{"id":4464,"date":"2023-02-14T05:46:39","date_gmt":"2023-02-14T05:46:39","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4464"},"modified":"2023-02-14T05:48:49","modified_gmt":"2023-02-14T05:48:49","slug":"ncert-solutions-class-9-maths-chapter-9-areas-of-parallelograms-and-triangles-ex-9-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-9-areas-of-parallelograms-and-triangles-ex-9-2\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2"},"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; 9 (Areas of Parallelograms and Triangles)\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; 9 Areas of Parallelograms and Triangles <\/strong>Exercise 9.2 has been provided here to help the students in solving the questions from this exercise.<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 9: Areas of Parallelograms and Triangles<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-9-areas-of-parallelograms-and-triangles-ex-9-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 9.1<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-9-areas-of-parallelograms-and-triangles-ex-9-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 9.3<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-9-areas-of-parallelograms-and-triangles-ex-9-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 9.4<\/a><\/li>\n<\/ul>\n<div>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 9.2<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. In Fig. 9.15, ABCD is a parallelogram, AE \u22a5 DC and CF \u22a5 AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4474\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Q1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"334\" height=\"218\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Q1.png 334w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Q1-300x196.png 300w\" sizes=\"auto, (max-width: 334px) 100vw, 334px\" \/><br \/>\n<\/strong><strong>Answer &#8211; Given:<\/strong> AB = 16 cm, AE = 8 cm, CF = 10 cm<\/span><br \/>\n<span style=\"color: #000000;\">In parallelogram ABCD, CD = AB = 16 cm\u00a0 \u00a0 \u00a0[Opposite sides of a parallelogram\u00a0are equal]<\/span><br \/>\n<span style=\"color: #000000;\">We know that area of a parallelogram = Base \u00d7 Corresponding altitude<\/span><br \/>\n<span style=\"color: #000000;\">Therefore,\u00a0Area of a parallelogram\u00a0ABCD = CD \u00d7 AE = AD \u00d7 CF<\/span><br \/>\n<span style=\"color: #000000;\">Here CD and AD act as the base and AE \u22a5 CD, CF \u22a5 AD are the corresponding altitudes.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AD = (CD \u00d7 AE) \/ CF<\/span><br \/>\n<span style=\"color: #000000;\">Substituting the values for CD, AE, and CF, we get AD = (16 \u00d7 8) \/10 cm = 12.8 cm<\/span><br \/>\n<span style=\"color: #000000;\">Thus, the length of AD is 12.8 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{2}}\" alt=\"\\mathbf{\\frac{1}{2}}\" align=\"absmiddle\" \/> ar(ABCD).<br \/>\n<\/strong><strong>Answer &#8211; <\/strong>If a triangle and parallelogram are on the same base and between the same parallel lines, then the area of the triangle will be half of the area of a parallelogram.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4475\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"288\" height=\"163\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans2.png 707w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans2-300x170.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans2-480x272.png 480w\" sizes=\"auto, (max-width: 288px) 100vw, 288px\" \/><br \/>\n<strong>Given :\u00a0<\/strong>E, F, G and H are the mid-points of the sides of a parallelogram ABCD, respectively.<br \/>\n<strong>To Prove :\u00a0<\/strong>ar (EFGH) = \u00bd ar(ABCD)<br \/>\n<strong>Construction :<\/strong>\u00a0H and F are joined.<br \/>\nAD || BC and AD = BC (Opposite sides of a parallelogram)<br \/>\n\u21d2 \u00bd AD = \u00bd BC<br \/>\nAlso,<br \/>\nAH || BF and and DH || CF<br \/>\n\u21d2 AH = BF and DH = CF (H and F are mid points)<br \/>\n\u2234 ABFH and HFCD are parallelograms.<br \/>\nNow,<br \/>\nWe know that, \u0394EFH and parallelogram ABFH, both lie on the same FH the common base and in-between the same parallel lines AB and HF.