{"id":126471,"date":"2023-09-03T12:30:00","date_gmt":"2023-09-03T07:00:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126471"},"modified":"2023-12-07T10:56:11","modified_gmt":"2023-12-07T05:26:11","slug":"rd-sharma-class-9-solutions-chapter-10-exercise-10-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-1\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126478\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-10-Exercise-10.1.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-10-Exercise-10.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-10-Exercise-10.1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1:<\/strong> Class 9 Maths is not actually difficult if you are studying it from the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a>. You can easily clear your doubts and learn the concepts in the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 10<\/a> Exercise 10.1. You can download the PDF from the link given in this blog. To know more, read the whole blog.&nbsp;<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69e791d8897a2\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69e791d8897a2\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-1\/#download-rd-sharma-class-9-solutions-chapter-10-exercise-101-pdf\" title=\"Download RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1 PDF\">Download RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1 PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-1\/#access-answers-of-rd-sharma-class-9-solutions-chapter-10-exercise-101\" title=\"Access answers of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1\">Access answers of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-1\/#faqs-on-rd-sharma-class-9-solutions-chapter-10-exercise-101\" title=\"FAQs on RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1\">FAQs on RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-1\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-10-exercise-101\" title=\"How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1?\">How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-1\/#can-i-access-the-rd-sharma-class-9-solutions-chapter-10-exercise-101-pdf-offline\" title=\"Can I access the RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1\u00a0PDF offline?\">Can I access the RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1\u00a0PDF offline?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-1\/#from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-10-exercise-101\" title=\"From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1?\">From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-10-exercise-101-pdf\"><\/span><strong>Download RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-10-Ex-10.1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-10-Ex-10.1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-class-9-solutions-chapter-10-exercise-101\"><\/span><strong>Access answers of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<br>Write the complement of each of the following angles:<br>(i) 20\u00b0<br>(ii) 35\u00b0<br>(iii) 90\u00b0<br>(iv) 77\u00b0<br>(v) 30\u00b0<br>Solution:<br>We know that two angles complement each other if their sum is 90\u00b0. Therefore,<br>(i) Complement of 20\u00b0 is (90\u00b0 \u2013 20\u00b0) = 70\u00b0<br>(ii) Complement of 35\u00b0 is (90\u00b0 \u2013 35\u00b0) = 55\u00b0<br>(iii) Complement of 90\u00b0 is (90\u00b0 \u2013 90\u00b0) = 0\u00b0<br>(iv) Complement of 77\u00b0 is (90\u00b0 \u2013 77\u00b0) = 13\u00b0<br>(v) Complement of 30\u00b0 is (90\u00b0 \u2013 30\u00b0) = 60\u00b0<\/p>\n<p>Question 2.