{"id":608,"date":"2020-12-08T04:11:47","date_gmt":"2020-12-07T22:41:47","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=608"},"modified":"2020-12-08T11:21:23","modified_gmt":"2020-12-08T05:51:23","slug":"pythagoras-theorem","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/pythagoras-theorem\/","title":{"rendered":"Pythagoras Theorem"},"content":{"rendered":"<h2><strong>Pythagoras Theorem<\/strong><\/h2>\n<p><img loading=\"lazy\" class=\"size-full wp-image-12223 aligncenter\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/08\/pythagoras-theorem.jpg\" alt=\"\" width=\"560\" height=\"422\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/08\/pythagoras-theorem.jpg 560w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2016\/08\/pythagoras-theorem-300x226.jpg 300w\" sizes=\"(max-width: 560px) 100vw, 560px\" \/><\/p>\n<p><strong>Theorem 1:<\/strong> In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.<br \/>\n<strong>Given:<\/strong> A right-angled triangle ABC in which B = \u222090\u00ba.<br \/>\n<strong>To Prove:<\/strong> (Hypotenuse)<sup>2<\/sup> = (Base)<sup>2<\/sup> + (Perpendicular)<sup>2<\/sup>.<br \/>\ni.e., AC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup><br \/>\n<strong>Construction:<\/strong> From B draw BD \u22a5\u00a0AC.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127162\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-1.png\" alt=\"Pythagoras Theorem 1\" width=\"264\" height=\"145\" \/><br \/>\n<strong>Proof:<\/strong> In triangle ADB and ABC, we have<br \/>\n\u2220ADB = \u2220ABC \u00a0 \u00a0 \u00a0 \u00a0 [Each equal to 90\u00ba]<br \/>\nand, \u2220A = \u2220A \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Common]<br \/>\nSo, by AA-similarity criterion, we have<br \/>\n\u2206ADB ~ \u2206ABC<br \/>\n\\(\\Rightarrow \\frac{AD}{AB}=\\frac{AB}{AC}\\) \u00a0 [\u2235 In similar triangles corresponding sides are proportional]<br \/>\n\u21d2 AB<sup>2<\/sup> = AD \u00d7 AC\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 &#8230;.(i)<br \/>\nIn triangles BDC and ABC, we have<br \/>\n\u2220CDB = \u2220ABC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Each equal to 90\u00ba]<br \/>\nand, \u2220C = \u2220C \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Common]<br \/>\nSo, by AA-similarity criterion, we have<br \/>\n\u2206BDC ~ \u2206ABC<br \/>\n\\(\\Rightarrow \\frac{DC}{BC}=\\frac{BC}{AC}\\) \u00a0 \u00a0 \u00a0 \u00a0 [\u2235 In similar triangles corresponding sides are proportional]<br \/>\n\u21d2 BC<sup>2<\/sup> = AC \u00d7 DC\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 &#8230;.(ii)<br \/>\nAdding equation (i) and (ii), we get<br \/>\nAB<sup>2<\/sup> + BC<sup>2<\/sup>\u00a0= AD \u00d7 AC + AC \u00d7 DC<br \/>\n\u21d2 AB<sup>2<\/sup> + BC<sup>2<\/sup> = AC (AD + DC)<br \/>\n\u21d2 AB<sup>2<\/sup> + BC<sup>2<\/sup> = AC \u00d7 AC<br \/>\n\u21d2\u00a0AC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup><br \/>\nHence, AC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup><br \/>\nThe converse of the above theorem is also true as proved below.<\/p>\n<p><iframe loading=\"lazy\" title=\"Pythagorean Theorem\" width=\"1200\" height=\"900\" src=\"https:\/\/www.youtube.com\/embed\/JDmglbduKvE?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p><strong>Read also:<\/strong><\/p>\n<ul>\n<li><a href=\"https:\/\/cbselibrary.com\/pythagorean-theorem-algebra\/\">Pythagorean Theorem in Algebra<\/a><\/li>\n<li><a href=\"https:\/\/cbselibrary.com\/pythagorean-theorem\/\">Pythagorean Theorem<\/a><\/li>\n<\/ul>\n<p><strong>Theorem 2:<\/strong> (Converse of Pythagoras Theorem).<br \/>\nIn a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle\u00a0opposite to the side is a right angle.