{"id":141,"date":"2017-08-23T09:08:26","date_gmt":"2017-08-23T09:08:26","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/mathematics\/?page_id=141"},"modified":"2017-09-18T18:31:31","modified_gmt":"2017-09-18T18:31:31","slug":"area-of-triangles-and-special-quadrilaterals","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/mathematics\/area-of-triangles-and-special-quadrilaterals\/","title":{"rendered":"Area of Triangles and Special Quadrilaterals"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/special-triangles\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/geometry-spatial-reasoning\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/the-pythagorean-theorem\">Next Lesson \u27a1<\/a><\/div>\n<p><!-- CONTENT BEGINS HERE --><\/p>\n<h1 id=\"title\">Area of Triangles and Special Quadrilaterals<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson, you will study how to find the area of triangles and special quadrilaterals.<\/p>\n<h4>Previously Covered:<\/h4>\n<ul>\n<li>Two triangles are <strong><em>congruent\u00a0<\/em><\/strong>if there is a correspondence between them such that every pair of\u00a0corresponding sides is congruent and every pair of angles is\u00a0congruent.<\/li>\n<li>A line segment with endpoints\u00a0that lie on two nonconsecutive vertices of the polygon is called\u00a0a <em><strong>diagonal<\/strong><\/em>.<\/li>\n<li>A <strong><em>trapezoid\u00a0<\/em><\/strong>is a quadrilateral in which exactly one pair of sides is\u00a0parallel.<\/li>\n<li>A <strong><em>parallelogram<\/em><\/strong> is a\u00a0quadrilateral in which both pairs of opposite sides are parallel.<\/li>\n<\/ul>\n<section>\n<h3><strong>How do we find the area of a parallelogram?<\/strong><\/h3>\n<p>The formula for the area of a rectangle is the product of its\u00a0base and height. This can also be used to find the area of any\u00a0parallelogram.<\/p>\n<p><span style=\"text-decoration: none;\">In any parallelogram <\/span><em><span style=\"text-decoration: none;\">ABCD<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0draw the segments from <\/span><em><span style=\"text-decoration: none;\">C\u00a0and D\u00a0<\/span><\/em><span style=\"text-decoration: none;\">perpendicular to the line containing <\/span><em><span style=\"text-decoration: none;\">AB<\/span><\/em><span style=\"text-decoration: none;\">. <\/span><\/p>\n<p style=\"text-decoration: none;\" align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20025.JPG\" alt=\"Parallelogram\" width=\"167\" height=\"117\" name=\"graphics3\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"text-decoration: none;\">Call these points of\u00a0intersection <\/span><em><span style=\"text-decoration: none;\">E <\/span><\/em><span style=\"text-decoration: none;\">and\u00a0<\/span><em><span style=\"text-decoration: none;\">F,\u00a0<\/span><\/em><span style=\"text-decoration: none;\">respectively. The length of either of the segments\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p2_html_372ae795.gif\" width=\"28\" height=\"21\" name=\"graphics16\" align=\"BOTTOM\" border=\"0\" \/>\u00a0<\/span><span style=\"text-decoration: none;\">or\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p2_html_m356afd3a.gif\" width=\"27\" height=\"23\" name=\"graphics17\" align=\"BOTTOM\" border=\"0\" \/>\u00a0is the height, <\/span><em><span style=\"text-decoration: none;\">h<\/span><\/em><span style=\"text-decoration: none;\">.\u00a0By the HL Theorem, triangle\u00a0<\/span><em><span style=\"text-decoration: none;\">ADE<\/span><\/em><span style=\"text-decoration: none;\"> and triangle\u00a0<\/span><em><span style=\"text-decoration: none;\">BCF<\/span><\/em><span style=\"text-decoration: none;\"> are congruent,\u00a0and they therefore have the same area. From the figure, we see\u00a0that the area of <\/span><em><span style=\"text-decoration: none;\">ABCD\u00a0<\/span><\/em><span style=\"text-decoration: none;\">is equal to the area of the rectangle <em>CDEF<\/em>. So we have\u00a0shown:<\/span><\/p>\n<p><em><span style=\"text-decoration: none;\">The area of a\u00a0parallelogram is equal to the product of its base and its height:\u00a0A=bh. <\/span><\/em><\/p>\n<h4><strong>How do we find the area of a triangle?<\/strong><\/h4>\n<p>Previously, we reviewed the following:<\/p>\n<p><em><span style=\"text-decoration: none;\">Each diagonal\u00a0separates a parallelogram into two congruent triangles. <\/span><\/em>Using this information and the formula for the area of a\u00a0parallelogram, we can derive the formula for the area of a\u00a0triangle:<\/p>\n<p><em><span style=\"text-decoration: none;\">The area of a\u00a0triangle is one-half the product of its base and its height:\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s11_p2_clip_image002.gif\" width=\"44\" height=\"34\" name=\"graphics4\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0.