{"id":6053,"date":"2023-08-17T03:01:34","date_gmt":"2023-08-17T03:01:34","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6053"},"modified":"2023-08-17T03:01:34","modified_gmt":"2023-08-17T03:01:34","slug":"ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-5","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-5\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 6 Triangles Ex 6.5"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"font-family: Georgia, Palatino;\"><strong><span style=\"color: #000000;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><\/span><br \/>\n<span style=\"font-family: Georgia, Palatino;\"><strong><span style=\"color: #000000;\">Chapter &#8211; 6 (Triangles)\u00a0<\/span><\/strong><\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 6 Triangles <\/strong>Exercise 6.5 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter : 6 Triangles<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.3<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-4\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.4<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-6\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.6<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"font-family: Georgia, Palatino;\"><strong><span style=\"color: #000000;\">Exercise &#8211; 6.5<\/span><\/strong><\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. \u00a0Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.<br \/>\n<\/strong>(i) 7 cm, 24 cm, 25 cm<br \/>\n(ii) 3 cm, 8 cm, 6 cm<br \/>\n(iii) 50 cm, 80 cm, 100 cm<br \/>\n(iv) 13 cm, 12 cm, 5 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.<br \/>\n<\/strong>Squaring the lengths of the sides of the, we will get 49, 576, and 625.<br \/>\n49 + 576 = 625<br \/>\n(7)<sup>2<\/sup>\u00a0+ (24)<sup>2<\/sup>\u00a0= (25)<sup>2<br \/>\n<\/sup>Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is <strong>right angled triangle<\/strong>.<br \/>\nLength of Hypotenuse = 25 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.<br \/>\n<\/strong>Squaring the lengths of these sides, we will get 9, 64, and 36.<br \/>\nClearly, 9 + 36 \u2260 64<br \/>\nOr, 3<sup>2<\/sup>\u00a0+ 6<sup>2<\/sup>\u00a0\u2260 8<sup>2<br \/>\n<\/sup>Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.<br \/>\nHence, the given triangle does <strong>not satisfies Pythagoras theorem<\/strong>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii)\u00a0Given, sides of triangle\u2019s are 50 cm, 80 cm, and 100 cm.<br \/>\n<\/strong>Squaring the lengths of these sides, we will get 2500, 6400, and 10000.<br \/>\nHowever, 2500 + 6400 \u2260 10000<br \/>\nOr, 50<sup>2<\/sup>\u00a0+ 80<sup>2<\/sup>\u00a0\u2260 100<sup>2<br \/>\n<\/sup>As you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.<br \/>\nTherefore, the given triangle does <strong>not satisfies Pythagoras theorem<\/strong>.<br \/>\nHence, it is not a right triangle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv) Given, sides are 13 cm, 12 cm, and 5 cm.<br \/>\n<\/strong>Squaring the lengths of these sides, we will get 169, 144, and 25.<br \/>\nThus, 144 +25 = 169<br \/>\nOr, 12<sup>2<\/sup>\u00a0+ 5<sup>2<\/sup>\u00a0= 13<sup>2<br \/>\n<\/sup>The sides of the given triangle are satisfying Pythagoras theorem.