<br \/>\n<strong>Given:<\/strong> A triangle ABC such that AC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127166\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-2.png\" alt=\"Pythagoras Theorem 2\" width=\"280\" height=\"134\" \/><br \/>\n<strong>Construction:<\/strong> Construct a triangle DEF such that DE = AB, EF = BC and \u2220E = 90\u00ba,<br \/>\n<strong>Proof:<\/strong> In order to prove that B = \u222090\u00ba, it is sufficient to show that \u2206ABC ~ \u2206DEF.<br \/>\nFor this we proceed as follows :<br \/>\nSince \u00a0\u2206DEF is a right angled triangle with right angle at E. Therefore, by Pythagoras theorem, we have<br \/>\nDF<sup>2<\/sup> = DE<sup>2<\/sup> + EF<sup>2<\/sup><br \/>\n\u21d2 DF<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0[\u2235\u00a0 DE = AB and EF = BC \u00a0 \u00a0 \u00a0 \u00a0\u00a0(By construction)]<br \/>\n\u21d2 DF<sup>2<\/sup> = AC<sup>2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [\u2235\u00a0 AB<sup>2<\/sup> + BC<sup>2<\/sup> = AC<sup>2<\/sup> (Given)]<br \/>\n\u21d2 DF = AC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(i)<br \/>\nThus, in \u00a0\u2206ABC and \u00a0\u2206DEF, we have<br \/>\nAB = DE, BC = EF \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [By construction]<br \/>\nand, AC = DF \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [From equation (i)]<br \/>\n\u2234\u00a0\u2206ABC ~\u00a0\u2206DEF<br \/>\n\u21d2 \u2220B = \u2220E = 90\u00ba<br \/>\nHence, \u2206ABC is a right triangle right angled at B.<\/p>\n<h2><strong>Pythagoras Theorem With Examples<\/strong><\/h2>\n<p><strong>Example 1:<\/strong> \u00a0 \u00a0Side of a triangle is given, determine it is a right triangle.<br \/>\n(2a \u2013 1) cm, \\(2\\sqrt { 2a } \\) cm, \u00a0and (2a + 1) cm<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let p = (2a \u2013 1) cm, q = \\(2\\sqrt { 2a } \\)\u00a0cm and r = (2a + 1) cm.<br \/>\nThen, (p<sup>2<\/sup> + q<sup>2<\/sup>) = (2a \u2013 1)<sup>2<\/sup> cm<sup>2<\/sup> + (2 )<sup>2<\/sup> cm<sup>2<\/sup><br \/>\n= {(4a<sup>2<\/sup> + 1\u2013 4a) + 8a}cm<sup>2<\/sup><br \/>\n= (4a<sup>2<\/sup> + 4a + 1)cm<sup>2<\/sup><br \/>\n= (2a + 1)<sup>2<\/sup> cm<sup>2<\/sup> = r<sup>2<\/sup>.<br \/>\n(p<sup>2<\/sup> + q<sup>2<\/sup>) = r<sup>2<\/sup>.<br \/>\nHence, the given triangle is right angled.<\/p>\n<p><strong>Example 2:<\/strong>\u00a0 \u00a0 A man goes 10 m due east and then 24 m due north. Find the distance from the starting point.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>Let the initial position of the man be O and his final position be B. Since the man goes \u00a0\u00a010 m due east and then 24 m due north. Therefore, \u2206AOB is a right triangle right-angled at A such that OA = 10 m and AB = 24 m.<br \/>\n<img loading=\"lazy\" class=\"alignnone  wp-image-127168\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-3.png\" alt=\"Pythagoras Theorem 3\" width=\"452\" height=\"367\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-3.png 626w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-3-300x243.png 300w\" sizes=\"(max-width: 452px) 100vw, 452px\" \/><br \/>\nBy Phythagoras theorem, we have<br \/>\nOB<sup>2<\/sup> = OA<sup>2<\/sup> + AB<sup>2<\/sup><br \/>\n\u21d2 OB<sup>2<\/sup> = 10<sup>2<\/sup> + 24<sup>2<\/sup> = 100 + 576 = 676<br \/>\n\u21d2 OB = \\(\\sqrt { 676 } \\)\u00a0= 26 m<br \/>\nHence, the man is at a distance of 26 m from the starting point.<\/p>\n<p><strong>Example 3:<\/strong>\u00a0 \u00a0\u00a0Two towers of heights 10 m and 30 m stand on a plane ground. If the distance between their feet is 15 m, find the distance between their tops.<br \/>\n<strong>Sol.