{"id":4434,"date":"2023-02-10T07:38:16","date_gmt":"2023-02-10T07:38:16","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4434"},"modified":"2023-02-10T07:39:06","modified_gmt":"2023-02-10T07:39:06","slug":"ncert-solutions-class-9-maths-chapter-8-quadrilaterals-ex-8-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-8-quadrilaterals-ex-8-2\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 8 (Quadrilaterals)\u00a0<\/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 9 Mathematics\u00a0<strong>Chapter &#8211; 8 Quadrilaterals <\/strong>Exercise 8.2 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 8: Quadrilaterals<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-8-quadrilaterals-ex-8-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 8.1<\/a><\/li>\n<\/ul>\n<div>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 8.2<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4453\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Q1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"241\" height=\"263\" \/><br \/>\n(i) <\/strong>SR || AC and SR = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>\u00a0 AC<br \/>\n<strong>(ii) <\/strong>PQ = SR<strong><br \/>\n(iii) <\/strong>PQRS is a parallelogram.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong>We will use the\u00a0mid-point theorem\u00a0here. It\u00a0that states that\u00a0the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it.<br \/>\n<strong>(i) SR || AC and SR = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> AC<br \/>\n<\/strong>In \u0394ADC, S and R are the mid-points of sides AD and CD respectively. Thus, by using the mid-point theorem <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 SR || AC and SR = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00a0<\/strong>AC\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (1)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) PQ = SR<br \/>\n<\/strong>In \u0394ABC, P and Q are mid-points of sides AB and BC. Therefore, by using the mid-point theorem,<\/span><br \/>\n<span style=\"color: #000000;\">PQ || AC and PQ = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/strong>AC\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (2)<\/span><br \/>\n<span style=\"color: #000000;\">Using Equations (1) and (2), we obtain PQ || SR and PQ = SR &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (3) <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 PQ = SR<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) PQRS is a parallelogram.<br \/>\n<\/strong>From Equation (3), we obtained PQ || SR and PQ = SR<\/span><br \/>\n<span style=\"color: #000000;\">Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal. Hence, PQRS is a parallelogram.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.<br \/>\n<\/strong><strong>Answer &#8211; <\/strong>We will use the mid-point theorem\u00a0here. It states that the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4454\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"326\" height=\"186\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans2.png 685w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans2-300x171.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans2-480x273.png 480w\" sizes=\"auto, (max-width: 326px) 100vw, 326px\" \/><br \/>\nIn \u0394ABC, P and Q are the\u00a0mid-points\u00a0of sides AB and BC respectively.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 PQ || AC and PQ = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/strong>AC (Using mid-point theorem) &#8212;&#8212;&#8212;&#8212;- (1)<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ADC,<\/span><br \/>\n<span style=\"color: #000000;\">R and S are the mid-points of CD and AD respectively.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 RS || AC and RS = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/strong>AC (Using mid-point theorem) &#8212;&#8212;&#8212;&#8212;- (2)<\/span><br \/>\n<span style=\"color: #000000;\">From Equations (1) and (2), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">PQ || RS and PQ = RS<\/span><br \/>\n<span style=\"color: #000000;\">Since in quadrilateral PQRS, one pair of opposite sides is equal and\u00a0parallel\u00a0to each other, it is a\u00a0parallelogram.<\/span><br \/>\n<span style=\"color: #000000;\">Since the sides of a rhombus are equal, AB = BC<\/span><br \/>\n<span style=\"color: #000000;\"><strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/strong>\u00d7 AB = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/strong>\u00d7 BC<\/span><br \/>\n<span style=\"color: #000000;\">PB = BQ (P and Q are the mid-points of sides AB and BC respectively)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220QPB = \u2220PQB (Sides opposite to equal angles are equal) &#8212;&#8212;&#8212;&#8212;- (3)<\/span><br \/>\n<span style=\"color: #000000;\">In\u00a0\u0394APS\u00a0and\u00a0\u0394CQR,\u00a0\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">AP = CQ\u00a0(P and Q are the mid-points of sides AB and BC respectively)<\/span><br \/>\n<span style=\"color: #000000;\">AS = CR\u00a0(S\u00a0and R\u00a0are the mid-points of sides AD\u00a0and CD\u00a0respectively)<\/span><br \/>\n<span style=\"color: #000000;\">PS = QR (Opposite sides of a parallelogram are equal)<\/span><br \/>\n<span style=\"color: #000000;\">BY SSS congruency,\u00a0\u0394APS\u00a0\u2245 \u0394CQR<\/span><br \/>\n<span style=\"color: #000000;\">So, \u2220APS = \u2220CQR (By CPCT) &#8212;&#8212;&#8212;&#8212;- (4)<\/span><br \/>\n<span style=\"color: #000000;\">Since AB is a straight line,\u00a0\u2220APS +\u00a0\u2220SPQ +\u00a0\u2220QPB = 180<sup>\u00b0<br \/>\n<\/sup>Since BC is a\u00a0straight line,\u00a0\u2220PQB +\u00a0\u2220PQR +\u00a0\u2220CQR = 180<sup>\u00b0<br \/>\n<\/sup>\u2220APS +\u00a0\u2220SPQ +\u00a0\u2220QPB =\u00a0\u2220PQB +\u00a0\u2220PQR +\u00a0\u2220CQR<\/span><br \/>\n<span style=\"color: #000000;\">\u2220APS +\u00a0\u2220SPQ +\u00a0\u2220QPB =\u00a0\u2220QPB +\u00a0\u2220PQR +\u00a0\u2220APS (By equations (3) and (4))<\/span><br \/>\n<span style=\"color: #000000;\">\u2220SPQ = \u2220PQR &#8212;&#8212;&#8212;&#8212;- (5)<\/span><br \/>\n<span style=\"color: #000000;\">Since \u2220SPQ and\u00a0\u2220PQR are interior angles on the same side of the transversal PQ, they form a pair of supplementary angles.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220SPQ +\u00a0\u2220PQR =\u00a0180<sup>\u00b0<br \/>\n<\/sup>2 \u2220SPQ = 180<sup>\u00b0<\/sup>\u00a0[From (5)]<\/span><br \/>\n<span style=\"color: #000000;\">\u2220SPQ = 90<sup>\u00b0<br \/>\n<\/sup>Clearly, PQRS is a parallelogram having one of its interior angles as 90\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">Hence, PQRS is a rectangle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus.<\/strong><br \/>\n<strong>Answer &#8211;\u00a0<\/strong>We will use the\u00a0mid-point theorem\u00a0here. It states that the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4455\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"274\" height=\"161\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans3.png 1022w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans3-300x176.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans3-768x451.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans3-480x282.png 480w\" sizes=\"auto, (max-width: 274px) 100vw, 274px\" \/><br \/>\nLet us join AC and BD. In \u25b3ABC,<\/span><br \/>\n<span style=\"color: #000000;\">P and Q are the\u00a0mid-points\u00a0of AB and BC respectively.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 PQ || AC and PQ = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/><\/strong> AC (Mid-point theorem) &#8212;&#8212;&#8212;&#8212;- (1)<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, in \u25b3ADC,<\/span><br \/>\n<span style=\"color: #000000;\">SR || AC and SR = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/strong>AC (Mid-point theorem) &#8212;&#8212;&#8212;&#8212;- (2)<\/span><br \/>\n<span style=\"color: #000000;\">Clearly, PQ || SR and PQ = SR [From equation (1) and (2)]<\/span><br \/>\n<span style=\"color: #000000;\">Since in quadrilateral PQRS, one pair of opposite sides are equal and parallel to each other, it is a\u00a0parallelogram.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 PS || QR and PS = QR (Opposite sides of the parallelogram) &#8212;&#8212;&#8212;&#8212;- (3)<\/span><br \/>\n<span style=\"color: #000000;\">In \u25b3BCD, Q and R are the mid-points of side BC and CD respectively.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 QR || BD and QR = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/strong>BD (Mid-point theorem) &#8212;&#8212;&#8212;&#8212;- (4)<\/span><br \/>\n<span style=\"color: #000000;\">However, the diagonals of a rectangle are equal.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AC = BD &#8212;&#8212;&#8212;&#8212;-\u00a0 (5)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, QR = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/strong>AC<\/span><br \/>\n<span style=\"color: #000000;\">Also, in\u00a0\u25b3BAD<\/span><br \/>\n<span style=\"color: #000000;\">PS || BD and PS = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/strong>BD<\/span><br \/>\n<span style=\"color: #000000;\">Thus, QR = PS &#8212;&#8212;&#8212;&#8212;- (6)<\/span><br \/>\n<span style=\"color: #000000;\">By using Equations (1), (2), (3), (4), and (5), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">PQ = QR = SR = PS<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, PQRS is a rhombus.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4456\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"265\" height=\"240\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong><strong>Given: <\/strong>ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.<br \/>\n<strong>To prove :\u00a0<\/strong>F is the mid-point of BC.<br \/>\nBD intersected EF at G.<br \/>\nIn \u0394BAD,<br \/>\nE is the mid point of AD and also EG || AB.<br \/>\nThus, G is the mid point of BD (Converse of mid point theorem)<br \/>\nNow,<br \/>\nIn \u0394BDC,<br \/>\nG is the mid point of BD and also GF || AB || DC.<br \/>\nThus, F is the mid point of BC (Converse of mid point theorem)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD, respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4457\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Q5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"265\" height=\"244\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong><strong>Given : <\/strong>ABCD is a parallelogram. E and F are the mid-points of sides AB and CD, respectively.