{"id":4465,"date":"2023-02-14T05:46:46","date_gmt":"2023-02-14T05:46:46","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4465"},"modified":"2023-02-14T05:48:16","modified_gmt":"2023-02-14T05:48:16","slug":"ncert-solutions-class-9-maths-chapter-9-areas-of-parallelograms-and-triangles-ex-9-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-9-areas-of-parallelograms-and-triangles-ex-9-3\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3"},"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 9.3 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-9-2\/\" 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-9-4\/\" 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; 9.3<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. In Fig.9.23, E is any point on the median AD of a \u0394ABC. Show that ar (ABE) = ar(ACE).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4482\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"239\" height=\"226\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q1.png 324w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q1-300x283.png 300w\" sizes=\"auto, (max-width: 239px) 100vw, 239px\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0Given : <\/strong>AD is the median of \u0394ABC.<br \/>\n\u2234, it will divide \u0394ABC into two triangles of equal area.<br \/>\n\u2234 Area (ABD) = Area (ACD)\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;\u2014 (i)<br \/>\nalso,<br \/>\nED is the median of \u0394ABC.<br \/>\n\u2234 Area (EBD) = Area (ECD)\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;\u2014 (ii)<br \/>\nSubtracting (ii) from (i),<br \/>\nArea (ABD) \u2013 Area (EBD) = Area (ACD) \u2013 Area (ECD)<br \/>\n\u21d2 Area (ABE) = Area (ACE)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. In a triangle ABC, E is the mid-point of median AD. Show that ar(BED) = \u00bc ar(ABC).<\/strong><br \/>\n<strong>Answer &#8211;\u00a0<\/strong>Let ABC be a triangle and AD is the median of\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\">\u0394<\/span><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">C, <\/span><\/span><\/span><\/span>E is the mid point of AD.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4483\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"182\" height=\"232\" \/><br \/>\n<strong>To prove : <\/strong>Area <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mopen\">(<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">E<\/span><span class=\"mord mathnormal\">D<\/span><span class=\"mclose\">) <\/span><span class=\"mrel\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{4}\" alt=\"\\frac{1}{4}\" align=\"absmiddle\" \/> Area <\/span><\/span><span class=\"base\"><span class=\"mopen\">(<\/span><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">C<\/span><span class=\"mclose\">)<br \/>\n<\/span><\/span><\/span><\/span>In\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\">\u0394<\/span><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">C<\/span><\/span><\/span><\/span>, <\/span><br \/>\n<span style=\"color: #000000;\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mopen\">Area (<\/span><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">D<\/span><span class=\"mclose\">) <\/span><span class=\"mrel\">= Area <\/span><\/span><span class=\"base\"><span class=\"mopen\">(<\/span><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">C<\/span><span class=\"mord mathnormal\">D<\/span><span class=\"mclose\">)<\/span><\/span><\/span><\/span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (1)<\/span><br \/>\n<span style=\"color: #000000;\">In\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\">\u0394<\/span><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">D<\/span><\/span><\/span><\/span>, BE is the median <\/span><br \/>\n<span style=\"color: #000000;\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mopen\">Area (<\/span><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">E<\/span><span class=\"mclose\">) <\/span><span class=\"mrel\">= Area <\/span><\/span><span class=\"base\"><span class=\"mopen\">(<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">E<\/span><span class=\"mord mathnormal\">D<\/span><span class=\"mclose\">)<\/span><\/span><\/span><\/span>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (2)<\/span><br \/>\n<span style=\"color: #000000;\">Now,\u00a0<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">a<\/span><span class=\"mord mathnormal\">r <\/span><span class=\"mopen\">(<\/span><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">D<\/span><span class=\"mclose\">) <\/span><span