<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127197\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-4.png\" alt=\"Pythagoras Theorem 4\" width=\"242\" height=\"318\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-4.png 242w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-4-228x300.png 228w\" sizes=\"(max-width: 242px) 100vw, 242px\" \/><br \/>\nBy Phythagoras theorem, we have<br \/>\nAC<sup>2<\/sup> = CE<sup>2<\/sup> + AE<sup>2<\/sup><br \/>\n\u21d2 AC<sup>2<\/sup> = 15<sup>2<\/sup> + 20<sup>2<\/sup> = 225 + 400 = 625<br \/>\n\u21d2 AC = \\(\\sqrt { 625 } \\)\u00a0= 25 m.<\/p>\n<p><strong>Example 4:<\/strong>\u00a0 \u00a0 \u00a0In Fig., \u2206ABC is an obtuse triangle, obtuse angled at B. If AD \u22a5\u00a0CB, prove that<br \/>\nAC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup> + 2BC \u00d7 BD<br \/>\n<strong>Sol. \u00a0 \u00a0Given:<\/strong> An obtuse triangle ABC, obtuse-angled at B and AD is perpendicular to CB produced.<br \/>\n<strong>To Prove:<\/strong> AC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup> + 2BC \u00d7 BD<br \/>\n<strong>Proof:<\/strong> Since \u2206ADB is a right triangle right angled at D. Therefore, by Pythagoras theorem, we have\u00a0AB<sup>2<\/sup> = AD<sup>2<\/sup> + DB<sup>2<\/sup> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(i)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127200\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-5.png\" alt=\"Pythagoras Theorem 5\" width=\"194\" height=\"171\" \/><br \/>\nAgain \u2206ADC is a right triangle right angled at D.<br \/>\nTherefore, by Phythagoras theorem, we have<br \/>\nAC<sup>2<\/sup> = AD<sup>2<\/sup> + DC<sup>2<\/sup><br \/>\n\u21d2 AC<sup>2<\/sup> = AD<sup>2<\/sup> + (DB + BC)<sup>2<\/sup><br \/>\n\u21d2 AC<sup>2<\/sup> = AD<sup>2<\/sup> + DB<sup>2<\/sup> + BC<sup>2<\/sup> + 2BC \u2022 BD<br \/>\n\u21d2 AC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup> + 2BC \u2022 BD \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Using (i)]<br \/>\nHence, AC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup> + 2BC \u2022 BD<\/p>\n<p><strong>Example 5:<\/strong>\u00a0 \u00a0 \u00a0In figure, \u2220B of \u2206ABC is an acute angle and AD \u22a5\u00a0BC, prove that<br \/>\nAC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup>\u00a0\u2013 2BC \u00d7 BD<br \/>\n<strong>Sol. \u00a0 \u00a0Given:<\/strong> A \u2206ABC in which \u2220B is an acute angle and AD \u22a5\u00a0BC.<br \/>\n<strong>To Prove:<\/strong> AC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup> \u2013 2BC \u00d7 BD.<br \/>\n<strong>Proof:<\/strong> Since \u2206ADB is a right triangle right-angled at D. So, by Pythagoras theorem, we have<br \/>\nAB<sup>2<\/sup> = AD<sup>2<\/sup> + BD<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(i)<br \/>\nAgain \u2206ADC is a right triangle right angled at D.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127202\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-6.png\" alt=\"Pythagoras Theorem 6\" width=\"300\" height=\"213\" \/><br \/>\nSo, by Pythagoras theorem, we have<br \/>\nAC<sup>2<\/sup> = AD<sup>2<\/sup> + DC<sup>2<\/sup><br \/>\n\u21d2 AC<sup>2<\/sup> = AD<sup>2<\/sup> + (BC\u00a0\u2013\u00a0BD)<sup>2<\/sup><br \/>\n\u21d2 AC<sup>2<\/sup> = AD<sup>2<\/sup> + (BC<sup>2<\/sup>\u00a0+\u00a0BD<sup>2<\/sup>\u00a0\u2013 2BC \u2022 BD)<br \/>\n\u21d2 AC<sup>2<\/sup> = (AD<sup>2<\/sup> + BD<sup>2<\/sup>) +\u00a0BC<sup>2<\/sup>\u00a0\u2013 2BC \u2022 BD<br \/>\n\u21d2 AC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup>\u00a0\u2013 2BC \u2022 BD \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Using (i)]<br \/>\nHence, AC<sup>2<\/sup> = AB<sup>2<\/sup> + BC<sup>2<\/sup>\u00a0\u2013 2BC \u2022 BD<\/p>\n<p><strong>Example 6:<\/strong>\u00a0 \u00a0 \u00a0If ABC is an equilateral triangle of side a, prove that its altitude = \\(\\frac { \\sqrt { 3 } \u00a0}{ 2 } a\\).