{"id":4466,"date":"2023-02-14T05:46:52","date_gmt":"2023-02-14T05:46:52","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4466"},"modified":"2023-02-14T05:48:03","modified_gmt":"2023-02-14T05:48:03","slug":"ncert-solutions-class-9-maths-chapter-9-areas-of-parallelograms-and-triangles-ex-9-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-9-areas-of-parallelograms-and-triangles-ex-9-4\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4"},"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 9.4 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 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-9-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 9.2<\/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<\/ul>\n<div>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 9.4<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0Given:<\/strong> The\u00a0parallelogram\u00a0and the rectangle have the same base and equal areas, therefore, they\u00a0will also lie between the same parallels.<\/span><br \/>\n<span style=\"color: #000000;\">Consider the parallelogram ABCD and rectangle ABEF as follows.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4501\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"356\" height=\"198\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans1.png 809w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans1-300x167.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans1-768x427.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans1-480x267.png 480w\" sizes=\"auto, (max-width: 356px) 100vw, 356px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Here, it can be observed that parallelogram ABCD and rectangle ABEF are lying between the same parallels AB and CF. <\/span><br \/>\n<span style=\"color: #000000;\">It is given\u00a0that the opposite sides of a parallelogram and a rectangle are of equal lengths.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, AB = EF (Opposite sides of a\u00a0rectangle\u00a0are equal.)<\/span><br \/>\n<span style=\"color: #000000;\">AB = CD (Opposite sides of a parallelogram are equal.)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 CD = EF<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AB + CD = AB + EF\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;-(i)<\/span><br \/>\n<span style=\"color: #000000;\">Now, in a\u00a0right-angled\u00a0triangle AFD, AD is a hypotenuse.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AF &lt; AD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;-(ii)<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, in a\u00a0right angled triangle\u00a0EBC, EB is altitude and BC is a\u00a0hypotenuse<\/span><\/p>\n<p><span style=\"color: #000000;\">\u2234 BE &lt; BC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;-(iii)<\/span><br \/>\n<span style=\"color: #000000;\">Adding equation (ii) and (iii),<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AF + BE &lt; AD + BC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;-(iv)<\/span><br \/>\n<span style=\"color: #000000;\">Now, from equations (i) and (iv), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">AB + EF + AF + BE &lt; AD + BC + AB + CD<\/span><br \/>\n<span style=\"color: #000000;\">Perimeter of rectangle\u00a0ABEF &lt; Perimeter of parallelogram ABCD.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. In Fig. 9.30, D and E are two points on BC such that BD = DE = EC. <\/strong><strong>Show that ar (ABD) = ar (ADE) = ar (AEC).<br \/>\n<\/strong><strong>Can you now answer the question that you have left in the \u2018Introduction\u2019 of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4502\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"337\" height=\"260\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q2.png 337w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q2-300x231.png 300w\" sizes=\"auto, (max-width: 337px) 100vw, 337px\" \/><br \/>\n<\/strong><strong>[Remark: <\/strong>Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into\u00a0<em>n\u00a0<\/em>equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide DABC into\u00a0<em>n\u00a0<\/em>triangles of equal areas.