<br>Write the supplement of each of the following angles:<br>(i) 54\u00b0<br>(ii) 132\u00b0<br>(iii) 138\u00b0<br>Solution:<br>We know that two angles are supplementary to each other if their sum is 180\u00b0. Therefore,<br>(i) Supplement of 54\u00b0 is (180\u00b0 \u2013 54\u00b0) = 126\u00b0<br>(ii) Supplement of 132\u00b0 is (180\u00b0 \u2013 132\u00b0) = 48\u00b0<br>(iii) Supplement of 138\u00b0 is (180\u00b0 \u2013 138\u00b0) = 42\u00b0<\/p>\n<p>Question 3.<br>If an angle is 28\u00b0 less than its complement, find its measure.<br>Solution:<br>Let the required angle = x, then<br>Its complement = x + 28\u00b0<br>\u2234&nbsp; x + x + 28\u00b0 = 90\u00b0 \u21d2&nbsp; 2x = 90\u00b0 \u2013 28\u00b0 = 62\u00b0<br>\u2234 x =&nbsp;<span id=\"MathJax-Element-1-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-1\" class=\"math\"><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"mfrac\"><span id=\"MathJax-Span-4\" class=\"msubsup\"><span id=\"MathJax-Span-5\" class=\"texatom\"><span id=\"MathJax-Span-6\" class=\"mrow\"><span id=\"MathJax-Span-7\" class=\"mn\">62<\/span><\/span><\/span><span id=\"MathJax-Span-8\" class=\"texatom\"><span id=\"MathJax-Span-9\" class=\"mrow\"><span id=\"MathJax-Span-10\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-11\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 31\u00b0<br>\u2234 Required angle = 31\u00b0<\/p>\n<p>Question 4.<br>If an angle is 30\u00b0 more than one-half of its complement, find the measure of the angle.<br>Solution:<br>Let the measure of the required angle = x<br>\u2234&nbsp; Its complement =&nbsp; 90\u00b0 \u2013 x<br>\u2234&nbsp; x =&nbsp;<span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-12\" class=\"math\"><span id=\"MathJax-Span-13\" class=\"mrow\"><span id=\"MathJax-Span-14\" class=\"mfrac\"><span id=\"MathJax-Span-15\" class=\"mn\">1<\/span><span id=\"MathJax-Span-16\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;(90\u00b0 \u2013 x) + 30\u00b0<br>2x = 90\u00b0 \u2013 x + 60\u00b0<br>\u21d2 2x + x = 90\u00b0 + 60\u00b0<br>\u21d2&nbsp; 3x = 150\u00b0<br>\u21d2 x = &nbsp;<span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-17\" class=\"math\"><span id=\"MathJax-Span-18\" class=\"mrow\"><span id=\"MathJax-Span-19\" class=\"mfrac\"><span id=\"MathJax-Span-20\" class=\"msubsup\"><span id=\"MathJax-Span-21\" class=\"texatom\"><span id=\"MathJax-Span-22\" class=\"mrow\"><span id=\"MathJax-Span-23\" class=\"mn\">150<\/span><\/span><\/span><span id=\"MathJax-Span-24\" class=\"texatom\"><span id=\"MathJax-Span-25\" class=\"mrow\"><span id=\"MathJax-Span-26\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-27\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp; = 50\u00b0<br>\u2234 Required angle = 50\u00b0<\/p>\n<p>Question 5.<br>Two supplementary angles are in the ratio 4: 5. Find the angles.<br>Solution:<br>Ratio in two supplementary angles = 4 : 5<br>Let first angle = 4x<br>The second angle = 5x<br>\u2234&nbsp; 4x + 5x = 180<br>\u21d2&nbsp; 9x = 180\u00b0<br>\u2234 x&nbsp; =&nbsp;<span id=\"MathJax-Element-4-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-28\" class=\"math\"><span id=\"MathJax-Span-29\" class=\"mrow\"><span id=\"MathJax-Span-30\" class=\"mfrac\"><span id=\"MathJax-Span-31\" class=\"msubsup\"><span id=\"MathJax-Span-32\" class=\"texatom\"><span id=\"MathJax-Span-33\" class=\"mrow\"><span id=\"MathJax-Span-34\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-35\" class=\"texatom\"><span id=\"MathJax-Span-36\" class=\"mrow\"><span id=\"MathJax-Span-37\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-38\" class=\"mn\">9<\/span><\/span><\/span><\/span><\/span>&nbsp;= 20\u00b0<br>\u2234&nbsp; First angle = 4x = 4 x 20\u00b0 = 80\u00b0<br>and second angle = 5x<br>= 5 x 20\u00b0 = 100\u00b0<\/p>\n<p>Question 6.<br>Two supplementary angles differ by 48\u00b0. Find the angles.