<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>\u2206ABD is an equilateral triangle.<br \/>\nWe are given that AB = BC = CA = a.<br \/>\nAD is the altitude, i.e., AD \u22a5\u00a0BC.<br \/>\nNow, in right angled triangles ABD and ACD, we have<br \/>\nAB = AC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(Given)<br \/>\nand AD = AD \u00a0 \u00a0 \u00a0 \u00a0 (Common side)<br \/>\n\u2206ABD \u2245\u00a0\u2206ACD \u00a0 \u00a0 (By RHS congruence)<br \/>\n\u21d2\u00a0BD = CD \u21d2\u00a0BD = DC = \\(\\frac { 1 }{ 2 }BC\\) = \\(\\frac { a }{ 2 }\\)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127207\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-7.png\" alt=\"Pythagoras Theorem 7\" width=\"218\" height=\"216\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-7.png 218w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-7-150x150.png 150w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-7-100x100.png 100w\" sizes=\"(max-width: 218px) 100vw, 218px\" \/><br \/>\nFrom right triangle ABD.<br \/>\nAB<sup>2<\/sup> = AD<sup>2<\/sup> + BD<sup>2<\/sup><br \/>\n\\(\\Rightarrow {{a}^{2}}=A{{D}^{2}}+{{\\left( \\frac{a}{2} \\right)}^{2}} \\)<br \/>\n\\(\\Rightarrow A{{D}^{2}}={{a}^{2}}-\\frac{{{a}^{2}}}{4}=\\frac{3}{4}{{a}^{2}}\\)<br \/>\n\\(\\Rightarrow AD=\\frac{\\sqrt{3}}{2}a \\)<\/p>\n<p><strong>Example 7:<\/strong>\u00a0 \u00a0 \u00a0ABC is a right-angled triangle, right-angled at A. A circle is inscribed in it. The lengths of the two sides containing the right angle are 5 cm and 12 cm. Find the radius of the circle.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>Given that \u2206ABC is right angled at A.<br \/>\nAC = 5 cm and AB = 12 cm<br \/>\nBC<sup>2<\/sup> = AC<sup>2<\/sup> + AB<sup>2<\/sup> = 25 + 144 = 169<br \/>\nBC = 13 cm<br \/>\nJoin OA, OB, OC<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127208\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-8.png\" alt=\"Pythagoras Theorem 8\" width=\"395\" height=\"303\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-8.png 395w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-8-300x230.png 300w\" sizes=\"(max-width: 395px) 100vw, 395px\" \/><br \/>\nLet the radius of the inscribed circle be r<br \/>\nArea of \u2206ABC = Area of \u2206OAB\u00a0+ Area of \u2206OBC + Area of \u2206OCA<br \/>\n\u21d2 1\/2 \u00d7 AB \u00d7 AC<br \/>\n\\(=\\frac{1}{2}\\left( 12\\text{ }\\times \\text{ }r \\right)\\text{ }+\\frac{1}{2}\\left( 13\\text{ }\\times \\text{ }r \\right)\\text{ }+\\frac{1}{2}\\left( 5\\text{ }\\times \\text{ }r \\right)\\)<br \/>\n\u21d2 \u00a012 \u00d7 5 = r \u00d7 {12 + 13 + 5}<br \/>\n\u21d2 \u00a060 = r \u00d7 30 \u21d2 \u00a0\u00a0r = 2 cm<\/p>\n<p><strong>Example 7:<\/strong>\u00a0 \u00a0 \u00a0ABCD is a rhombus. Prove that<br \/>\nAB<sup>2<\/sup> + BC<sup>2<\/sup> + CD<sup>2<\/sup> + DA<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup><br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>Let the diagonals AC and BD of rhombus ABCD intersect at O.<br \/>\nSince the diagonals of a rhombus bisect each other at right angles.<br \/>\n\u2234\u00a0\u2220AOB = \u2220BOC = \u2220COD = \u2220DOA = 90\u00ba<br \/>\nand AO = CO, BO = OD.<br \/>\nSince \u2206AOB is a right triangle right-angle at O.