{"id":6054,"date":"2023-08-17T03:01:40","date_gmt":"2023-08-17T03:01:40","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6054"},"modified":"2023-08-17T03:01:40","modified_gmt":"2023-08-17T03:01:40","slug":"ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-6","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-6\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 6 Triangles Ex 6.6"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 6 (Triangles)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 6 Triangles <\/strong>Exercise 6.6 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-6-5\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.5<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 6.6<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. In Figure, PS is the bisector of \u2220 QPR of \u2206 PQR. Prove that QS\/PQ = SR\/PR<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6404\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Que1.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"320\" height=\"136\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Que1.png 320w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Que1-300x128.png 300w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given : <\/strong>PS is the bisector of \u2220QPR of \u2206PQR.\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Now, Draw RT SP || to meet QP produced in T.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6405\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Ans1.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"136\" height=\"200\" \/><\/span><br \/>\n<span style=\"color: #000000;\"><strong>Proof :<br \/>\n<\/strong>\u2235 RT SP || and transversal PR intersects them<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u22201 = \u22202\u00a0(Alternate interior angle) &#8212;&#8212;&#8212;&#8212;&#8212; (i)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 RT SP || and transversalQT intersects them<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u22203 = \u22204\u00a0(Corresponding angle)\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212; (ii)<\/span><br \/>\n<span style=\"color: #000000;\">But \u22201 = \u22203\u00a0(Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u22202 = \u22204\u00a0[From Eqs. (i) and (ii)]<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 PT = PR \u2026(iii)\u00a0(\u2235 Sides opposite to equal angles of a triangle are equal)<\/span><br \/>\n<span style=\"color: #000000;\">Now, in \u2206QRT, <\/span><br \/>\n<span style=\"color: #000000;\">PS || RT\u00a0(By construction) <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 QS\/SR = PQ\/PT\u00a0(By basic proportionally theorem) <\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 QS\/SR = PQ\/PR\u00a0[From Eq. (iii)]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. In Fig. 6.57, D is a point on hypotenuse AC of \u2206ABC, such that BD \u22a5AC, DM \u22a5 BC and DN \u22a5 AB. Prove that:<br \/>\n<\/strong>(i) DM<sup>2<\/sup>\u00a0= DN . MC<br \/>\n(ii) DN<sup>2<\/sup> = DM . AN.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6406\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"197\" height=\"200\" \/><br \/>\n<\/strong><strong>Given<\/strong> : D is a point on hypotenuse AC of \u2206ABC, DM \u22a5 BC and DN \u22a5 AB.