<br>Solution:<br>Let first angle = x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \u201d<br>The second angle = x + 48\u00b0<br>\u2234&nbsp; x + x + 48\u00b0 = 180\u00b0\u21d2&nbsp; 2x + 48\u00b0 = 180\u00b0<br>\u21d2&nbsp; 2x = 180\u00b0 \u2013 48\u00b0 = 132\u00b0<br>x=&nbsp;<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-39\" class=\"math\"><span id=\"MathJax-Span-40\" class=\"mrow\"><span id=\"MathJax-Span-41\" class=\"mfrac\"><span id=\"MathJax-Span-42\" class=\"msubsup\"><span id=\"MathJax-Span-43\" class=\"texatom\"><span id=\"MathJax-Span-44\" class=\"mrow\"><span id=\"MathJax-Span-45\" class=\"mn\">132<\/span><\/span><\/span><span id=\"MathJax-Span-46\" class=\"texatom\"><span id=\"MathJax-Span-47\" class=\"mrow\"><span id=\"MathJax-Span-48\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-49\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;=66\u00b0<br>\u2234&nbsp; First angle = 66\u00b0<br>and second angle = x + 48\u00b0 = 66\u00b0 + 48\u00b0 = 114\u00b0<br>\u2234 Angles are 66\u00b0, 114\u00b0<\/p>\n<p>Question 7.<br>An angle is equal to 8 times its complement. Determine its measure.<br>Solution:<br>Let the required angle = x<br>Then its complement angle = 90\u00b0 \u2013 x<br>\u2234 x = 8(90\u00b0 \u2013 x)<br>\u21d2 x = 720\u00b0 \u2013 8x \u21d2&nbsp; x + 8x = 720\u00b0<br>\u21d2 9x = 720\u00b0&nbsp;\u21d2 x =&nbsp;&nbsp;<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-50\" class=\"math\"><span id=\"MathJax-Span-51\" class=\"mrow\"><span id=\"MathJax-Span-52\" class=\"mfrac\"><span id=\"MathJax-Span-53\" class=\"msubsup\"><span id=\"MathJax-Span-54\" class=\"texatom\"><span id=\"MathJax-Span-55\" class=\"mrow\"><span id=\"MathJax-Span-56\" class=\"mn\">720<\/span><\/span><\/span><span id=\"MathJax-Span-57\" class=\"texatom\"><span id=\"MathJax-Span-58\" class=\"mrow\"><span id=\"MathJax-Span-59\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-60\" class=\"mn\">9<\/span><\/span><\/span><\/span><\/span>&nbsp;= 80\u00b0<br>\u2234&nbsp; Required angle = 80\u00b0<\/p>\n<p>Question 8.<br>If the angles (2x \u2013 10)\u00b0 and (x \u2013 5)\u00b0 are complementary angles, find x.<br>Solution:<br>First complementary angle = (2x \u2013 10\u00b0) and second = (x \u2013 5)\u00b0<br>\u2234 2x \u2013 10\u00b0 + x \u2013 5\u00b0 = 90\u00b0<br>\u21d2 3x \u2013 15\u00b0 = 90\u00b0 \u21d2&nbsp; 3x = 90\u00b0 + 15\u00b0 = 105\u00b0<br>\u2234 x =&nbsp;<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-61\" class=\"math\"><span id=\"MathJax-Span-62\" class=\"mrow\"><span id=\"MathJax-Span-63\" class=\"mfrac\"><span id=\"MathJax-Span-64\" class=\"msubsup\"><span id=\"MathJax-Span-65\" class=\"texatom\"><span id=\"MathJax-Span-66\" class=\"mrow\"><span id=\"MathJax-Span-67\" class=\"mn\">105<\/span><\/span><\/span><span id=\"MathJax-Span-68\" class=\"texatom\"><span id=\"MathJax-Span-69\" class=\"mrow\"><span id=\"MathJax-Span-70\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-71\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;= 35\u00b0<br>\u2234&nbsp; First angle = 2x \u2013 10\u00b0 = 2 x 35\u00b0 \u2013 10\u00b0<br>= 70\u00b0 \u2013 10\u00b0 = 60\u00b0<br>and second angle = x \u2013 5 = 35\u00b0 \u2013 5 = 30\u00b0<\/p>\n<p>Question 9.<br>If an angle differs from its complement by 10\u00b0, find the angle.