{"id":6083,"date":"2023-09-18T04:30:36","date_gmt":"2023-09-18T04:30:36","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6083"},"modified":"2023-09-18T04:30:55","modified_gmt":"2023-09-18T04:30:55","slug":"ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-1\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 11 Constructions Ex 11.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 11 (Constructions)\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 10 Mathematics <strong>Chapter &#8211; 11 Constructions <\/strong>Exercise 11.1 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 : 11 Constructions<\/strong><\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-11-constructions-ex-11-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 11.2<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 11.1<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>In each of the following, give the justification of the construction also:<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.<br \/>\n<\/strong><b>Solution &#8211;\u00a0<\/b>Steps of Construction:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6583 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans1-300x161.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"161\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans1-300x161.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans1.png 492w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/> <\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Step I &#8211;<\/strong> Draw AB = 7.6 cm<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Step II &#8211;<\/strong> Draw ray AX, making an acute angle with AB<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Step III &#8211;<\/strong> Mark 13\u00a0(i.e, 5\u00a0+ 8) points as A\u2081, A<sub>\u2082<\/sub> ,\u2026. A<sub>\u2081\u2083<\/sub>\u00a0on AX such that\u00a0AA<sub>\u2081<\/sub>\u00a0= A<sub>\u2081<\/sub>A<sub>\u2082<\/sub>\u00a0= A<sub>\u2082<\/sub>A<sub>\u2083<\/sub> = &#8230;&#8230; A<sub>\u2081\u2082<\/sub>A<sub>\u2081\u2083<br \/>\n<\/sub><strong>Step IV &#8211; <\/strong>Join BA<sub>\u2081\u2083<br \/>\n<\/sub><strong>Step V &#8211; <\/strong>Through A<sub>\u2085<\/sub>\u00a0(since we need 5\u00a0parts to 8 parts) draw CA<sub>\u2085<\/sub>\u00a0parallel to BA<sub>\u2081\u2083<\/sub>\u00a0where C lies on AB.<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Now AC : CB = 5 : 8 <\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">By measurement, we find that AC = 2.9 cm and CB = 4.7 cm <\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Proof : <\/strong><\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">CA<sub>\u2085<\/sub>\u00a0is parallel to BA<sub>\u2081\u2083<br \/>\n<\/sub>By Basic Proportionality theorem, in \u0394AA<sub>\u2081\u2083<\/sub>B <\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">AC\/BC = AA<sub>\u2085<\/sub>\/A<sub>\u2085<\/sub>A<sub>\u2081\u2083<\/sub> = 5\/8 (By Construction) <\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\">Thus, C divides AB in the ratio 5 : 8.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>2.\u00a0 Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose\u00a0sides are 2\/3 of the corresponding sides of the first triangle.<br \/>\n<b>Solution &#8211;\u00a0<\/b><\/strong>Steps of Construction:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6588\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"207\" height=\"256\" \/><br \/>\n<strong>Step I &#8211; <\/strong>Draw BC = 6 cm. With B and C as centres and radii, 5 cm and 4 cm respectively draw arcs to intersect at A. \u0394ABC is obtained.