<\/span><br \/>\n<span style=\"color: #000000;\">Now, join NM. <\/span><br \/>\n<span style=\"color: #000000;\">Let BD and NM intersect at O. <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) In \u2206DMC and \u2206NDM,<br \/>\n<\/strong>\u2220DMC = \u2220NDM\u00a0(Each equal to 90\u00b0) <\/span><br \/>\n<span style=\"color: #000000;\">\u2220MCD = \u2220DMN <\/span><br \/>\n<span style=\"color: #000000;\">Let MCD = \u22201 <\/span><br \/>\n<span style=\"color: #000000;\">Then, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220MDC = 90\u00b0 \u2212 (90\u00b0-\u22201) <\/span><br \/>\n<span style=\"color: #000000;\">= \u22201\u00a0(\u2235\u2220MCD + \u2220MDC + \u2220DMC = 180\u00b0)\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220ODM = 90\u00b0 \u2212 (90\u00b0 \u2212 \u22201)<\/span><br \/>\n<span style=\"color: #000000;\">= \u22201 <\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 \u2220DMN = \u22201 <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2206DMO ~ \u2206NDM\u00a0\u00a0(AA similarity criterion) <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 DM\/ND = MC\/DM (Corresponding sides of the similar triangles are proportional) <\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 DM<sup>2<\/sup>\u00a0= MC ND<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) In \u2206DNM and \u2206NAD,<br \/>\n<\/strong>\u2220NDM = \u2220AND\u00a0(Each equal to 90\u00b0)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DNM = \u2220NAD<\/span><br \/>\n<span style=\"color: #000000;\">Let \u2220NAD\u00a0= \u22202<\/span><br \/>\n<span style=\"color: #000000;\">Then, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220NDA = 90\u00b0 \u2212 \u22202 <\/span><br \/>\n<span style=\"color: #000000;\">\u2235\u2220NDA + \u2220DAN + \u2220DNA = 180\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220ODN = 90\u00b0 \u2212 (90\u00b0 \u2212 \u22202) = \u22202 <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220DNO = \u22202<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2206DNM ~ \u2206NAD\u00a0\u00a0(AA similarity criterion)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 DN\/NA = DM\/ND<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 DN\/AN = DM\/DN <\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 DN<sup>2<\/sup>\u00a0=\u00a0DM\u00d7AN<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. In Figure, ABC is a triangle in which \u2220ABC &gt; 90\u00b0 and AD \u22a5 CB produced. Prove that<br \/>\n<\/strong><strong>AC<sup>2<\/sup>= AB<sup>2<\/sup>+ BC<sup>2<\/sup>+ 2 BC.BD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6407\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Que3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"162\" \/><br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given<\/strong> : ABC is a triangle in which \u2220ABC&gt; 90\u00b0 and AD \u22a5 CB produced.\u00a0<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Proof : <\/strong>In right ADC,\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220D = 90\u00b0\u00a0 <\/span><br \/>\n<span style=\"color: #000000;\">AC<sup>2<\/sup>\u00a0= AD<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup>\u00a0\u00a0(By Pythagoras theorem)\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">= AD<sup>2<\/sup>\u00a0+ (BD + BC)<sup>2<\/sup>\u00a0[\u2235DC = DB + BC]\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">= (AD<sup>2<\/sup>\u00a0+ DB<sup>2<\/sup>) + BC<sup>2<\/sup>\u00a0+ 2BD.BC\u00a0[\u2235 (a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0+ 2ab]\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">= AB<sup>2<\/sup>\u00a0+ BC<sup>2<\/sup> + 2BC.BD\u00a0\u00a0[\u2235In right ADB with \u2220D = 90\u00b0, AB<sup>2<\/sup>\u00a0= AD<sup>2<\/sup>\u00a0+ DB<sup>2<\/sup>] (By Pythagoras theorem)\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. In Figure, ABC is a triangle in which \u2220 ABC &lt; 90\u00b0 and AD \u22a5 BC. Prove that <\/strong><strong>AC<sup>2<\/sup>= AB<sup>2<\/sup>+ BC<sup>2<\/sup>\u00a0\u2013 2 BC.BD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6408\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Que4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"158\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given<\/strong> : ABC is a triangle in which \u2220ABC &lt; 90\u00b0 and AD \u22a5 BC.