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127211\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-9.png\" alt=\"Pythagoras Theorem 9\" width=\"342\" height=\"253\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-9.png 342w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-9-300x222.png 300w\" sizes=\"(max-width: 342px) 100vw, 342px\" \/><br \/>\n\u2234\u00a0AB<sup>2<\/sup> = OA<sup>2<\/sup> + OB<sup>2<\/sup><br \/>\n\\(\\Rightarrow A{{B}^{2}}={{\\left( \\frac{1}{2}AC \\right)}^{2}}+{{\\left( \\frac{1}{2}BD \\right)}^{2}}\\) \u00a0 \u00a0 \u00a0 \u00a0 [\u2235 OA = OC and OB = OD]<br \/>\n\u21d2 4AB<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(i)<br \/>\nSimilarly, we have<br \/>\n4BC<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(ii)<br \/>\n4CD<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(iii)<br \/>\nand, 4AD<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(iv)<br \/>\nAdding all these results, we get<br \/>\n4(AB<sup>2<\/sup> + BC<sup>2<\/sup> + AD<sup>2<\/sup>) = 4(AC<sup>2<\/sup> + BD<sup>2<\/sup>)<br \/>\n\u21d2 AB<sup>2<\/sup> + BC<sup>2<\/sup> + AD<sup>2<\/sup> + DA<sup>2<\/sup> = AC<sup>2<\/sup> + BD<sup>2<\/sup><\/p>\n<p><strong>Example 8:<\/strong>\u00a0 \u00a0 \u00a0P and Q are the mid-points of the sides CA and CB respectively of a \u2206ABC, right angled at C. Prove that:<br \/>\n(i) 4AQ<sup>2<\/sup> = 4AC<sup>2<\/sup> + BC<sup>2<\/sup><br \/>\n(ii) 4BP<sup>2<\/sup> = 4BC<sup>2<\/sup> + AC<sup>2<\/sup><br \/>\n(iii) (4AQ<sup>2<\/sup> + BP<sup>2<\/sup>) = 5AB<sup>2<\/sup><br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127214\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-10.png\" alt=\"Pythagoras Theorem 10\" width=\"202\" height=\"216\" \/><br \/>\n<strong>(i)\u00a0 <\/strong>Since \u2206AQC is a right triangle right-angled at C.<br \/>\n\u2234\u00a0AQ<sup>2<\/sup> = AC<sup>2<\/sup> + QC<sup>2<\/sup><br \/>\n\u21d2 4AQ<sup>2<\/sup> = 4AC<sup>2<\/sup> + 4QC<sup>2 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/sup>[Multiplying both sides by 4]<br \/>\n\u21d2 4AQ<sup>2<\/sup> = 4AC<sup>2<\/sup> + (2QC)<sup>2<\/sup><br \/>\n\u21d2 4AQ<sup>2<\/sup> = 4AC<sup>2<\/sup> + BC<sup>2<\/sup> [\u2235\u00a0 BC = 2QC]<br \/>\n<strong>(ii) \u00a0<\/strong>Since \u2206BPC is a right triangle right-angled at C.<br \/>\n\u2234\u00a0BP<sup>2<\/sup> = BC<sup>2<\/sup> + CP<sup>2<\/sup><br \/>\n\u21d2 4BP<sup>2<\/sup> = 4BC<sup>2<\/sup> + 4CP<sup>2 \u00a0 \u00a0 \u00a0<\/sup>[Multiplying both sides by 4]<br \/>\n\u21d2 4BP<sup>2<\/sup> = 4BC<sup>2<\/sup> + (2CP)<sup>2<\/sup><br \/>\n\u21d2 4BP<sup>2<\/sup> = 4BC<sup>2<\/sup> + AC<sup>2<\/sup> [\u2235\u00a0 AC = 2CP]<br \/>\n<strong>(iii) \u00a0<\/strong>From (i) and (ii), we have<br \/>\n4AQ<sup>2<\/sup> = 4AC<sup>2<\/sup> + BC<sup>2<\/sup> and, 4BC<sup>2<\/sup> = 4BC<sup>2<\/sup> + AC<sup>2<\/sup><br \/>\n\u2234 4AQ<sup>2<\/sup> + 4BP<sup>2<\/sup> = (4AC<sup>2<\/sup> + BC<sup>2<\/sup>) + (4BC<sup>2<\/sup> + AC<sup>2<\/sup>)<br \/>\n\u21d2 4(AQ<sup>2<\/sup> + BP<sup>2<\/sup>) = 5 (AC<sup>2<\/sup> + BC<sup>2<\/sup>)<br \/>\n\u21d2 4(AQ<sup>2<\/sup> + BP<sup>2<\/sup>) = 5 AB<sup>2<\/sup><br \/>\n[In \u2206ABC, we have AB<sup>2<\/sup> = AC<sup>2<\/sup> + BC<sup>2<\/sup>]<\/p>\n<p><strong>Example 9:<\/strong>\u00a0 \u00a0 \u00a0From a point O in the interior of a \u2206ABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that :<br \/>\n(i) AF<sup>2<\/sup> + BD<sup>2<\/sup> + CE<sup>2<\/sup> = OA<sup>2<\/sup> + OB<sup>2<\/sup> + OC<sup>2<\/sup> \u2013 OD<sup>2<\/sup> \u2013 OE<sup>2<\/sup> \u2013 OF<sup>2<\/sup><br \/>\n(ii) AF<sup>2<\/sup> + BD<sup>2<\/sup> + CE<sup>2<\/sup> = AE<sup>2<\/sup> + CD<sup>2<\/sup> + BF<sup>2<\/sup><br \/>\n<strong>Sol. \u00a0 \u00a0\u00a0<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127217\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-11.png\" alt=\"Pythagoras Theorem 11\" width=\"216\" height=\"230\" \/><br \/>\nLet O be a point in the interior of \u2206ABC and let OD \u22a5\u00a0BC, OE \u22a5\u00a0CA and OF \u22a5\u00a0AB.