{"id":4573,"date":"2023-02-16T05:33:08","date_gmt":"2023-02-16T05:33:08","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4573"},"modified":"2023-02-16T05:34:03","modified_gmt":"2023-02-16T05:34:03","slug":"ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/","title":{"rendered":"NCERT Solutions Class 9 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 9 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 9 Mathematics\u00a0<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><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 11.2<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 11.1<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Construct an angle of 90\u00b0 at the initial point of a given ray and justify the construction.<br \/>\nAnswer &#8211;<\/strong>\u00a0Steps of Construction :<br \/>\n1. Draw a ray PQ.<br \/>\n2. Take P as a centre with any radius, draw an arc TSR is that cuts PQ at R.<br \/>\n3. With R as a centre with the same radius, mark a point S on the arc TSR.<br \/>\n4. With S as a centre and the same radius, mark a point T on the arc TSR.<br \/>\n5. Take S and T as centre, draw two arcs which intersect each other with the same radius at U.<br \/>\n6. Finally, the ray PU is joined which makes an angle 90\u00b0 with PU is formed.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4612\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"374\" height=\"281\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1.png 374w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1-300x225.png 300w\" sizes=\"auto, (max-width: 374px) 100vw, 374px\" \/><br \/>\n<strong>Justification<br \/>\n<\/strong>To prove \u2220UPQ = 90\u00b0<br \/>\n\u2220SPQ = 60\u00b0<br \/>\n\u2220TPS = 60\u00b0<br \/>\n\u2220UPS = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220TPS = 30\u00b0<br \/>\n\u2220UPQ = \u2220UPS + \u2220SPR<br \/>\n\u2220UPQ = 30\u00b0 + 60\u00b0 = 90\u00b0<br \/>\nHence, justified.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Construct an angle of 45\u00b0 at the initial point of a given ray and justify the construction.<br \/>\nAnswer &#8211;\u00a0<\/strong>Steps of Construction :<br \/>\n1. Draw a ray PQ.<br \/>\n2. Take U as a centre with any radius, draw an arc TSR that cuts PQ at R.<br \/>\n3. With R as a centre with the same radius, mark a point S on the arc TSR.<br \/>\n4. With R as a centre and the same radius, mark a point T on the arc TSR.<br \/>\n5. Take S and T as centre, draw two arcs which intersect each other with the same radius at U.<br \/>\n6. Finally, the ray PU is joined which makes an angle 90\u00b0 with PU is formed.<br \/>\n7. Take R and V as centre draw the perpendicular bisector which intersects at the point W.<br \/>\n8. Draw a line that joins the point P and W.<br \/>\n9. So, the angle formed \u2220WPQ = 45\u00b0<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4613\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"400\" height=\"296\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans2.png 400w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans2-300x222.png 300w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><br \/>\n<strong>Justification<br \/>\n<\/strong>\u2220UPQ = 90\u00b0<br \/>\n\u2220WPQ = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220POA<\/span><\/p>\n<p>\u2220WPQ = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 90\u00b0 = 45\u00b0<br \/>\nHence, justified.<\/p>\n<p><span style=\"color: #000000;\"><strong>3. Construct the angles of the following measurements:<\/strong><\/span><br \/>\n<span style=\"color: #000000;\"><strong>(i) 30\u00b0\u00a0<\/strong> <\/span><br \/>\n<span style=\"color: #000000;\"><strong>(ii) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{22\\frac{1}{2}^{\\circ}}\" alt=\"\\mathbf{22\\frac{1}{2}^{\\circ}}\" align=\"absmiddle\" \/>\u00a0<\/strong> <\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iii) 15\u00b0 <\/strong><\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong><strong>(i) 30\u00b0\u00a0<\/strong> \u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Steps of Construction :<\/span><br \/>\n<span style=\"color: #000000;\">1. Draw a ray PQ.<br \/>\n2. Take P as a centre with any radius, draw an arc RS which cuts PQ at R.<br \/>\n3. With R and S as centres, draw two arcs which intersect each other at the point T and the perpendicular bisector is drawn.<br \/>\n4. Thus, \u2220TPQ is the required angle making 30\u00b0 with PQ.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4614\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans3i.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"327\" height=\"239\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans3i.png 327w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans3i-300x219.png 300w\" sizes=\"auto, (max-width: 327px) 100vw, 327px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong><img decoding=\"async\" class=\"\" title=\"Ncert solutions class 9 chapter 11-6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/07\/ncert-solutions-class-9-chapter-11-6.png\" alt=\"Ncert solutions class 9 chapter 11-6\" \/><br \/>\n<\/strong>Steps of Construction :<\/span><br \/>\n<span style=\"color: #000000;\">1. Draw a ray PQ.<br \/>\n2. To construct an angle of 60\u00b0, with P as a center and any\u00a0radius, draw a wide arc to intersect PQ at R.\u00a0 With R as a center and same radius draw an arc to intersect the initial arc at S. Then, \u2220SPR = 60\u00b0<br \/>\n3. To construct an\u00a0adjacent angle of 60\u00b0, with S as the center and the same radius as before, draw an arc to intersect the initial arc at T. Then, \u2220TPS = 60\u00b0<br \/>\n4. To bisect \u2220TPS, with T and S as centers and radius greater than half of TS , draw arcs to intersect each other at Z. Join P and Z. Then, \u2220ZPQ = 90\u00b0e)\u00a0To bisect \u2220ZPQ, with R and U as centers and radius more than half of RU, draw arcs to intersect each other at V. Join P and V. Then, \u2220VPQ = 45\u00b0<br \/>\n5. To bisect \u2220VPQ = 45\u00b0, with W and R as centers and radius greater than half of WR, draw arcs to intersect each other at X. Join P and X. PX bisects \u2220VPQ.