class=\"mrel\">= Area <\/span><\/span><span class=\"base\"><span class=\"mopen\">(<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">E<\/span><span class=\"mord mathnormal\">D<\/span><span class=\"mclose\">)<br \/>\n<\/span><\/span><\/span><\/span>= 2 Area (BED)\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;\">Area (ABC) = Area (ABD) + Area (ACD)<\/span><br \/>\n<span style=\"color: #000000;\">Area (ABD) = 2 Area (ABD)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 using (1)<\/span><br \/>\n<span style=\"color: #000000;\">Area (ABC) = 2 \u00d7 2 Area (BED)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 using (2)<\/span><br \/>\n<span style=\"color: #000000;\">Area (ABC) = 4 Area (BED)<\/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 Area <\/span><\/span><span class=\"base\"><span class=\"mopen\">(<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">E<\/span><span class=\"mord mathnormal\">D<\/span><span class=\"mclose\">) <\/span><span class=\"mrel\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{4}\" alt=\"\\frac{1}{4}\" align=\"absmiddle\" \/> <\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist-s\">\u200bArea <\/span><\/span><\/span><\/span><\/span><span class=\"mopen\">(<\/span><span class=\"mord mathnormal\">A<\/span><span class=\"mord mathnormal\">B<\/span><span class=\"mord mathnormal\">C<\/span><span class=\"mclose\">)<br \/>\n<\/span><\/span><\/span><\/span>Hence it is proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>We know that the diagonals of a parallelogram bisect each other. <\/span><br \/>\n<span style=\"color: #000000;\">Also, the\u00a0median\u00a0of a triangle divides it into two\u00a0triangles of equal areas. By the use of these observations, we can get the required result. <\/span><br \/>\n<span style=\"color: #000000;\">Let&#8217;s draw a diagram according to the question statement.\u00a0<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4484\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"344\" height=\"179\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans3.png 977w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans3-300x156.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans3-768x399.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans3-480x250.png 480w\" sizes=\"auto, (max-width: 344px) 100vw, 344px\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">O is the midpoint of AC and BD. (Diagonals bisect each other)<br \/>\nIn \u0394ABC, BO is the median.<br \/>\n\u2234 Area (AOB) = Area (BOC)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;\u2014 (i)<br \/>\nalso,<br \/>\nIn \u0394BCD, CO is the median.<br \/>\n\u2234 Area (BOC) = Area (COD)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;\u2014 (ii)<br \/>\nIn \u0394ACD, OD is the median.<br \/>\n\u2234 Area (AOD) = Area (COD)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;\u2014 (iii)<br \/>\nIn \u0394ABD, AO is the median.<br \/>\n\u2234 Area (AOD) = Area (AOB)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;\u2014 (iv)<br \/>\nFrom equations (i), (ii), (iii) and (iv), we get<br \/>\nArea (BOC) = Area (COD) = Area (AOD) = Area (AOB)<br \/>\nHence, we get the diagonals of a parallelogram that divide it into four triangles of equal area.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. In Fig. 9.24, ABC and ABD are two triangles on the same base AB. If the line-segment CD is bisected by AB at O, show that ar(ABC) = ar(ABD).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4485\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"234\" height=\"246\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>AB bisects CD at O\u00a0signifies that O is the mid-point of CD. AO and BO are\u00a0medians of triangles ADC and BDC. <\/span><br \/>\n<span style=\"color: #000000;\">Also, the median divides the triangle into two\u00a0triangles\u00a0of equal areas.\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Consider \u0394ACD. <\/span><br \/>\n<span style=\"color: #000000;\">Line-segment CD is\u00a0bisected by AB at O. Therefore, AO is the median of \u0394ACD. <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (\u0394ACO) = Area (\u0394ADO)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;(1)<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, Considering \u0394BCD , BO is the\u00a0median.