<br \/>\n<strong>(i)<\/strong>\u00a0 In right triangles \u2206OFA, \u2206ODB and \u2206OEC, we have<br \/>\nOA<sup>2<\/sup> = AF<sup>2<\/sup> + OF<sup>2<\/sup><br \/>\nOB<sup>2<\/sup> = BD<sup>2<\/sup> + OD<sup>2<\/sup><br \/>\nand, OC<sup>2<\/sup> = CE<sup>2<\/sup> + OE<sup>2<\/sup><br \/>\nAdding all these results, we get<br \/>\nOA<sup>2<\/sup> + OB<sup>2<\/sup> + OC<sup>2<\/sup> = AF<sup>2<\/sup> + BD<sup>2<\/sup> + CE<sup>2<\/sup> + OF<sup>2<\/sup> + OD<sup>2<\/sup> + OE<sup>2<\/sup><br \/>\n\u21d2 AF<sup>2<\/sup> + BD<sup>2<\/sup> + CE<sup>2<\/sup> = OA<sup>2<\/sup> + OB<sup>2<\/sup> + OC<sup>2<\/sup> \u2013 OD<sup>2<\/sup> \u2013 OE<sup>2<\/sup> \u2013 OF<sup>2<\/sup><br \/>\n<strong>(ii)<\/strong>\u00a0 In right triangles \u2206ODB and \u2206ODC, we have<br \/>\nOB<sup>2<\/sup> = OD<sup>2<\/sup> + BD<sup>2<\/sup><br \/>\nand, OC<sup>2<\/sup> = OD<sup>2<\/sup> + CD<sup>2<\/sup><br \/>\nOB<sup>2<\/sup> \u2013 OC<sup>2<\/sup> = (OD<sup>2<\/sup> + BD<sup>2<\/sup>) \u2013 (OD<sup>2<\/sup> + CD<sup>2<\/sup>)<br \/>\n\u21d2 OB<sup>2<\/sup> \u2013 OC<sup>2<\/sup> = BD<sup>2<\/sup> \u2013 CD<sup>2 \u00a0 \u00a0 \u00a0\u00a0<\/sup> &#8230;.(i)<br \/>\nSimilarity, we have<br \/>\nOC<sup>2<\/sup> \u2013 OA<sup>2<\/sup> = CE<sup>2<\/sup> \u2013 AE<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(ii)<br \/>\nand, OA<sup>2<\/sup> \u2013 OB<sup>2<\/sup> = AF<sup>2<\/sup> \u2013 BF<sup>2<\/sup> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(iii)<br \/>\nAdding (i), (ii) and (iii), we get<br \/>\n(OB<sup>2<\/sup> \u2013 OC<sup>2<\/sup>) + (OC<sup>2<\/sup> \u2013 OA<sup>2<\/sup>) + (OA<sup>2<\/sup> \u2013 OB<sup>2<\/sup>)<br \/>\n= (BD<sup>2<\/sup> \u2013 CD<sup>2<\/sup>) + (CE<sup>2<\/sup> \u2013 AE<sup>2<\/sup>) + (AF<sup>2<\/sup> \u2013 BF<sup>2<\/sup>)<br \/>\n\u21d2 (BD<sup>2<\/sup> + CE<sup>2<\/sup> + AF<sup>2<\/sup>) \u2013 (AE<sup>2<\/sup> + CD<sup>2<\/sup> + BF<sup>2<\/sup>) = 0<br \/>\n\u21d2 AF<sup>2<\/sup> + BD<sup>2<\/sup> + CE<sup>2<\/sup> = AE<sup>2<\/sup> + CD<sup>2<\/sup> + BF<sup>2<\/sup><\/p>\n<p><strong>Example 10:<\/strong>\u00a0 \u00a0 \u00a0In a right triangle ABC right-angled at C, P and Q are the points on the sides CA and CB respectively, which divide these sides in the ratio 2 : 1. Prove that<br \/>\n(i) 9 AQ<sup>2<\/sup> = 9 AC<sup>2<\/sup> + 4 BC<sup>2<\/sup><br \/>\n(ii) 9 BP<sup>2<\/sup> = 9 BC<sup>2<\/sup> + 4 AC<sup>2<\/sup><br \/>\n(iii) 9 (AQ<sup>2<\/sup> + BP<sup>2<\/sup>) = 13 AB<sup>2<\/sup><br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>It is given that P divides CA in the ratio 2 : 1. Therefore,<br \/>\n\\(CP=\\frac { 2 }{ 3 }AC\\)\u00a0 \u00a0 \u00a0 &#8230;.(i)<br \/>\nAlso, Q divides CB in the ratio 2 : 1.<br \/>\n\u2234 \\(QC=\\frac { 2 }{ 3 }BC\\) \u00a0 \u00a0 \u00a0&#8230;.(ii)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127220\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-12.png\" alt=\"Pythagoras Theorem 12\" width=\"183\" height=\"185\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-12.png 183w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-12-100x100.png 100w\" sizes=\"(max-width: 183px) 100vw, 183px\" \/><br \/>\n<strong>(i)<\/strong>\u00a0 Applying pythagoras theorem in right-angled triangle ACQ, we have<br \/>\nAQ<sup>2<\/sup> = QC<sup>2<\/sup> + AC<sup>2<\/sup><br \/>\n\u21d2 AQ<sup>2<\/sup> = \\(\\frac { 4 }{ 9 }\\) BC<sup>2<\/sup> + AC<sup>2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/sup> [Using (ii)]<br \/>\n\u21d2 9 AQ<sup>2<\/sup> = 4\u00a0BC<sup>2<\/sup> + 9 AC<sup>2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/sup> &#8230;.