<strong>]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong><strong>Given :\u00a0<\/strong>BD = DE = EC<br \/>\n<strong>To prove :\u00a0<\/strong>Area (\u25b3ABD) = Area (\u25b3ADE) = Area (\u25b3AEC)<br \/>\nProof,<br \/>\nIn (\u25b3ABE), AD is median [since, BD = DE, given]<br \/>\nWe know that, the median of a triangle divides it into two parts of equal areas<br \/>\nArea (\u25b3ABD) = Area (\u25b3AED)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;-\u2014(i)<br \/>\nSimilarly,<br \/>\nIn (\u25b3ADC), AE is median [since, DE = EC, given]<br \/>\nArea (ADE) = Area (AEC)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8211;\u2014(ii)<br \/>\nFrom the equation (i) and (ii), we get<br \/>\nArea (ABD) = Area (ADE) = Area (AEC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4503\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"303\" height=\"232\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q3.png 303w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q3-300x230.png 300w\" sizes=\"auto, (max-width: 303px) 100vw, 303px\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong><strong>Given :\u00a0<\/strong>ABCD, DCFE and ABFE are parallelograms<br \/>\n<strong>To prove :\u00a0<\/strong>ar (\u25b3ADE) = ar (\u25b3BCF)<br \/>\nIn \u25b3ADE and \u25b3BCF,<br \/>\nAD = BC [Since they are the opposite sides of the parallelogram ABCD]<br \/>\nDE = CF [Since they are the opposite sides of the parallelogram DCFE]<br \/>\nAE = BF [Since they are the opposite sides of the parallelogram ABFE]<br \/>\n\u25b3ADE \u2245 \u25b3BCF [Using SSS Congruence theorem]<br \/>\nArea (\u25b3ADE) = Area (\u25b3BCF)\u00a0 \u00a0 \u00a0 \u00a0[By CPCT]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4504\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"275\" height=\"359\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q4.png 275w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q4-230x300.png 230w\" sizes=\"auto, (max-width: 275px) 100vw, 275px\" \/><br \/>\n<\/strong><strong>[Hint : <\/strong>Join AC<strong>.]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; Given:\u00a0<\/strong>ABCD is a parallelogram<br \/>\nAD = CQ<br \/>\n<strong>To prove:<\/strong> Area (\u25b3BPC) = Area (\u25b3DPQ)<br \/>\nIn \u25b3ADP and \u25b3QCP,<br \/>\n\u2220APD = \u2220QPC [Vertically Opposite Angles]<br \/>\n\u2220ADP = \u2220QCP [Alternate Angles]<br \/>\nAD = CQ [given]<br \/>\n\u25b3ABO \u2245 \u25b3ACD [AAS congruency]<br \/>\nDP = CP [CPCT]<br \/>\nIn \u25b3CDQ, QP is median. [Since, DP = CP]<br \/>\nSince, median of a triangle divides it into two parts of equal areas.<br \/>\nArea (\u25b3DPQ) = Area (\u25b3QPC)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;\u2014(i)<br \/>\nIn \u25b3PBQ, PC is median. [Since, AD = CQ and AD = BC \u21d2 BC = QC]<br \/>\nSince, median of a triangle divides it into two parts of equal areas.<br \/>\nArea (\u25b3QPC) = Area (\u25b3BPC)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;(ii)<br \/>\nFrom the equation (i) and (ii), we get<br \/>\nArea (\u25b3BPC) = Area (\u25b3DPQ)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4505\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"321\" height=\"360\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q5.png 374w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q5-267x300.png 267w\" sizes=\"auto, (max-width: 321px) 100vw, 321px\" \/><br \/>\n<\/strong><strong>(i) <\/strong>ar (BDE) =1\/4 ar (ABC)<br \/>\n<strong>(ii) <\/strong>ar (BDE) = \u00bd ar (BAE)<br \/>\n<strong>(iii) <\/strong>ar (ABC) = 2 ar (BEC)<br \/>\n<strong>(iv) <\/strong>ar (BFE) = ar (AFD)<br \/>\n<strong>(v) <\/strong>ar (BFE) = 2 ar (FED)<br \/>\n<strong>(vi) <\/strong>ar (FED) = 1\/8 ar (AFC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong><strong>(i) ar (BDE) =1\/4 ar (ABC)<br \/>\n<\/strong>Let G and H be the mid-points of side AB and AC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4506\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"345\" height=\"346\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans5.png 374w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans5-300x300.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans5-150x150.png 150w\" sizes=\"auto, (max-width: 345px) 100vw, 345px\" \/><br \/>\nLine segment GH is joining the mid-points and is parallel to the third side. Therefore, GH will be half of the length of BC (mid-point theorem).