{"id":2890,"date":"2023-05-16T19:10:44","date_gmt":"2023-05-16T19:10:44","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&#038;p=2890"},"modified":"2025-08-26T03:40:32","modified_gmt":"2025-08-26T03:40:32","slug":"area-and-circumference-learn-it-2","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/area-and-circumference-learn-it-2\/","title":{"raw":"Area and Circumference: Learn It 2","rendered":"Area and Circumference: Learn It 2"},"content":{"raw":"<h2>Find the Area and Perimeter of a Triangle<\/h2>\r\n<p>We now know how to find the area of a rectangle. We can use this fact to help us visualize the formula for the area of a triangle. In the rectangle below, we\u2019ve labeled the length [latex]b[\/latex] and the width [latex]h[\/latex], so its area is [latex]bh[\/latex].<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"151\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223912\/CNX_BMath_Figure_09_04_035.png\" alt=\"A rectangle with the side labeled h and the bottom labeled b. The center says A equals bh.\" width=\"151\" height=\"89\" \/> Figure 1. The area, A = bh[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p><br \/>\r\nWe can divide this rectangle into two congruent triangles (see the image below). Triangles that are congruent have identical side lengths and angles, and so their areas are equal. The area of each triangle is one-half the area of the rectangle, or [latex]\\Large\\frac{1}{2}\\normalsize bh[\/latex]. This example helps us see why the formula for the area of a triangle is [latex]A=\\Large\\frac{1}{2}\\normalsize bh[\/latex].<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"323\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223913\/CNX_BMath_Figure_09_04_036.png\" alt=\"A rectangle with a diagonal line drawn from the upper left corner to the bottom right corner. The side of the rectangle is labeled h and the bottom is labeled b. Each triangle says one-half bh. To the right of the rectangle, it says &quot;Area of each triangle A = one-half bh&quot;. \" width=\"323\" height=\"107\" \/> Figure 2. The area of each triangle, half the rectangle, is A = 1\/2(bh)[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>To find the area of the triangle, you need to know its base and height. The base is the length of one side of the triangle, usually the side at the bottom. The height is the length of the line that connects the base to the opposite vertex, and makes a [latex]\\text{90}^ \\circ[\/latex] angle with the base. The image below\u00a0shows three triangles with the base and height of each marked.<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"563\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223914\/CNX_BMath_Figure_09_04_037.png\" alt=\"Three triangles. The triangle on the left is a right triangle. The bottom is labeled b and the side is labeled h. The middle triangle is an acute triangle. The bottom is labeled b. There is a dotted line from the top vertex to the base of the triangle, forming a right angle with the base. That line is labeled h. The triangle on the right is an obtuse triangle. The bottom of the triangle is labeled b. The base has a dotted line extended out and forms a right angle with a dotted line to the top of the triangle. The vertical line is labeled h.\" width=\"563\" height=\"107\" \/> Figure 3. These triangles have their heights and bases labeled[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>triangle properties<\/h3>\r\n<p>For any triangle [latex]\\Delta ABC[\/latex], the sum of the measures of the angles is [latex]\\text{180}^ \\circ[\/latex].<\/p>\r\n<p>&nbsp;<\/p>\r\n<p style=\"text-align: center;\">[latex]m\\angle{A}+m\\angle{B}+m\\angle{C}=180^\\circ [\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>The <strong>perimeter<\/strong> of a triangle is the sum of the lengths of the sides.<\/p>\r\n<p>&nbsp;<\/p>\r\n<p style=\"text-align: center;\">[latex]P=a+b+c[\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>The <strong>area<\/strong> of a triangle is one-half the base, [latex]b[\/latex], times the height, [latex]h[\/latex].<\/p>\r\n<p>&nbsp;<\/p>\r\n<p style=\"text-align: center;\">[latex]A={\\Large\\frac{1}{2}}bh[\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<br \/>\r\n<center><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223917\/CNX_BMath_Figure_09_04_038_img.png\" alt=\"A triangle, with vertices labeled A, B, and C. The sides are labeled a, b, and c. There is a vertical dotted line from vertex B at the top of the triangle to the base of the triangle, meeting the base at a right angle. The dotted line is labeled h.