<br \/>\n\u2234 area of EFH = \u00bd area of ABFH\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;-\u2014 (i)<br \/>\nAnd, area of GHF = \u00bd area of HFCD &#8212;&#8212;&#8212;&#8212;-\u2014 (ii)<br \/>\nAdding (i) and (ii),<br \/>\narea of \u0394EFH\u00a0+ area of \u0394GHF = \u00bd area of ABFH\u00a0+ \u00bd area of HFCD<br \/>\n\u21d2 area of EFGH = \u00bd (area of ABFH + area of HFCD)<br \/>\n\u2234 ar (EFGH) = \u00bd ar(ABCD)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. P and Q are any two points lying on the sides DC and AD, respectively of a parallelogram ABCD.<br \/>\n<\/strong><strong>Show that ar(APB) = ar(BQC).<br \/>\n<\/strong><strong>Answer &#8211; <\/strong>If a triangle and a parallelogram are on the same base and between the same parallel lines, the area of the triangle will be half of the\u00a0area of the\u00a0parallelogram.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4476\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"303\" height=\"154\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans3.png 668w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans3-300x152.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans3-480x244.png 480w\" sizes=\"auto, (max-width: 303px) 100vw, 303px\" \/><br \/>\n\u0394APB and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.<br \/>\nar(\u0394APB) = \u00bd ar(parallelogram ABCD)\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;\u2014 (i)<br \/>\nSimilarly,<br \/>\nar(\u0394BQC) = \u00bd ar(parallelogram ABCD)\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;\u2014 (ii)<br \/>\nFrom (i) and (ii), we have<br \/>\nar(\u0394APB) = ar(\u0394BQC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4477\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"255\" height=\"166\" \/><br \/>\n<\/strong><strong>(i) <\/strong>ar(APB) + ar(PCD) = \u00bd ar(ABCD)<br \/>\n<strong>(ii) <\/strong>ar(APD) + ar(PBC) = ar(APB) + ar(PCD)<br \/>\n<strong>[Hint : <\/strong>Through P, draw a line parallel to AB.<strong>]<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>If a triangle and parallelogram are on the same base and between the same\u00a0parallel lines, then the\u00a0area of the triangle\u00a0will be half of the area of the parallelogram.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4478\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"241\" height=\"135\" \/><br \/>\n<strong>(i) ar(APB) + ar(PCD) = \u00bd ar(ABCD)<br \/>\n<\/strong>A line GH is drawn parallel to AB passing through P.<br \/>\nIn a parallelogram,<br \/>\nAB || GH (by construction)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;\u2014 (i)<br \/>\n\u2234 AD || BC \u21d2 AG || BH\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;\u2014\u00a0 (ii)<br \/>\nFrom equations (i) and (ii),<br \/>\nABHG is a parallelogram.<br \/>\nNow,<br \/>\n\u0394APB and parallelogram ABHG are lying on the same base AB and in-between the same parallel lines AB and GH.<br \/>\n\u2234 ar(\u0394APB) = \u00bd ar(ABHG)\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;\u2014 (iii)<br \/>\nalso,<br \/>\n\u0394PCD and parallelogram CDGH are lying on the same base CD and in-between the same parallel lines CD and GH.<br \/>\n\u2234 ar(\u0394PCD) = \u00bd ar(CDGH)\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;\u2014 (iv)<br \/>\nAdding equations (iii) and (iv),<br \/>\nar(\u0394APB) + ar(\u0394PCD) = \u00bd [ar(ABHG)+ar(CDGH)]<br \/>\n\u21d2 ar(APB)+ ar(PCD) = \u00bd ar(ABCD)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)<br \/>\n<\/strong>A line EF is drawn parallel to AD passing through P.<br \/>\nIn the parallelogram,<br \/>\nAD || EF (by construction)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;\u2014 (i)<br \/>\n\u2234\u00a0AB || CD \u21d2 AE || DF\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;\u2014 (ii)<br \/>\nFrom equations (i) and (ii),<br \/>\nAEDF is a parallelogram.<br \/>\nNow,<br \/>\n\u0394APD and parallelogram AEFD are lying on the same base AD and in-between the same parallel lines AD and EF.<br \/>\n\u2234 ar(\u0394APD) = \u00bd ar(AEFD)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;\u2014 (iii)<br \/>\nalso,<br \/>\n\u0394PBC and parallelogram BCFE are lying on the same base BC and in-between the same parallel lines BC and EF.