<br \/>\n<strong>Step II &#8211; <\/strong>Draw ray BX making an acute angle with BC.<br \/>\n<strong>Step III &#8211; <\/strong>Mark 3 (since, 3 &gt; 2 in the ratio 2\/3) points B\u2081, B\u2082, B\u2083 on BX such that BB\u2081 = B\u2081B\u2082 = B\u2082B\u2083.<br \/>\n<strong>Step IV &#8211; <\/strong>Join B\u2083C and draw a line through B\u2082 (second point where 2 &lt; 3 in the ratio) parallel to B\u2083C meeting BC at C&#8217;.<br \/>\n<strong>Step V &#8211; <\/strong>Draw a line thorough C&#8217; parallel to CA to meet BA at A\u2019. Now \u0394A&#8217;BC&#8217; is the required triangle similar to \u0394ABC where BC&#8217;\/BC = BA&#8217;\/BA = C&#8217;A&#8217;\/CA = 2\/3<br \/>\n<strong>Proof:<br \/>\n<\/strong>In \u0394BB\u2083C, B\u2082C&#8217; is parallel to B\u2083C.<br \/>\nHence by Basic proportionality theorem,<br \/>\nB\u2082B\u2083\/BB\u2082 = C&#8217;C\/BC&#8217; = 1\/2<br \/>\nAdding 1 to both the sides of C&#8217;C\/BC&#8217; = 1\/2,<br \/>\nC&#8217;C\/BC&#8217; + 1 = 1\/2 + 1<br \/>\n(C&#8217;C + BC&#8217;)\/BC&#8217; = 3\/2<br \/>\nBC\/BC&#8217; = 3\/2<br \/>\nor BC&#8217;\/BC = 2\/3\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;- (i)<br \/>\nConsider \u0394BA&#8217;C&#8217; and \u0394BAC<br \/>\n\u2220A&#8217;BC&#8217; = \u2220ABC (Common)<br \/>\n\u2220BA&#8217;C&#8217; = \u2220BAC (Corresponding angles \u2235 C&#8217; A&#8217; ||CA)<br \/>\nHence by AA similarity, \u0394BA&#8217;C&#8217; ~ \u0394BAC<br \/>\nCorresponding sides are proportional<br \/>\nBA&#8217;\/BA = C&#8217;A&#8217;\/CA = BC&#8217;\/BC = 2\/3 [From equation (i)] <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7\/5 of the corresponding sides of the first triangle.<br \/>\n<\/strong><strong><b>Solution &#8211;\u00a0<\/b><\/strong>Steps of Construction:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6585\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"473\" height=\"407\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans3.png 473w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans3-300x258.png 300w\" sizes=\"auto, (max-width: 473px) 100vw, 473px\" \/> <\/span><br \/>\n<span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Step I &#8211; <\/strong>Draw BC = 7cm with B and C as centres and radii 5 cm and 6 cm respectively. Draw arcs to intersect at A. \u0394ABC is obtained.<br \/>\n<strong>Step II &#8211; <\/strong>Draw ray BX making \u2220CBX an acute angle.<br \/>\n<strong>Step III &#8211; <\/strong>Mark 7 points (greater of 7 and 5 in 7\/5 ) B\u2081, B\u2082,\u2026\u2026\u2026B\u2087 on BX such that BB\u2081 = B\u2081B\u2082 =&#8230;&#8230;&#8230;&#8230;&#8230; = B\u2086B\u2087<br \/>\n<strong>Step IV &#8211; <\/strong>Join B\u2085\u00a0(smaller of 7 and 5 in 7\/5 which is\u00a0the 5<sup>th<\/sup> point) to C and draw B\u2087C&#8217; parallel to B\u2085C intersecting the extension of BC at C&#8217;.<br \/>\n<strong>Step V &#8211; <\/strong>Through C&#8217; draw C&#8217;A&#8217; parallel to CA to meet the extension of BA at A\u2019. Now, \u0394A&#8217; B&#8217;C&#8217; is the required triangle similar to \u0394ABC where BA&#8217;\/BA = C&#8217;A&#8217;\/CA = BC&#8217;\/BC = 7\/5<br \/>\n<strong>Proof :<br \/>\n<\/strong>In \u0394BB\u2087C&#8217;, B\u2083C is parallel to B<span style=\"font-size: 12pt;\"><sub>\u2087<\/sub><\/span>C&#8217;<br \/>\nHence by Basic proportionality theorem,<br \/>\nB\u2085B\u2087\/BB\u2085 = CC&#8217;\/BC = 2\/5<br \/>\nAdding 1 to both the sides of CC&#8217;\/BC = 2\/5,<br \/>\nCC&#8217;\/BC + 1 = 2\/5 + 1<br \/>\n(BC + CC&#8217;)\/BC = 7\/5<br \/>\nBC&#8217;\/BC = 7\/5<br \/>\nConsider \u0394BAC and \u0394BA&#8217;C&#8217;<br \/>\n\u2220ABC = \u2220A&#8217;BC&#8217; (Common)<br \/>\n\u2220BCA = \u2220BC&#8217;A&#8217; (Corresponding angles \u2235 CA || C&#8217;A&#8217;)<br \/>\nBy AA criteria, \u0394BAC \u223c \u0394BA&#8217;C&#8217;<br \/>\n\u2234 Corresponding sides are proportional<br \/>\nHence,<br \/>\nBA&#8217;\/BA = C&#8217;A&#8217;\/A = BC&#8217;\/BC = 7\/5<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.<br \/>\n<\/strong><strong><b>Solution &#8211;\u00a0<\/b><\/strong>Steps of Construction:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6599 size-full\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"480\" height=\"409\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans6.png 480w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans6-300x256.png 300w\" sizes=\"auto, (max-width: 480px) 100vw, 480px\" \/><br \/>\n<strong>Step I &#8211; <\/strong>Draw BC = 8cm.<br \/>\n<strong>Step II &#8211; <\/strong>Through D, the mid-point of BC, draw the perpendicular to BC and cut an arc from D on it such that DA = 4cm. Join BA and CA. \u0394ABC is obtained.<br \/>\n<strong>Step III &#8211; <\/strong>Draw the ray BX so that \u2220CBX is acute.<br \/>\n<strong>Step IV &#8211; <\/strong>Mark 3 (since, 3 &gt; 2 in <span class=\"equation\"><span id=\"MathJax-Element-2-Frame\" class=\"mjx-chtml MathJax_CHTML\" style=\"box-sizing: inherit; margin: 0px; padding: 1px 0px; border: 0px; font-style: normal; font-variant: inherit; font-weight: normal; font-stretch: inherit; font-size: 18.08px; line-height: 0; font-optical-sizing: inherit; font-kerning: inherit; font-feature-settings: inherit; font-variation-settings: inherit; vertical-align: baseline; display: inline-block; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;\" tabindex=\"0\" role=\"presentation\" data-mathml=\"&lt;math xmlns=&quot;http:\/\/www.w3.org\/1998\/Math\/MathML&quot;&gt;&lt;mn&gt;1&lt;\/mn&gt;&lt;mrow class=&quot;MJX-TeXAtom-ORD&quot;&gt;&lt;mstyle mathsize=&quot;1.44em&quot;&gt;&lt;mfrac&gt;&lt;mn&gt;1&lt;\/mn&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/mfrac&gt;&lt;\/mstyle&gt;&lt;\/mrow&gt;&lt;\/math&gt;\"><span id=\"MJXc-Node-11\" class=\"mjx-math\" aria-hidden=\"true\"><span id=\"MJXc-Node-12\" class=\"mjx-mrow\"><span id=\"MJXc-Node-13\" class=\"mjx-mn\"><span class=\"mjx-char MJXc-TeX-main-R\">1<\/span><\/span><span id=\"MJXc-Node-14\" class=\"mjx-texatom\"><span id=\"MJXc-Node-15\" class=\"mjx-mrow\"><span id=\"MJXc-Node-16\" class=\"mjx-mstyle\"><span id=\"MJXc-Node-17\" class=\"mjx-mrow\"><span id=\"MJXc-Node-18\" class=\"mjx-mfrac\"><span class=\"mjx-box MJXc-stacked\"><span class=\"mjx-numerator\"><span id=\"MJXc-Node-19\" class=\"mjx-mn\"><span class=\"mjx-char MJXc-TeX-main-R\">1<\/span><\/span><\/span><span class=\"mjx-denominator\"><span id=\"MJXc-Node-20\" class=\"mjx-mn\"><span class=\"mjx-char MJXc-TeX-main-R\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"MJX_Assistive_MathML\" role=\"presentation\">112<\/span><\/span><\/span> = 3\/2) points B<sub>1<\/sub>, B<sub>2<\/sub>, B<sub>3<\/sub> on B<sub>X<\/sub> such that BB<sub>1<\/sub> = B<sub>1<\/sub>B<sub>2<\/sub> = B<sub>2<\/sub>B<sub>3<br \/>\n<\/sub><strong>Step V &#8211; <\/strong>Join B<sub>2<\/sub> (2<sup>nd<\/sup> point \u2235 2 &lt; 3) to C and draw B\u2083C&#8217; parallel to B\u2082C, intersecting BC extended at C\u2019.