\" width=\"190\" height=\"160\" \/><\/center>\r\n<p>&nbsp;<\/p>\r\n<\/div>\r\n<\/section>\r\n<section class=\"textbox example\">Find the area of a triangle whose base is [latex]11[\/latex] inches and whose height is [latex]8[\/latex] inches.<br \/>\r\n[reveal-answer q=\"247910\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"247910\"]\r\n\r\n<table id=\"eip-id1168468457178\" class=\"unnumbered unstyled\" summary=\"Step 1 says, \">\r\n<tbody>\r\n<tr>\r\n<td>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223918\/CNX_BMath_Figure_09_04_073_img-01.png\" alt=\"A triangle with the base labeled 11 in and a dotted vertical line from the top vertex to the base to form a right angle. This dotted line is labeled 8 in.\" width=\"318\" height=\"202\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td>the area of the triangle<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\r\n<td>let <em>A<\/em> = area of the triangle<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 4.<strong>Translate.<\/strong><\/p>\r\n<p>Write the appropriate formula.<\/p>\r\n<p>Substitute.<\/p>\r\n<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223920\/CNX_BMath_Figure_09_04_073_img-02.png\" alt=\"The equation A = one half times b times h. The equation is written again with 11 substituted for b and 8 substituted for h.\" width=\"318\" height=\"110\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td>[latex]A=44[\/latex] square inches.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 6. <strong>Check.<\/strong><\/p>\r\n<\/td>\r\n<td>\r\n<p>[latex]A=\\frac{1}{2}bh[\/latex]<\/p>\r\n<p>[latex]44\\stackrel{?}{=}\\frac{1}{2}(11)8[\/latex]<\/p>\r\n<p>[latex]44=44\\quad\\checkmark[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td>The area is [latex]44[\/latex] square inches.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question height=\"300\" hide_question_numbers=1]6992[\/ohm2_question]<\/section>\r\n<section class=\"textbox example\">The perimeter of a triangular garden is [latex]24[\/latex] feet. The lengths of two sides are [latex]4[\/latex] feet and [latex]9[\/latex] feet. How long is the third side?<br \/>\r\n[reveal-answer q=\"371512\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"371512\"]\r\n\r\n<table id=\"eip-id1168466081900\" class=\"unnumbered unstyled\" summary=\"Step 1 says, \">\r\n<tbody>\r\n<tr>\r\n<td>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223923\/CNX_BMath_Figure_09_04_074_img-01.png\" alt=\"An acute triangle with one side labeled 4 feet, the second side labeled 9 feet, and the third side labeled c. Beneath the triangle, it says P = 24 feet.\" width=\"317\" height=\"188\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td>length of the third side of a triangle<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\r\n<td>Let <em>c<\/em> = the third side<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 4.<strong>Translate.<\/strong><\/p>\r\n<p>Write the appropriate formula.<\/p>\r\n<p>Substitute in the given information.<\/p>\r\n<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223925\/CNX_BMath_Figure_09_04_074_img-02.png\" alt=\"The equation P = a + b + c. The equation is written again with 24 substituted in for P, 4 substituted in for a, and 9 substituted in for b.\" width=\"317\" height=\"67\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td>\r\n<p>[latex]24=13+c[\/latex]<\/p>\r\n<p>[latex]11=c[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 6. <strong>Check.<\/strong><\/p>\r\n<\/td>\r\n<td>\r\n<p>[latex]P=a+b+c[\/latex]<\/p>\r\n<p>[latex]24\\stackrel{?}{=}4+9+11[\/latex]<\/p>\r\n<p>[latex]24=24\\checkmark[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td>The third side is [latex]11[\/latex] feet long.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]6993[\/ohm2_question]<\/section>\r\n<section class=\"textbox example\">The area of a triangular church window is [latex]90[\/latex] square meters. The base of the window is [latex]15[\/latex] meters. What is the window\u2019s height?<br \/>\r\n[reveal-answer q=\"632571\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"632571\"]\r\n\r\n<table id=\"eip-id1168467155173\" class=\"unnumbered unstyled\" summary=\"Step 1 says, \">\r\n<tbody>\r\n<tr>\r\n<td>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223929\/CNX_BMath_Figure_09_04_075_img-01.png\" alt=\"A triangle with base labeled 15 m and a dotted line down the center from the top vertex, forming a right angle with the base. The line is labeled h.\" width=\"225\" height=\"222\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td>height of a triangle<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\r\n<td>Let <em>h<\/em> = the height<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 4.<strong>Translate.<\/strong><\/p>\r\n<p>Write the appropriate formula.