<br \/>\nTherefore, it is a <strong>right triangle.<\/strong><br \/>\nHence, length of the hypotenuse of this triangle is 13 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. PQR is a triangle right angled at P and M is a point on QR such that PM \u22a5 QR. Show that PM<sup>2<\/sup> = QM\u00a0\u00d7\u00a0MR.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong><strong>Given :<\/strong>\u00a0\u0394PQR is right angled at P is a point on QR such that PM\u00a0\u22a5QR<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6385\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"112\" \/><br \/>\nWe have to prove, PM<sup>2<\/sup>\u00a0= QM\u00a0\u00d7\u00a0MR<br \/>\nIn \u0394PQM, by Pythagoras theorem<br \/>\nPQ<sup>2<\/sup>\u00a0= PM<sup>2<\/sup>\u00a0+ QM<sup>2<br \/>\n<\/sup>Or, PM<sup>2<\/sup>\u00a0= PQ<sup>2<\/sup>\u00a0\u2013 QM<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<strong><br \/>\n<\/strong>In \u0394PMR, by Pythagoras theorem<br \/>\nPR<sup>2<\/sup>\u00a0= PM<sup>2<\/sup>\u00a0+ MR<sup>2<br \/>\n<\/sup>Or,<br \/>\nPM<sup>2<\/sup>\u00a0= PR<sup>2<\/sup>\u00a0\u2013 MR<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (ii)<br \/>\nAdding\u00a0equation,\u00a0(i)\u00a0and\u00a0(ii), we get,<br \/>\n2PM<sup>2<\/sup>\u00a0=\u00a0(PQ<sup>2<\/sup>\u00a0+ PM<sup>2<\/sup>) \u2013 (QM<sup>2<\/sup>\u00a0+ MR<sup>2<\/sup>)<br \/>\n= QR<sup>2<\/sup>\u00a0\u2013 QM<sup>2<\/sup>\u00a0\u2013 MR<sup>2 \u00a0 \u00a0 \u00a0\u00a0<\/sup>\u00a0[\u2234 QR<sup>2<\/sup>\u00a0= PQ<sup>2<\/sup>\u00a0+ PR<sup>2<\/sup>]<br \/>\n= (QM\u00a0+ MR)<sup>2<\/sup>\u00a0\u2013 QM<sup>2<\/sup>\u00a0\u2013 MR<sup>2<br \/>\n<\/sup>= 2QM \u00d7 MR<br \/>\n\u2234 PM<sup>2<\/sup>\u00a0= QM\u00a0\u00d7\u00a0MR<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. In Figure, ABD is a triangle right angled at A\u00a0and AC \u22a5 BD. Show that<br \/>\n<\/strong>(i) AB<sup>2<\/sup>\u00a0= BC \u00d7\u00a0BD<br \/>\n(ii) AC<sup>2<\/sup>\u00a0= BC \u00d7\u00a0DC<br \/>\n(iii) AD<sup>2<\/sup>\u00a0= BD\u00a0\u00d7\u00a0CD<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6386\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Que3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"193\" height=\"200\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>(i) In \u0394ADB and \u0394CAB,<br \/>\n<\/strong>\u2220DAB = \u2220ACB (Each 90\u00b0)<br \/>\n\u2220ABD = \u2220CBA (Common angles)<br \/>\n\u2234 \u0394ADB ~ \u0394CAB [AA similarity criterion]<br \/>\n\u21d2 AB\/CB = BD\/AB<br \/>\n\u21d2 AB<sup>2<\/sup>\u00a0= CB \u00d7\u00a0BD<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) Let \u2220CAB = x<br \/>\n<\/strong>In \u0394CBA,<br \/>\n\u2220CBA = 180\u00b0 \u2013 90\u00b0 \u2013 x<br \/>\n\u2220CBA = 90\u00b0 \u2013 x<br \/>\nSimilarly, in \u0394CAD<br \/>\n\u2220CAD = 90\u00b0 \u2013 \u2220CBA<br \/>\n= 90\u00b0 \u2013<em>\u00a0<\/em>x<br \/>\n\u2220CDA = 180\u00b0 \u2013 90\u00b0 \u2013 (90\u00b0 \u2013\u00a0x)<br \/>\n\u2220CDA =\u00a0x<br \/>\nIn \u0394CBA and \u0394CAD, we have<br \/>\n\u2220CBA = \u2220CAD<br \/>\n\u2220CAB = \u2220CDA<br \/>\n\u2220ACB = \u2220DCA (Each 90\u00b0)<br \/>\n\u2234 \u0394CBA ~ \u0394CAD [AAA similarity criterion]<br \/>\n\u21d2 AC\/DC = BC\/AC<br \/>\n\u21d2 AC<sup>2<\/sup>\u00a0= \u00a0DC \u00d7 BC<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) In \u0394DCA and \u0394DAB,<br \/>\n<\/strong>\u2220DCA = \u2220DAB (Each 90\u00b0)<br \/>\n\u2220CDA = \u2220ADB (common angles)<br \/>\n\u2234 \u0394DCA ~ \u0394DAB [AA similarity criterion]<br \/>\n\u21d2 DC\/DA = DA\/DA<br \/>\n\u21d2 AD<sup>2<\/sup>\u00a0= BD\u00a0\u00d7\u00a0CD<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. ABC is an isosceles triangle right angled at C. Prove that AB<sup>2<\/sup>\u00a0= 2AC<sup>2<\/sup>\u00a0.