<br \/>\n<strong>Step VI &#8211; <\/strong>Through C\u2019 draw C&#8217;A&#8217; parallel to CA to intersect BA extended to A\u2019. Now, \u0394A&#8217;BC&#8217; is the required triangle similar to \u0394ABC where BA&#8217;\/BA = C&#8217;A&#8217;\/CA = BC&#8217;\/BC = 3\/2<br \/>\n<strong>Proof:<br \/>\n<\/strong>In \u0394BB<sub>3<\/sub>C&#8217;, B<sub>2<\/sub>C || B<sub>3<\/sub>C&#8217;,<br \/>\nHence by Basic proportionality theorem,<br \/>\nB<sub>2<\/sub>B<sub>3<\/sub>\/BB<sub>2<\/sub> = CC&#8217;\/BC = 1\/2<br \/>\nAdding 1 to CC&#8217;\/BC = 1\/2<br \/>\nCC&#8217;\/BC + 1 = 1\/2 + 1<br \/>\n(BC+CC&#8217;)\/BC = 3\/2<br \/>\nBC&#8217;\/BC = 3\/2<br \/>\nConsider \u0394BAC and \u0394BA&#8217;C&#8217;<br \/>\n\u2220ABC = \u2220A&#8217;BC&#8217; (Common)<br \/>\n\u2220BCA = \u2220BC&#8217;A&#8217; (Corresponding angles \u2235 CA || C&#8217;A&#8217;)<br \/>\n\u2220BAC = \u2220BA&#8217;C&#8217; (Corresponding angles)<br \/>\nBy AAA axiom, \u0394BAC ~ \u0394BA&#8217;C&#8217;<br \/>\n\u2234 Corresponding sides are proportional<br \/>\nHence,<br \/>\nBA&#8217;\/BA = BC&#8217;\/BC= C&#8217;A&#8217;\/CA = 3\/2\u00a0 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and \u2220ABC = 60\u00b0. Then construct a triangle whose sides are 3\/4 of the corresponding sides of the triangle ABC.<br \/>\n<\/strong><strong><b>Solution &#8211;\u00a0<\/b><\/strong>Steps of Construction:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6596\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"493\" height=\"408\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans5.png 493w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans5-300x248.png 300w\" sizes=\"auto, (max-width: 493px) 100vw, 493px\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step I &#8211; <\/strong>Draw a line BC = 6 cm.<br \/>\n<strong>Step II &#8211; <\/strong>At B, make \u2220C = 60\u00b0 and cut an arc at A on the same line so that BA = 5 cm. Join AC, \u0394ABC is obtained.<br \/>\n<strong>Step III &#8211; <\/strong>Draw the ray BX such that \u2220CBX is acute.<br \/>\n<strong>Step IV &#8211; <\/strong>Mark 4 (since 4 &gt; in 3\/4) points B\u2081, B\u2082, B\u2083, B\u2084 on BX such that BB\u2081 = B\u2081B\u2082 = B\u2082B\u2083 = B\u2083B\u2084<br \/>\n<strong>Step V &#8211; <\/strong>Join B\u2084 to C and draw B\u2083C&#8217; parallel to B\u2084C to intersect BC at C&#8217;.<br \/>\n<strong>Step VI &#8211; <\/strong>Draw C&#8217;A&#8217; parallel to CA to intersect BA at A\u2019.<br \/>\nNow, \u0394A&#8217;BC&#8217; is the required triangle similar to \u0394ABC where BA&#8217;\/BA = BC&#8217;\/BC = C&#8217;A&#8217;\/CA = 3\/4<br \/>\n<strong>Proof:<br \/>\n<\/strong>In \u0394BB\u2084C , B\u2083C&#8217; || B\u2084C<br \/>\nHence by Basic proportionality theorem,<br \/>\nB\u2083B\u2084\/BB\u2083 = C&#8217;C\/BC&#8217; = 1\/3<br \/>\nC&#8217;C \/BC&#8217; + 1 = 1\/3 + 1<br \/>\n(C&#8217;C + BC&#8217;)\/BC&#8217; = 4\/3<br \/>\nBC\/BC&#8217; = 4\/3 or BC&#8217;\/BC = 3\/4<br \/>\nConsider \u0394BA&#8217;C&#8217; and \u0394BAC<br \/>\n\u2220A&#8217;BC&#8217; = \u2220ABC = 60\u00b0<br \/>\n\u2220BCA&#8217; = \u2220BCA (Corresponding angles \u2235 C&#8217;A&#8217;||CA)<br \/>\n\u2220BA&#8217;C&#8217; = \u2220BAC (Corresponding angles)<br \/>\nBy AAA axiom, \u0394BA&#8217;C&#8217; ~ \u0394BAC<br \/>\nTherefore, corresponding sides are proportional,<br \/>\nBC&#8217;\/BC = BA&#8217;\/BA = C&#8217;A&#8217;\/CA = 3\/4<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. Draw a triangle ABC with side BC = 7 cm, \u2220B = 45\u00b0, \u2220A = 105\u00b0. Then, construct a triangle whose sides are 4\/3 times the corresponding sides of \u0394 ABC.