<br \/>\n\u2234 ar(\u0394PBC) = \u00bd ar(BCFE)\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;\u2014 (iv)<br \/>\nAdding equations (iii) and (iv),<br \/>\nar (\u0394APD)+ ar (\u0394PBC) = \u00bd [ar (AEFD) + ar (BCFE)]<br \/>\n\u21d2 ar(APD)+ar(PBC) = ar(APB)+ar(PCD)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4479\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Q5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"220\" height=\"166\" \/><br \/>\n(i) <\/strong>ar (PQRS) = ar (ABRS)<br \/>\n<strong>(ii) <\/strong>ar (AXS) = \u00bd ar (PQRS)\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong>If a triangle and a\u00a0parallelogram\u00a0are on the same base and between the same parallel lines, the area of the triangle will be half of that of a parallelogram or if two parallelograms are on the same and between two parallel lines then their area will be equal.\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) ar (PQRS) = ar (ABRS)<br \/>\n<\/strong>Parallelogram PQRS and ABRS lie on the same base SR and in-between the same parallel lines SR and PB.<br \/>\n\u2234 ar(PQRS) = ar(ABRS)\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;-\u2014 (i)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) ar (AXS) = \u00bd ar (PQRS)\u00a0<\/strong><br \/>\n\u0394AXS and parallelogram ABRS are lying on the same base AS and in-between the same parallel lines AS and BR.<br \/>\n\u2234 ar(\u0394AXS) = \u00bd ar(ABRS)\u00a0 \u00a0&#8212;&#8212;&#8212;-\u2014 (ii)<br \/>\nFrom (i) and (ii), we find that,<br \/>\nar (\u0394AXS) = \u00bd ar (PQRS)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. A farmer had a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts is the fields divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it? <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong>Let&#8217;s draw the figure as shown below.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4480\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"300\" height=\"157\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans6.png 658w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans6-300x158.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.2-Ans6-480x252.png 480w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">From the figure, it can be observed that point A divides the parallelogram-shaped field into three parts. These parts are triangular in shape: \u00a0\u0394PSA, \u0394PAQ, and \u0394QRA<\/span><br \/>\n<span style=\"color: #000000;\">From the figure, we can observe that:<\/span><br \/>\n<span style=\"color: #000000;\">Area of \u0394PSA + Area of \u0394PAQ + Area of \u0394QRA =\u00a0Area of parallelogram PQRS\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (1)<\/span><br \/>\n<span style=\"color: #000000;\">We know that if a\u00a0parallelogram\u00a0and a triangle are on the same base and between the same set of\u00a0parallel lines, then the area of the triangle is half the area of the parallelogram.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (\u0394PAQ) = 1\/2 Area (PQRS)\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#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;\">Area (\u0394PSA) + Area (\u0394QRA) = 1\/2 Area (PQRS)\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (3)<\/span><br \/>\n<span style=\"color: #000000;\">Clearly, it can be observed that the farmer must sow wheat in the triangular part \u0394PAQ and pulses in the other two triangular parts &#8211; \u0394PSA and \u0394QRA, or wheat in triangular parts &#8211; \u0394PSA and \u0394QRA and pulses in the triangular part\u00a0\u0394PAQ.<\/span><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 9 (Areas of Parallelograms and Triangles)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 9 Areas of Parallelograms and Triangles Exercise 9.2 has been provided here to help the students in solving the questions from this exercise. Chapter 9: Areas of Parallelograms and Triangles NCERT [&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":[717,702,186,718,701,5],"class_list":["post-4464","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-9-areas-of-parallelograms-and-triangles-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-9-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 9 Areas of Parallelograms and Triangles Ex 9.2 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 9 Maths\u00a0Chapter - 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