{"id":581566,"date":"2022-02-24T03:48:38","date_gmt":"2022-02-24T03:48:38","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=581566"},"modified":"2022-02-25T09:43:11","modified_gmt":"2022-02-25T09:43:11","slug":"maharashtra-board-solutions-class-9-maths-part-2-chapter-5-1-quadrilaterals","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-5-1-quadrilaterals\/","title":{"rendered":"Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.1- Quadrilaterals"},"content":{"rendered":"\n<p>Class 9: Maths Chapter 5.1 solutions. Complete Class 9 Maths Chapter 5.1 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-maharashtra-board-solutions-class-9-maths-part-2-chapter-5-1-quadrilaterals\">Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.1- Quadrilaterals<\/h2>\n\n\n\n<p>Maharashtra Board 9th Maths Chapter 5.1, Class 9 Maths Chapter 5.1 solutions<\/p>\n\n\n\n<p><strong>Question 1.<br>Diagonals of a parallelogram WXYZ intersect each other at point O. If \u2220XYZ\u2220 = 135\u00b0, then measure of \u2220XWZ and \u2220YZW? If l(OY) = 5 cm, then l(WY) = ?<br>Solution:<\/strong><br><img loading=\"lazy\" decoding=\"async\" width=\"173\" height=\"113\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/02\/Maharashtra-Board-Class-9-Maths-Solutions-Chapter-5-Quadrilaterals-Practice-Set-5.1-1.jpg\" alt=\"Maharashtra Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.1 1\"><br>i. \u2220XYZ = 135\u00b0<br>\u25a1WXYZ is a parallelogram.<br>\u2220XWZ = \u2220XYZ<br>\u2234 \u2220XWZ = 135\u00b0 \u2026..(i)<\/p>\n\n\n\n<p>ii. \u2220YZW + \u2220XYZ = 180\u00b0 [Adjacent angles of a parallelogram are supplementary]<br>\u2234 \u2220YZW + 135\u00b0= 180\u00b0 [From (i)]<br>\u2234 \u2220YZW = 180\u00b0- 135\u00b0<br>\u2234 \u2220YZW = 45\u00b0<\/p>\n\n\n\n<p>iii. l(OY) = 5 cm [Given]<br>l(OY) = 12 l(WY) [Diagonals of a parallelogram bisect each other]<br>\u2234 l(WY) = 2 x l(OY)<br>= 2 x 5<br>\u2234 l(WY) = 10 cm<br>\u2234\u2220XWZ = 135\u00b0, \u2220YZW = 45\u00b0, l(WY) = 10 cm<\/p>\n\n\n\n<p><strong>Question 2.<br>In a parallelogram ABCD, if \u2220A = (3x + 12)\u00b0, \u2220B = (2x \u2013 32)\u00b0, then liptl the value of x and the measures of \u2220C and \u2220D.<br>Solution:<\/strong><br><img loading=\"lazy\" decoding=\"async\" width=\"168\" height=\"114\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/02\/Maharashtra-Board-Class-9-Maths-Solutions-Chapter-5-Quadrilaterals-Practice-Set-5.1-2.jpg\" alt=\"Maharashtra Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.1 2\"><br>\u25a1ABCD is a parallelogram. [Given]<br>\u2234 \u2220A + \u2220B = 180\u00b0 [Adjacent angles of a parallelogram are supplementary],<br>\u2234 (3x + 12)\u00b0 + (2x-32)\u00b0 = 180\u00b0<br>\u2234 3x + 12 + 2x \u2013 32 = 180<br>\u2234 5x \u2013 20 = 180<br>\u2234 5x= 180 + 20<br>\u2234 5x = 200<br>\u2234 x = 2005<br>\u2234 x = 40<\/p>\n\n\n\n<p>ii. \u2220A = (3x + 12)\u00b0<br>= [3(40) + 12]\u00b0<br>=(120 +12)\u00b0= 132\u00b0<br>\u2220B = (2x \u2013 32)\u00b0<br>= [2(40) \u2013 32]\u00b0<br>= (80 \u2013 32)\u00b0 = 48\u00b0<br>\u2234 \u2220C = \u2220A = 132\u00b0<br>\u2220D = \u2220B = 48\u00b0 [Opposite angles of a parallelogram]<br>\u2234 The value of x is 40, and the measures of \u2220C and \u2220D are 132\u00b0 and 48\u00b0 respectively.<\/p>\n\n\n\n<p><strong>Question 3.<br>Perimeter of a parallelogram is 150 cm. One of its sides is greater than the other side by 25 cm. Find the lengths of all sides.