\u00a0<\/span><\/em><\/p>\n<p><span style=\"text-decoration: none;\">Given triangle\u00a0<\/span><em><span style=\"text-decoration: none;\">ABC<\/span><\/em><span style=\"text-decoration: none;\">, with base <\/span><em><span style=\"text-decoration: none;\">AB\u00a0<\/span><\/em><span style=\"text-decoration: none;\">= <\/span><em><span style=\"text-decoration: none;\">b\u00a0<\/span><\/em><span style=\"text-decoration: none;\">and height <\/span><em><span style=\"text-decoration: none;\">h<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0draw the line parallel to\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p2_html_786a0686.gif\" width=\"27\" height=\"23\" name=\"graphics18\" align=\"absmiddle\" border=\"0\" \/>\u00a0containing the point <\/span><em><span style=\"text-decoration: none;\">A<\/span><\/em><span style=\"text-decoration: none;\">. Choose a point <\/span><em><span style=\"text-decoration: none;\">D\u00a0<\/span><\/em><span style=\"text-decoration: none;\">on this line so that <\/span><em><span style=\"text-decoration: none;\">DC\u00a0<\/span><\/em><span style=\"text-decoration: none;\">= <\/span><em><span style=\"text-decoration: none;\">AB\u00a0<\/span><\/em><span style=\"text-decoration: none;\">= <\/span><em><span style=\"text-decoration: none;\">b<\/span><\/em><span style=\"text-decoration: none;\">. By construction, <\/span><em><span style=\"text-decoration: none;\">ABCD\u00a0<\/span><\/em><span style=\"text-decoration: none;\">is a parallelogram and\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p2_html_m76b936f1.gif\" width=\"28\" height=\"23\" name=\"graphics19\" align=\"absmiddle\" border=\"0\" \/>\u00a0is a diagonal. Therefore, triangle\u00a0<\/span><em><span style=\"text-decoration: none;\">ADC \u00a0<\/span><\/em><span style=\"text-decoration: none;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p2_html_11aa9d18.gif\" width=\"15\" height=\"13\" name=\"graphics15\" align=\"BOTTOM\" border=\"0\" \/>triangle\u00a0<\/span><em><span style=\"text-decoration: none;\">CBA<\/span><\/em><span style=\"text-decoration: none;\"> and the areas of these triangles are equal. If we call this area <\/span><em><span style=\"text-decoration: none;\">A<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0we have that:<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"text-decoration: none;\">2<\/span><em><span style=\"text-decoration: none;\">A\u00a0= area(ABCD)\u00a0= bh. So we have proved <img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s11_p2_clip_image003.gif\" width=\"44\" height=\"34\" name=\"graphics5\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0.<\/span><\/em><em><span style=\"text-decoration: none;\"><br \/>\n<\/span><\/em><\/p>\n<p align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20026.JPG\" alt=\"Proof of triangle area formula\" width=\"167\" height=\"117\" name=\"graphics6\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<div>\n<p>What is the area of an equilateral triangle with side length 3?<\/p>\n<ol>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s11_p2_clip_image005.gif\" width=\"9\" height=\"34\" name=\"graphics7\" align=\"BOTTOM\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s11_p2_clip_image007.gif\" width=\"29\" height=\"38\" name=\"graphics8\" align=\"BOTTOM\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s11_p2_clip_image009.gif\" width=\"29\" height=\"38\" name=\"graphics9\" align=\"BOTTOM\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s11_p2_clip_image011.gif\" width=\"26\" height=\"18\" name=\"graphics10\" align=\"BOTTOM\" border=\"0\" \/><\/li>\n<\/ol>\n<\/div>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>The correct answer is C. To use the area formula, we need to\u00a0determine the height of the triangle. From the previous section,\u00a0we know that\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s11_p2_clip_image013.gif\" width=\"135\" height=\"41\" name=\"graphics11\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0.\u00a0This gives area\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p2_html_m1f83e79.gif\" width=\"171\" height=\"53\" name=\"graphics12\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<\/div>\n<\/section>\n<h3><strong>How do we find the area of a trapezoid?