<br \/>\n<strong>Proof <\/strong>: In right \u25b3ADC, \u2220D = 90\u00b0<br \/>\nAC<sup>2<\/sup>\u00a0= AD<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup>\u00a0\u00a0(By Pythagoras theorem) \u00a0<\/span><br \/>\n<span style=\"color: #000000;\">= AD<sup>2<\/sup>\u00a0+ (BC &#8211; BD)<sup>2<\/sup> \u00a0[\u2235 BC = BD + DC]<br \/>\n= AD<sup>2<\/sup>\u00a0+ (BC &#8211; BD)<sup>2<\/sup>\u00a0\u00a0(BC = BD + DC)<br \/>\n= AD<sup>2<\/sup>\u00a0+ BC<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0&#8211; 2BC.BD\u00a0[\u2235 (a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup> + 2ab]<br \/>\n= (AD<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>) + BC<sup>2<\/sup> &#8211; 2BC . BD<br \/>\n= AB<sup>2<\/sup>\u00a0+ BC<sup>2<\/sup> &#8211; 2BC . BD {In right \u25b3ADB with \u2220D = 90\u00b0, AB<sup>2<\/sup>\u00a0= AD<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>} (By Pythagoras theorem)\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. In Figure, AD is a median of a triangle ABC and AM \u22a5 BC. Prove that :<br \/>\n<\/strong>(i) AC<sup>2<\/sup>\u00a0= AD<sup>2<\/sup>\u00a0+ BC.DM + 2 (BC\/2)<sup>\u00a02<br \/>\n<\/sup>(ii) AB<sup>2<\/sup>\u00a0= AD<sup>2<\/sup>\u00a0\u2013 BC.DM + 2 (BC\/2)<sup>\u00a02<br \/>\n<\/sup>(iii) AC<sup>2<\/sup>\u00a0+ AB<sup>2<\/sup>\u00a0= 2 AD<sup>2<\/sup>\u00a0+ \u00bd BC<sup>2<\/sup><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6409\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Que5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"150\" \/><br \/>\n<strong>Solution &#8211;<br \/>\n<\/strong><strong>Given<\/strong> : AD is a median of a \u2206ABC and AM \u22a5 BC.<br \/>\n<strong>Proof :<br \/>\n(i)<\/strong> In right \u2206AMC, \u2220M = 90\u00b0<br \/>\nAC<sup>2 <\/sup>= AM<sup>2<\/sup>\u00a0+ MC<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (By Pythagoras theorem)<br \/>\nAC<sup>2 <\/sup>=\u00a0AM<sup>2<\/sup>\u00a0+ (MD + DC)<sup>\u00a02<\/sup>\u00a0 \u00a0 \u00a0(\u2235 MC = MD + DC)<br \/>\nAC<sup>2 <\/sup>= AM<sup>2<\/sup>\u00a0+ MD<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup> + 2MD.DC\u00a0 \u00a0 \u00a0(\u2235 (a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup> + 2ab)<br \/>\nAC<sup>2 <\/sup>= AD<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup> + 2MD.DC<br \/>\n[\u2235 In right \u25b3AMD with \u2220M = 90\u00b0, AD<sup>2<\/sup> = AM<sup>2<\/sup> + MD<sup>2<\/sup>} (By Pythagoras theorem]<br \/>\nSince, DC = BC\/2,<br \/>\nthus, we get,<br \/>\nAC<sup>2 <\/sup>= AD<sup>2\u00a0<\/sup>+ (BC\/2)<sup>\u00a02\u00a0<\/sup>+ 2MD.(BC\/2)<sup>\u00a02<\/sup><br \/>\nAC<sup>2 <\/sup>= AD<sup>2\u00a0<\/sup>+ (BC\/2)<sup>\u00a02\u00a0<\/sup>+ 2MD \u00d7 BC<br \/>\n<strong>Hence, proved.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> In right \u2206AMB, \u2220M = 90\u00b0<br \/>\nAB<sup>2<\/sup>\u00a0= AM<sup>2<\/sup>\u00a0+ MB<sup>2<br \/>\n<\/sup>= (AD<sup>2<\/sup>\u00a0\u2212 DM<sup>2<\/sup>) + MB<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (By Pythagoras theorem)<br \/>\n= (AD<sup>2<\/sup>\u00a0\u2212 DM<sup>2<\/sup>) + (BD \u2212 MD)<sup>\u00a02<\/sup>\u00a0 \u00a0(\u2235 BD = BM + MD)<br \/>\n= AD<sup>2<\/sup>\u00a0\u2212 DM<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0+ MD<sup>2<\/sup> \u2212 2BD \u00d7 MD\u00a0 \u00a0 (\u2235 (a &#8211; b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup> &#8211; 2ab)<br \/>\n= AD<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0\u2212 2BD \u00d7 MD<br \/>\nSince, 2BD = BC, AD is a median of \u2206ABC<br \/>\nthus, we get,<br \/>\n= AD<sup>2\u00a0<\/sup>+ (BC\/2)<sup>2\u00a0<\/sup>\u2013 2(BC\/2) MD<br \/>\n= AD<sup>2\u00a0<\/sup>+ (BC\/2)<sup>2\u00a0<\/sup>\u2013 BC MD<br \/>\nHence, proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) <\/strong>By applying Pythagoras Theorem in \u2206ABM, we get,<strong><br \/>\n<\/strong>AM<sup>2<\/sup>\u00a0+ MB<sup>2<\/sup>\u00a0= AB<sup>2<\/sup>\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)<br \/>\nBy applying Pythagoras Theorem in \u2206AMC, we get,<br \/>\nAM<sup>2<\/sup>\u00a0+ MC<sup>2<\/sup>\u00a0= AC<sup>2<\/sup> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii)<br \/>\nAdding both the equations (i) and (ii), we get,<br \/>\n2AM<sup>2<\/sup>\u00a0+ MB<sup>2<\/sup>\u00a0+ MC<sup>2<\/sup>\u00a0= AB<sup>2<\/sup>\u00a0+ AC<sup>2<br \/>\n<\/sup>2AM<sup>2<\/sup>\u00a0+ (BD \u2212 DM)<sup>\u00a02<\/sup>\u00a0+ (MD + DC)<sup>\u00a02<\/sup>\u00a0= AB<sup>2<\/sup>\u00a0+ AC<sup>2<br \/>\n<\/sup>2AM<sup>2<\/sup>+BD<sup>2<\/sup>\u00a0+ DM<sup>2<\/sup>\u00a0\u2212 2BD.DM + MD<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup>\u00a0+ 2MD.DC = AB<sup>2<\/sup>\u00a0+ AC<sup>2<br \/>\n<\/sup>2AM<sup>2<\/sup>\u00a0+ 2MD<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup>\u00a0+ 2MD (\u2212 BD + DC) = AB<sup>2<\/sup>\u00a0+ AC<sup>2<br \/>\n<\/sup>2(AM<sup>2<\/sup>+ MD<sup>2<\/sup>) + (BC\/2)<sup>\u00a02<\/sup>\u00a0+ (BC\/2)<sup>\u00a02<\/sup>\u00a0+ 2MD (-BC\/2 + BC\/2)<sup>\u00a02<\/sup>\u00a0= AB<sup>2<\/sup>\u00a0+ AC<sup>2<br \/>\n<\/sup>2AD<sup>2\u00a0<\/sup>+ BC<sup>2<\/sup>\/2 = AB<sup>2<\/sup>\u00a0+ AC<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides. <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong><br \/>\n<strong>Given<\/strong> : ABCD is a parallelogram whose diagonals are AC and BD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6410\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Ans6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"141\" \/><br \/>\nNow, draw AM\u22a5DC and BN\u22a5D\u00a0\u00a0<b>(produced)<\/b>.<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Proof : <\/strong>In right \u2206AMD and \u2206BNC, <\/span><br \/>\n<span style=\"color: #000000;\">AD = BC (Opposite sides of a parallelogram)<br \/>\nAM = BN (Both are altitudes of the same parallelogram to the same base) ,<br \/>\n\u25b3AMD \u2a6d \u25b3BNC (RHS congruence criterion)<br \/>\nMD = NC (CPCT) &#8212;&#8212;&#8212;&#8212;&#8211; (i)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>In right \u25b3BND, \u2220N = 90\u00b0<br \/>\n<\/strong>BD<sup>2<\/sup>\u00a0= BN<sup>2<\/sup>\u00a0+ DN<sup>2<\/sup>(By Pythagoras theorem)<br \/>\n= BN<sup>2<\/sup>\u00a0+ (DC + CN)<sup>2<\/sup>\u00a0(\u2235 DN = DC + CN)<br \/>\n= BN<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup>\u00a0+ CN<sup>2<\/sup>\u00a0+ 2DC.CN\u00a0[\u2235 (a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0+ 2ab]<br \/>\n= (BN<sup>2<\/sup>\u00a0+ CN<sup>2<\/sup>) + DC<sup>2<\/sup>\u00a0+ 2DC.CN<br \/>\n= BC<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup> + 2DC.CN\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8211; (ii)\u00a0(\u2235In right \u25b3BNC with \u2220N = 90\u00b0)<br \/>\nBN<sup>2<\/sup>\u00a0+ CN<sup>2<\/sup>\u00a0= BC<sup>2<\/sup>\u00a0 (By Pythagoras theorem)<\/span><\/p>\n<p><strong>In right \u25b3AMC, \u2220M = 90\u00b0<br \/>\n<\/strong>AC<sup>2<\/sup>\u00a0= AM<sup>2<\/sup>\u00a0+ MC<sup>2<\/sup>(\u2235MC = DC &#8211; DM)<br \/>\n= AM<sup>2<\/sup>\u00a0+ (DC &#8211; DM)<sup>2<\/sup>\u00a0[\u2235 (a + b)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0+ 2ab]<br \/>\n= AM<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup>\u00a0+ DM<sup>2<\/sup>\u00a0&#8211; 2DC.DM<br \/>\n= (AM<sup>2<\/sup>\u00a0+ DM<sup>2<\/sup>) + DC<sup>2<\/sup>\u00a0&#8211; 2DC.DM<br \/>\n= AD<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup> &#8211; 2DC.DM [\u2235 In\u00a0 right triangle AMD with \u2220M = 90\u00b0, AD<sup>2<\/sup>\u00a0= AM<sup>2<\/sup>\u00a0+ DM<sup>2<\/sup>\u00a0(By Pythagoras theorem)]<br \/>\n= AD<sup>2<\/sup>\u00a0+ AB<sup>2<\/sup> = 2DC.CN\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8211; (iii) [\u2235 DC = AB, opposite sides of parallelogram and BM = CN from eq (i)]<\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Now, on adding Eqs. (iii) and (ii), we get<br \/>\nAC<sup>2<\/sup>\u00a0+ BD<sup>2<\/sup>\u00a0= (AD<sup>2<\/sup>\u00a0+ AB<sup>2<\/sup>) + (BC<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup>)<br \/>\n= AB<sup>2<\/sup>\u00a0+ BC<sup>2<\/sup>\u00a0+ CD<sup>2<\/sup>\u00a0+ DA<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. In Figure, two chords AB and CD intersect each other at the point P. Prove that :<br \/>\n<\/strong>(i) \u2206APC ~ \u2206 DPB<br \/>\n(ii) AP . PB = CP . DP<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6411\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Que7.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"188\" height=\"200\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given<\/strong> :\u00a0 two chords AB and CD intersects each other at the point P.<br \/>\n<strong>Proof:<br \/>\n<\/strong><strong>(i) \u2206APC and \u2206DPB<br \/>\n<\/strong>\u2220APC =\u00a0\u2220DPB (Vertically opposite angles)<br \/>\n\u2220CAP = \u2220BDP (Angles in the same segment for chord CB)<br \/>\nTherefore,<br \/>\n\u2206APC\u00a0\u223c\u00a0\u2206DPB (AA similarity criterion)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) \u2206APC ~ \u2206DPB<br \/>\n<\/strong>In the above, we have proved that \u2206APC \u223c \u2206DPB<br \/>\nWe know that the corresponding sides of similar triangles are proportional.<br \/>\n\u2234 AP\/DP = PC\/PB = CA\/BD<br \/>\n\u21d2 AP\/DP = PC\/PB<br \/>\n\u2234 AP. PB = PC. DP<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:<br \/>\n<\/strong>(i) \u2206 PAC ~ \u2206 PDB<br \/>\n(ii) PA . PB = PC . PD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6412\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Que8.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"182\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given<\/strong> : two chords AB and CD of a circle intersect each other at the point P (when produced) out the circle.<br \/>\n<strong>Proof :<br \/>\n(i) In \u2206PAC and \u2206PDB,<br \/>\n<\/strong>\u2220P = \u2220P (Common Angles)<br \/>\nAs we know, exterior angle of a cyclic quadrilateral is \u2220PCA and \u2220PBD is opposite interior angle, which are both equal.<br \/>\n\u2220PAC = \u2220PDB<br \/>\nThus, \u2206PAC \u223c \u2206PDB(AA similarity criterion)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) \u2206PAC ~ \u2206PDB<br \/>\n<\/strong>We have already proved above,<br \/>\n\u2206APC \u223c \u2206DPB<br \/>\nWe know that the corresponding sides of similar triangles are proportional.<br \/>\nTherefore,<br \/>\nAP\/DP = PC\/PB = CA\/BD<br \/>\nAP\/DP = PC\/PB<br \/>\n\u2234 AP. PB = PC. DP<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. In Figure, D is a point on side BC of \u2206 ABC such that BD\/CD = AB\/AC. Prove that AD is the bisector of \u2220 BAC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6413\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Que9.