<br \/>\n\u2234 GH = 1\/2 BC and GH || BD<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 GH = BD = DC and GH || BD (D is the mid-point of BC)<\/span><br \/>\n<span style=\"color: #000000;\">Similarly,<\/span><br \/>\n<span style=\"color: #000000;\">GD = HC = HA<\/span><br \/>\n<span style=\"color: #000000;\">HD = AG = BG<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, clearly, triangle ABC is divided into 4 equal equilateral triangles viz \u0394BGD, \u0394AGH, \u0394DHC, and \u0394GHD. <\/span><br \/>\n<span style=\"color: #000000;\">In other words,<\/span><br \/>\n<span style=\"color: #000000;\">\u0394BGD = 1\/4 \u0394ABC<\/span><br \/>\n<span style=\"color: #000000;\">Now consider \u0394BDG and \u0394BDE<\/span><br \/>\n<span style=\"color: #000000;\">BD = BD (Common base)<\/span><br \/>\n<span style=\"color: #000000;\">As both triangles are equilateral triangle, we can say BG = BE<\/span><br \/>\n<span style=\"color: #000000;\">DG = DE<\/span><br \/>\n<span style=\"color: #000000;\">Therefore,<\/span><br \/>\n<span style=\"color: #000000;\">\u0394BDG \u2245 \u0394BDE [By SSS congruency]<\/span><br \/>\n<span style=\"color: #000000;\">The SSS rule states that: If three sides of one triangle are equal to three sides of another triangle, then the\u00a0triangles are congruent.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, area (\u0394BDG) = area (\u0394BDE)<\/span><br \/>\n<span style=\"color: #000000;\">ar (\u0394BDE) = 1\/4 ar (\u0394ABC) <\/span><br \/>\n<span style=\"color: #000000;\">Hence proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) ar (BDE) = \u00bd ar (BAE)<br \/>\n<\/strong>Join points A and D.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4507\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans5ii.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"316\" height=\"324\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans5ii.png 374w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans5ii-293x300.png 293w\" sizes=\"auto, (max-width: 316px) 100vw, 316px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394BDE) = Area (\u0394AED) (Common base DE and DE || AB)<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394BDE) &#8211; Area (\u0394FED) = Area (\u0394AED) &#8211; Area (\u0394FED)<\/span><br \/>\n<span style=\"color: #000000;\">Now Area (\u0394BEF) = Area (\u0394AFD)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211;(1)<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394ABD) = Area (\u0394ABF) + Area (\u0394AFD)<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394ABD) = Area (\u0394ABF) + Area (\u0394BEF) [From equation (1)]<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394ABD) = Area (\u0394ABE)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;(2)<\/span><br \/>\n<span style=\"color: #000000;\">AD is the\u00a0median\u00a0in \u0394ABC (D is the midpoint of BC)<\/span><br \/>\n<span style=\"color: #000000;\">ar (\u0394ABD) = 1\/2 ar (\u0394ABC)\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">ar (\u0394BDE) = 1\/4 ar (\u0394ABC)\u00a0[As proved earlier in (i)]<\/span><br \/>\n<span style=\"color: #000000;\">ar(\u0394ABD) = 2 ar (\u0394BDE)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;(3)<\/span><br \/>\n<span style=\"color: #000000;\">From (2) and (3), we obtain,<\/span><br \/>\n<span style=\"color: #000000;\">2 Area (\u0394BDE) = Area (\u0394ABE)<\/span><br \/>\n<span style=\"color: #000000;\">Area (BDE) = 1\/2 Area (BAE)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) ar (ABC) = 2 ar (BEC)<br \/>\n<\/strong>Join points C and E<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4508\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans5iii.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"374\" height=\"372\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans5iii.png 374w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans5iii-300x298.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans5iii-150x150.png 150w\" sizes=\"auto, (max-width: 374px) 100vw, 374px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394ABE) = Area (\u0394BEC) (Common base BE and BE || AC)<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394ABF) + Area (\u0394BEF) = Area (\u0394BEC)<\/span><br \/>\n<span style=\"color: #000000;\">Using equation (1), we obtain,<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394ABF) + Area (\u0394AFD) = Area (\u0394BEC)<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394ABD) = Area (\u0394BEC)<\/span><br \/>\n<span style=\"color: #000000;\">1\/2 Area (\u0394ABC) = Area (\u0394BEC)<\/span><br \/>\n<span style=\"color: #000000;\">Area(\u0394ABC) = 2 Area (\u0394BEC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) ar (BFE) = ar (AFD)<br \/>\n<\/strong>\u0394BDE and \u0394AED lie on the same base (DE) and between the parallels DE and AB.