<\/strong><\/h3>\n<p>An <abbr title=\"any segment that intersects both bases of a trapezoid at right angles \">altitude\u00a0of a trapezoid<\/abbr> is any segment perpendicular to the\u00a0bases . Its length is the <abbr title=\" The perpendicular distance between the bases of a trapezoid\">height<\/abbr>.<\/p>\n<blockquote><p><em><span style=\"text-decoration: none;\">The area of a\u00a0trapezoid is equal to one-half the product of its height and the\u00a0sum of the lengths of its bases, b<sub>1<\/sub> and b<sub>2<\/sub>:<br \/>\n<\/span><\/em><span style=\"text-decoration: none;\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s11_p3_clip_image002.gif\" width=\"88\" height=\"34\" name=\"graphics3\" align=\"ABSMIDDLE\" border=\"0\" \/><\/span><em><span style=\"text-decoration: none;\">.<br \/>\n<\/span><\/em><\/p><\/blockquote>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20027.JPG\" alt=\"Trapezoid\" width=\"167\" height=\"117\" name=\"graphics4\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<div>\n<p>What is area of the trapezoid shown below?<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20028.JPG\" alt=\"Trapezoid with legs 2 and 8\" width=\"167\" height=\"117\" name=\"graphics5\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<ol>\n<li>15<\/li>\n<li>20<\/li>\n<li>25<\/li>\n<li>30<\/li>\n<\/ol>\n<\/div>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>T<span style=\"text-decoration: none;\">he correct answer is A .\u00a0If we take the altitude shown below, then we can use that the\u00a0resulting triangle is a 45-45-90 triangle along with the given\u00a0information to get that <\/span><em><span style=\"text-decoration: none;\">h\u00a0<\/span><\/em><span style=\"text-decoration: none;\">= 3. Using the\u00a0formula, we get\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s11_p3_clip_image004.gif\" width=\"119\" height=\"37\" name=\"graphics6\" align=\"ABSMIDDLE\" border=\"0\" \/>.<br \/>\n<\/span><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/Math%20Mod%204.3%20Art%20029.JPG\" alt=\" Trapezoid area\" name=\"graphics7\" \/><\/p>\n<\/div>\n<\/section>\n<h3>Review of New Vocabulary and Concepts<\/h3>\n<ul>\n<li>If the sum of the measures of\u00a0two angles is 180\u00b0, then the angles are called <strong><em>supplementary<\/em><\/strong>,\u00a0and each angle is called the <strong><em>supplement\u00a0<\/em><\/strong>of the other.<\/li>\n<li>If the sum of the measures of\u00a0two angles is 90\u00b0, then the angles are called <em><strong>complementary\u00a0<\/strong><\/em>and each angle is called the <em><strong>complement <\/strong><\/em>of\u00a0the other.<\/li>\n<li>The <strong><em>angle bisector\u00a0<\/em><\/strong><span style=\"text-decoration: none;\">of angle\u00a0<\/span><em><span style=\"text-decoration: none;\">BAC<\/span><\/em><span style=\"text-decoration: none;\"> is the ray\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p4_html_mb09fd5c.gif\" width=\"28\" height=\"21\" name=\"graphics14\" align=\"BOTTOM\" border=\"0\" \/>\u00a0<\/span><em><span style=\"text-decoration: none;\"><span style=\"font-style: normal;\">s<\/span><\/span><\/em><span style=\"text-decoration: none;\">uch\u00a0that angle<\/span><em><span style=\"text-decoration: none;\"> BAD\u00a0<\/span><\/em><span style=\"text-decoration: none;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p4_html_11aa9d18.gif\" width=\"15\" height=\"13\" name=\"graphics8\" align=\"BOTTOM\" border=\"0\" \/>\u00a0angle<\/span><em><span style=\"text-decoration: none;\"> DAC<\/span><\/em><span style=\"text-decoration: none;\">.<br \/>\n<\/span><\/li>\n<li>Two lines that are not coplanar\u00a0are called <strong><em>skew<\/em><\/strong> lines.<\/li>\n<li>A triangle is called <em><strong>scalene<\/strong><\/em>,\u00a0<em><strong>isosceles<\/strong><\/em>, or <strong><em>equilateral\u00a0<\/em><\/strong>according to whether it has no congruent sides, at least two\u00a0congruent sides, or three congruent sides, respectively.<\/li>\n<li>A triangle is called an <strong><em>acute<\/em>,\u00a0<em>right<\/em><\/strong>, or <em><strong>obtuse triangle\u00a0<\/strong><\/em>according to whether it includes three acute angles, one right\u00a0angle, or one obtuse angle, respectively.<\/li>\n<li>In a right triangle, the side\u00a0opposite the right angle is the <strong><em>hypotenuse<\/em><\/strong>,\u00a0and the other two sides are the <strong><em>legs<\/em>. <\/strong><\/li>\n<li>A <strong><em>median\u00a0<\/em><\/strong>of a triangle is the segment from any vertex to the midpoint of\u00a0the opposite side.