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"151\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given<\/strong> : D is a point on side BC of \u2206ABC such that BD\/CD = AB\/AC <\/span><br \/>\n<span style=\"color: #000000;\">Now, from BA produce cut off AE = A. JoinCE.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6414\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Ans9.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"171\" height=\"246\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">By using the converse of basic proportionality theorem, we get,<br \/>\nAD || PC<br \/>\n\u2220BAD = \u2220APC (Corresponding angles)\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212; (i)<br \/>\nAnd,<br \/>\n\u2220DAC = \u2220ACP (Alternate interior angles) &#8212;&#8212;&#8212;&#8212;&#8212; (ii)<br \/>\nBy the new figure, we have;<br \/>\nAP = AC<br \/>\n\u21d2 \u2220APC = \u2220ACP\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212; (iii)<br \/>\nOn comparing equations (i), (ii), and (iii), we get,<br \/>\n\u2220BAD = \u2220APC<br \/>\nTherefore, AD is the bisector of the angle BAC.<br \/>\nHence, proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6415\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Que10.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"320\" height=\"254\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Que10.png 320w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Que10-300x238.png 300w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Length of the string that she has out<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6416\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Ana10i.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"172\" height=\"183\" \/><br \/>\nTo find AC, we have to use Pythagoras theorem in \u2206ABC, is such way;<br \/>\nAC<sup>2<\/sup>\u00a0= AB<sup>2<\/sup>+ BC<sup>2<br \/>\n<\/sup>AC<sup>2\u00a0<\/sup>= (1.8 m)<sup>\u00a02<\/sup>\u00a0+ (2.4 m)<sup> 2<br \/>\n<\/sup>AC<sup>2\u00a0<\/sup>= (3.24 + 5.76) m<sup>2<br \/>\n<\/sup>AC<sup>2<\/sup>\u00a0= 9.00 m<sup>2<br \/>\n<\/sup>\u27f9 AC = \u221a9 m = 3m<br \/>\nThus, the length of the string out is 3 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Length of string out after 12 seconds is AD.<br \/>\nAD = AC \u2212 String pulled by Nazima in 12 seconds<br \/>\n= (3.00 \u221212 \u00d7 5 cm)<br \/>\n= (3.00 \u2212 60 cm)<br \/>\n= (3.00 \u2212 0.6) m<br \/>\n= 2.4 m<br \/>\nIn \u2206ADB, by Pythagoras Theorem,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6417\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.6-Ana10ii.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"263\" height=\"160\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">AD<sup>2 <\/sup>= AB<sup>2<\/sup>\u00a0+ BD<sup>2<br \/>\n<\/sup>(2.4 m)<sup> 2\u00a0<\/sup>= (1.8 m) <sup>2<\/sup>\u00a0+ BD<sup>2<\/sup><br \/>\nBD<sup>2<\/sup>\u00a0= (5.76 \u2212 3.24) m<sup>2<\/sup>\u00a0= 2.52 m<sup>2<br \/>\n<\/sup>BD = 1.587 m<br \/>\nHorizontal distance of fly from Nazima after 12 sec = BD + 1.2 m<br \/>\n= (1.587 + 1.2) m = 2.787 m<br \/>\n= 2.79 m (Approximate)<\/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 6.6 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":[1045,1079,1085,1044,1049,1048],"class_list":["post-6054","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-6-6-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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