<br>Solution:<br><img loading=\"lazy\" decoding=\"async\" width=\"162\" height=\"120\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/02\/Maharashtra-Board-Class-9-Maths-Solutions-Chapter-5-Quadrilaterals-Practice-Set-5.1-3.jpg\" alt=\"Maharashtra Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.1 3\"><\/strong><br>i. Let \u25a1ABCD be the parallelogram and the length of AD be x cm.<br>One side is greater than the other by 25 cm.<br>\u2234 AB = x + 25 cm<br>AD = BC = x cm<br>AB = DC = (x + 25) cm [Opposite angles of a parallelogram]<\/p>\n\n\n\n<p>ii. Perimeter of \u25a1ABCD = 150 cm [Given]<br>\u2234 AB + BC + DC + AD = 150<br>\u2234 (x + 25) +x + (x + 25) + x \u2013 150<br>\u2234 4x + 50 = 150<br>\u2234 4x = 150 \u2013 50<br>\u2234 4x = 100<br>\u2234 x = 1004<br>\u2234 x = 25<\/p>\n\n\n\n<p>iii. AD = BC = x = 25 cm<br>AB = DC = x + 25 = 25 + 25 = 50 cm<br>\u2234 The lengths of the sides of the parallelogram are 25 cm, 50 cm, 25 cm and 50 cm.<\/p>\n\n\n\n<p><strong>Question 4.<br>If the ratio of measures of two adjacent angles of a parallelogram is 1 : 2, find the measures of all angles of the parallelogram.<br>Solution:<\/strong><br><img loading=\"lazy\" decoding=\"async\" width=\"168\" height=\"130\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/02\/Maharashtra-Board-Class-9-Maths-Solutions-Chapter-5-Quadrilaterals-Practice-Set-5.1-4.jpg\" alt=\"Maharashtra Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.1 4\"><br>i. Let \u25a1ABCD be the parallelogram.<br>The ratio of measures of two adjacent angles of a parallelogram is 1 : 2.<br>Let the common multiple be x.<br>\u2234 \u2220A = x\u00b0 and \u2220B = 2x\u00b0<br>\u2220A + \u2220B = 180\u00b0 [Adjacent angles of a parallelogram are supplementary]<br>\u2234 x + 2x = 180<br>\u2234 3x = 180<br>\u2234 x = 1803<br>\u2234 x = 60<\/p>\n\n\n\n<p>ii. \u2220A = x\u00b0 = 60\u00b0<br>\u2220B = 2x\u00b0 = 2 x 60\u00b0 = 120\u00b0<br>\u2220A = \u2220C = 60\u00b0<br>\u2220B = \u2220D= 120\u00b0 [Opposite angles of a parallelogram]<br>\u2234 The measures of the angles of the parallelogram are 60\u00b0, 120\u00b0, 60\u00b0 and 120\u00b0.<\/p>\n\n\n\n<p><strong>Question 5.<br>Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO show that \u25a1ABCD is a rhombus.<\/strong><br><img loading=\"lazy\" decoding=\"async\" width=\"169\" height=\"115\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/02\/Maharashtra-Board-Class-9-Maths-Solutions-Chapter-5-Quadrilaterals-Practice-Set-5.1-5.jpg\" alt=\"Maharashtra Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.1 5\"><br>Given: AO = 5, BO = 12 and AB = 13.<br>To prove: \u25a1ABCD is a rhombus.<br>Solition:<br>Proof:<br>AO = 5, BO = 12, AB = 13 [Given]<br>AO<sup>2<\/sup> + BO<sup>2<\/sup> = 5<sup>2<\/sup> + 12<sup>2<\/sup><br>= 25 + 144<br>\u2234 AO<sup>2<\/sup> + BO<sup>2<\/sup> = 169 \u2026..(i)<br>AB<sup>2<\/sup> = 13<sup>2<\/sup> = 169 \u2026.(ii)<br>\u2234 AB<sup>2<\/sup> = AO<sup>2<\/sup> + BO<sup>2<\/sup> [From (i) and (ii)]<br>\u2234 \u2206AOB is a right-angled triangle. [Converse of Pythagoras theorem]<br>\u2234 \u2220AOB = 90\u00b0<br>\u2234 seg AC \u22a5 seg BD \u2026..(iii) [A-O-C]<br>\u2234 In parallelogram ABCD,<br>\u2234 seg AC \u22a5 seg BD [From (iii)]<br>\u2234 \u25a1ABCD is a rhombus. [A parallelogram is a rhombus perpendicular to each other]<\/p>\n\n\n\n<p><strong>Question 6.<br>In the adjoining figure, \u25a1PQRS and \u25a1ABCR are two parallelograms. If \u2220P = 110\u00b0, then find the measures of all the angles of \u25a1ABCR.