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given :<\/strong>\u00a0\u0394ABC is an isosceles triangle right angled at C.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6387\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"192\" height=\"195\" \/><br \/>\nIn \u0394ACB, \u2220C = 90\u00b0<br \/>\nAC = BC (By isosceles triangle property)<br \/>\nAB<sup>2<\/sup>\u00a0= AC<sup>2<\/sup>\u00a0+ BC<sup>2<\/sup>\u00a0[By Pythagoras theorem]<br \/>\n= AC<sup>2<\/sup>\u00a0+\u00a0AC<sup>2<\/sup>\u00a0[Since, AC = BC]<br \/>\nAB<sup>2<\/sup>\u00a0= 2AC<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5.\u00a0ABC is an isosceles triangle with AC = BC. If\u00a0AB<sup>2<\/sup>\u00a0= 2AC<sup>2<\/sup>, prove that ABC is a right triangle.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given <\/strong>:\u00a0 \u0394ABC is an isosceles triangle having AC = BC and AB<sup>2<\/sup>\u00a0= 2AC<sup>2<br \/>\n<\/sup><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6387\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"192\" height=\"195\" \/><br \/>\nIn \u0394ACB,<br \/>\nAC = BC<br \/>\nAB<sup>2<\/sup>\u00a0= 2AC<sup>2<br \/>\n<\/sup>AB<sup>2<\/sup>\u00a0= AC<sup>2\u00a0<\/sup>+ AC<sup>2<br \/>\n<\/sup>= AC<sup>2<\/sup>\u00a0+ BC<sup>2\u00a0<\/sup>[Since, AC = BC]<br \/>\nHence, by Pythagoras theorem\u00a0\u0394ABC is right angle triangle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6.\u00a0ABC is an equilateral triangle of side 2a. Find each of its altitudes<\/strong>.<br \/>\n<strong>Solution &#8211;<br \/>\n<\/strong><strong>Given :<\/strong>\u00a0ABC is an equilateral triangle of side 2a.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6388\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"205\" height=\"241\" \/><br \/>\nDraw, AD \u22a5 BC<br \/>\nIn \u0394ADB and \u0394ADC,<br \/>\nAB = AC<br \/>\nAD = AD<br \/>\n\u2220ADB = \u2220ADC [Both are 90\u00b0]<br \/>\nTherefore, \u0394ADB \u2245 \u0394ADC by RHS congruence.<br \/>\nHence, BD = DC [by CPCT]<br \/>\nIn right angled \u0394ADB,<br \/>\nAB<sup>2<\/sup>\u00a0= AD<sup>2\u00a0<\/sup>+ BD<sup>2<br \/>\n<\/sup>(2<em>a<\/em>)<sup>2<\/sup>\u00a0= AD<sup>2\u00a0<\/sup>+\u00a0<em>a<\/em><sup>2<br \/>\n<\/sup>\u21d2\u00a0AD<sup>2 <\/sup>=\u00a04<em>a<\/em><sup>2<\/sup>\u00a0\u2013\u00a0<em>a<\/em><sup>2<br \/>\n<\/sup>\u21d2\u00a0AD<sup>2 =<\/sup>\u00a03<em>a<\/em><sup>2<br \/>\n<\/sup>\u21d2\u00a0AD<sup>\u00a0=<\/sup>\u00a0\u221a3a<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals. <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given :<\/strong>\u00a0ABCD is a rhombus whose diagonals AC and BD intersect at O.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6389\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans7.