<br \/>\n<\/strong><strong><b>Solution &#8211;\u00a0 <\/b><\/strong>Sum of all side of triangle = 180\u00b0<br \/>\n\u2234 \u2220A \u00a0+ \u2220B \u00a0+ \u2220C \u00a0= 180\u00b0<br \/>\n\u2220C = 180\u00b0 &#8211; 150\u00b0 = 30\u00b0<br \/>\nSteps of Construction:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6599\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"437\" height=\"372\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans6.png 480w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans6-300x256.png 300w\" sizes=\"auto, (max-width: 437px) 100vw, 437px\" \/><br \/>\n<strong>Step I &#8211; <\/strong>Draw BC = 7cm and at B, make an angle \u2220CBY = 45\u00b0 and at C, make \u2220BCZ = 30\u00b0 [Since, 180\u00b0 &#8211; (45\u00b0 + 105\u00b0 )]. Both BY and CZ intersect at A and thus \u0394ABC is\u00a0constructed.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step II &#8211; <\/strong>Draw the ray BX so that \u2220CBX is acute.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step III &#8211; <\/strong>Mark 4 (since 4 &gt; 3 in 4\/3) points B\u2081, B\u2082, B\u2083, B\u2084\u00a0on BX such that\u00a0BB\u2081\u00a0= B\u2081B\u2082\u00a0= B\u2082B\u2083\u00a0= B\u2083B\u2084<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step IV &#8211; <\/strong>Join B\u2083\u00a0third point on BX, (since 3 &lt; 4 in 4\/3) to C and draw B C&#8217; parallel to BC such that C\u2019 lies on the extension of BC.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step V &#8211; <\/strong>Draw C\u2019A\u2019 parallel to CA to intersect the extension of BA at A\u2019. Now, triangle A&#8217;BC&#8217; is the required triangle similar to \u0394ABC where,\u00a0BA&#8217;\/BA = BC&#8217;\/BC = C&#8217;A&#8217;\/CA = 4\/3<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Proof:<br \/>\n<\/strong>In \u0394BB\u2084C&#8217;, B\u2083C || B\u2084C&#8217; <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence by Basic proportionality theorem, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">B\u2083B\u2084\/BB\u2083 = CC&#8217;\/BC = 1\/3 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">CC&#8217;\/BC + 1 = 1\/3 + 1 (Adding 1) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(BC + CC&#8217;)\/BC = 4\/3 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">BC&#8217;\/BC = 4\/3 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Consider \u0394BA&#8217;C&#8217; and \u0394BAC <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220A&#8217;BC&#8217; = \u2220ABC = 45\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220BC&#8217;A&#8217; = \u2220BCA = 30\u00b0 (Corresponding angles as CA|| C&#8217;A&#8217;) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220BA&#8217;C&#8217; = \u2220BAC = 105\u00b0 (Corresponding angles) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">By AAA axiom, \u0394BA&#8217;C&#8217; ~ \u0394BAC <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence corresponding sides are proportional <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">BA&#8217;\/BA = BC&#8217;\/BC = C&#8217;A&#8217;\/CA = 4\/3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5\/3 times the corresponding sides of the given triangle.<br \/>\nSolution &#8211;\u00a0<\/strong>Steps of Construction:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6600\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans7.