<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394BDE) = Area (\u0394AED)<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394BDE) &#8211; Area (\u0394FED) = Area (\u0394AED) &#8211; Area (\u0394FED) [Subtracting ar (\u0394FED) on both the sides]<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394BFE) = Area (\u0394AFD)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) ar (BFE) = 2 ar (FED)<br \/>\n<\/strong>Let h be the height of vertex E, corresponding to the side BD in \u0394BDE<\/span><br \/>\n<span style=\"color: #000000;\">Let H be the height of vertex A, corresponding to the side BC in \u0394ABC.<\/span><br \/>\n<span style=\"color: #000000;\">Area (ABC) = 1\/2 \u00d7 BC \u00d7 H<\/span><br \/>\n<span style=\"color: #000000;\">Area (BDE) = 1\/2 \u00d7 1\/2(BC) \u00d7 h<\/span><br \/>\n<span style=\"color: #000000;\">In (i), it was shown that ar (BDE) = 1\/4 Area (ABC)<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, H = 2h<\/span><br \/>\n<span style=\"color: #000000;\">Now, <\/span><br \/>\n<span style=\"color: #000000;\">Area (AFD) = 1\/2 \u00d7 FD \u00d7 2h<\/span><br \/>\n<span style=\"color: #000000;\">Area (FED) = 1\/2 \u00d7 (FD) \u00d7 h<\/span><br \/>\n<span style=\"color: #000000;\">Hence, Area (AFD) = 2 \u00d7 Area (FED)<\/span><br \/>\n<span style=\"color: #000000;\">In (iv), it was shown that Area (\u0394BFE) = Area (\u0394AFD)<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, Area (\u0394BFE) = Area (\u0394AFD) = 2 Area (\u0394FED)<\/span><br \/>\n<span style=\"color: #000000;\">Hence proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi) ar (FED) = 1\/8 ar (AFC)<br \/>\n<\/strong>Area (\u0394AFC) = Area (\u0394AFD) + Area (\u0394ADC)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 Area (\u0394FED) + 1\/2 Area (\u0394ABC) [using (v)]<\/span><br \/>\n<span style=\"color: #000000;\">= 2 Area (\u0394FED) + 1\/2 [4 \u00d7 Area (\u0394BDE)] [Using result of part (i)]<\/span><br \/>\n<span style=\"color: #000000;\">= 2 Area (\u0394FED) + 2 Area (\u0394BDE)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 Area (\u0394FED) + 2 Area (\u0394AED)<\/span><br \/>\n<span style=\"color: #000000;\">[\u0394BDE and \u0394AED are on the same base and between same parallels]<\/span><br \/>\n<span style=\"color: #000000;\">= 2 Area (\u0394FED) + 2 [Area (\u0394AFD) + Area (\u0394FED)]<\/span><br \/>\n<span style=\"color: #000000;\">= 2 Area (\u0394FED) + 2 Area (\u0394AFD) + 2 Area (\u0394FED)\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">= 4 Area (\u0394FED) + 4 Area (\u0394FED) [using result of (v)]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (\u0394AFC) = 8 Area (\u0394FED)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (\u0394FED) = 1\/8 Area (\u0394AFC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that<br \/>\n<\/strong><strong>ar (APB)\u00d7ar (CPD) = ar (APD)\u00d7ar (BPC).<br \/>\n<\/strong><strong>[Hint : <\/strong>From A and C, draw perpendiculars to BD.<strong>]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; Given:\u00a0<\/strong>The diagonal AC and BD of the quadrilateral ABCD, intersect each other at point E.<br \/>\n<strong>Construction:\u00a0<\/strong>From A, draw AM perpendicular to BD<br \/>\nFrom C, draw CN perpendicular to BD<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4509\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"360\" height=\"316\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans6.png 767w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans6-300x264.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans6-480x422.png 480w\" sizes=\"auto, (max-width: 360px) 100vw, 360px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>To Prove : <\/strong>Area (\u0394AED) Area (\u0394BEC) = Area (\u0394ABE) \u00d7 Area (\u0394CDE)<br \/>\nArea (\u0394ABE) = \u00bd \u00d7 BE \u00d7 AM\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;(i)<br \/>\nArea (\u0394AED) = \u00bd \u00d7 DE \u00d7 AM\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;(ii)<br \/>\nDividing eq. (ii) by (i) , we get,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4510\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans6i.png\" alt=\"\" width=\"197\" height=\"78\" \/><br \/>\nArea (AED)\/Area (ABE) = DE\/BE\u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;(iii)<br \/>\nSimilarly,<br \/>\nArea (CDE)\/Area (BEC) = DE\/BE\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (iv)<br \/>\nFrom eq. (iii) and (iv) , we get<br \/>\nArea (AED)\/Area (ABE) = Area (CDE)\/Area (BEC)<br \/>\nArea (\u0394AED) \u00d7 Area (\u0394BEC) = Area (\u0394ABE) \u00d7 Area (\u0394CDE)<br \/>\nHence proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7.