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"264\" height=\"201\" \/><br \/>\nWe have to prove, as per the question,<br \/>\nAB<sup>2\u00a0<\/sup>+ BC<sup>2\u00a0<\/sup>+ CD<sup>2<\/sup>\u00a0+ AD<sup>2\u00a0<\/sup>= AC<sup>2\u00a0<\/sup>+ BD<sup>2<br \/>\n<\/sup>Since, the diagonals of a rhombus bisect each other at right angles.<br \/>\nTherefore, AO = CO and BO = DO<br \/>\nIn \u0394AOB,<br \/>\n\u2220AOB =\u00a090\u00b0<br \/>\nAB<sup>2<\/sup>\u00a0= AO<sup>2\u00a0<\/sup>+ BO<sup>2\u00a0<\/sup>\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (i)\u00a0[By Pythagoras theorem]<br \/>\nSimilarly,<br \/>\nAD<sup>2<\/sup>\u00a0= AO<sup>2\u00a0<\/sup>+ DO<sup>2\u00a0 \u00a0\u00a0<\/sup>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (ii)<br \/>\nDC<sup>2<\/sup>\u00a0= DO<sup>2\u00a0<\/sup>+ CO<sup>2\u00a0 \u00a0 <\/sup>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (iii)<br \/>\nBC<sup>2<\/sup>\u00a0= CO<sup>2\u00a0<\/sup>+ BO<sup>2\u00a0\u00a0<\/sup>\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (iv)<br \/>\nAdding equations\u00a0(i) + (ii)\u00a0+ (iii)\u00a0+ (iv), we get,<br \/>\nAB<sup>2\u00a0<\/sup>+ AD<sup>2\u00a0<\/sup>+<sup>\u00a0<\/sup>DC<sup>2\u00a0<\/sup>+<sup>\u00a0<\/sup>BC<sup>2<\/sup>\u00a0= 2(AO<sup>2\u00a0<\/sup>+ BO<sup>2\u00a0<\/sup>+ DO<sup>2\u00a0<\/sup>+ CO<sup>2<\/sup>)<br \/>\n= 4AO<sup>2\u00a0<\/sup>+ 4BO<sup>2\u00a0<\/sup>[Since, AO = CO and BO =DO]<br \/>\n= (2AO)<sup>2\u00a0<\/sup>+ (2BO)<sup>2<\/sup>\u00a0= AC<sup>2\u00a0<\/sup>+ BD<sup>2<br \/>\n<\/sup>AB<sup>2\u00a0<\/sup>+ AD<sup>2\u00a0<\/sup>+<sup>\u00a0<\/sup>DC<sup>2\u00a0<\/sup>+<sup>\u00a0<\/sup>BC<sup>2<\/sup>\u00a0= AC<sup>2\u00a0<\/sup>+ BD<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>8. In Fig. 6.54, O is a point in the interior of a triangle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6390\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Que8.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"151\" height=\"200\" \/><br \/>\n<\/strong><strong>ABC, OD \u22a5 BC, OE\u00a0\u22a5\u00a0AC and OF \u22a5 AB. Show that:<br \/>\n<\/strong>(i) OA<sup>2<\/sup>\u00a0+ OB<sup>2<\/sup>\u00a0+ OC<sup>2<\/sup>\u00a0\u2013 OD<sup>2<\/sup>\u00a0\u2013 OE<sup>2<\/sup>\u00a0\u2013 OF<sup>2<\/sup>\u00a0= AF<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0+ CE<sup>2<\/sup>\u00a0,<br \/>\n(ii) AF<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0+ CE<sup>2<\/sup>\u00a0= AE<sup>2<\/sup>\u00a0+ CD<sup>2<\/sup>\u00a0+ BF<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given :<\/strong>\u00a0 in \u0394ABC, O is a point in the interior of a triangle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6391\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans9.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"172\" \/><br \/>\nAnd OD \u22a5 BC, OE \u22a5 AC and OF \u22a5 AB.