<br>Solution:<br>Let required angle = x\u00b0<br>Then its complement angle = 90\u00b0 \u2013 x\u00b0<br>\u2234 x \u2013 (90\u00b0 \u2013 x) = 10<br>\u21d2&nbsp; x \u2013 90\u00b0 + x = 10\u00b0\u21d2&nbsp; 2x = 10\u00b0 + 90\u00b0 = 100\u00b0 100\u00b0<br>\u21d2 x =&nbsp;&nbsp;<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-72\" class=\"math\"><span id=\"MathJax-Span-73\" class=\"mrow\"><span id=\"MathJax-Span-74\" class=\"mfrac\"><span id=\"MathJax-Span-75\" class=\"msubsup\"><span id=\"MathJax-Span-76\" class=\"texatom\"><span id=\"MathJax-Span-77\" class=\"mrow\"><span id=\"MathJax-Span-78\" class=\"mn\">100<\/span><\/span><\/span><span id=\"MathJax-Span-79\" class=\"texatom\"><span id=\"MathJax-Span-80\" class=\"mrow\"><span id=\"MathJax-Span-81\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-82\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 50\u00b0<br>\u2234 Required angle = 50\u00b0<\/p>\n<p>Question 10.<br>If the supplement of an angle is two-thirds of itself Determine the angle and its supplement.<br>Solution:<br>Let required angle = x<br>Then its supplement angle = 180\u00b0 \u2013 x<br>\u2234&nbsp; (180\u00b0-x)=&nbsp;<span id=\"MathJax-Element-9-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-83\" class=\"math\"><span id=\"MathJax-Span-84\" class=\"mrow\"><span id=\"MathJax-Span-85\" class=\"mfrac\"><span id=\"MathJax-Span-86\" class=\"mn\">2<\/span><span id=\"MathJax-Span-87\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>x<br>540\u00b0 \u2013 3x = 2x \u21d2 2x + 3x = 540\u00b0<br>\u21d2 5x = 540\u00b0\u21d2&nbsp; x =&nbsp;<span id=\"MathJax-Element-10-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-88\" class=\"math\"><span id=\"MathJax-Span-89\" class=\"mrow\"><span id=\"MathJax-Span-90\" class=\"mfrac\"><span id=\"MathJax-Span-91\" class=\"msubsup\"><span id=\"MathJax-Span-92\" class=\"texatom\"><span id=\"MathJax-Span-93\" class=\"mrow\"><span id=\"MathJax-Span-94\" class=\"mn\">540<\/span><\/span><\/span><span id=\"MathJax-Span-95\" class=\"texatom\"><span id=\"MathJax-Span-96\" class=\"mrow\"><span id=\"MathJax-Span-97\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-98\" class=\"mn\">5<\/span><\/span><\/span><\/span><\/span>&nbsp;= 108\u00b0<br>-. Supplement angle = 180\u00b0 \u2013 108\u00b0 = 72\u00b0<\/p>\n<p>Question 11.<br>An angle is 14\u00b0 more than its complementary angle. What is its measure?<br>Solution:<br>Let required angle = x<br>Then its complementary angle = 90\u00b0 \u2013 x<br>\u2234&nbsp; x + 14\u00b0 = 90\u00b0 \u2013 x<br>x + x = 90\u00b0 \u2013 14\u00b0 \u21d2&nbsp; 2x = 76\u00b0<br>\u21d2 x =&nbsp;&nbsp;<span id=\"MathJax-Element-11-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-99\" class=\"math\"><span id=\"MathJax-Span-100\" class=\"mrow\"><span id=\"MathJax-Span-101\" class=\"mfrac\"><span id=\"MathJax-Span-102\" class=\"msubsup\"><span id=\"MathJax-Span-103\" class=\"texatom\"><span id=\"MathJax-Span-104\" class=\"mrow\"><span id=\"MathJax-Span-105\" class=\"mn\">76<\/span><\/span><\/span><span id=\"MathJax-Span-106\" class=\"texatom\"><span id=\"MathJax-Span-107\" class=\"mrow\"><span id=\"MathJax-Span-108\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-109\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 38\u00b0<br>\u2234&nbsp; Required angle = 38\u00b0<\/p>\n<p>Question 12.<br>The measure of an angle is twice the measure of its supplementary angle. Find its measure.<br>Solution:<br>Let the required angle = x<br>\u2234&nbsp; Its supplementary angle = 180\u00b0 \u2013 x<br>\u2234&nbsp; x = 2(180\u00b0-x) = 360\u00b0-2x<br>\u21d2&nbsp; x +&nbsp;2x = 360\u00b0<br>\u21d2 3x = 360\u00b0<br>\u21d2&nbsp; x =&nbsp;<span id=\"MathJax-Element-12-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-110\" class=\"math\"><span id=\"MathJax-Span-111\" class=\"mrow\"><span id=\"MathJax-Span-112\" class=\"mfrac\"><span id=\"MathJax-Span-113\" class=\"msubsup\"><span id=\"MathJax-Span-114\" class=\"texatom\"><span id=\"MathJax-Span-115\" class=\"mrow\"><span id=\"MathJax-Span-116\" class=\"mn\">360<\/span><\/span><\/span><span id=\"MathJax-Span-117\" class=\"texatom\"><span id=\"MathJax-Span-118\" class=\"mrow\"><span id=\"MathJax-Span-119\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-120\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;= 120\u00b0<br>\u2234&nbsp; Required angle = 120\u00b0<\/p>\n<p>Question 13.