{"id":6052,"date":"2023-08-17T03:01:26","date_gmt":"2023-08-17T03:01:26","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6052"},"modified":"2023-08-17T03:01:26","modified_gmt":"2023-08-17T03:01:26","slug":"ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-4\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 6 Triangles Ex 6.4"},"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 6.4 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-6-5\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.5<\/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;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 6.4<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. Let \u0394ABC ~ \u0394DEF and their areas be, respectively, 64 cm<sup>2<\/sup>\u00a0and 121 cm<sup>2<\/sup>. If EF = 15.4 cm, find BC.<br \/>\n<\/strong><strong>Solution &#8211; \u00a0<\/strong>Given, \u0394ABC ~ \u0394DEF,<br \/>\nArea of \u0394ABC = 64 cm<sup>2<br \/>\n<\/sup>Area of \u0394DEF =\u00a0121 cm<sup>2<br \/>\n<\/sup>EF = 15.4 cm<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6372 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans1-300x190.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"190\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans1-300x190.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans1.png 477w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\nTherefore,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{ar(\\Delta&amp;space;ABC)}{ar(\\Delta&amp;space;DEF)}&amp;space;=&amp;space;\\frac{AB^2}{DE^2}&amp;space;=&amp;space;\\frac{AC^2}{DF^2}&amp;space;=&amp;space;\\frac{BC^2}{EF^2}\" alt=\"\\frac{ar(\\Delta ABC)}{ar(\\Delta DEF)} = \\frac{AB^2}{DE^2} = \\frac{AC^2}{DF^2} = \\frac{BC^2}{EF^2}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{64}{121}&amp;space;=&amp;space;\\frac{BC^2}{(15.4)^2}\" alt=\"\\frac{64}{121} = \\frac{BC^2}{(15.4)^2}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 (BC)\u00b2 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(15.4)^2&amp;space;\\times&amp;space;8^2}{11^2}\" alt=\"\\frac{(15.4)^2 \\times 8^2}{11^2}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 BC = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{15.4&amp;space;\\times&amp;space;8}{11}\" alt=\"\\frac{15.4 \\times 8}{11}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 BC = 11.2 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD. <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong>Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6373 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"224\" height=\"150\" \/><br \/>\nIn \u0394AOB and \u0394COD, we have<br \/>\n\u22201 = \u22202 (Alternate angles)<br \/>\n\u22203 = \u22204 (Alternate angles)<br \/>\n\u22205 = \u22206 (Vertically opposite angle)<br \/>\n\u2234 \u0394AOB ~ \u0394COD [AAA similarity criterion]<br \/>\nAs we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,<br \/>\nArea of (\u0394AOB)\/Area of (\u0394COD) = AB<sup>2<\/sup>\/CD<sup>2<br \/>\n<\/sup>= (2CD)<sup>2<\/sup>\/CD<sup>2<\/sup>\u00a0[\u2234 AB = 2CD]<br \/>\n\u2234 Area of (\u0394AOB)\/Area of (\u0394COD)<br \/>\n= 4CD<sup>2<\/sup>\/CD<sup>2<\/sup>\u00a0= 4\/1<br \/>\nHence, the required ratio of the area of \u0394AOB and \u0394COD = 4 : 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>3. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (\u0394ABC)\/area (\u0394DBC) = AO\/DO.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6374\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Que3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"178\" height=\"200\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><strong>Given :<\/strong> ABC and DBC are triangles on the same base BC. Ad intersects BC at O.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>To Prove :<\/strong> area (\u0394ABC)\/area (\u0394DBC) = AO\/DO.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Construction :<\/strong> Let us draw two perpendiculars AP and DM on line BC.