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (\u0394BCO) = Area (\u0394BDO)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212; (2)<\/span><br \/>\n<span style=\"color: #000000;\">Adding equation (1) and equation (2), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394ACO) +\u00a0Area (\u0394BCO) =\u00a0Area (\u0394ADO) +\u00a0Area (\u0394BDO)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Area (\u0394ABC) = Area (\u0394ABD)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. D, E and F are, respectively, the midpoints of the sides BC, CA and AB of a \u0394ABC. Show that<br \/>\n<\/strong><strong>(i) <\/strong>BDEF is a parallelogram<strong><br \/>\n<\/strong><strong>(ii) <\/strong>ar(DEF) = \u00bc ar(ABC)<strong><br \/>\n<\/strong><strong>(iii) <\/strong>ar (BDEF) = \u00bd ar(ABC)<br \/>\n<strong>Answer &#8211;\u00a0<\/strong>The midpoint theorem states that the line joining the mid-points of two sides of a triangle is parallel to the third side and half of its length. <\/span><br \/>\n<span style=\"color: #000000;\">If one pair of the opposite side in a quadrilateral is parallel and equal to each other then it is a parallelogram. Diagonals separate the parallelogram into two triangles of equal\u00a0areas. <\/span><br \/>\n<span style=\"color: #000000;\">Let&#8217;s construct a diagram according to the given question.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4486\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"207\" height=\"148\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans5.png 706w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans5-300x215.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans5-480x344.png 480w\" sizes=\"auto, (max-width: 207px) 100vw, 207px\" \/><\/span><br \/>\n<span style=\"color: #000000;\"><strong>(i) BDEF is a parallelogram<br \/>\n<\/strong>In \u0394ABC,<br \/>\nEF || BC and EF = \u00bd BC (by the midpoint theorem)<br \/>\nalso,<br \/>\nBD = \u00bd BC (D is the midpoint)<br \/>\nSo, BD = EF<br \/>\nalso,<br \/>\nBF and DE are parallel and equal to each other.<br \/>\n\u2234 the pair of opposite sides are equal in length and parallel to each other.<br \/>\n\u2234 BDEF is a parallelogram.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) ar(DEF) = \u00bc ar(ABC)<br \/>\n<\/strong>Proceeding from the result of (i),<br \/>\nBDEF, DCEF, and AFDE are parallelograms.<br \/>\nA diagonal of a parallelogram divides it into two triangles of equal area.<br \/>\n\u2234 Area (\u0394BFD) = Area (\u0394DEF) (For parallelogram BDEF)\u00a0 \u00a0 \u00a0&#8212;&#8212;&#8211;\u2014 (i)<br \/>\nalso,<br \/>\nArea (\u0394AFE) = Area (\u0394DEF) (For parallelogram DCEF)\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;\u2014 (ii)<br \/>\nArea (\u0394CDE) = Area (\u0394DEF) (For parallelogram AFDE)\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8211;\u2014 (iii)<br \/>\nFrom (i), (ii) and (iii)<br \/>\nArea (\u0394BFD) = Area (\u0394AFE) = Area (\u0394CDE) = Area (\u0394DEF)<br \/>\n\u21d2 Area (\u0394BFD) + Area (\u0394AFE) + Area (\u0394CDE) + Area (\u0394DEF) = Area (\u0394ABC)<br \/>\n\u21d2 4 Area (\u0394DEF) = Area (\u0394ABC)<br \/>\n\u21d2 Area (DEF) = \u00bc Area (ABC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) ar (BDEF) = \u00bd ar(ABC)<br \/>\n<\/strong>Area (parallelogram BDEF) = Area (\u0394DEF) + Area (\u0394BDE)<br \/>\n\u21d2 Area (parallelogram BDEF) = Area (\u0394DEF) + Area (\u0394DEF)<br \/>\n\u21d2 Area (parallelogram BDEF) = 2\u00d7 Area (\u0394DEF)<br \/>\n\u21d2 Area (parallelogram BDEF) = 2\u00d7 \u00bc Area (\u0394ABC)<br \/>\n\u21d2 Area (parallelogram BDEF) = \u00bd Area (\u0394ABC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. In Fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O, such that OB = OD.<br \/>\nIf AB = CD, then show that<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4487\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"340\" height=\"290\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q6.png 340w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q6-300x256.png 300w\" sizes=\"auto, (max-width: 340px) 100vw, 340px\" \/><br \/>\n<\/strong><strong>(i) <\/strong>ar (DOC) = ar (AOB)<br \/>\n<strong>(ii) <\/strong>ar (DCB) = ar (ACB)<br \/>\n<strong>(iii) <\/strong>DA || CB or ABCD is a parallelogram.<br \/>\n<strong>[Hint: <\/strong>From D and B, draw perpendiculars to AC.<strong>]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>We can draw a perpendicular from vertices B and D on diagonal AC which will help us to make congruent\u00a0triangles\u00a0and we know that congruent triangles are always equal in\u00a0areas. <\/span><br \/>\n<span style=\"color: #000000;\">If two triangles have the same base and also have equal areas, then these triangles must lie between the same parallels.