{"id":1929,"date":"2022-11-04T08:29:33","date_gmt":"2022-11-04T08:29:33","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=1929"},"modified":"2022-11-04T08:29:33","modified_gmt":"2022-11-04T08:29:33","slug":"ncert-solutions-class-7-maths-chapter-6-the-triangle-and-its-properties-ex-6-5","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-6-the-triangle-and-its-properties-ex-6-5\/","title":{"rendered":"NCERT Solutions Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 7 Mathematics\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 6 (The Triangle and its Properties)<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 7 Mathematics <strong>Chapter &#8211; 6 The Triangle and its Properties <\/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; font-family: georgia, palatino, serif;\"><strong>Chapter : 6 The Triangles and its Properties<\/strong><\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-6-the-triangle-and-its-properties-ex-6-1\" target=\"_blank\" rel=\"noopener\"><span style=\"font-family: georgia, palatino, serif;\">NCERT Solution Class 7 Maths Exercise &#8211; 6.1<\/span><\/a><\/span><\/li>\n<li><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-6-the-triangle-and-its-properties-ex-6-2\" target=\"_blank\" rel=\"noopener\"><span style=\"font-family: georgia, palatino, serif;\">NCERT Solution Class 7 Maths Exercise &#8211; 6.2<\/span><\/a><\/span><\/span><\/li>\n<li><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-6-the-triangle-and-its-properties-ex-6-3\" target=\"_blank\" rel=\"noopener\"><span style=\"font-family: georgia, palatino, serif;\">NCERT Solution Class 7 Maths Exercise &#8211; 6.3<\/span><\/a><\/span><\/span><\/li>\n<li><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-6-the-triangle-and-its-properties-ex-6-4\" target=\"_blank\" rel=\"noopener\"><span style=\"font-family: georgia, palatino, serif;\">NCERT Solution Class 7 Maths Exercise &#8211; 6.4<\/span><\/a><\/span><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 6.5\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>1. PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong>Let us draw a rough sketch of right-angled triangle<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1984\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-6.5-Q1.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"237\" height=\"135\" \/><br \/>\nIn right angled triangle PQR, we have<br \/>\nBy the rule of Pythagoras Theorem,<br \/>\nQR<sup>2<\/sup>\u00a0= PQ<sup>2<\/sup>\u00a0+ PR<sup>2<\/sup><br \/>\n\u21d2 (10)<sup>2<\/sup>\u00a0+ (24)<sup>2<\/sup><br \/>\n\u21d2 100 + 576 = 676<br \/>\n\u2234 QR = \u221a<span id=\"MathJax-Element-1-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-1\" class=\"math\"><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"msqrt\"><span id=\"MathJax-Span-4\" class=\"mrow\"><span id=\"MathJax-Span-5\" class=\"mn\">676 <\/span><\/span><\/span><\/span><\/span><\/span>= 26 cm<br \/>\nThe, the required length of QR = 26 cm.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>2. ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong>Let us draw a rough sketch of right-angled triangle<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1985\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-6.5-Q2.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"249\" height=\"135\" \/><br \/>\nIn right angled \u2206ABC, we have<br \/>\nBy the rule of Pythagoras Theorem,<br \/>\nBC<sup>2<\/sup>\u00a0+ (7)<sup>2<\/sup>\u00a0= (25)<sup>2<\/sup><br \/>\n\u21d2 BC<sup>2<\/sup> + 49 = 625<br \/>\n\u21d2 BC<sup>2<\/sup> = 625 \u2013 49<br \/>\n\u21d2 BC<sup>2<\/sup> = 576<br \/>\n\u2234 BC = <span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-6\" class=\"math\"><span id=\"MathJax-Span-7\" class=\"mrow\"><span id=\"MathJax-Span-8\" class=\"msqrt\">\u221a<\/span><\/span><\/span><\/span><span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-6\" class=\"math\"><span id=\"MathJax-Span-7\" class=\"mrow\"><span id=\"MathJax-Span-8\" class=\"msqrt\"><span id=\"MathJax-Span-9\" class=\"mrow\"><span id=\"MathJax-Span-10\" class=\"mn\">576<\/span><\/span><\/span><\/span><\/span><\/span>\u00a0= 24 cm<br \/>\nThus, the required length of BC = 24 cm.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1986\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-6.5-Q3.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"225\" height=\"198\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong>Here, the ladder forms a right angled triangle.