<br \/>\nJoin OA, OB and OC<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) By Pythagoras theorem in \u0394AOF, we have<br \/>\n<\/strong>OA<sup>2<\/sup>\u00a0= OF<sup>2<\/sup>\u00a0+\u00a0AF<sup>2<br \/>\n<\/sup>Similarly, in\u00a0\u0394BOD<br \/>\nOB<sup>2<\/sup>\u00a0= OD<sup>2<\/sup>\u00a0+ BD<sup>2<br \/>\n<\/sup>Similarly,\u00a0in\u00a0\u0394COE<br \/>\nOC<sup>2<\/sup>\u00a0= OE<sup>2<\/sup>\u00a0+ EC<sup>2<br \/>\n<\/sup>Adding these equations,<br \/>\nOA<sup>2<\/sup>\u00a0+ OB<sup>2<\/sup>\u00a0+ OC<sup>2<\/sup>\u00a0= OF<sup>2<\/sup>\u00a0+ AF<sup>2<\/sup>\u00a0+ OD<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0+ OE<sup>2\u00a0<\/sup>+ EC<sup>2<br \/>\n<\/sup>OA<sup>2<\/sup>\u00a0+ OB<sup>2<\/sup>\u00a0+ OC<sup>2<\/sup>\u00a0\u2013 OD<sup>2<\/sup>\u00a0\u2013 OE<sup>2<\/sup>\u00a0\u2013 OF<sup>2<\/sup>\u00a0= AF<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0+ CE<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) AF<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0+ EC<sup>2<\/sup>\u00a0= (OA<sup>2<\/sup>\u00a0\u2013 OE<sup>2<\/sup>)\u00a0+ (OC<sup>2<\/sup>\u00a0\u2013 OD<sup>2<\/sup>)\u00a0+ (OB<sup>2<\/sup>\u00a0\u2013 OF<sup>2<\/sup>)<br \/>\n<\/strong>\u2234 AF<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0+ CE<sup>2<\/sup>\u00a0= AE<sup>2<\/sup>\u00a0+ CD<sup>2<\/sup>\u00a0+ BF<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given:<\/strong> a ladder 10 m long reaches a window 8 m above the ground.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6392\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans09.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"178\" \/><br \/>\nLet BA be the wall and AC be the ladder,<br \/>\nTherefore, by Pythagoras theorem,<br \/>\nAC<sup>2<\/sup>\u00a0=<sup>\u00a0<\/sup>AB<sup>2<\/sup>\u00a0+ BC<sup>2<br \/>\n<\/sup>10<sup>2<\/sup>\u00a0= 8<sup>2<\/sup>\u00a0+ BC<sup>2<br \/>\n<\/sup>BC<sup>2\u00a0<\/sup>= 100 \u2013 64<br \/>\nBC<sup>2\u00a0<\/sup>= 36<br \/>\nBC<sup>\u00a0<\/sup>= 6m<br \/>\nTherefore, the distance of the foot of the ladder from the base of the wall is\u00a06 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong><br \/>\n<strong>Given :<\/strong>\u00a0a guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6393\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans10.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"194\" \/><br \/>\nLet AB be the pole and AC be the wire.<br \/>\nBy Pythagoras theorem,<br \/>\nAC<sup>2<\/sup>\u00a0=<sup>\u00a0<\/sup>AB<sup>2<\/sup>\u00a0+ BC<sup>2<br \/>\n<\/sup>24<sup>2<\/sup>\u00a0= 18<sup>2<\/sup>\u00a0+ BC<sup>2<br \/>\n<\/sup>BC<sup>2\u00a0<\/sup>= 576 \u2013 324<br \/>\nBC<sup>2\u00a0<\/sup>= 252<br \/>\nBC<sup>\u00a0<\/sup>= 6\u221a7m<br \/>\nTherefore, the distance from the base is 6\u221a7m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{1\\frac{1}{2}}\" alt=\"\\mathbf{1\\frac{1}{2}}\" align=\"absmiddle\" \/><\/strong><strong> hours?\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given : <\/strong>Speed of first aeroplane = 1000 km\/hr<br \/>\nDistance covered by first aeroplane flying due north in <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?1\\frac{1}{2}\" alt=\"1\\frac{1}{2}\" align=\"absmiddle\" \/><strong>\u00a0<\/strong> hours (OA) = 1000 \u00d7 3\/2 km = 1500 km<br \/>\nSpeed of second aeroplane = 1200 km\/hr<br \/>\nDistance covered by second aeroplane flying due west in <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?1\\frac{1}{2}\" alt=\"1\\frac{1}{2}\" align=\"absmiddle\" \/> hours (OB) = 1200 \u00d7 3\/2 km = 1800 km<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6394\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans11.