\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Let us construct DN \u22a5 AC and BM \u22a5 AC.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4488\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"340\" height=\"242\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans6.png 340w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans6-300x214.png 300w\" sizes=\"auto, (max-width: 340px) 100vw, 340px\" \/><\/span><br \/>\n<span style=\"color: #000000;\"><strong>Given: <\/strong>OB = OD and AB = CD<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) ar (DOC) = ar (AOB)<br \/>\n<\/strong>In \u0394DOE and \u0394BOF,<br \/>\n\u2220DEO = \u2220BFO (Perpendiculars)<br \/>\n\u2220DOE = \u2220BOF (Vertically opposite angles)<br \/>\nOD = OB (Given)<br \/>\n\u2234 \u0394DOE \u2245 \u0394BOF by AAS congruence condition.<br \/>\n\u2234 DE = BF (By CPCT)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8211;\u2014 (i)<br \/>\nalso,<br \/>\nArea (\u0394DOE) = Area (\u0394BOF) (Congruent triangles)\u00a0 &#8212;&#8212;&#8212;&#8211;\u2014 (ii)<br \/>\nNow,<br \/>\nIn \u0394DEC and \u0394BFA,<br \/>\n\u2220DEC = \u2220BFA (Perpendiculars)<br \/>\nCD = AB (Given)<br \/>\nDE = BF (From i)<br \/>\n\u2234 \u0394DEC \u2245 \u0394BFA by RHS congruence condition.<br \/>\n\u2234 Area (\u0394DEC) = Area (\u0394BFA) (Congruent triangles)\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8211;\u2014 (iii)<br \/>\nAdding (ii) and (iii),<br \/>\nArea (\u0394DOE) + Area (\u0394DEC) = Area (\u0394BOF) + Area (\u0394BFA)<br \/>\n\u21d2 Area (DOC) = Area (AOB)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) ar (DCB) = ar (ACB)<br \/>\n<\/strong>Area (\u0394DOC) = Area (\u0394AOB)<br \/>\nAdding Area (\u0394OCB) in LHS and RHS, we get<br \/>\n\u21d2 Area (\u0394DOC) + Area (\u0394OCB) = Area (\u0394AOB) + Area (\u0394OCB)<br \/>\n\u21d2 Area (\u0394DCB) = Area (\u0394ACB)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) DA || CB or ABCD is a parallelogram.<br \/>\n<\/strong>When two triangles have the same base and equal areas, the triangles will be in between the same parallel lines<br \/>\nArea (\u0394DCB) = Area (\u0394ACB)<br \/>\nDA || BC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;\u2014 (iv)<br \/>\nFor quadrilateral ABCD, one pair of opposite sides are equal (AB = CD), and the other pair of opposite sides are parallel.<br \/>\n\u2234 ABCD is parallelogram.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. D and E are points on sides AB and AC, respectively, of \u0394ABC such that ar(DBC) = ar(EBC). Prove that DE || BC.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>If two\u00a0triangles\u00a0are on a common base and have equal\u00a0areas, then they will lie between the same\u00a0parallel lines. <\/span><br \/>\n<span style=\"color: #000000;\">Let&#8217;s draw the given triangle ABC.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4489\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans7.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"230\" height=\"167\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans7.png 692w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans7-300x218.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans7-480x349.png 480w\" sizes=\"auto, (max-width: 230px) 100vw, 230px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">\u0394DBC and \u0394EBC are on the same base BC and also have equal areas.<br \/>\n\u2234 they will lie between the same parallel lines.<br \/>\n\u2234 DE || BC<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that<br \/>\nar(\u0394ABE) = ar(\u0394ACF)<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>If a\u00a0triangle\u00a0and parallelogram are lying on the same base and between the same\u00a0parallel lines, the area of the triangle will be equal\u00a0to half of the area of a parallelogram.<\/span><br \/>\n<span style=\"color: #000000;\">Also, if two parallelograms are lying on the same base and between the same pair of parallel lines then both of them will have equal\u00a0area. <\/span><br \/>\n<span style=\"color: #000000;\">Let\u2019s draw points X\u00a0and Y, intersected by line EF\u00a0on sides AB and AC respectively.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4490\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans8.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"390\" height=\"194\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans8.png 1120w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans8-300x149.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans8-1024x508.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans8-768x381.