<br>Solution:<\/strong><br><img loading=\"lazy\" decoding=\"async\" width=\"262\" height=\"108\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/02\/Maharashtra-Board-Class-9-Maths-Solutions-Chapter-5-Quadrilaterals-Practice-Set-5.1-6.jpg\" alt=\"Maharashtra Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.1 6\"><br>\u25a1PQRS is a parallelogram. [Given]<br>\u2234 \u2220R = \u2220P [Opposite angles of a parallelogram]<br>\u2234 \u2220R = 110\u00b0 \u2026..(iii)<br>\u25a1ABCR is a parallelogram. [Given]<br>\u2234 \u2220A + \u2220R= 180\u00b0 [Adjacent angles of a parallelogram are supplementary]<br>\u2234 \u2220A+ 110\u00b0= 180\u00b0 [From (i)]<br>\u2234 \u2220A= 180\u00b0- 110\u00b0<br>\u2234 \u2220A = 70\u00b0<br>\u2234 \u2220C = \u2220A = 70\u00b0<br>\u2234 \u2220B = \u2220R= 110\u00b0 [Opposite angles of a parallelogram]<br>\u2234 \u2220A = 70\u00b0, \u2220B = 110\u00b0,<br>\u2234 \u2220C = 70\u00b0, \u2220R = 110\u00b0<\/p>\n\n\n\n<p><strong>Question 7.<br>In the adjoining figure, \u25a1ABCD is a parallelogram. Point E is on the ray AB such that BE = AB, then prove that line ED bisects seg BC at point F.<\/strong><br><img loading=\"lazy\" decoding=\"async\" width=\"281\" height=\"120\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/02\/Maharashtra-Board-Class-9-Maths-Solutions-Chapter-5-Quadrilaterals-Practice-Set-5.1-7.jpg\" alt=\"Maharashtra Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.1 7\"><br>Given: \u25a1ABCD is a parallelogram.<br>BE = AB<br>To prove: Line ED bisects seg BC at point F i.e. FC = FB<br>Solution:<br>Proof:<br>\u25a1ABCD is a parallelogram. [Given]<br>\u2234 seg AB \u2245 seg DC \u2026\u2026.(i) [Opposite angles of a parallelogram]<br>seg AB \u2245 seg BE \u2026\u2026..(ii) [Given]<br>seg DC \u2245 seg BE \u2026\u2026..(iii) [From (i) and (ii)]<br>side DC || side AB [Opposite sides of a parallelogram]<br>i.e. side DC || seg AE and seg DE is their transversal. [A-B-E]<br>\u2234 \u2220CDE \u2245 \u2220AED<br>\u2234 \u2220CDF \u2245 \u2220BEF \u2026..(iv) [D-F-E, A-B-E]<br>In \u2206DFC and \u2206EFB,<br>seg DC = seg EB [From (iii)]<br>\u2220CDF \u2245 \u2220BEF [From (iv)]<br>\u2220DFC \u2245 \u2220EFB [Vertically opposite angles]<br>\u2234 \u2206DFC \u2245 \u2206EFB [SAA test]<br>\u2234 FC \u2245 FB [c.s.c.t]<br>\u2234 Line ED bisects seg BC at point F.<\/p>\n\n\n\n<p><strong>Question 1.<br>Write the following pairs considering \u25a1ABCD. (Textbook pg. no 57)<\/strong><br><img loading=\"lazy\" decoding=\"async\" width=\"173\" height=\"148\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/02\/Maharashtra-Board-Class-9-Maths-Solutions-Chapter-5-Quadrilaterals-Practice-Set-5.1-8.jpg\" alt=\"Maharashtra Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.1 8\"><br>Pairs of adjacent sides:<br>i. AB, AD<br>ii. AD, DC<br>iii. DC, BC<br>iv. BC, AB<\/p>\n\n\n\n<p>Pairs of adjacent angles:<br>i. \u2220A, \u2220B<br>ii. \u2220C, \u2220D<br>iii. \u2220B, \u2220C<br>iv. \u2220D, \u2220A<\/p>\n\n\n\n<p>Pairs of opposite sides:<br>i. AB, DC<br>ii. AD, BC<\/p>\n\n\n\n<p>Pairs of opposite angles:<br>i. \u2220A, \u2220C<br>ii. \u2220B, \u2220D<\/p>\n\n\n\n<p><strong>Question 2.<br>Complete the following tree diagram. (Textbook pg. no 57)<br><img loading=\"lazy\" decoding=\"async\" width=\"718\" height=\"431\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/02\/Maharashtra-Board-Class-9-Maths-Solutions-Chapter-5-Quadrilaterals-Practice-Set-5.1-9.jpg\" alt=\"Maharashtra Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.1 9\"><\/strong><\/p>\n\n\n\n<p><strong>Question 3.