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"161\" \/><br \/>\nIn right angle \u0394AOB, by Pythagoras Theorem,<br \/>\nAB<sup>2<\/sup>\u00a0=<sup>\u00a0<\/sup>AO<sup>2<\/sup>\u00a0+ OB<sup>2<br \/>\n<\/sup>\u21d2 AB<sup>2<\/sup>\u00a0=<sup>\u00a0<\/sup>(1500)<sup>2<\/sup>\u00a0+ (1800)<sup>2<br \/>\n<\/sup>\u21d2 AB = \u221a(2250000\u00a0+ 3240000)<br \/>\n= \u221a5490000<br \/>\n\u21d2 AB = 300\u221a61\u00a0km<br \/>\nHence, the distance between two aeroplanes will be 300\u221a61\u00a0km.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops. <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given :<\/strong> Two poles of heights 6 m and 11 m stand on a plane ground.<br \/>\nDistance between the feet of the poles is 12 m.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6395\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans12.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"320\" height=\"233\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans12.png 320w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans12-300x218.png 300w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><br \/>\nLet AB and CD be the poles of height 6m and 11m.<br \/>\nTherefore, CP = 11 \u2013 6 = 5m<br \/>\nFrom the figure, it can be observed that AP = 12m<br \/>\nBy Pythagoras theorem for \u0394APC, we get,<br \/>\nAP<sup>2<\/sup>\u00a0=<sup>\u00a0<\/sup>PC<sup>2<\/sup>\u00a0+ AC<sup>2<br \/>\n<\/sup>(12m)<sup>2<\/sup>\u00a0+ (5m)<sup>2<\/sup>\u00a0= (AC)<sup>2<br \/>\n<\/sup>AC<sup>2<\/sup>\u00a0= (144+25) m<sup>2<\/sup>\u00a0= 169 m<sup>2<br \/>\n<\/sup>AC = 13m<br \/>\nTherefore, the distance between their tops is 13 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0= AB<sup>2<\/sup>\u00a0+ DE<sup>2<\/sup>.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given :<\/strong>\u00a0D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6396\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans13.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"192\" height=\"196\" \/><br \/>\nBy Pythagoras theorem in \u0394ACE, we get<br \/>\nAC<sup>2<\/sup>\u00a0+<sup>\u00a0<\/sup>CE<sup>2<\/sup>\u00a0= AE<sup>2<\/sup>\u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)<br \/>\nIn \u0394BCD, by Pythagoras theorem, we get<br \/>\nBC<sup>2<\/sup>\u00a0+<sup>\u00a0<\/sup>CD<sup>2<\/sup>\u00a0= BD<sup>2<\/sup>\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii)<br \/>\nFrom equations\u00a0(i)\u00a0and\u00a0(ii), we get,<br \/>\nAC<sup>2<\/sup>\u00a0+<sup>\u00a0<\/sup>CE<sup>2<\/sup>\u00a0+ BC<sup>2<\/sup>\u00a0+<sup>\u00a0<\/sup>CD<sup>2<\/sup>\u00a0= AE<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (iii)<br \/>\nIn \u0394CDE, by Pythagoras theorem, we get<br \/>\nDE<sup>2<\/sup>\u00a0=<sup>\u00a0<\/sup>CD<sup>2<\/sup>\u00a0+ CE<sup>2<br \/>\n<\/sup>In \u0394ABC, by Pythagoras theorem, we get<br \/>\nAB<sup>2<\/sup>\u00a0=<sup>\u00a0<\/sup>AC<sup>2<\/sup>\u00a0+ CB<sup>2<br \/>\n<\/sup>Putting the above two values in equation\u00a0(iii), we get<br \/>\nDE<sup>2<\/sup>\u00a0+ AB<sup>2<\/sup>\u00a0= AE<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>14. The perpendicular from A on side BC of a\u00a0\u0394 ABC intersects BC at D such that DB = 3CD\u00a0(see Figure). Prove that 2AB<sup>2<\/sup>\u00a0= 2AC<sup>2<\/sup>\u00a0+ BC<sup>2<\/sup>.