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans8-480x238.png 480w\" sizes=\"auto, (max-width: 390px) 100vw, 390px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Let&#8217;s consider BCYE<\/span><br \/>\n<span style=\"color: #000000;\">It is given that, XY || BC so,\u00a0EY || BC<\/span><br \/>\n<span style=\"color: #000000;\">Also,\u00a0BE || AC so, BE || CY<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, BCYE is a parallelogram.<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, In BCFX<\/span><br \/>\n<span style=\"color: #000000;\">It is given that, XY || BC so,\u00a0XF\u00a0|| BC<\/span><br \/>\n<span style=\"color: #000000;\">Since, CF || AB so, CF || BX<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, BCFX\u00a0is a parallelogram.<\/span><br \/>\n<span style=\"color: #000000;\">Parallelograms\u00a0BCYE and BCFX\u00a0are lying on the same base BC and between the same parallels BC and EF.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, According to Theorem 9.1: Parallelograms on the same base and between the same parallels are equal in area.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (BCYE) = Area (BCFX)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- (i)<\/span><br \/>\n<span style=\"color: #000000;\">Now, Consider parallelogram BCYE and \u0394AEB <\/span><br \/>\n<span style=\"color: #000000;\">They are lying on the same base BE and are between the same parallels BE and AC. <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (\u0394ABE) = 1\/2 Area (BCYE)\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- (ii)<\/span><br \/>\n<span style=\"color: #000000;\">Also, parallelogram BCFX\u00a0and \u0394ACF are lying on the same base CF and existing between the same parallels CF and AB.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Area (\u0394ACF) = 1\/2 Area (BCFX)\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- (iii)<\/span><br \/>\n<span style=\"color: #000000;\">From Equations (i), (ii), and (iii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394ABE) = Area (\u0394ACF) proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, and then parallelogram PBQR is completed (see Fig. 9.26). Show that<br \/>\nar(ABCD) = ar(PBQR).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4491\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q9.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"183\" height=\"195\" \/><br \/>\n[Hint: <\/strong>Join AC and PQ. Now, compare ar(ACQ) and ar(APQ).<strong>]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong>Let us join AC and PQ.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4492\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans9.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"183\" height=\"170\" \/><br \/>\nArea (\u25b3ACQ) = Area (\u25b3APQ) (On the same base AQ and between the same parallel lines AQ and CP)<br \/>\n\u21d2 Area (\u25b3ACQ) &#8211; Area (\u25b3ABQ) = Area (\u25b3APQ) &#8211; Area (\u25b3ABQ)<br \/>\n\u21d2 Area (\u25b3ABC) = Area (\u25b3QBP)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;\u2014 (i)<br \/>\nAC and QP are diagonals ABCD and PBQR.<br \/>\n\u2234 Area (ABC) = \u00bd Area (ABCD)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;-\u2014 (ii)<br \/>\nArea (QBP) = \u00bd Area (PBQR)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;-\u2014 (iii)<br \/>\nFrom (ii) and (ii),<br \/>\n\u00bd Area (ABCD) = \u00bd Area (PBQR)<br \/>\n\u21d2 Area (ABCD) = Area (PBQR)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).<\/strong><br \/>\n<strong>Answer &#8211;\u00a0<\/strong>Let&#8217;s draw the given trapezium ABCD<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4493\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans10.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"271\" height=\"161\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans10.png 659w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans10-300x178.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans10-480x286.png 480w\" sizes=\"auto, (max-width: 271px) 100vw, 271px\" \/><br \/>\n\u25b3DAC and \u25b3DBC lie on the same base DC and between the same parallels AB and CD.<br \/>\nArea (\u25b3DAC) = Area (\u25b3DBC)<br \/>\n\u21d2 Area (\u25b3DAC) \u2013 Area (\u25b3DOC) = Area (\u25b3DBC) \u2013 Area (\u25b3DOC)<br \/>\n\u21d2 Area (\u25b3AOD) = Area (\u25b3BOC)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>11. In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.<br \/>\n<\/strong><strong>Show that<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4494\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q11.