<\/p>\r\n<p>Substitute in the given information.<\/p>\r\n<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223931\/CNX_BMath_Figure_09_04_075_img-02.png\" alt=\"The equation A = one half times b times h. The equation is written again with 90 substituted in for A and 15 substituted in for b.\" width=\"303\" height=\"99\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td>\r\n<p>[latex]90={\\Large\\frac{1}{2}}\\normalsize(15)h[\/latex]<\/p>\r\n<p>[latex]12=h[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 6. <strong>Check.<\/strong><\/p>\r\n<\/td>\r\n<td>\r\n<p>[latex]A={\\Large\\frac{1}{2}}\\normalsize bh[\/latex]<\/p>\r\n<p>[latex]90\\stackrel{?}{=}{\\Large\\frac{1}{2}}\\normalsize\\cdot 15\\cdot 12[\/latex]<\/p>\r\n<p>[latex]90=90\\quad\\checkmark[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td>The height of the triangle is [latex]12[\/latex] meters.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]6994[\/ohm2_question]<\/section>\r\n<h3>Isosceles and Equilateral Triangles<\/h3>\r\n<p>Besides the right triangle, some other triangles have special names. A triangle with two sides of equal length is called an <strong>isosceles triangle<\/strong>. A triangle that has three sides of equal length is called an <strong>equilateral triangle<\/strong>. The image below\u00a0shows both types of triangles.<\/p>\r\n<center>\r\n[caption id=\"\" align=\"aligncenter\" width=\"367\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223935\/CNX_BMath_Figure_09_04_045.png\" alt=\"Two triangles. All three sides of the triangle on the right are labeled s. It is labeled equilateral triangle. The triangle on the left is labeled isosceles triangle and just two of the sides are labeled s.\" width=\"367\" height=\"231\" \/> Figure 4. The isosceles triangle on the left has two equal sides. The equilateral triangle on the right has three equal sides.[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>isosceles and equilateral triangles<\/h3>\r\n<p>An <strong>isosceles<\/strong> triangle has two sides the same length.<\/p>\r\n<p>&nbsp;<\/p>\r\n<p>An <strong>equilateral<\/strong> triangle has three sides of equal length.<\/p>\r\n<\/div>\r\n<\/section>\r\n<section class=\"textbox example\">Arianna has [latex]156[\/latex] inches of beading to use as trim around a scarf. The scarf will be an isosceles triangle with a base of [latex]60[\/latex] inches. How long can she make the two equal sides?<br \/>\r\n[reveal-answer q=\"327649\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"327649\"]\r\n\r\n<table id=\"eip-id1168466073183\" class=\"unnumbered unstyled\" summary=\"Step 1 says, \">\r\n<tbody>\r\n<tr>\r\n<td>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\r\n<td>\r\n<p><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223942\/CNX_BMath_Figure_09_04_077_img-01.png\" alt=\"A triangle with two sides labeled s and one side labeled 60 inches\" width=\"189\" height=\"115\" \/><\/p>\r\n<p><em>P<\/em> = [latex]156[\/latex] in.<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\r\n<td>the lengths of the two equal sides<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\r\n<td>Let <em>s<\/em> = the length of each side<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 4.<strong>Translate.<\/strong><\/p>\r\n<p>Write the appropriate formula.<\/p>\r\n<p>Substitute in the given information.<\/p>\r\n<\/td>\r\n<td><img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223943\/CNX_BMath_Figure_09_04_077_img-02.png\" alt=\"The equation P = a + b + c. The equation is written again with 156 substituted in for P, 60 substituted in for b, and s substituted in for both a and c.\" width=\"300\" height=\"51\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\r\n<td>\r\n<p>[latex]156=2s=60[\/latex]<\/p>\r\n<p>[latex]96=2s[\/latex]<\/p>\r\n<p>[latex]48=s[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p>Step 6. <strong>Check.<\/strong><\/p>\r\n<\/td>\r\n<td>\r\n<p>[latex]P=a+b+c[\/latex]<\/p>\r\n<p>[latex]156\\stackrel{?}{=}48+60+48[\/latex]<\/p>\r\n<p>[latex]156=156\\quad\\checkmark[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\r\n<td>Arianna can make each of the two equal sides [latex]48[\/latex] inches long.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]6995[\/ohm2_question]<\/section>","rendered":"<h2>Find the Area and Perimeter of a Triangle<\/h2>\n<p>We now know how to find the area of a rectangle. We can use this fact to help us visualize the formula for the area of a triangle. In the rectangle below, we\u2019ve labeled the length [latex]b[\/latex] and the width [latex]h[\/latex], so its area is [latex]bh[\/latex].