{"id":4356,"date":"2023-02-09T06:58:40","date_gmt":"2023-02-09T06:58:40","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4356"},"modified":"2023-02-09T07:00:39","modified_gmt":"2023-02-09T07:00:39","slug":"ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-4\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 7 Triangles Ex 7.4"},"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; 7 (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; 7 Triangles <\/strong>Exercise 7.4 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 7: Triangles<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.1<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.2<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.3<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-5\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.5<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 7.4\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Show that in a right-angled triangle, the hypotenuse is the longest side.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Let us consider a\u00a0right-angled triangle\u00a0ABC,\u00a0right-angled\u00a0at B.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4412\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"177\" height=\"159\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans1.png 2100w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans1-300x270.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans1-1024x921.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans1-768x690.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans1-1536x1381.png 1536w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans1-2048x1841.png 2048w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans1-480x432.png 480w\" sizes=\"auto, (max-width: 177px) 100vw, 177px\" \/><br \/>\nIn \u2206ABC,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A + \u2220B + \u2220C = 180\u00b0 (Angle sum property of a triangle)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A + 90<sup>o\u00a0<\/sup>+ \u2220C = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A + \u2220C = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Hence, the other two angles have to be acute (i.e., less than 90<sup>o<\/sup>\u00a0).<\/span><br \/>\n<span style=\"color: #000000;\">Thus, \u2220B is the largest angle in \u2206ABC.<\/span><br \/>\n<span style=\"color: #000000;\">So, \u2220B &gt; \u2220A and \u2220B &gt; \u2220C<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, AC &gt; BC and AC &gt; AB [Using theorem 7.7 of triangles, in any triangle, the side opposite to the larger (greater) angle is longer.] <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, AC is the largest side in \u2206ABC.<\/span><br \/>\n<span style=\"color: #000000;\">However, AC is the\u00a0hypotenuse\u00a0of \u2206ABC.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, the hypotenuse is the longest side in a right-angled triangle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. In Fig. 7.48, sides AB and AC of \u0394ABC are extended to points P and Q, respectively. Also, \u2220PBC &lt; \u2220QCB. Show that AC &gt; AB.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4413\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Q2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"185\" height=\"197\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 \u00a0<\/strong>In the given figure,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABC + \u2220PBC = 180\u00b0 [Linear pair of angles]<\/span><br \/>\n<span style=\"color: #000000;\">Also, \u2220ABC = 180\u00b0 &#8211; \u2220PBC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;- (1)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ACB + \u2220QCB = 180\u00b0 [Linear pair]<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ACB = 180\u00b0 &#8211; \u2220QCB\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;-\u00a0 (2)<\/span><br \/>\n<span style=\"color: #000000;\">As \u2220PBC &lt; \u2220QCB (given),<\/span><br \/>\n<span style=\"color: #000000;\">180<sup>o<\/sup>\u00a0&#8211; \u2220PBC &gt; 180<sup>o<\/sup>\u00a0&#8211; \u2220QCB<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABC &gt; \u2220ACB [From Equations (1) and (2)]<\/span><br \/>\n<span style=\"color: #000000;\">Thus, AC &gt; AB (Side opposite to the larger angle is larger).<\/span><br \/>\n<span style=\"color: #000000;\">Hence proved, AC &gt; AB.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. In Fig. 7.49, \u2220B &lt; \u2220A and \u2220C &lt; \u2220D. Show that AD &lt; BC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4414\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Q3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"220\" height=\"258\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong><strong>Given:\u00a0<\/strong>\u2220B &lt; \u2220A and \u2220C &lt; \u2220D<\/span><br \/>\n<span style=\"color: #000000;\"><strong>To prove:\u00a0<\/strong>AD &lt; BC <\/span><br \/>\n<span style=\"color: #000000;\">We can use the fact that in any\u00a0triangle, the side opposite to the larger (greater) angle is longer.<br \/>\nIn \u0394AOB, \u2220B &lt; \u2220A (given)<\/span><br \/>\n<span style=\"color: #000000;\">AO &lt; OB (The side opposite to the smaller angle is smaller)\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212; (1)<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394COD, \u2220C &lt; \u2220D<\/span><br \/>\n<span style=\"color: #000000;\">OD &lt; OC (The side opposite to the smaller angle is smaller)\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212; (2)<\/span><br \/>\n<span style=\"color: #000000;\">On adding Equations (1) and (2), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">AO + OD &lt; BO + OC<\/span><br \/>\n<span style=\"color: #000000;\">AD &lt; BC<\/span><br \/>\n<span style=\"color: #000000;\">Hence, proved<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). <\/strong><strong>Show that \u2220A &gt; \u2220C and \u2220B &gt; \u2220D.