{"id":11469,"date":"2022-07-12T12:41:20","date_gmt":"2022-07-12T07:11:20","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11469"},"modified":"2022-07-12T12:54:38","modified_gmt":"2022-07-12T07:24:38","slug":"prove-that-the-sum-of-the-squares-of-the-sides-of-a-rhombus-is-equal-to-the-sum-of-the-squares-of-its-diagonals","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-the-sum-of-the-squares-of-the-sides-of-a-rhombus-is-equal-to-the-sum-of-the-squares-of-its-diagonals\/","title":{"rendered":"Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals."},"content":{"rendered":"<h2>Solution :<\/h2>\n<p>Let the diagonals BD and AC of the rhombus ABCD intersect each other at O.<img decoding=\"async\" loading=\"lazy\" class=\" wp-image-11489 alignright\" src=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-7.jpeg?resize=190%2C143&#038;ssl=1\" alt=\"rhombus\" width=\"190\" height=\"143\" srcset=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-7.jpeg?resize=300%2C225&amp;ssl=1 300w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-7.jpeg?resize=1024%2C768&amp;ssl=1 1024w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-7.jpeg?resize=768%2C576&amp;ssl=1 768w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-7.jpeg?resize=1536%2C1152&amp;ssl=1 1536w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-7.jpeg?w=1700&amp;ssl=1 1700w\" sizes=\"(max-width: 190px) 100vw, 190px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Since the diagonals of the rhombus bisect each other at right angles.<\/p>\n<p>\\(\\therefore\\)\u00a0 \\(\\angle\\) AOB = \\(\\angle\\) BOC = \\(\\angle\\) COD = \\(\\angle\\) DOA = 90<\/p>\n<p>and\u00a0 AO = CO, BO = OD<\/p>\n<p>Since AOB is a right triangle angled at O, therefore,<\/p>\n<p>\\({AB}^2\\) = \\({OA}^2\\) + \\({OB}^2\\)<\/p>\n<p>Since AO = CO, BO = OD,<\/p>\n<p>\\(\\implies\\) \\({AB}^2\\) = \\(({1\\over 2}{AB})^2\\) + \\(({1\\over 2}{BD})^2\\)<\/p>\n<p>\\(\\implies\\)\u00a0 \\(4{AB}^2\\) = \\({AC}^2\\) + \\({BD}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;&#8230;&#8230;.(1)<\/p>\n<p>Similarly, we have :<\/p>\n<p>\\(4{BC}^2\\) = \\({AC}^2\\) + \\({BD}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;&#8230;..(2)<\/p>\n<p>\\(4{CD}^2\\) = \\({AC}^2\\) + \\({BD}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;&#8230;..(3)<\/p>\n<p>\\(4{AD}^2\\) = \\({AC}^2\\) + \\({BD}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;&#8230;&#8230;(4)<\/p>\n<p>Adding all the equations above, we get<\/p>\n<p><strong>\\({AB}^2\\) + \\({BC}^2\\) + \\({CD}^2\\) + \\({DA}^2\\) = \\({AC}^2\\) + \\({BD}^2\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Solution : Let the diagonals BD and AC of the rhombus ABCD intersect each other at O. Since the diagonals of the rhombus bisect each other at right angles. \\(\\therefore\\)\u00a0 \\(\\angle\\) AOB = \\(\\angle\\) BOC = \\(\\angle\\) COD = \\(\\angle\\) DOA = 90 and\u00a0 AO = CO, BO = OD Since AOB is a right &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathemerize.com\/prove-that-the-sum-of-the-squares-of-the-sides-of-a-rhombus-is-equal-to-the-sum-of-the-squares-of-its-diagonals\/\"> <span class=\"screen-reader-text\">Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,912],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. - Mathemerize<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/prove-that-the-sum-of-the-squares-of-the-sides-of-a-rhombus-is-equal-to-the-sum-of-the-squares-of-its-diagonals\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : Let the diagonals BD and AC of the rhombus ABCD intersect each other at O. Since the diagonals of the rhombus bisect each other at right angles. (therefore)\u00a0 (angle) AOB = (angle) BOC = (angle) COD = (angle) DOA = 90 and\u00a0 AO = CO, BO = OD Since AOB is a right &hellip; Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. 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