<br \/>\nBy the rule of Pythagoras Theorem,<br \/>\n\u2234 a<sup>2<\/sup>\u00a0+ (12)<sup>2<\/sup>\u00a0= (15)<sup>2<\/sup><br \/>\n\u21d2 a<sup>2<\/sup>+ 144 = 225<br \/>\n\u21d2 a2 = 225 \u2013 144<br \/>\n\u21d2 a<sup>2<\/sup>\u00a0= 81<br \/>\n\u2234 a = <span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-11\" class=\"math\"><span id=\"MathJax-Span-12\" class=\"mrow\"><span id=\"MathJax-Span-13\" class=\"msqrt\">\u221a<\/span><\/span><\/span><\/span>\u00a0<span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-11\" class=\"math\"><span id=\"MathJax-Span-12\" class=\"mrow\"><span id=\"MathJax-Span-13\" class=\"msqrt\"><span id=\"MathJax-Span-14\" class=\"mrow\"><span id=\"MathJax-Span-15\" class=\"mn\">81 <\/span><\/span><\/span><\/span><\/span><\/span>= 9 m<br \/>\nThus, the distance of the foot from the ladder = 9m<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>4. Which of the following can be the sides of a right triangle?<br \/>\n<\/strong><strong>(i) 2.5 cm, 6.5 cm, 6 cm.<br \/>\n<\/strong><strong>(ii) 2 cm, 2 cm, 5 cm.<br \/>\n<\/strong><strong>(iii) 1.5 cm, 2cm, 2.5 cm.<br \/>\n<\/strong><strong>In the case of right-angled triangles, identify the right angles.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(i) 2.5 cm, 6.5 cm, 6 cm.<br \/>\n<\/strong>Square of the longer side = (6.5)<sup>2<\/sup>\u00a0= 42.25 cm.<br \/>\nSum of the square of other two sides<br \/>\n\u21d2 (2.5)<sup>2<\/sup>\u00a0+ (6)<sup>2<\/sup><br \/>\n\u21d2 6.25 + 36<br \/>\n\u21d2 42.25 cm.<br \/>\nSince, the square of the longer side in a triangle is equal to the sum of the squares of other two sides.<br \/>\n\u2234 The given sides form a right triangle.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(ii) 2 cm, 2 cm, 5 cm.<br \/>\n<\/strong>Square of the longer side = (5)<sup>2<\/sup>\u00a0= 25 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Sum of the square of other two sides<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u21d2 (2)<sup>2<\/sup>\u00a0+ (2)<sup>2<\/sup>\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u21d2 4 + 4 <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u21d2 8 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Since 25 cm \u2260 8 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u2234 The given sides do not form a right triangle.<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(iii) 1.5 cm, 2 cm, 2.5 cm<br \/>\n<\/strong>Square of the longer side = (2.5)<sup>2<\/sup>\u00a0= 6.25 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Sum of the square of other two sides<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u21d2 (1.5)<sup>2<\/sup>\u00a0+ (2)<sup>2<\/sup>\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u21d2 2.25 + 4<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u21d2 6.25 cm<br \/>\nSince the square of longer side in a triangle is equal to the sum of square of other two sides.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u2234 The given sides form a right triangle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>5. A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong>Let ABC is the triangle and B is the point where tree is broken at the height 5 m from the ground.<br \/>\nTree top touches the ground at a distance of AC = 12 m from the base of the tree,<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1987\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-6.5-Q5.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"263\" height=\"138\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">By observing the figure we came to conclude that right angle triangle is formed at A.<br \/>\nFrom the rule of Pythagoras theorem,<br \/>\nBC<sup>2<\/sup>\u00a0= AB<sup>2<\/sup>\u00a0+ AC<sup>2<br \/>\n<\/sup>BC<sup>2<\/sup>\u00a0= 5<sup>2<\/sup>\u00a0+ 12<sup>2<br \/>\n<\/sup>BC<sup>2<\/sup>\u00a0= 25 + 144<br \/>\nBC<sup>2<\/sup>\u00a0= 169<br \/>\nBC = \u221a169<br \/>\nBC = 13 m<br \/>\nThen, the original height of the tree = AB + BC<br \/>\n= 5 + 13<br \/>\n= 18 m<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>6. Angles Q and R of a \u0394PQR are 25<sup>o<\/sup>\u00a0and 65<sup>o<\/sup>. <\/strong><strong>Write which of the following is true:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1988 alignright\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-6.5-Q6.