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"221\" height=\"193\" \/><br \/>\n<\/strong><strong>(i) <\/strong>ar(\u25b3ACB) = ar(\u25b3ACF)<strong><br \/>\n<\/strong><strong>(ii) <\/strong>ar(AEDF) = ar(ABCDE)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n(i) ar(\u25b3ACB) = ar(\u25b3ACF)<br \/>\n<\/strong>\u25b3ACB and \u25b3ACF lie on the same base AC and between the same parallels AC and BF.<br \/>\n\u2234 Area (\u25b3ACB) = Area (\u25b3ACF)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) ar(AEDF) = ar(ABCDE)<br \/>\n<\/strong>\u21d2 Area (\u25b3ACB) + Area (ACDE) = Area (\u25b3ACF) + Area (ACDE)<br \/>\n\u21d2\u00a0 Area (ABCDE) = Area (AEDF)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>12. A villager, Itwaari, has a plot of land in the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given an equal amount of land in lieu of the land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0\u00a0<\/strong>Let\u00a0ABCD\u00a0be the real\u00a0shape of\u00a0the\u00a0quadrilateral\u00a0field.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4495\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans12.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"484\" height=\"176\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans12.png 1022w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans12-300x109.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans12-768x280.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans12-480x175.png 480w\" sizes=\"auto, (max-width: 484px) 100vw, 484px\" \/><br \/>\nThe proposal may be implemented as follows. <\/span><br \/>\n<span style=\"color: #000000;\">Join the BD diagonal and construct\u00a0a line parallel to BD through point A.<\/span><br \/>\n<span style=\"color: #000000;\">Let that line meet further to the extended side CD of ABCD at point E.<\/span><br \/>\n<span style=\"color: #000000;\">Join BE and let BE and AD intersect each other at O.<\/span><br \/>\n<span style=\"color: #000000;\">Then, part\u00a0\u2206AOB is cut from the original field which is taken by the Gram Panchayat.<\/span><br \/>\n<span style=\"color: #000000;\">The new shape of the field will be \u0394BCE. (See figure).<\/span><br \/>\n<span style=\"color: #000000;\">We have to prove that the area of \u0394AOB is equal to the area of \u0394DEO.<\/span><br \/>\n<span style=\"color: #000000;\">We can observe that \u2206DEB and \u2206DAB are lying\u00a0on the same base BD and existing between the same parallels BD and AE.<\/span><br \/>\n<span style=\"color: #000000;\">According to Theorem 9.2: Two\u00a0triangles\u00a0on the same base (or equal bases) and between the same parallels are equal in\u00a0area. <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, Area (\u0394DEB) = Area (\u0394DAB)<\/span><br \/>\n<span style=\"color: #000000;\">Now, subtract Area (\u0394DOB) from both side<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394DEB) &#8211; Area (\u0394DOB) = Area (\u0394DAB) &#8211; Area (\u0394DOB)<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394DEO) = Area (\u0394AOB)<\/span><br \/>\n<span style=\"color: #000000;\">Hence\u00a0\u0394AOB is the area of the plot taken by the gram Panchayat and an equivalent area of the plot that is\u00a0\u0394DEO is given to Itwaari\u00a0so that the overall plot is triangular in shape that is\u00a0\u0394BCE.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (\u25b3ADX) = ar (\u25b3ACY).<br \/>\n<\/strong><strong>[Hint : Join CX.]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong>Let&#8217;s draw\u00a0the given\u00a0trapezium\u00a0ABCD with AB || DC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4496\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans13.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"309\" height=\"193\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans13.png 830w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans13-300x188.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans13-768x480.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans13-480x300.png 480w\" sizes=\"auto, (max-width: 309px) 100vw, 309px\" \/><br \/>\nWe observe that \u0394ADX and \u0394ACX are lying\u00a0on the same base AX and are existing between the same\u00a0parallels\u00a0AB and DC.<\/span><br \/>\n<span style=\"color: #000000;\">According to Theorem 9.2: Two\u00a0triangles\u00a0on the same base (or equal bases) and between the same parallels are equal in area.