<br>In the above theorem, to prove \u2220DAB \u2245 \u2220BCD, is any change in the construction needed? If so, how will you write the proof making the change? (Textbook pg. no. 60)<br><img loading=\"lazy\" decoding=\"async\" width=\"203\" height=\"123\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/02\/Maharashtra-Board-Class-9-Maths-Solutions-Chapter-5-Quadrilaterals-Practice-Set-5.1-10.jpg\" alt=\"Maharashtra Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.1 10\"><br>Solution:<\/strong><br>Yes<br>Construction: Draw diagonal BD.<br>Proof:<br>side AB || side CD and diagonal BD is their transversal. [Given]<br>\u2234 \u2220ABD \u2245 \u2220CDB \u2026\u2026..(i) [Alternate angles]<br>side BC || side AD and diagonal BD is their transversal. [Given]<br>\u2234 \u2220ADB \u2245 \u2220CBD \u2026\u2026..(ii) [Alternate angles]<br>In \u2206DAB and \u2206BCD,<br>\u2220ABD \u2245 \u2220CDB [From (i)]<br>seg BD \u2245 seg DB [Common side]<br>\u2234 \u2220ADB \u2245 \u2220CBD [From (ii)]<br>\u2234 \u2206DAB \u2245 \u2206BCD [ASA test]<br>\u2234 \u2220DAB \u2245 \u2220BCD [c.a.c.t.]<br>Note: \u2220DAB s \u2220BCD can be proved using the same construction as in the above theorem.<br>\u2220BAC \u2245 \u2220DCA \u2026..(i)<br>\u2220DAC \u2245 \u2220BCA \u2026\u2026(ii)<br>\u2234 \u2220BAC + \u2220DAC \u2245 \u2220DCA + \u2220BCA [Adding (i) and (ii)]<br>\u2234 \u2220DAB \u2245 \u2220BCD [Angle addition property]<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"download-pdf\"><strong>Download PDF<\/strong><\/h2>\n\n\n\n<p>Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.1- Quadrilaterals<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/docs\/d18606c1-441a-41bd-b285-51b036ec5127\" target=\"_blank\" rel=\"noreferrer noopener\"><strong>Download PDF<\/strong>: Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.1- Quadrilaterals PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"block-2da0bf3e-e6a7-4a61-aec1-a1dc708a500d\"><strong>Chapterwise Maharashtra Board Solutions Class 9 Maths :<\/strong><\/h2>\n\n\n\n<p id=\"block-9fd708a1-440b-436d-bb5e-2f94067b7b02\"><strong>Part 2<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\" id=\"block-85c5bba7-0b7d-46dd-8961-c54d822793b1\"><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-1-1-basic-concepts-in-geometry\/\">Chapter 1.1- Basic Concepts in Geometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-1-2-basic-concepts-in-geometry\/\">Chapter 1.2- Basic Concepts in Geometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-1-3-basic-concepts-in-geometry\/\">Chapter 1.3- Basic Concepts in Geometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-2-1-parallel-lines\/\">Chapter 2.1- Parallel Lines<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-2-2-parallel-lines\/\">Chapter 2.2- Parallel Lines<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-3-1-triangles\/\">Chapter 3.1- Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-3-2-triangles\/\">Chapter 3.2- Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-3-3-triangles\/\">Chapter 3.3- Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-3-4-triangles\/\">Chapter 3.4- Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-3-5-triangles\/\">Chapter 3.5- Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-4-1-constructions-of-triangles\/\">Chapter 4.1- Constructions of Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-4-2-constructions-of-triangles\/\">Chapter 4.2- Constructions of Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-4-3-constructions-of-triangles\/\">Chapter 4.3- Constructions of Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-5-1-quadrilaterals\/\">Chapter 5.1- Quadrilaterals<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-5-2-quadrilaterals\/\">Chapter 5.2- Quadrilaterals<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-5-3-quadrilaterals\/\">Chapter 