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"410\" height=\"438\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans7.png 410w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans7-281x300.png 281w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex11.1-Ans7-300x320.png 300w\" sizes=\"auto, (max-width: 410px) 100vw, 410px\" \/><br \/>\n<strong>Step I &#8211; <\/strong>Draw BC = 4. At B, make an angle \u2220CBY\u00a0= 90\u00b0 and mark A on BY\u00a0such that BA = 3 cm. Join A to C. Thus \u0394ABC is constructed.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step II &#8211; <\/strong>Draw the ray BX so that \u2220CBX is acute.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step III &#8211; <\/strong>Mark 5 (since, 5 &gt; 3 in 5\/3) points B\u2081, B\u2082, B\u2083, B\u2084, B\u2085\u00a0on BX so that BB\u2081\u00a0= B\u2081B\u2082\u00a0= B\u2082B\u2083\u00a0= B\u2083B\u2084\u00a0= B\u2084B\u2085<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step IV &#8211; <\/strong>Join B\u2083 (3<sup>rd<\/sup>\u00a0point on BX as 3 &lt; 5) to C and draw B\u2085C&#8217; parallel to B\u2083C so that C&#8217; lies on the extension of BC.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Step V &#8211; <\/strong>Draw C&#8217;A&#8217; parallel to CA to intersect of the extension of BA at A\u2019. Now \u0394BA&#8217;C&#8217; is the required triangle similar to \u0394BAC where\u00a0BA&#8217;\/BA = BC&#8217;\/BC = C&#8217;A&#8217;\/CA = 5\/3<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Proof:<br \/>\n<\/strong>In \u0394BB\u2085C&#8217;, B\u2083C || B\u2083C&#8217; <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence by Basic proportionality theorem, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">B\u2083B\u2085\/BB\u2083 = CC&#8217;\/BC = 2\/3 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">CC&#8217;\/BC + 1 = 2\/3 + 1 (Adding 1) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(CC&#8217; + BC)\/BC = 5\/3 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">BC&#8217;\/BC = 5\/3 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Consider \u0394BAC and \u0394BA&#8217;C&#8217; <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220ABC = \u2220A&#8217;BC&#8217; = 90\u00b0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220BCA = \u2220BC&#8217;A&#8217; (Corresponding angles as CA || C&#8217;A&#8217;) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220BAC = \u2220BA&#8217;C&#8217; <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">By AAA axiom, \u0394BAC ~ \u0394BA&#8217;C&#8217; <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, corresponding sides are proportional, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">BA&#8217;\/BA = BC&#8217;\/BC = C&#8217;A&#8217;\/CA = 5\/3<\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000ff;\"><b><i>Go Back To Chapters<\/i><\/b><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 11 (Constructions)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 11 Constructions Exercise 11.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 11 Constructions NCERT Class 10 Maths Solution Ex &#8211; 11.2 Exercise &#8211; 11.1 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1102,1103,1044,1049,1048],"class_list":["post-6083","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-11-constructions-solutions","tag-ncert-class-10-mathematics-exercise-11-1-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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