<br \/>\nHence, \u2220XPQ = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220WPQ = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 45\u00b0 =\u00a0<span class=\"equation\"><span id=\"MathJax-Element-4-Frame\" class=\"mjx-chtml MathJax_CHTML\" style=\"box-sizing: inherit; font-family: 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; 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;22&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-31\" class=\"mjx-math\" aria-hidden=\"true\"><span id=\"MJXc-Node-32\" class=\"mjx-mrow\"><span id=\"MJXc-Node-33\" class=\"mjx-mn\"><span class=\"mjx-char MJXc-TeX-main-R\">22<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> <\/span><\/span><\/span><\/span><\/span><\/span>\u00b0<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4615\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans3ii.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"394\" height=\"291\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans3ii.png 394w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans3ii-300x222.png 300w\" sizes=\"auto, (max-width: 394px) 100vw, 394px\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 15\u00b0<br \/>\n<\/strong>Steps of Construction :<br \/>\n1. Draw a ray PQ.<br \/>\n2. To construct an angle of 60\u00b0, with P as a center and any radius, draw a wide arc to intersect PQ at R. With R as a center and same radius draw an arc to intersect the initial arc at S. Then, \u2220SPR = 60\u00b0<br \/>\n3. Now, we will bisect \u2220SPR. With R and S as centers and radius greater than half of RS, draw arcs to intersect each other at T. Join P and T. So,\u00a0PT is the\u00a0angle bisector\u00a0of \u2220SPR.<br \/>\n\u2220TPQ = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220SPR = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 60\u00b0\u00a0= 30\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">4. To bisect \u2220TPQ, With R and W as centers and radius greater than half of arc RW, draw arcs to intersect each other at U. Join P and U. PU is the angle bisector of \u2220TPQ .<\/span><br \/>\n<span style=\"color: #000000;\">\u2220UPQ = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220TPQ = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 30\u00b0 = 15\u00b0<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4616\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans3iii.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"313\" height=\"248\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans3iii.png 313w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans3iii-300x238.png 300w\" sizes=\"auto, (max-width: 313px) 100vw, 313px\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Construct the following angles and verify by measuring them by a protractor:<br \/>\n<\/strong><strong>(i) 75\u00b0<br \/>\n(ii) 105\u00b0<br \/>\n(iii) 135\u00b0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong><strong>(i) 75\u00b0<br \/>\n<\/strong>Steps of Construction :<br \/>\n1. Draw a ray PQ.<br \/>\n2. To construct an angle of 60\u00b0, with P as a center and any radius, draw a wide arc to intersect PQ at R. With R as a center and same radius, draw an arc to intersect the initial arc at S. Thus,\u2220SPR = 60\u00b0<br \/>\n3. To construct an\u00a0 adjacent angle of 60\u00b0 with S as a center and the same radius, draw an arc that intersects the initial arc at T.<br \/>\n4. To bisect \u2220SPT, with T and S as centers and radius more than half of TS draw arcs to bisect each other at U. Join U and P.<br \/>\n\u2220UPS = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220TPS = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 60\u00b0 = 30\u00b0<br \/>\n5. To bisect \u2220UPS, with V and S as centers and\u00a0radius\u00a0greater than half of VS draw arcs to intersect each other at X.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220XPS = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00a0\u2220UPS = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 30\u00b0 = 15\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220XPQ = \u2220XPS + \u2220SPQ <\/span><br \/>\n<span style=\"color: #000000;\">= 15\u00b0 + 60\u00b0 = 75\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0\u2220XPQ =\u00a075\u00b0<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4617\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans4i.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"366\" height=\"293\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans4i.png 366w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans4i-300x240.png 300w\" sizes=\"auto, (max-width: 366px) 100vw, 366px\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 105\u00b0<br \/>\n<\/strong>Steps of Construction :<br \/>\n1. Draw a ray PQ.<br \/>\n2. To construct an angle of 60\u00b0 With P as a center and any radius, draw a wide arc to intersect PQ at R. With R as a center and same radius, draw an arc to intersect the initial arc at S. Thus, \u2220SPR = 60\u00b0.<br \/>\n3. To construct an adjacent angle with S as the center and the same radius as before, draw an arc to intersect the initial arc at T. \u2220TPS = 60\u00b0.<br \/>\n4. To bisect \u2220TPS, with T and S as centers and radius greater than TS draw arcs to bisect each other at U. Join U and P.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220UPS = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220TPS = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 60\u00b0 = 30\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220UPT =\u00a0\u2220UPS =\u00a030\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">5.\u00a0To bisect \u2220UPT, with T and V as centers and radius greater than half of TV, draw arcs to intersect each other at W. Join P and W.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220WPU = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00a0\u2220UPT = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 30\u00b0 = 15\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220WPR = \u2220WPU + \u2220UPS + \u2220SPR<\/span><br \/>\n<span style=\"color: #000000;\">= 15\u00b0 + 30\u00b0 + 60\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">= 105\u00b0<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4618\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans4ii.