5.3- Quadrilaterals<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-5-4-quadrilaterals\/\">Chapter 5.4- Quadrilaterals<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-5-5-quadrilaterals\/\">Chapter 5.5- Quadrilaterals<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-6-1-circle\/\">Chapter 6.1- Circle<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-6-2-circle\/\">Chapter 6.2- Circle<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-6-3-circle\/\">Chapter 6.3- Circle<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-7-1-co-ordinate-geometry\/\">Chapter 7.1- Co-ordinate Geometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-7-2-co-ordinate-geometry\/\">Chapter 7.2- Co-ordinate Geometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-8-1-trigonometry\/\">Chapter 8.1- Trigonometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-8-2-trigonometry\/\">Chapter 8.2- Trigonometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-9-1-surface-area-and-volume\/\">Chapter 9.1- Surface Area and Volume<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-9-2-surface-area-and-volume\/\">Chapter 9.2- Surface Area and Volume<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-9-3-surface-area-and-volume\/\">Chapter 9.3- Surface Area and Volume<\/a><\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-faqs\">FAQs<\/h2>\n\n\n\n<div class=\"schema-faq wp-block-yoast-faq-block\"><div class=\"schema-faq-section\" id=\"faq-question-1637607587822\"><strong class=\"schema-faq-question\">Where do I get the Maharashtra State Board Books PDF For free download?<\/strong> <p class=\"schema-faq-answer\">You can download the Maharashtra State Board Books from the eBalbharti official website, i.e. cart.ebalbharati.in or from this article.<\/p> <\/div> <div class=\"schema-faq-section\" id=\"faq-question-1636085525999\"><strong class=\"schema-faq-question\">How to Download Maharashtra State Board Books?<\/strong> <p class=\"schema-faq-answer\">Students can get the Maharashtra Books for primary, secondary, and senior secondary classes from here.\u00a0 You can view or download the\u00a0<strong>Maharashtra State Board Books<\/strong>\u00a0from this page or from the official website for free of cost. Students can follow the detailed steps below to visit the official website and download the e-books for all subjects or a specific subject in different mediums.<br\/><strong>Step 1:<\/strong>\u00a0Visit the official website\u00a0<em><a rel=\"noreferrer noopener\" href=\"https:\/\/ebalbharati.in\/main\/publicHome.aspx\" target=\"_blank\">ebalbharati.in<\/a><\/em><br\/><strong>Step 2:<\/strong>\u00a0On the top of the screen, select &#8220;Download PDF textbooks&#8221;\u00a0<br\/><strong>Step 3:\u00a0<\/strong>From the &#8220;Classes&#8221;\u00a0section, select your class.<br\/><strong>Step 4:\u00a0<\/strong>From &#8220;Medium&#8221;, select the medium suitable to you.<br\/><strong>Step 5:\u00a0<\/strong>All Maharashtra board books for your class will now be displayed on the right side.\u00a0<br\/>Step 6:\u00a0Click on the &#8220;Download&#8221;\u00a0option to download the PDF book.<\/p> <\/div> <div class=\"schema-faq-section\" id=\"faq-question-1637607561423\"><strong class=\"schema-faq-question\">Who developed the Maharashtra State board books?<\/strong> <p class=\"schema-faq-answer\">As of now, the MSCERT and Balbharti are responsible for the syllabus and textbooks of Classes 1 to 8, while Classes 9 and 10 are under the Maharashtra State Board of Secondary and Higher Secondary Education (MSBSHSE).