<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Proof :<\/strong> We know that area of a triangle = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 Base \u00d7 Height<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6375\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Que3i.png\" alt=\"\" width=\"229\" height=\"62\" \/><br \/>\nIn \u0394APO and \u0394DMO,<br \/>\n\u2220APO = \u2220DMO (Each equals to 90\u00b0)<br \/>\n\u2220AOP = \u2220DOM (Vertically opposite angles)<br \/>\n\u2234 \u0394APO ~ \u0394DMO (By AA similarity criterion)<br \/>\n\u2234 AP\/DM = AO\/DO<br \/>\n\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{area(\\bigtriangleup&amp;space;ABC)}{area(\\bigtriangleup&amp;space;DBC)}&amp;space;=&amp;space;\\frac{AO}{DO}\" alt=\"\\frac{area(\\bigtriangleup ABC)}{area(\\bigtriangleup DBC)} = \\frac{AO}{DO}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. If the areas of two similar triangles are equal, prove that they are congruent.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>Say \u0394ABC and \u0394PQR are two similar triangles and equal in area<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6376\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"320\" height=\"130\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans4.png 320w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans4-300x122.png 300w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><br \/>\nNow let us prove \u0394ABC \u2245 \u0394PQR.<br \/>\nSince, \u0394ABC ~ \u0394PQR<br \/>\n\u2234 Area of (\u0394ABC)\/Area of (\u0394PQR) = BC<sup>2<\/sup>\/QR<sup>2<br \/>\n<\/sup>\u21d2 BC<sup>2<\/sup>\/QR<sup>2<\/sup> = 1 [Since, Area(\u0394ABC) = (\u0394PQR)<br \/>\n\u21d2 BC<sup>2<\/sup>\/QR<sup>2<br \/>\n<\/sup>\u21d2 BC = QR<br \/>\nSimilarly, we can prove that<br \/>\nAB = PQ and AC = PR<br \/>\nThus, \u0394ABC \u2245 \u0394PQR [SSS criterion of congruence]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. D, E and F are respectively the mid-points of sides AB, BC and CA of \u0394ABC. Find the ratio of the area of \u0394DEF and \u0394ABC.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6377\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"166\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Given :<\/strong> D, E and F are the mid-points of the sides AB, BC and CA respectively of the \u0394ABC.<br \/>\n<strong>To Find :<\/strong> area(\u0394DEF) and area(\u0394ABC)<br \/>\n<strong>Solution :<\/strong> In \u0394ABC, we have<br \/>\nF is the mid point of AB (Given)<br \/>\nE is the mid-point of AC (Given)<br \/>\nSo, by the mid-point theorem, we have<br \/>\nFE || BC and FE = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> BC<br \/>\n\u21d2 FE || BC and FE || BD [BD = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> BC]<br \/>\n\u2234 BDEF is parallelogram [Opposite sides of parallelogram are equal and parallel]<br \/>\nSimilarly in \u0394FBD and \u0394DEF, we have<br \/>\nFB = DE (Opposite sides of parallelogram BDEF)<br \/>\nFD = FD (Common)<br \/>\nBD = FE (Opposite sides of parallelogram BDEF)<br \/>\n\u2234 \u0394 FBD \u2245 \u0394 DEF<br \/>\nSimilarly, we can prove that<br \/>\n\u0394AFE \u2245 \u0394DEF<br \/>\n\u0394EDC \u2245 \u0394DEF<br \/>\nIf triangles are congruent,then they are equal in area.<br \/>\nSo, area (\u0394FBD) = area (\u0394DEF)\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;- (i)<br \/>\narea (\u0394AFE) = area (\u0394DEF)\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;- (ii)<br \/>\nand, area (\u0394EDC) = area (\u0394DEF) &#8212;&#8212;&#8212;&#8212;- (iii)<br \/>\nNow,<br \/>\narea (\u0394ABC) = area (\u0394FBD) + area (\u0394DEF) + area (\u0394AFE) + area (\u0394EDC) &#8212;&#8212;&#8212;&#8212;- (iv)<br \/>\narea (\u0394ABC) = area (\u0394DEF) + area (\u0394DEF) + area (\u0394DEF) + area (\u0394DEF)<br \/>\n\u21d2 area (\u0394DEF) = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{4}\" alt=\"\\frac{1}{4}\" align=\"absmiddle\" \/> area (\u0394ABC) [From (i), (ii) and (iii)]<br \/>\n\u21d2 area (\u0394DEF)\/area (\u0394ABC) = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{4}\" alt=\"\\frac{1}{4}\" align=\"absmiddle\" \/><br \/>\nHence,<br \/>\narea (\u0394DEF) : area (\u0394ABC) = 1 : 4<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong><strong>Given :<\/strong> AM and DN are the medians of triangles ABC and DEF respectively and \u0394ABC ~ \u0394DEF.