<\/strong>\u00a0<strong>P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that:<br \/>\n<\/strong><strong>(i) <\/strong>ar (PRQ) = \u00bd ar (ARC)<strong><br \/>\n<\/strong><strong>(ii) <\/strong>ar (RQC) = (3\/8) ar (ABC)<strong><br \/>\n<\/strong><strong>(iii) <\/strong>ar (PBQ) = ar (ARC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong>Let&#8217;s draw the given triangle ABC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4511\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans7.jpg\" alt=\"NCERT Class 9 Solutions Maths\" width=\"321\" height=\"181\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans7.jpg 2560w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans7-300x169.jpg 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans7-1024x576.jpg 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans7-768x432.jpg 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans7-1536x864.jpg 1536w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans7-2048x1152.jpg 2048w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Ans7-480x270.jpg 480w\" sizes=\"auto, (max-width: 321px) 100vw, 321px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) ar (PRQ) = \u00bd ar (ARC)<br \/>\n<\/strong>PC is the\u00a0median\u00a0of \u0394ABC.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (\u0394BPC) = Area (\u0394APC)\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;- (i)<\/span><br \/>\n<span style=\"color: #000000;\">RC is the median\u00a0of \u0394APC.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (\u0394ARC) = 1\/2 Area (\u0394APC) &#8212;&#8212;&#8212;&#8212;&#8212;- (ii)<\/span><br \/>\n<span style=\"color: #000000;\">[Median divides the triangle into two triangles of equal area]<\/span><br \/>\n<span style=\"color: #000000;\">PQ is the median of \u0394BPC.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (\u0394PQC) = 1\/2 Area (\u0394BPC)\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- (iii)<\/span><br \/>\n<span style=\"color: #000000;\">From equation\u00a0(i) and (iii), we get,<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394PQC) = 1\/2 Area (\u0394APC)\u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (iv)<\/span><br \/>\n<span style=\"color: #000000;\">From equation\u00a0(ii) and (iv), we get,<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394PQC) = Area (\u0394ARC)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (v)<\/span><br \/>\n<span style=\"color: #000000;\">We are given that P and Q are the mid-points of AB and BC, respectively.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 PQ || AC and PQ = 1\/2 AC<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (\u0394APQ) = Area (\u0394PQC)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (vi) [triangles between same parallel are equal in area ]<\/span><br \/>\n<span style=\"color: #000000;\">From equation\u00a0(v) and (vi), we get<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394APQ) = Area (\u0394ARC)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (vii)<\/span><br \/>\n<span style=\"color: #000000;\">R is the mid-point of AP. Therefore, RQ is the median of \u0394APQ.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (\u0394PRQ) = 1\/2 Area (\u0394APQ)\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (viii)<\/span><br \/>\n<span style=\"color: #000000;\">From (vii) and (viii), we get,<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (\u0394PRQ) = 1\/2 Area (\u0394ARC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) ar (RQC) = (3\/8) ar (ABC)<br \/>\n<\/strong>PQ is the median of \u0394BPC.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (\u0394PQC) = 1\/2 Area (\u0394BPC) = 1\/2 \u00d7 1\/2 Area (\u0394ABC)\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (ix)<\/span><br \/>\n<span style=\"color: #000000;\">Also,<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394PRC) = 1\/2 Area (\u0394APC) [Using (iv)]<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394PRC) = 1\/2 \u00d7 1\/2 Area (\u0394ABC) = 1\/4 Area (\u0394ABC)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (x)<\/span><br \/>\n<span style=\"color: #000000;\">Adding equation\u00a0(ix) and (x), we get,<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394PQC) + Area (\u0394PRC) = ( 1\/4 + 1\/4 ) Area (\u0394ABC)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (quadrilateral PQCR) = 1\/2 Area (\u0394ABC)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (xi)<\/span><br \/>\n<span style=\"color: #000000;\">Subtracting ar (\u0394PRQ) from the both sides,<\/span><br \/>\n<span style=\"color: #000000;\">Area (quadrilateral PQCR) &#8211; Area (\u0394PRQ) = 