<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 151px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223912\/CNX_BMath_Figure_09_04_035.png\" alt=\"A rectangle with the side labeled h and the bottom labeled b. The center says A equals bh.\" width=\"151\" height=\"89\" \/><figcaption class=\"wp-caption-text\">Figure 1. The area, A = bh<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>\nWe can divide this rectangle into two congruent triangles (see the image below). Triangles that are congruent have identical side lengths and angles, and so their areas are equal. The area of each triangle is one-half the area of the rectangle, or [latex]\\Large\\frac{1}{2}\\normalsize bh[\/latex]. This example helps us see why the formula for the area of a triangle is [latex]A=\\Large\\frac{1}{2}\\normalsize bh[\/latex].<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 323px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223913\/CNX_BMath_Figure_09_04_036.png\" alt=\"A rectangle with a diagonal line drawn from the upper left corner to the bottom right corner. The side of the rectangle is labeled h and the bottom is labeled b. Each triangle says one-half bh. To the right of the rectangle, it says &quot;Area of each triangle A = one-half bh&quot;.\" width=\"323\" height=\"107\" \/><figcaption class=\"wp-caption-text\">Figure 2. The area of each triangle, half the rectangle, is A = 1\/2(bh)<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>To find the area of the triangle, you need to know its base and height. The base is the length of one side of the triangle, usually the side at the bottom. The height is the length of the line that connects the base to the opposite vertex, and makes a [latex]\\text{90}^ \\circ[\/latex] angle with the base. The image below\u00a0shows three triangles with the base and height of each marked.<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 563px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223914\/CNX_BMath_Figure_09_04_037.png\" alt=\"Three triangles. The triangle on the left is a right triangle. The bottom is labeled b and the side is labeled h. The middle triangle is an acute triangle. The bottom is labeled b. There is a dotted line from the top vertex to the base of the triangle, forming a right angle with the base. That line is labeled h. The triangle on the right is an obtuse triangle. The bottom of the triangle is labeled b. The base has a dotted line extended out and forms a right angle with a dotted line to the top of the triangle. The vertical line is labeled h.\" width=\"563\" height=\"107\" \/><figcaption class=\"wp-caption-text\">Figure 3. These triangles have their heights and bases labeled<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>triangle properties<\/h3>\n<p>For any triangle [latex]\\Delta ABC[\/latex], the sum of the measures of the angles is [latex]\\text{180}^ \\circ[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]m\\angle{A}+m\\angle{B}+m\\angle{C}=180^\\circ[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The <strong>perimeter<\/strong> of a triangle is the sum of the lengths of the sides.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]P=a+b+c[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The <strong>area<\/strong> of a triangle is one-half the base, [latex]b[\/latex], times the height, [latex]h[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]A={\\Large\\frac{1}{2}}bh[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223917\/CNX_BMath_Figure_09_04_038_img.png\" alt=\"A triangle, with vertices labeled A, B, and C. The sides are labeled a, b, and c. There is a vertical dotted line from vertex B at the top of the triangle to the base of the triangle, meeting the base at a right angle. The dotted line is labeled h.\" width=\"190\" height=\"160\" \/><\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Find the area of a triangle whose base is [latex]11[\/latex] inches and whose height is [latex]8[\/latex] inches.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q247910\">Show Solution<\/button><\/p>\n<div id=\"q247910\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"eip-id1168468457178\" class=\"unnumbered unstyled\" summary=\"Step 1 says,\">\n<tbody>\n<tr>\n<td>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223918\/CNX_BMath_Figure_09_04_073_img-01.png\" alt=\"A triangle with the base labeled 11 in and a dotted vertical line from the top vertex to the base to form a right angle. This dotted line is labeled 8 in.\" width=\"318\" height=\"202\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>the area of the triangle<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>let <em>A<\/em> = area of the triangle<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 4.<strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula.<\/p>\n<p>Substitute.<\/p>\n<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223920\/CNX_BMath_Figure_09_04_073_img-02.png\" alt=\"The equation A = one half times b times h. The equation is written again with 11 substituted for b and 8 substituted for h.\" width=\"318\" height=\"110\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>[latex]A=44[\/latex] square inches.<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 6. <strong>Check.<\/strong><\/p>\n<\/td>\n<td>\n[latex]A=\\frac{1}{2}bh[\/latex]<br \/>\n[latex]44\\stackrel{?}{=}\\frac{1}{2}(11)8[\/latex]<br \/>\n[latex]44=44\\quad\\checkmark[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The area is [latex]44[\/latex] square inches.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6992\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6992&theme=lumen&iframe_resize_id=ohm6992&source=tnh\" width=\"100%\" height=\"300\"><\/iframe><\/section>\n<section class=\"textbox example\">The perimeter of a triangular garden is [latex]24[\/latex] feet. The lengths of two sides are [latex]4[\/latex] feet and [latex]9[\/latex] feet. How long is the third side?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q371512\">Show Solution<\/button><\/p>\n<div id=\"q371512\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"eip-id1168466081900\" class=\"unnumbered unstyled\" summary=\"Step 1 says,\">\n<tbody>\n<tr>\n<td>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223923\/CNX_BMath_Figure_09_04_074_img-01.png\" alt=\"An acute triangle with one side labeled 4 feet, the second side labeled 9 feet, and the third side labeled c. Beneath the triangle, it says P = 24 feet.\" width=\"317\" height=\"188\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>length of the third side of a triangle<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>Let <em>c<\/em> = the third side<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 4.<strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula.<\/p>\n<p>Substitute in the given information.<\/p>\n<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223925\/CNX_BMath_Figure_09_04_074_img-02.png\" alt=\"The equation P = a + b + c. The equation is written again with 24 substituted in for P, 4 substituted in for a, and 9 substituted in for b.\" width=\"317\" height=\"67\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>\n[latex]24=13+c[\/latex]<br \/>\n[latex]11=c[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 6. <strong>Check.<\/strong><\/p>\n<\/td>\n<td>\n[latex]P=a+b+c[\/latex]<br \/>\n[latex]24\\stackrel{?}{=}4+9+11[\/latex]<br \/>\n[latex]24=24\\checkmark[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The third side is [latex]11[\/latex] feet long.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6993\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6993&theme=lumen&iframe_resize_id=ohm6993&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox example\">The area of a triangular church window is [latex]90[\/latex] square meters. The base of the window is [latex]15[\/latex] meters. What is the window\u2019s height?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q632571\">Show Solution<\/button><\/p>\n<div id=\"q632571\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"eip-id1168467155173\" class=\"unnumbered unstyled\" summary=\"Step 1 says,\">\n<tbody>\n<tr>\n<td>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223929\/CNX_BMath_Figure_09_04_075_img-01.png\" alt=\"A triangle with base labeled 15 m and a dotted line down the center from the top vertex, forming a right angle with the base. The line is labeled h.\" width=\"225\" height=\"222\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>height of a triangle<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>Let <em>h<\/em> = the height<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 4.<strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula.<\/p>\n<p>Substitute in the given information.<\/p>\n<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223931\/CNX_BMath_Figure_09_04_075_img-02.png\" alt=\"The equation A = one half times b times h. The equation is written again with 90 substituted in for A and 15 substituted in for b.\" width=\"303\" height=\"99\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>\n[latex]90={\\Large\\frac{1}{2}}\\normalsize(15)h[\/latex]<br \/>\n[latex]12=h[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 6. <strong>Check.<\/strong><\/p>\n<\/td>\n<td>\n[latex]A={\\Large\\frac{1}{2}}\\normalsize bh[\/latex]<br \/>\n[latex]90\\stackrel{?}{=}{\\Large\\frac{1}{2}}\\normalsize\\cdot 15\\cdot 12[\/latex]<br \/>\n[latex]90=90\\quad\\checkmark[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>The height of the triangle is [latex]12[\/latex] meters.