<\/li>\n<li>An <em><strong>altitude\u00a0<\/strong><\/em>of a triangle is the segment from any vertex that is\u00a0perpendicular to the line containing the opposite side.<\/li>\n<li><strong>Exterior Angle Theorem: <\/strong><span style=\"text-decoration: none;\">An\u00a0exterior angle of a triangle is equal to the sum of its remote\u00a0interior angles. <\/span><\/li>\n<li>Two\u00a0sides of a triangle are congruent if and only if the angles\u00a0opposite those sides are congruent.<\/li>\n<li>A\u00a0triangle has three congruent sides if and only if it has three\u00a0congruent angles.<\/li>\n<li>If\u00a0angle ABC is an isosceles triangle with congruent sides\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p4_html_299c8a9a.gif\" width=\"27\" height=\"21\" name=\"graphics10\" align=\"BOTTOM\" border=\"0\" \/>\u00a0and\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p4_html_m76b936f1.gif\" width=\"28\" height=\"23\" name=\"graphics11\" align=\"BOTTOM\" border=\"0\" \/>,\u00a0then the median from A to\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p4_html_786a0686.gif\" width=\"27\" height=\"23\" name=\"graphics12\" align=\"BOTTOM\" border=\"0\" \/>\u00a0is the altitude from A to\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p4_html_786a0686.gif\" width=\"27\" height=\"23\" name=\"graphics13\" align=\"BOTTOM\" border=\"0\" \/>.<\/li>\n<li><strong>HL Theorem : <\/strong>If\u00a0the hypotenuse and one leg of a right triangle are congruent to\u00a0corresponding sides of another right triangle, then the triangles\u00a0are congruent.<\/li>\n<li>Ina 30-60-90 triangle, the length of the hypotenuse is twice the\u00a0length of the shorter leg and the length of the longer leg equals\u00a0the shorter leg times\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/s11_p4_html_7b57d20.gif\" width=\"24\" height=\"24\" name=\"graphics9\" align=\"BOTTOM\" border=\"0\" \/><\/li>\n<li>In\u00a0a 45-45-90 triangle, the length of the hypotenuse is\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/sqroo.gif\" width=\"25\" height=\"22\" name=\"graphics3\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0times the shorter leg.<\/li>\n<li><span style=\"text-decoration: none;\">The\u00a0area of a parallelogram is equal to the product of its base and\u00a0its height: <\/span><em><span style=\"text-decoration: none;\">A=bh<\/span><\/em><span style=\"text-decoration: none;\">.<br \/>\n<\/span><\/li>\n<li>The area\u00a0of a triangle is one-half the product of its base and its height:\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s11_p4_clip_image002.gif\" width=\"44\" height=\"34\" name=\"graphics4\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/li>\n<li>The area of a\u00a0trapezoid is equal to one-half the product of its height and the\u00a0sum of the lengths of its bases, <em><span style=\"text-decoration: none;\">b<sub>1\u00a0<\/sub><\/span><\/em><span style=\"text-decoration: none;\">and <\/span><em><span style=\"text-decoration: none;\">b<sub>2<\/sub><\/span><\/em><span style=\"text-decoration: none;\">:\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/4\/images\/s11_p4_clip_image004.gif\" width=\"88\" height=\"34\" name=\"graphics5\" align=\"ABSMIDDLE\" border=\"0\" \/>.<br \/>\n<\/span><\/li>\n<\/ul>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/special-triangles\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/geometry-spatial-reasoning\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/the-pythagorean-theorem\">Next Lesson \u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson \u27a1 Area of Triangles and Special Quadrilaterals Objective In this lesson, you will study how to find the area of triangles and special quadrilaterals. Previously Covered: Two triangles are congruent\u00a0if there is a correspondence between them such that every pair of\u00a0corresponding sides is congruent and every pair of angles is\u00a0congruent. [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-141","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/141","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/comments?post=141"}],"version-history":[{"count":9,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/141\/revisions"}],"predecessor-version":[{"id":760,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/141\/revisions\/760"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/media?parent=141"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}