<\/p> <\/div> <div class=\"schema-faq-section\" id=\"faq-question-1637607721404\"><strong class=\"schema-faq-question\">How many state boards are there in Maharashtra?<\/strong> <p class=\"schema-faq-answer\">The Maharashtra State Board of Secondary &amp; Higher Secondary Education, conducts the HSC and SSC Examinations in the state of Maharashtra through its\u00a0<strong>nine<\/strong>\u00a0Divisional Boards located at Pune, Mumbai, Aurangabad, Nasik, Kolhapur, Amravati, Latur, Nagpur and Ratnagiri.<\/p> <\/div> <\/div>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-about-maharashtra-state-board-msbshse\">About Maharashtra State Board (<strong>MSBSHSE<\/strong>)<\/h2>\n\n\n\n<p>The Maharashtra State Board of Secondary and Higher Secondary Education or MSBSHSE (Marathi: \u092e\u0939\u093e\u0930\u093e\u0937\u094d\u091f\u094d\u0930 \u0930\u093e\u091c\u094d\u092f \u092e\u093e\u0927\u094d\u092f\u092e\u093f\u0915 \u0906\u0923\u093f \u0909\u091a\u094d\u091a \u092e\u093e\u0927\u094d\u092f\u092e\u093f\u0915 \u0936\u093f\u0915\u094d\u0937\u0923 \u092e\u0902\u0921\u0933), is an&nbsp;<strong>autonomous and statutory body established in 1965<\/strong>. The board was amended in the year 1977 under the provisions of the Maharashtra Act No. 41 of 1965.<\/p>\n\n\n\n<p>The Maharashtra State Board of Secondary &amp; Higher Secondary Education (MSBSHSE), Pune is an independent body of the Maharashtra Government. There are more than 1.4 million students that appear in the examination every year. The Maha State Board conducts the board examination twice a year. This board conducts the examination for SSC and HSC.&nbsp;<\/p>\n\n\n\n<p>The Maharashtra government established the Maharashtra State Bureau of Textbook Production and Curriculum Research, also commonly referred to as Ebalbharati, in 1967 to take up the responsibility of providing quality textbooks to students from all classes studying under the Maharashtra State Board. MSBHSE prepares and updates the curriculum to provide holistic development for students. It is designed to tackle the difficulty in understanding the concepts with simple language with simple illustrations. Every year around 10 lakh students are enrolled in schools that are affiliated with the Maharashtra State Board.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Class 9: Maths Chapter 5.1 solutions. Complete Class 9 Maths Chapter 5.1 Notes. Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.1- Quadrilaterals Maharashtra Board 9th Maths Chapter 5.1, Class 9 Maths Chapter 5.1 solutions Question 1.Diagonals of a parallelogram WXYZ intersect each other at point O. If \u2220XYZ\u2220 = 135\u00b0, then measure of \u2220XWZ [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":581568,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,921,22],"tags":[2078,2178],"boards":[1318],"class_list":["post-581566","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-9","category-maharashtra","tag-english-medium","tag-msbshse-maths-class-9","boards-msbshse","entry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.0 (Yoast SEO v27.1.1) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>Maharashtra Board for Class 9, Maths Chapter 5.1 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.1- Quadrilaterals | MSBSHSE Class 9 IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-5-1-quadrilaterals\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.1- Quadrilaterals\" \/>\n<meta property=\"og:description\" content=\"Class 9: Maths Chapter 5.1 solutions. 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