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6994\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6994&theme=lumen&iframe_resize_id=ohm6994&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h3>Isosceles and Equilateral Triangles<\/h3>\n<p>Besides the right triangle, some other triangles have special names. A triangle with two sides of equal length is called an <strong>isosceles triangle<\/strong>. A triangle that has three sides of equal length is called an <strong>equilateral triangle<\/strong>. The image below\u00a0shows both types of triangles.<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 367px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223935\/CNX_BMath_Figure_09_04_045.png\" alt=\"Two triangles. All three sides of the triangle on the right are labeled s. It is labeled equilateral triangle. The triangle on the left is labeled isosceles triangle and just two of the sides are labeled s.\" width=\"367\" height=\"231\" \/><figcaption class=\"wp-caption-text\">Figure 4. The isosceles triangle on the left has two equal sides. The equilateral triangle on the right has three equal sides.<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>isosceles and equilateral triangles<\/h3>\n<p>An <strong>isosceles<\/strong> triangle has two sides the same length.<\/p>\n<p>&nbsp;<\/p>\n<p>An <strong>equilateral<\/strong> triangle has three sides of equal length.<\/p>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Arianna has [latex]156[\/latex] inches of beading to use as trim around a scarf. The scarf will be an isosceles triangle with a base of [latex]60[\/latex] inches. How long can she make the two equal sides?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q327649\">Show Solution<\/button><\/p>\n<div id=\"q327649\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"eip-id1168466073183\" class=\"unnumbered unstyled\" summary=\"Step 1 says,\">\n<tbody>\n<tr>\n<td>Step 1. <strong>Read<\/strong> the problem. Draw the figure and label it with the given information.<\/td>\n<td>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223942\/CNX_BMath_Figure_09_04_077_img-01.png\" alt=\"A triangle with two sides labeled s and one side labeled 60 inches\" width=\"189\" height=\"115\" \/><\/p>\n<p><em>P<\/em> = [latex]156[\/latex] in.<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>Step 2. <strong>Identify<\/strong> what you are looking for.<\/td>\n<td>the lengths of the two equal sides<\/td>\n<\/tr>\n<tr>\n<td>Step 3. <strong>Name.<\/strong> Choose a variable to represent it.<\/td>\n<td>Let <em>s<\/em> = the length of each side<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 4.<strong>Translate.<\/strong><\/p>\n<p>Write the appropriate formula.<\/p>\n<p>Substitute in the given information.<\/p>\n<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24223943\/CNX_BMath_Figure_09_04_077_img-02.png\" alt=\"The equation P = a + b + c. The equation is written again with 156 substituted in for P, 60 substituted in for b, and s substituted in for both a and c.\" width=\"300\" height=\"51\" \/><\/td>\n<\/tr>\n<tr>\n<td>Step 5. <strong>Solve<\/strong> the equation.<\/td>\n<td>\n[latex]156=2s=60[\/latex]<br \/>\n[latex]96=2s[\/latex]<br \/>\n[latex]48=s[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p>Step 6. <strong>Check.<\/strong><\/p>\n<\/td>\n<td>\n[latex]P=a+b+c[\/latex]<br \/>\n[latex]156\\stackrel{?}{=}48+60+48[\/latex]<br \/>\n[latex]156=156\\quad\\checkmark[\/latex]\n<\/td>\n<\/tr>\n<tr>\n<td>Step 7. <strong>Answer<\/strong> the question.<\/td>\n<td>Arianna can make each of the two equal sides [latex]48[\/latex] inches long.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6995\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6995&theme=lumen&iframe_resize_id=ohm6995&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":19,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":71,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2890"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":43,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2890\/revisions"}],"predecessor-version":[{"id":15650,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2890\/revisions\/15650"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/parts\/71"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2890\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=2890"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=2890"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=2890"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=2890"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}