(iii)<br \/>\n<strong>(ii) \u00a0<\/strong>Applying pythagoras theorem in right triangle BCP, we have<br \/>\nBP<sup>2<\/sup> = BC<sup>2<\/sup> + CP<sup>2<\/sup><br \/>\n\u21d2 BP<sup>2<\/sup> = BC<sup>2<\/sup> + AC<sup>2<\/sup> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Using (i)]<br \/>\n\u21d2 9 BP<sup>2<\/sup> = 9 BC<sup>2<\/sup> + 4 AC<sup>2<\/sup> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(iv)<br \/>\n<strong>(iii) \u00a0<\/strong>Adding (iii) and (iv), we get<br \/>\n9 (AQ<sup>2<\/sup> + BP<sup>2<\/sup>) = 13 (BC<sup>2<\/sup> + AC<sup>2<\/sup>)<br \/>\n\u21d2 9 (AQ<sup>2<\/sup> + BP<sup>2<\/sup>) = 13 AB<sup>2<\/sup> [\u2235\u00a0 BC<sup>2<\/sup> = AC<sup>2<\/sup> + AB<sup>2<\/sup>]<\/p>\n<p><strong>Example 11:<\/strong>\u00a0 \u00a0 \u00a0In a \u2206ABC, AD \u22a5\u00a0BC and AD<sup>2<\/sup> = BC \u00d7 CD. Prove \u2206ABC is a right triangle.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127279\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-13.png\" alt=\"Pythagoras Theorem 13\" width=\"245\" height=\"219\" \/><br \/>\nIn right triangles ADB and ADC, we have<br \/>\nAB<sup>2<\/sup> = AD<sup>2<\/sup> + BD<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(i)<br \/>\nand, AC<sup>2<\/sup> = AD<sup>2<\/sup> + DC<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(ii)<br \/>\nAdding (i) and (ii), we get<br \/>\nAB<sup>2<\/sup> + AC<sup>2<\/sup> = 2 AD<sup>2<\/sup> \u00d7 BD<sup>2<\/sup> + DC<sup>2<\/sup><br \/>\n\u21d2 AB<sup>2<\/sup> + AC<sup>2<\/sup> = 2BD \u00d7 CD + BD<sup>2<\/sup> + DC<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0[\u2235\u00a0 AD<sup>2<\/sup> = BD \u00d7 CD (Given)]<br \/>\n\u21d2 AB<sup>2<\/sup> + AC<sup>2<\/sup> = (BD + CD)<sup>2<\/sup> = BC<sup>2<\/sup><br \/>\nThus, in \u2206ABC, we have<br \/>\nAB<sup>2<\/sup> = AC<sup>2<\/sup> + BC<sup>2<\/sup><br \/>\nHence, \u2206ABC, is a right triangle right-angled at A.<\/p>\n<p><strong>Example 12:<\/strong>\u00a0 \u00a0 \u00a0 The perpendicular AD on the base BC of a \u2206ABC intersects BC at D so that DB = 3 CD.\u00a0Prove that 2AB<sup>2<\/sup> = 2AC<sup>2<\/sup> + BC<sup>2<\/sup>.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>We have,<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127280\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-14.png\" alt=\"Pythagoras Theorem 14\" width=\"216\" height=\"215\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-14.png 216w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-14-150x150.png 150w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-14-100x100.png 100w\" sizes=\"(max-width: 216px) 100vw, 216px\" \/><br \/>\nDB = 3CD<br \/>\nBC = BD + DC<br \/>\nThe perpendicular AD on the base BC of a \u2206ABC intersects BC at D so that DB = 3 CD. Prove that<br \/>\n2AC2 + BC2.<br \/>\nWe have,<br \/>\nDB = 3CD<br \/>\n\u2234\u00a0BC = BD + DC<br \/>\n\u21d2 BC = 3 CD + CD<br \/>\n\u21d2 BD = 4 CD \u21d2 \u00a0CD = \\(\\frac { 1 }{ 4 }\\)\u00a0BC<br \/>\n\u2234\u00a0CD = \\(\\frac { 1 }{ 4 }\\) BC\u00a0 and BD = 3CD = \\(\\frac { 1 }{ 4 }\\)\u00a0BC\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(i)<br \/>\nSince \u2206ABD is a right triangle right-angled at D.<br \/>\n\u2234\u00a0AB<sup>2<\/sup> = AD<sup>2<\/sup> + BD<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(ii)<br \/>\nSimilarly, \u2206ACD is a right triangle right angled at D.