1\/2 Area (\u0394ABC) &#8211; Area (\u0394PRQ)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (\u0394RQC) = 1\/2 Area (\u0394ABC) &#8211; 1\/2 Area (\u0394ARC) [Using result (i)]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (\u0394RQC) = 1\/2 Area (\u0394ABC) &#8211; 1\/2 \u00d7 1\/2 Area (\u0394APC)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (\u0394RQC) = 1\/2 Area (\u0394ABC) &#8211; 1\/4 Area (\u0394APC)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (\u0394RQC) = 1\/2 Area (\u0394ABC) &#8211; 1\/4 \u00d7 1\/2 Area (\u0394ABC) [PC is median of DABC]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (\u0394RQC) = 1\/2 Area (\u0394ABC) &#8211; 1\/8 Area (\u0394ABC)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (\u0394RQC) = (1\/2 &#8211; 1\/8) \u00d7 Area (\u0394ABC)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (\u0394RQC) = 3\/8 \u00d7 Area (\u0394ABC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) ar (PBQ) = ar (ARC)<br \/>\n<\/strong>Area (\u0394PRQ) = 1\/2 Area (\u0394ARC) [Using result (i)]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 2 Area (\u0394PRQ) = Area (\u0394ARC)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (xii)<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394PRQ) = 1\/2 Area (\u0394APQ) [RQ is the medium of \u0394APQ]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;- (xiii)<\/span><br \/>\n<span style=\"color: #000000;\">But Area (\u0394APQ) = Area (\u0394PQC) [Using reason of eq. (vi)]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212; (xiv)<\/span><br \/>\n<span style=\"color: #000000;\">From eq. (xiii) and (xiv), we get,<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394PRQ) = 1\/2 Area (\u0394PRQ)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- (xv)<\/span><br \/>\n<span style=\"color: #000000;\">But,<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394BPQ) = Area (\u0394PQC) [PQ is the median of DBPC]\u00a0 &#8212;&#8212;&#8212;&#8212;&#8211; (xvi)<\/span><br \/>\n<span style=\"color: #000000;\">From eq. (xv) and (xvi), we get,<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394PRQ) = 1\/2 Area (\u0394BPQ)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211;\u00a0 (xvii)<\/span><br \/>\n<span style=\"color: #000000;\">Now from (xii) and (xvii), we get,<\/span><br \/>\n<span style=\"color: #000000;\">2 \u00d7 1\/2 Area (\u0394BPQ) = Area (ARC)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (\u0394BPQ) = Area (\u0394ARC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB, respectively. Line segment AX ^ DE meets BC at Y. Show that:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4512\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q8.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"281\" height=\"365\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q8.png 402w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.4-Q8-231x300.png 231w\" sizes=\"auto, (max-width: 281px) 100vw, 281px\" \/><br \/>\n<\/strong><strong>(i)<\/strong> \u0394MBC \u2245 \u0394ABD<br \/>\n<strong>(ii)<\/strong> ar(BYXD) = 2ar(MBC)<br \/>\n<strong>(iii)<\/strong> ar(BYXD) = ar(ABMN)<br \/>\n<strong>(iv)<\/strong> \u0394FCB \u2245 \u0394ACE<br \/>\n<strong>(v)<\/strong> ar(CYXE) = 2ar(FCB)<br \/>\n<strong>(vi)<\/strong> ar(CYXE) = ar(ACFG)<br \/>\n<strong>(vii)<\/strong> ar(BCED) = ar(ABMN)+ar(ACFG)<br \/>\n<strong>Note : <\/strong>Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong>We can use a suitable congruency rule to show the triangles congruent. Also, we can use some theorem or properties like triangle and parallelogram on the same base and between the same parallels. The Triangle area will be half of the\u00a0parallelogram\u00a0area.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) \u0394MBC \u2245 \u0394ABD<br \/>\n<\/strong>We know that each angle of a square is 90\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">Hence,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABM = \u2220DBC = 90<sup>o<br \/>\n<\/sup>\u2234 \u2220ABM + \u2220ABC = \u2220DBC + \u2220ABC<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220MBC = \u2220ABD<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394MBC and \u0394ABD,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220MBC = \u2220ABD (Proved above)<\/span><br \/>\n<span style=\"color: #000000;\">MB = AB (Sides of square ABMN)<\/span><br \/>\n<span style=\"color: #000000;\">BC = BD (Sides of square BCED)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394MBC \u2245 \u0394ABD (By SAS\u00a0congruence rule)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) ar(BYXD) = 2ar(MBC)<br \/>\n<\/strong>We have\u00a0\u0394MBC \u2245 \u0394ABD<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 ar (\u0394MBC) = ar (\u0394ABD)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;-(1)<\/span><br \/>\n<span style=\"color: #000000;\">It is given that AX \u22a5 DE and BD \u22a5 DE (Adjacent sides of square BDEC)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 BD|| AX (Two lines perpendicular to the same line are parallel to each other)<\/span><br \/>\n<span style=\"color: #000000;\">\u2206ABD and parallelogram BYXD are on the same base BD and between the same parallels BD and AX.