<br \/>\n<strong>To Prove :\u00a0<\/strong>AF and EC trisect the diagonal BD.<br \/>\nABCD is a parallelogram, AB || CD<br \/>\nalso, AE || FC<br \/>\nNow,<br \/>\nAB = CD (Opposite sides of parallelogram ABCD)<br \/>\n\u21d2 <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/strong>AB = <strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/strong>CD<br \/>\n\u21d2 AE = FC (E and F are midpoints of side AB and CD)<br \/>\nAECF is a parallelogram (AE and CF are parallel and equal to each other)<br \/>\nAF || EC (Opposite sides of a parallelogram)<br \/>\nNow,<br \/>\nIn \u0394DQC,<br \/>\nF is mid point of side DC and FP || CQ (as AF || EC).<br \/>\nP is the mid-point of DQ (Converse of mid-point theorem)<br \/>\n\u21d2 DP = PQ\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;\u2014 (1)<br \/>\nSimilarly,<br \/>\nIn \u0394APB,<br \/>\nE is midpoint of side AB and EQ || AP (as AF || EC).<br \/>\nQ is the mid-point of PB (Converse of mid-point theorem)<br \/>\n\u21d2 PQ = QB\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;\u2014 (2)<br \/>\nFrom equations (1) and (2),<br \/>\nDP = PQ = BQ<br \/>\nHence, the line segments AF and EC trisect the diagonal BD.<br \/>\nHence Proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6<\/strong>.\u00a0<strong>Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.<\/strong><br \/>\n<strong>Answer &#8211; <\/strong>Let ABCD be a quadrilateral and P, Q, R and S the mid points of AB, BC, CD and DA, respectively.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4458\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"343\" height=\"320\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans6.png 418w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans6-300x280.png 300w\" sizes=\"auto, (max-width: 343px) 100vw, 343px\" \/><br \/>\nIn\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\">\u25b3<\/span><span class=\"mord\"><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">D<\/span><span class=\"mord mathnormal\">C<\/span><\/span><span class=\"mpunct\">,<\/span><span class=\"mord mathnormal\">S<\/span><\/span><\/span><\/span>\u00a0is the mid-point of\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">D<\/span><\/span><\/span><\/span>\u00a0and\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">R<\/span><\/span><\/span><\/span>\u00a0is the mid-point of\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">C<\/span><span class=\"mord mathnormal\">D<br \/>\n<\/span><\/span><\/span><\/span>In\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\">\u25b3<\/span><span class=\"mord\"><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">C<\/span><\/span><span class=\"mpunct\">,<\/span><span class=\"mord mathnormal\">P<\/span><\/span><\/span><\/span>\u00a0is the mid-point of\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">B<\/span><\/span><\/span><\/span>\u00a0and\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">Q<\/span><\/span><\/span><\/span>\u00a0is the mid-point of\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">C<br \/>\n<\/span><\/span><\/span><\/span>Line segments joining the mid-points of two sides of a triangle is parallel to the third side and is half of of it.<\/span><br \/>\n<span style=\"color: #000000;\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mrel amsrm\">\u2234 <\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">S<\/span><span class=\"mord mathnormal\">R || <\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">C<\/span><\/span><\/span><\/span>\u00a0and\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">S<\/span><span class=\"mord mathnormal\">R <\/span><span class=\"mrel\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">C<\/span><\/span><\/span><\/span>\u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mopen\">(<\/span><span class=\"mord\">1<\/span><span class=\"mclose\">)<\/span><\/span><\/span><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mrel amsrm\">\u2234 <\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><span class=\"mord mathnormal\">Q || <\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">C<\/span><\/span><\/span><\/span>\u00a0and\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><span class=\"mord mathnormal\">Q <\/span><span class=\"mrel\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00a0<\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">C<\/span><\/span><\/span><\/span>\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mopen\">(<\/span><span class=\"mord\">2<\/span><span class=\"mclose\">)<br \/>\n<\/span><\/span><\/span><\/span>From\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mopen\">(<\/span><span class=\"mord\">1<\/span><span class=\"mclose\">)<\/span><\/span><\/span><\/span>\u00a0and\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mopen\">(<\/span><span class=\"mord\">2<\/span><span class=\"mclose\">)<br \/>\n<\/span><\/span><\/span><\/span><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mrel\">\u21d2 <\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><span class=\"mord mathnormal\">Q <\/span><span class=\"mrel\">= <\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">S<\/span><span class=\"mord mathnormal\">R<\/span><\/span><\/span><\/span>\u00a0and\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><span class=\"mord mathnormal\">Q <\/span><span class=\"mrel\">|| <\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">S<\/span><span class=\"mord mathnormal\">R<br \/>\n<\/span><\/span><\/span><\/span>So, In <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><span class=\"mord mathnormal\">Q<\/span><span class=\"mord mathnormal\">R<\/span><span class=\"mord mathnormal\">S<\/span><span class=\"mpunct\">,<br \/>\n<\/span><\/span><\/span><\/span>one pair of opposite sides is parallel and equal.