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6397\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Que14.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"158\" \/><br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given <\/strong>: The perpendicular from A on side BC of a \u0394 ABC intersects BC at D such that;<br \/>\nDB = 3CD.<br \/>\nIn \u0394 ABC,<br \/>\nAD \u22a5 BC and BD = 3CD<br \/>\nIn right angle triangle, ADB and ADC, by Pythagoras theorem,<br \/>\nAB<sup>2<\/sup>\u00a0=<sup>\u00a0<\/sup>AD<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">AC<sup>2<\/sup>\u00a0=<sup>\u00a0<\/sup>AD<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (ii)<br \/>\nSubtracting equation\u00a0(ii)\u00a0from equation\u00a0(i), we get<br \/>\nAB<sup>2<\/sup>\u00a0\u2013 AC<sup>2<\/sup>\u00a0= BD<sup>2<\/sup>\u00a0\u2013 DC<sup>2<br \/>\n<\/sup>= 9CD<sup>2<\/sup>\u00a0\u2013 CD<sup>2<\/sup>\u00a0[Since, BD = 3CD]<br \/>\n= 8CD<sup>2<br \/>\n<\/sup>= 8(BC\/4)<sup>2\u00a0<\/sup>[Since, BC = DB\u00a0+ CD = 3CD\u00a0+ CD = 4CD]<br \/>\nTherefore, AB<sup>2<\/sup>\u00a0\u2013 AC<sup>2<\/sup>\u00a0= BC<sup>2<\/sup>\/2<br \/>\n\u21d2 2(AB<sup>2<\/sup>\u00a0\u2013 AC<sup>2<\/sup>) = BC<sup>2<br \/>\n<\/sup>\u21d2 2AB<sup>2<\/sup>\u00a0\u2013 2AC<sup>2<\/sup>\u00a0= BC<sup>2<br \/>\n<\/sup>\u2234 2AB<sup>2<\/sup>\u00a0= 2AC<sup>2<\/sup>\u00a0+ BC<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>15. \u00a0In an equilateral triangle ABC, D is a point on side BC such that BD = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{3}}\" alt=\"\\mathbf{\\frac{1}{3}}\" align=\"absmiddle\" \/> BC. Prove that 9AD<sup>2<\/sup>\u00a0= 7AB<sup>2<\/sup>.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given <\/strong>: ABC is an equilateral triangle.<br \/>\nD is a point on side BC such that BD = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> BC<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6398\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans15.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"184\" height=\"170\" \/><br \/>\nLet the side of the equilateral triangle be\u00a0<em>a<\/em>, and AE be the altitude of \u0394ABC.<br \/>\n\u2234 BE = EC = BC\/2 = a\/2<br \/>\nAnd, AE =\u00a0a\u221a3\/2<br \/>\nGiven, BD = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> BC<br \/>\n\u2234 BD = a\/3<br \/>\nDE = BE \u2013 BD =\u00a0a\/2 \u2013\u00a0a\/3 =\u00a0a\/6<br \/>\nIn \u0394ADE, by Pythagoras theorem,<br \/>\nAD<sup>2<\/sup>\u00a0= AE<sup>2<\/sup>\u00a0+ DE<sup>2<br \/>\n<\/sup>AD<sup>2<\/sup> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{a\\sqrt{3}}{2}&amp;space;\\right&amp;space;)^2&amp;space;+&amp;space;\\left&amp;space;(&amp;space;\\frac{a}{6}&amp;space;\\right&amp;space;)^2\" alt=\"\\left ( \\frac{a\\sqrt{3}}{2} \\right )^2 + \\left ( \\frac{a}{6} \\right )^2\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">AD<sup>2<\/sup> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3a^2}{4}&amp;space;+\\frac{a^2}{36}\" alt=\"\\frac{3a^2}{4} +\\frac{a^2}{36}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">AD<sup>2<\/sup> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{28a^2}{36}\" alt=\"\\frac{28a^2}{36}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7}{9}AB^2\" alt=\"\\frac{7}{9}AB^2\" align=\"absmiddle\" \/><br \/>\n\u21d2 9 AD<sup>2<\/sup>\u00a0= 7 AB<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes. <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given : <\/strong>An equilateral triangle say ABC,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6399\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans16.