<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u2206BYXD) = 2 Area (\u2206MBC) [Using equation (1)]\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- (2)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) ar(BYXD) = ar(ABMN)<br \/>\n<\/strong>\u0394MBC and parallelogram ABMN lie on the same base MB and between the same parallels MB and NC.<\/span><br \/>\n<span style=\"color: #000000;\">2 Area (\u0394MBC) = Area (ABMN)<\/span><br \/>\n<span style=\"color: #000000;\">Area (BYXD) = Area (ABMN) [Using equation (2)]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- (3)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iv)<\/strong> <strong>\u0394FCB \u2245 \u0394ACE<br \/>\n<\/strong>We know that each angle of a square is 90\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220FCA = \u2220BCE = 90<sup>o<br \/>\n<\/sup>\u2234 \u2220FCA + \u2220ACB = \u2220BCE + \u2220ACB<\/span><br \/>\n<span style=\"color: #000000;\">Adding \u2220ACB on both the sides<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220FCB = \u2220ACE<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394FCB and \u0394ACE,<\/span><br \/>\n<span style=\"color: #000000;\">FC = AC (Sides of square ACFG)<\/span><br \/>\n<span style=\"color: #000000;\">CB = CE (Sides of square BCED)<\/span><br \/>\n<span style=\"color: #000000;\">\u0394FCB \u2245 \u0394ACE (SAS congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">SAS Congruence Rule &#8211; If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent.<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(v) ar(CYXE) = 2ar(FCB)<br \/>\n<\/strong>It is given that AX ^ DE and CE ^ DE (Adjacent sides of square BDEC)<\/span><br \/>\n<span style=\"color: #000000;\">Hence, CE || AX (Two lines perpendicular to the same line are parallel to each other) Consider \u2206ACE and parallelogram CYXE<\/span><br \/>\n<span style=\"color: #000000;\">DACE and parallelogram CYXE are on the same base CE and between the same parallels CE and AX.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (\u0394CYXE) = 2 Area (\u0394ACE)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (4)<\/span><br \/>\n<span style=\"color: #000000;\">We had proved that<\/span><br \/>\n<span style=\"color: #000000;\">\u0394FCB \u2245 \u0394ACE<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394FCB) \u2245 Area (\u0394ACE)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (5)<\/span><br \/>\n<span style=\"color: #000000;\">On comparing equations (4) and (5), we obtain <\/span><br \/>\n<span style=\"color: #000000;\">Area (CYXE) = 2 Area (\u0394FCB)\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (6)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(vi) ar(CYXE) = ar(ACFG)<br \/>\n<\/strong>Consider \u2206FCB and parallelogram ACFG<\/span><br \/>\n<span style=\"color: #000000;\">\u0394FCBand parallelogram ACFG are lying on the same base CF and between the same parallels CF and BG.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (ACFG) = 2 Area (\u0394FCB)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (ACFG) = Area (CYXE) [Using equation (6)]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (7)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(vii) ar(BCED) = ar(ABMN)+ar(ACFG)<br \/>\n<\/strong>From the figure, it is evident that<\/span><br \/>\n<span style=\"color: #000000;\">Area (BCED) = Area (BYXD) + Area (CYXE)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (BCED) = Area (ABMN) + Area (ACFG) [Using equations (3) and (7)]<\/span><\/p>\n<p style=\"text-align: justify;\">\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 9.4 has been provided here to help the students in solving the questions from this exercise.\u00a0 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-4466","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 9.4 | 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