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, Area (\u0394ADX) = Area (\u0394ACX)\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;(i)<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, \u0394ACY and \u0394ACX are lying on the same base AC and are existing between the same parallels AC and XY.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, Area (\u0394ACY) = Area (\u0394ACX)\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;-(ii)<\/span><br \/>\n<span style=\"color: #000000;\">From Equations (i) and (ii), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">Area (\u0394ADX) = Area (\u0394ACY)<\/span><br \/>\n<span style=\"color: #000000;\">Henced proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>14. In Fig.9.28, AP || BQ || CR. Prove that ar(\u25b3AQC) = ar(\u25b3PBR).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4497\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q14.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"281\" height=\"206\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q14.png 403w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q14-300x220.png 300w\" sizes=\"auto, (max-width: 281px) 100vw, 281px\" \/><br \/>\n<\/strong><strong>Answer &#8211; Given :\u00a0<\/strong>AP || BQ || CR<br \/>\n<strong>To prove :\u00a0<\/strong>Area (AQC) = Area (PBR)<br \/>\nArea (\u25b3AQB) = Area (\u25b3PBQ)\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;\u2014 (i) (Since they are on the same base BQ and between the same parallels AP and BQ.)<br \/>\nalso,<br \/>\nArea (\u25b3BQC) = Area (\u25b3BQR)\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8211;\u2014 (ii) (Since they are on the same base BQ and between the same parallels BQ and CR.)<br \/>\nAdding (i) and (ii),<br \/>\nArea (\u25b3AQB) + Area (\u25b3BQC) = Area (\u25b3PBQ) + Area (\u25b3BQR)<br \/>\n\u21d2 Area (\u25b3AQC) = Area (\u25b3PBR)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(\u25b3AOD) = ar(\u25b3BOC). Prove that ABCD is a trapezium.<br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Let&#8217;s draw the given\u00a0quadrilateral\u00a0ABCD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4498\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans15.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"310\" height=\"180\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans15.png 1022w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans15-300x174.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans15-768x445.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Ans15-480x278.png 480w\" sizes=\"auto, (max-width: 310px) 100vw, 310px\" \/><br \/>\n<strong>Given :\u00a0<\/strong>ar(\u25b3AOD) = ar(\u25b3BOC)<br \/>\n<strong>To prove :\u00a0<\/strong>ABCD is a trapezium.<br \/>\nArea (\u25b3AOD) = Area (\u25b3BOC)<br \/>\n\u21d2 Area (\u25b3AOD) + Area (\u25b3AOB) = Area (\u25b3BOC) + Area (\u25b3AOB)<br \/>\n\u21d2 Area (\u25b3ADB) = Area (\u25b3ACB)<br \/>\nAreas of \u25b3ADB and \u25b3ACB are equal.<br \/>\n\u2234 they must lie between the same parallel lines.<br \/>\n\u2234 AB || CD<br \/>\n\u2234 ABCD is a trapezium.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>16. In Fig.9.29, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4499\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q16.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"312\" height=\"245\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q16.png 383w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-9.3-Q16-300x236.png 300w\" sizes=\"auto, (max-width: 312px) 100vw, 312px\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong><strong>Given :\u00a0<\/strong>ar(\u25b3DRC) = ar(\u25b3DPC),<br \/>\nar(\u25b3BDP) = ar(\u25b3ARC)<br \/>\n<strong>To prove :\u00a0<\/strong>ABCD and DCPR are trapeziums.<br \/>\nArea (\u25b3BDP) = Area (\u25b3ARC)<br \/>\n\u21d2 Area (\u25b3BDP) \u2013 Area (\u25b3DPC) = Area (\u25b3DRC)<br \/>\n\u21d2 Area (\u25b3BDC) = Area (\u25b3ADC)<br \/>\n\u2234 Area (\u25b3BDC) and Area (\u25b3ADC) are lying between the same parallel lines.<br \/>\n\u2234 AB || CD<br \/>\nABCD is a trapezium.<br \/>\nSimilarly,<br \/>\nArea (\u25b3DRC) = Area (\u25b3DPC).<br \/>\n\u2234 Area (\u25b3DRC) and Area (\u25b3DPC) are lying between the same parallel lines.<br \/>\n\u2234 DC || PR<br \/>\n\u2234 DCPR is a trapezium.<\/span><\/p>\n<p>&nbsp;<\/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 9.3 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-4465","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 9.3 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 9 Maths\u00a0Chapter - 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