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6378\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"320\" height=\"139\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans6.png 320w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans6-300x130.png 300w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><br \/>\n<strong>We have to prove :<\/strong>\u00a0Area(\u0394ABC)\/Area(\u0394DEF) = AM<sup>2<\/sup>\/DN<sup>2<br \/>\n<\/sup>Since, \u0394ABC ~ \u0394DEF (Given)<br \/>\n\u2234 Area(\u0394ABC)\/Area(\u0394DEF) = (AB<sup>2<\/sup>\/DE<sup>2<\/sup>) &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-(i)<strong><br \/>\n<\/strong>and, AB\/DE = BC\/EF = CA\/FD\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-(ii)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6379\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans6i.png\" alt=\"\" width=\"142\" height=\"65\" \/><br \/>\nIn \u0394ABM and \u0394DEN,<br \/>\nSince \u0394ABC ~ \u0394DEF<br \/>\n\u2234 \u2220B = \u2220E<br \/>\nAB\/DE = BM\/EN [Already Proved in equation\u00a0(i)]<br \/>\n\u2234 \u0394ABC ~ \u0394DEF [SAS similarity criterion]<br \/>\n\u21d2 AB\/DE = AM\/DN\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (iii)<br \/>\n\u2234 \u0394ABM ~ \u0394DEN<br \/>\nAs the areas of two similar triangles are proportional to the squares of the corresponding sides.<br \/>\n\u2234 area (\u0394ABC) \/area (\u0394DEF) = AB<sup>2<\/sup>\/DE<sup>2<\/sup>\u00a0= AM<sup>2<\/sup>\/DN<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one 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 square whose one diagonal is AC. \u0394APC and \u0394BQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6380\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans7.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"172\" \/><br \/>\nArea(\u0394BQC) = \u00bd Area(\u0394APC)<br \/>\nSince, \u0394APC and \u0394BQC are both equilateral triangles, as per given,<br \/>\n\u2234 \u0394APC ~ \u0394BQC [AAA similarity criterion]<br \/>\n\u2234 area(\u0394APC)\/area(\u0394BQC) = (AC<sup>2<\/sup>\/BC<sup>2<\/sup>) = AC<sup>2<\/sup>\/BC<sup>2<br \/>\n<\/sup>Since, Diagonal = \u221a2 side = \u221a2 BC = AC<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{\\sqrt{2}BC}{BC}&amp;space;\\right&amp;space;)^2&amp;space;=&amp;space;\\frac{2BC^2}{BC^2}&amp;space;=&amp;space;2\" alt=\"\\left ( \\frac{\\sqrt{2}BC}{BC} \\right )^2 = \\frac{2BC^2}{BC^2} = 2\" align=\"absmiddle\" \/><br \/>\n\u21d2 area (\u0394APC) = 2 \u00d7 area (\u0394BQC)<br \/>\n\u21d2 area (\u0394BQC) = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> area (\u0394APC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is<br \/>\n<\/strong>(A) 2 : 1<br \/>\n(B) 1 : 2<br \/>\n(C) 4 : 1<br \/>\n(D) 1 : 4<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong>Given<strong>,\u00a0<\/strong>\u0394ABC and \u0394BDE are two equilateral triangle. D is the midpoint of BC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6381\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans8.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"209\" height=\"188\" \/><br \/>\n\u2234 BD = DC = 1\/2 BC<br \/>\nLet each side of triangle is 2<em>a<\/em>.<br \/>\nAs, \u0394ABC ~ \u0394BDE<br \/>\n\u2234 Area(\u0394ABC)\/Area(\u0394BDE) = AB<sup>2<\/sup>\/BD<sup>2<\/sup>\u00a0= (2<em>a<\/em>)<sup>2<\/sup>\/(<em>a<\/em>)<sup>2<\/sup>\u00a0= 4<em>a<\/em><sup>2<\/sup>\/<em>a<\/em><sup>2<\/sup> = 4\/1 = 4 : 1<br \/>\nHence, the correct <strong>answer is (C)<\/strong>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio<br \/>\n<\/strong>(A) 2 : 3<br \/>\n(B) 4 : 9<br \/>\n(C) 81 : 16<br \/>\n(D) 16 : 81<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong>Given, Sides of two similar triangles are in the ratio 4 : 9.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6382\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans9.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"320\" height=\"144\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans9.png 320w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.4-Ans9-300x135.png 300w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><br \/>\nLet ABC and DEF are two similar triangles, such that,<br \/>\n\u0394ABC ~ \u0394DEF<br \/>\nAnd AB\/DE = AC\/DF = BC\/EF = 4\/9<br \/>\nAs, the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,<br \/>\n\u2234 Area(\u0394ABC)\/Area(\u0394DEF) = AB<sup>2<\/sup>\/DE<sup>2<br \/>\n<\/sup>\u2234 Area(\u0394ABC)\/Area(\u0394DEF) =\u00a0(4\/9)<sup>2\u00a0<\/sup>= 16\/81 = 16 : 81<br \/>\nHence, the correct <strong>answer is (D).<\/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 6.4 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":[1045,1079,1083,1044,1049,1048],"class_list":["post-6052","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-6-4-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 v27.5) - 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