohm_question]<\/section>","rendered":"

Using Heron’s Formula to Find the Area of a Triangle<\/h2>\n

We already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use Heron\u2019s formula<\/strong> instead of finding the height. Heron of Alexandria<\/strong> was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known.<\/p>\n

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Heron’s formula<\/h3>\n

Heron\u2019s formula finds the area of oblique triangles in which sides [latex]a,b[\/latex], and [latex]c[\/latex] are known.<\/p>\n

[latex]\\text{Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}[\/latex]<\/p>\n

where [latex]s=\\frac{\\left(a+b+c\\right)}{2}[\/latex] is one half of the perimeter of the triangle, sometimes called the semi-perimeter.<\/p>\n<\/section>\n

Find the area of the triangle using Heron\u2019s formula.\"A<\/p>\n