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"292\" height=\"130\" \/><br \/>\n<\/strong><strong>(i) PQ<sup>2<\/sup>\u00a0+ QR<sup>2<\/sup>\u00a0= RP<sup>2<br \/>\n<\/sup><\/strong><strong>(ii) PQ<sup>2<\/sup>\u00a0+ RP<sup>2<\/sup>\u00a0= QR<sup>2<br \/>\n<\/sup><\/strong><strong>(iii) RP<sup>2<\/sup>\u00a0+ QR<sup>2<\/sup>\u00a0= PQ<sup>2<\/sup><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">We know that<br \/>\n\u2220P + \u2220Q + \u2220R = 180\u00b0 (Angle sum property)<br \/>\n\u2220P + 25\u00b0 + 65\u00b0 = 180\u00b0<br \/>\n\u2220P + 90\u00b0 = 180\u00b0<br \/>\n\u2220P = 180\u00b0 \u2013 90\u00b0 = 90\u00b0<br \/>\n\u2206PQR is a right triangle, right angled at P<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(i) PQ<sup>2<\/sup>\u00a0+ QR<sup>2<\/sup>\u00a0= RP<sup>2<br \/>\n<\/sup><\/strong>Not True<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u2234 PQ<sup>2<\/sup>\u00a0+ QR<sup>2<\/sup>\u00a0\u2260 RP<sup>2<\/sup>\u00a0(By Pythagoras property)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(ii) PQ<sup>2<\/sup>\u00a0+ RP<sup>2<\/sup>\u00a0= QR<sup>2<br \/>\n<\/sup><\/strong>True<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u2234 PQ<sup>2<\/sup>\u00a0+ RP<sup>2<\/sup>\u00a0= QP<sup>2<\/sup>\u00a0(By Pythagoras property)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(iii) RP<sup>2<\/sup>\u00a0+ QR<sup>2<\/sup>\u00a0= PQ<sup>2<br \/>\n<\/sup><\/strong>Not True<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u2234 RP<sup>2<\/sup>\u00a0+ QR<sup>2<\/sup>\u00a0\u2260 PQ<sup>2<\/sup>\u00a0(By Pythagoras property)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1989\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-6.5-Q7.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"171\" height=\"131\" \/><br \/>\n<\/strong>Let ABCD be the rectangular plot.<br \/>\nThen, AB = 40 cm and AC = 41 cm<br \/>\nBC =?<br \/>\nAccording to Pythagoras theorem,<br \/>\nFrom right angle triangle ABC, we have:<br \/>\n\u21d2 AC<sup>2<\/sup>\u00a0= AB<sup>2<\/sup>\u00a0+ BC<sup>2<br \/>\n<\/sup>\u21d2 41<sup>2<\/sup>\u00a0= 40<sup>2<\/sup>\u00a0+ BC<sup>2<br \/>\n<\/sup>\u21d2 BC<sup>2<\/sup>\u00a0= 41<sup>2<\/sup>\u00a0\u2013 40<sup>2<br \/>\n<\/sup>\u21d2 BC<sup>2<\/sup>\u00a0= 1681 \u2013 1600<br \/>\n\u21d2 BC<sup>2<\/sup>\u00a0= 81<br \/>\n\u21d2 BC = \u221a81<br \/>\n\u21d2 BC = 9 cm<br \/>\nHence, the perimeter of the rectangle plot = 2 (length + breadth)<br \/>\nWhere, length = 40 cm, breadth = 9 cm<br \/>\nThen,<br \/>\n= 2(40 + 9)<br \/>\n= 2 \u00d7 49<br \/>\n= 98 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1990\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-6.5-Q8.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"195\" height=\"216\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Let ABCD be a rhombus whose diagonals intersect each other at O such that AC = 16 cm and BD = 30 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Since, the diagonals of a rhombus bisect each other at 90\u00b0.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u2234 OA = OC = 8 cm and OB = OD = 15 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">In right \u2206OAB,<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">According to Pythagoras theorem,<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u21d2 AB<sup>2<\/sup>\u00a0= OA<sup>2<\/sup>\u00a0+ OB<sup>2<\/sup> <\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u21d2 AB<sup>2<\/sup>\u00a0= (8)<sup>2<\/sup>+ (15)<sup>2<\/sup>\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u21d2 AB<sup>2<\/sup>\u00a0= 64 + 225<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u21d2 AB<sup>2<\/sup>\u00a0= 289<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u21d2 AB = <span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-26\" class=\"math\"><span id=\"MathJax-Span-27\" class=\"mrow\"><span id=\"MathJax-Span-28\" class=\"msqrt\">\u221a<\/span><\/span><\/span><\/span><span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-26\" class=\"math\"><span id=\"MathJax-Span-27\" class=\"mrow\"><span id=\"MathJax-Span-28\" class=\"msqrt\"><span id=\"MathJax-Span-29\" class=\"mrow\"><span id=\"MathJax-Span-30\" class=\"mn\">289<br \/>\n<\/span><\/span><\/span><\/span><\/span><\/span>\u21d2 AB = 17 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Since AB = BC = CD = DA (Property of rhombus)<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u2234 Required perimeter of rhombus<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 4 \u00d7 side = 4 \u00d7 17 = 68 cm.<\/span><\/p>\n<p style=\"text-align: justify;\">\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 7 Mathematics\u00a0 Chapter &#8211; 6 (The Triangle and its Properties) The NCERT Solutions in English Language for Class 7 Mathematics Chapter &#8211; 6 The Triangle and its Properties Exercise 6.5 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 6 The Triangles and its [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[219],"tags":[296,297,283,5],"class_list":["post-1929","post","type-post","status-publish","format-standard","hentry","category-class-7-maths","tag-ncert-class-7-maths-chapter-6-the-triangle-and-its-properties-in-english","tag-ncert-solutions-class-7-maths-chapter-6-in-english","tag-ncert-solutions-class-7-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - 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