<br>If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of the angle.<br>Solution:<br>Let required angle = x<br>Then its complement angle = 90\u00b0 \u2013 x<br>and supplement angle = 180\u00b0 \u2013 x<br>\u2234&nbsp; 3(90\u00b0 \u2013 x) = 180\u00b0 \u2013 x<br>\u21d2 270\u00b0 \u2013 3x = 180\u00b0 \u2013 x<br>\u21d2270\u00b0 \u2013 180\u00b0 = -x + 3x =&gt; 2x = 90\u00b0<br>\u21d2 x = 45\u00b0<br>\u2234&nbsp; Required angle = 45\u00b0<\/p>\n<p>Question 14.<br>If the supplement of an angle is three times its complement, find the angle.<br>Solution:<br>Let required angle = x<br>Then its complement = 90\u00b0-&nbsp; x<br>and supplement = 180\u00b0 \u2013 x<br>\u2234&nbsp; 180\u00b0-x = 3(90\u00b0-x)<br>\u21d2&nbsp; 180\u00b0 \u2013 x = 270\u00b0 \u2013 3x<br>\u21d2&nbsp; -x + 3x = 270\u00b0 \u2013 180\u00b0<br>\u21d2 2x = 90\u00b0 \u21d2 x =&nbsp;<span id=\"MathJax-Element-13-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-121\" class=\"math\"><span id=\"MathJax-Span-122\" class=\"mrow\"><span id=\"MathJax-Span-123\" class=\"mfrac\"><span id=\"MathJax-Span-124\" class=\"msubsup\"><span id=\"MathJax-Span-125\" class=\"texatom\"><span id=\"MathJax-Span-126\" class=\"mrow\"><span id=\"MathJax-Span-127\" class=\"mn\">90<\/span><\/span><\/span><span id=\"MathJax-Span-128\" class=\"texatom\"><span id=\"MathJax-Span-129\" class=\"mrow\"><span id=\"MathJax-Span-130\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-131\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;=45\u00b0<br>\u2234 Required angle = 45\u00b0<\/p>\n<p>This is the complete blog on RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1. To know more about the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 9 Maths exam, ask in the comments.&nbsp;<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-9-solutions-chapter-10-exercise-101\"><\/span><strong>FAQs on RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631263484709\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-10-exercise-101\"><\/span>How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631263506728\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-class-9-solutions-chapter-10-exercise-101-pdf-offline\"><\/span>Can I access the RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631263524612\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-10-exercise-101\"><\/span>From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1: Class 9 Maths is not actually difficult if you are studying it from the RD Sharma Solutions Class 9 Maths. You can easily clear your doubts and learn the concepts in the RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1. You can download the PDF &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-1\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":126478,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411],"tags":[3086,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126471"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=126471"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126471\/revisions"}],"predecessor-version":[{"id":499615,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126471\/revisions\/499615"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126478"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126471"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126471"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126471"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}