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"201\" height=\"177\" \/><br \/>\nLet the sides of the equilateral triangle be of length a, and AE be the altitude of \u0394ABC.<br \/>\n\u2234 BE = EC = BC\/2 =\u00a0a\/2<br \/>\nIn \u0394ABE, by Pythagoras Theorem, we get<br \/>\nAB<sup>2<\/sup>\u00a0= AE<sup>2<\/sup>\u00a0+ BE<sup>2<br \/>\n<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">a<sup>2<\/sup> = AE<sup>2<\/sup> + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{a}{2}&amp;space;\\right&amp;space;)^2\" alt=\"\\left ( \\frac{a}{2} \\right )^2\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">AE<sup>2<\/sup>\u00a0 =\u00a0 a<sup>2<\/sup>\u00a0 &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a^2}{4}\" alt=\"\\frac{a^2}{4}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">AE<sup>2<\/sup>\u00a0 =\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3a^2}{4}\" alt=\"\\frac{3a^2}{4}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">4AE<sup>2<\/sup>\u00a0= 3a<sup>2<br \/>\n<\/sup>\u21d2 4 \u00d7 (Square of altitude) = 3 \u00d7 (Square of one side)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>17. Tick the correct answer and justify: In \u0394ABC, AB = 6\u221a3\u00a0cm, AC = 12 cm and BC = 6 cm.<br \/>\nThe angle B is:<br \/>\n<\/strong>(A) 120\u00b0<br \/>\n(B) 60\u00b0<br \/>\n(C) 90\u00b0<br \/>\n(D) 45\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given : <\/strong>\u0394ABC, AB = 6\u221a3\u00a0cm, AC = 12 cm and BC = 6 cm.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6400\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.5-Ans17.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"252\" height=\"152\" \/><br \/>\nWe can observe that,<br \/>\nAB<sup>2<\/sup> = 108<br \/>\nAC<sup>2<\/sup> = 144<br \/>\nAnd,<br \/>\nBC<sup>2<\/sup> = 36<br \/>\nAB<sup>2<\/sup>\u00a0+ BC<sup>2<\/sup>\u00a0= AC<sup>2<br \/>\n<\/sup>The given triangle, \u0394ABC, is satisfying Pythagoras theorem.<br \/>\nTherefore, the triangle is a right triangle, right-angled at B.<br \/>\n\u2234 \u2220B = 90\u00b0<br \/>\nHence, the correct <strong>answer is (C).<\/strong><\/span><\/p>\n<p><span style=\"color: #0000ff; font-family: Georgia, Palatino;\"><b><i><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\">Go Back To Chapters<\/a><\/i><\/b><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 6 (Triangles)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 6 Triangles Exercise 6.5 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 6 Triangles NCERT Class 10 Maths Solution Ex &#8211; 6.1 NCERT Class 10 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1079,1084,1044,1049,1048],"class_list":["post-6053","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-6-triangles-solutions","tag-ncert-class-10-mathematics-exercise-6-5-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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