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4415\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"185\" height=\"275\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong><strong>Given:\u00a0<\/strong>AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD<\/span><br \/>\n<span style=\"color: #000000;\"><strong>To prove:\u00a0<\/strong>\u2220A &gt; \u2220C and \u2220B &gt; \u2220D<\/span><br \/>\n<span style=\"color: #000000;\">Let&#8217;s join vertex A to C and D to B as shown below.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4417\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Answer4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"197\" height=\"229\" \/><br \/>\nIn \u0394ABD, we see that<br \/>\nAB &lt; AD &lt; BD<br \/>\nSo, \u2220ADB &lt; \u2220ABD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;\u2014 (i) (Since the angle opposite to the longer side is always larger)<br \/>\nNow, in \u0394BCD,<br \/>\nBC &lt; DC &lt; BD<br \/>\nHence, it can be concluded that<br \/>\n\u2220BDC &lt; \u2220CBD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;\u2014 (ii)<br \/>\nNow, by adding equation (i) and equation (ii) we get,<br \/>\n\u2220ADB + \u2220BDC &lt; \u2220ABD + \u2220CBD<br \/>\n\u2220ADC &lt; \u2220ABC<br \/>\n\u2220B &gt; \u2220D<br \/>\nSimilarly, In triangle ABC,<br \/>\n\u2220ACB &lt; \u2220BAC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;\u2014 (iii) (Since the angle opposite to the longer side is always larger)<br \/>\nNow, In \u0394ADC,<br \/>\n\u2220DCA &lt; \u2220DAC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;\u2014 (iv)<br \/>\nBy adding equation (iii) and equation (iv), we get,<br \/>\n\u2220ACB + \u2220DCA &lt; \u2220BAC+\u2220DAC<br \/>\n\u21d2 \u2220BCD &lt; \u2220BAD<br \/>\n\u2234 \u2220A &gt; \u2220C<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. In Fig 7.51, PR &gt; PQ and PS bisect \u2220QPR. Prove that \u2220PSR &gt; \u2220PSQ.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4419\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Q5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"267\" height=\"251\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong><strong>Given:\u00a0<\/strong>PR &gt; PQ and PS bisects \u2220QPR<\/span><br \/>\n<span style=\"color: #000000;\"><strong>To prove:\u00a0<\/strong>\u2220PSR &gt; \u2220PSQ<\/span><br \/>\n<span style=\"color: #000000;\">As PR &gt; PQ, \u2220PQS &gt; \u2220PRS (Angle opposite to larger side is larger ) &#8212;&#8212;&#8212;&#8212; (1)<\/span><br \/>\n<span style=\"color: #000000;\">PS is the\u00a0angle bisector\u00a0of \u2220QPR.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220QPS = \u2220RPS &#8212;&#8212;&#8212;&#8212; (2) [Since,PS bisects \u2220QPR]<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PSR is the exterior angle of \u0394PQS.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PSR = \u2220PQS + \u2220QPS &#8212;&#8212;&#8212;&#8212; (3) [Using\u00a0 exterior angle sum property]<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PSQ is the exterior angle of \u0394PRS.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PSQ = \u2220PRS + \u2220RPS &#8212;&#8212;&#8212;&#8212; (4) [Using exterior angle sum property]<\/span><br \/>\n<span style=\"color: #000000;\">From Equations (1) and (2), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PQS\u00a0+ \u2220QPS &gt; \u2220PRS\u00a0+ \u2220RPS<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PSR &gt; \u2220PSQ [From Equations (3) and (4)]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong>We know that in a triangle if one angle is 90 degrees, then the other angles have to be acute.<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4420\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"473\" height=\"175\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans6.png 1146w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans6-300x111.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans6-1024x379.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans6-768x284.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.4-Ans6-480x178.png 480w\" sizes=\"auto, (max-width: 473px) 100vw, 473px\" \/><br \/>\nLet us take a line l and from point P, that is, not on line l, draw two\u00a0line segments PN and PM. Let PN be perpendicular\u00a0to line l and PM is drawn at some other angle.<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394PNM, \u2220N = 90<sup>o<br \/>\n<\/sup>\u2220P + \u2220N + \u2220M = 180<sup>o\u00a0<\/sup>(Angle sum property of a triangle)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220P + \u2220M = 90<sup>o<br \/>\n<\/sup>Clearly, \u2220M is an acute angle.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220M &lt; \u2220N<\/span><br \/>\n<span style=\"color: #000000;\">PN &lt; PM (The side opposite to the smaller angle is smaller)<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to all of them. Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.<\/span><\/p>\n<p style=\"text-align: justify;\">\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 7 (Triangles)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 7 Triangles Exercise 7.4 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 7: Triangles NCERT Solution Class 9 Maths Ex &#8211; 7.1 NCERT Solution Class 9 Maths [&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":[713,702,186,714,701,5],"class_list":["post-4356","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-7-triangles-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-7-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) - 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