<\/span><br \/>\n<span style=\"color: #000000;\">Hence,\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><span class=\"mord mathnormal\">Q<\/span><span class=\"mord mathnormal\">R<\/span><span class=\"mord mathnormal\">S<\/span><\/span><\/span><\/span>\u00a0is a parallelogram.<\/span><br \/>\n<span style=\"color: #000000;\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><span class=\"mord mathnormal\">R<\/span><\/span><\/span><\/span>\u00a0and\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">S<\/span><span class=\"mord mathnormal\">Q<\/span><\/span><\/span><\/span>\u00a0are diagonals of parallelogram\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><span class=\"mord mathnormal\">Q<\/span><span class=\"mord mathnormal\">R<\/span><span class=\"mord mathnormal\">S<br \/>\n<\/span><\/span><\/span><\/span>So, <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">O<\/span><span class=\"mord mathnormal\">P<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">O<\/span><span class=\"mord mathnormal\">R<\/span><\/span><\/span><\/span>\u00a0and\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">O<\/span><span class=\"mord mathnormal\">Q<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">O<\/span><span class=\"mord mathnormal\">S<\/span><\/span><\/span><\/span>\u00a0since diagonals of a parallelogram bisect\u00a0each other.<\/span><br \/>\n<span style=\"color: #000000;\">Hence proved.\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that<br \/>\n(i) <\/strong>D is the mid-point of AC<br \/>\n<strong>(ii) <\/strong>MD \u22a5 AC<strong><br \/>\n(iii) <\/strong>CM = MA = \u00bd AB<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong>By converse of\u00a0mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle that is parallel to another side bisects the third side.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4459\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans7.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"216\" height=\"285\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans7.png 586w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans7-228x300.png 228w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.2-Ans7-480x632.png 480w\" sizes=\"auto, (max-width: 216px) 100vw, 216px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) <\/strong>D is the mid-point of AC<br \/>\nIn \u0394ABC,<\/span><br \/>\n<span style=\"color: #000000;\">It is given that M is the mid-point of AB and MD || BC.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234\u00a0D is the mid-point of AC. [Converse of mid-point theorem]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) <\/strong>MD \u22a5 AC<br \/>\nAs DM || CB and AC is a transversal,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220MDC + \u2220DCB = 180\u00b0 [Co-interior angles]<\/span><br \/>\n<span style=\"color: #000000;\">\u2220MDC + 90\u00b0 = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220MDC = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 MD \u22a5 AC<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) <\/strong>CM = MA = \u00bd AB<br \/>\nJoin MC<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394AMD and \u0394CMD,<\/span><br \/>\n<span style=\"color: #000000;\">AD = CD (D is the mid-point of side AC)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ADM = \u2220CDM (Each 90\u00b0)<\/span><br \/>\n<span style=\"color: #000000;\">DM = DM (Common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394AMD \u2245 \u0394CMD (By\u00a0SAS congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, AM = CM (By\u00a0CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">However, we also know that AM = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> AB (M is the mid-point of AB)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 CM = AM = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> AB<\/span><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 8 (Quadrilaterals)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 8 Quadrilaterals Exercise 8.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 8: Quadrilaterals NCERT Solution Class 9 Maths Ex &#8211; 8.1 Exercise &#8211; 8.2 1. ABCD [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[715,702,186,716,701,5],"class_list":["post-4434","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-8-quadrilaterals-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-8-in-english","tag-ncert-solutions-class-9-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) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 9 Maths\u00a0Chapter - 8 (Quadrilaterals)\u00a0The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter - 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