<br \/>\n\u2234\u00a0AC<sup>2<\/sup> = AD<sup>2<\/sup> + CD<sup>2<\/sup> \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(iii)<br \/>\nSubtracting equation (iii) from equation (ii) \u00a0we get<br \/>\nAB<sup>2<\/sup> \u2013 AC<sup>2<\/sup>\u00a0= BD<sup>2<\/sup> \u2013 CD<sup>2<\/sup><br \/>\n\u21d2 AB<sup>2<\/sup> \u2013 AC<sup>2<\/sup> = \\({{\\left( \\frac{3}{4}BC \\right)}^{2}}-{{\\left( \\frac{1}{4}BC \\right)}^{2}}\\)\u00a0 \u00a0 \u00a0 \u00a0\\(\\left[ From\\ (i)\\ CD=\\frac{1}{4}BC,\\ BD=\\frac{3}{4}BC \\right]\\)<br \/>\n\u21d2 AB<sup>2<\/sup> \u2013 AC<sup>2<\/sup> = \\(\\frac { 9 }{ 16 }\\) BC<sup>2<\/sup> \u2013 \\(\\frac { 1 }{ 16 }\\) BC<sup>2<\/sup><br \/>\n\u21d2 AB<sup>2<\/sup> \u2013 AC<sup>2<\/sup> = \\(\\frac { 1 }{ 2 }\\) BC<sup>2<\/sup><br \/>\n\u21d2 2(AB<sup>2<\/sup> \u2013 AC<sup>2<\/sup>) = BC<sup>2<\/sup><br \/>\n\u21d2 2AB<sup>2<\/sup> = 2AC<sup>2<\/sup> + BC<sup>2<\/sup>.<\/p>\n<p><strong>Example 13:<\/strong>\u00a0 \u00a0 \u00a0 ABC is a right triangle right-angled at C. Let BC = a, CA = b, AB = c and let p be the length of perpendicular from C on AB, prove that<br \/>\n(i) cp = ab<br \/>\n(ii)\u00a0\\(\\frac{1}{{{p}^{2}}}=\\frac{1}{{{a}^{2}}}+\\frac{1}{{{b}^{2}}}\\)<br \/>\n<strong>Sol. \u00a0\u00a0 (i) \u00a0<\/strong>Let CD \u22a5\u00a0AB. Then, CD = p.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-127282\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-14-1.png\" alt=\"Pythagoras Theorem 14\" width=\"216\" height=\"215\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-14-1.png 216w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-14-1-150x150.png 150w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2017\/04\/Pythagoras-Theorem-14-1-100x100.png 100w\" sizes=\"(max-width: 216px) 100vw, 216px\" \/><br \/>\n\u2234\u00a0Area of \u2206ABC = \\(\\frac { 1 }{ 2 }\\)\u00a0(Base \u00d7 Height)<br \/>\n\u21d2 \u00a0Area of \u2206ABC = \\(\\frac { 1 }{ 2 }\\)\u00a0(AB \u00d7 CD) = \u00a0\\(\\frac { 1 }{ 2 }\\) cp<br \/>\nAlso,<br \/>\nArea of \u2206ABC = \\(\\frac { 1 }{ 2 }\\) (BC \u00d7 AC) = \\(\\frac { 1 }{ 2 }\\) ab<br \/>\n\u2234\u00a0\\(\\frac { 1 }{ 2 }\\) cp = \\(\\frac { 1 }{ 2 }\\) ab<br \/>\n\u21d2 \u00a0cp = ab<br \/>\n<strong>(ii) \u00a0<\/strong>Since \u2206ABC is right triangle right-angled at C.<br \/>\n\u2234\u00a0AB<sup>2<\/sup>\u00a0= BC<sup>2<\/sup>\u00a0+ AC<sup>2<\/sup><br \/>\n\u21d2 \u00a0c<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup><br \/>\n\\(\\Rightarrow {{\\left( \\frac{ab}{p} \\right)}^{2}}={{a}^{2}}+{{b}^{2}} \\) \u00a0 \u00a0 \u00a0 \u00a0\u00a0\\(\\left[ \\because \\ cp\\,=\\,ab\\ \\ \\therefore c=\\frac{ab}{p} \\right]\\)<br \/>\n\\(\\Rightarrow \\frac{{{a}^{2}}{{b}^{2}}}{{{p}^{2}}}={{a}^{2}}+{{b}^{2}} \\)<br \/>\n\\(\\Rightarrow \\frac{1}{{{p}^{2}}}=\\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}{{b}^{2}}}\\Rightarrow \\frac{1}{{{p}^{2}}}=\\frac{1}{{{b}^{2}}}+\\frac{1}{{{a}^{2}}} \\)<br \/>\n\\(\\Rightarrow \\frac{1}{{{p}^{2}}}=\\frac{1}{{{a}^{2}}}+\\frac{1}{{{b}^{2}}} \\)<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Pythagoras Theorem Theorem 1: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Given: A right-angled triangle ABC in which B = \u222090\u00ba. To Prove: (Hypotenuse)2 = (Base)2 + (Perpendicular)2. i.e., AC2 = AB2 + BC2 Construction: From B draw BD &#8230; <a title=\"Pythagoras Theorem\" class=\"read-more\" href=\"https:\/\/cbselibrary.com\/pythagoras-theorem\/\" aria-label=\"More on Pythagoras Theorem\">Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[73,72,71],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v18.4.1) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Pythagoras Theorem - CBSE Library<\/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:\/\/cbselibrary.com\/pythagoras-theorem\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Pythagoras Theorem\" \/>\n<meta property=\"og:description\" content=\"Pythagoras Theorem Theorem 1: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. 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