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"362\" height=\"291\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans4ii.png 362w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans4ii-300x241.png 300w\" sizes=\"auto, (max-width: 362px) 100vw, 362px\" \/><\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 135\u00b0<br \/>\n<\/strong>Steps of Construction :<br \/>\n1. Draw a ray PQ.<br \/>\n2. To construct an angle of 60\u00b0, with P as a center and any radius, draw a wide arc to intersect PQ at R. Thus, \u2220SPR = 60\u00b0.<br \/>\n3. To construct an adjacent angle of 60\u00b0, with S as the center and the same radius as before, draw an arc to intersect the initial arc at T. Thus, \u2220TPS = 60\u00b0.<br \/>\n4. To construct the second adjacent angle of 60\u00b0, with T as a center and the same radius as before, draw an arc to intersect the initial arc at U. Thus, \u2220UPT = 60\u00b0<br \/>\n5. To bisect \u2220UPT, with T and U as centers and radius greater than half of TU, draw two\u00a0arcs to intersect each other at V.<br \/>\n\u2220VPT = \u2220VPU = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220UPT = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 60\u00b0 = 30\u00b0<br \/>\n6.\u00a0To bisect \u2220VPT, with W and T as centers and radius greater than half of WT, draw arcs to intersect each other at X.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220XPT = \u2220XPV = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00a0\u2220VPT = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00a0\u00d7 30\u00b0 = 15\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220XPQ = \u2220XPT + \u2220TPS + \u2220SPR<\/span><br \/>\n<span style=\"color: #000000;\">= 15\u00b0 + 60\u00b0 + 60\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">= 135\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0\u2220XPQ =\u00a0135\u00b0<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4619\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans4iii.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"511\" height=\"266\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans4iii.png 511w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans4iii-300x156.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans4iii-480x250.png 480w\" sizes=\"auto, (max-width: 511px) 100vw, 511px\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Construct an equilateral triangle, given its side and justify the construction.<br \/>\nAnswer &#8211;\u00a0<\/strong>Steps of Construction:<br \/>\n1. Draw a ray AB.<\/span><br \/>\n<span style=\"color: #000000;\">2. With A as center and\u00a0radius equal to 3 cm, draw an arc to cut ray AB at C such that AC = 3 cm.<\/span><br \/>\n<span style=\"color: #000000;\">3. With C as the center and radius equal to AC (3cm), draw an arc to intersect the initial arc at D.<\/span><br \/>\n<span style=\"color: #000000;\">4. Join AD and DC.\u00a0 <\/span><br \/>\n<span style=\"color: #000000;\">5. Triangle ADC is an equilateral triangle with sides of 3cm each.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4620\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"395\" height=\"242\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans5.png 395w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans5-300x184.png 300w\" sizes=\"auto, (max-width: 395px) 100vw, 395px\" \/><\/span><br \/>\n<span style=\"color: #000000;\"><strong>Justification:<br \/>\n<\/strong>AC = AD (By construction since the radius of the arc is the same)<\/span><br \/>\n<span style=\"color: #000000;\">AC = CD (By construction since the same radius was used again)<\/span><br \/>\n<span style=\"color: #000000;\">Hence AC = AD = CD = 3cm<\/span><br \/>\n<span style=\"color: #000000;\">Thus, ADC is an equilateral triangle.<\/span><\/p>\n<p style=\"text-align: justify;\">\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 11 (Constructions)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#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 Solution Class 9 Maths Ex &#8211; 11.2 Exercise &#8211; 11.1 1. Construct [&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":[721,702,186,722,701,5],"class_list":["post-4573","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-11-constructions-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-11-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 11 Constructions Ex 11.1 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 9 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter - 11 Constructions Exercise 11.1 has been provided here to help the students in solving the questions from this exercise.\u00a0\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions Class 9 Maths Chapter 11 Constructions Ex 11.1\" \/>\n<meta property=\"og:description\" content=\"NCERT Solutions Class 9 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter - 11 Constructions Exercise 11.1 has been provided here to help the students in solving the questions from this exercise.\u00a0\" \/>\n<meta property=\"og:url\" content=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/\" \/>\n<meta property=\"og:site_name\" content=\"TheExamPillar NCERT\" \/>\n<meta property=\"article:published_time\" content=\"2023-02-16T05:33:08+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-02-16T05:34:03+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1.png\" \/>\n<meta name=\"author\" content=\"Admin\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@exampillar\" \/>\n<meta name=\"twitter:site\" content=\"@exampillar\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Admin\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"13 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/#article\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/\"},\"author\":{\"name\":\"Admin\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"headline\":\"NCERT Solutions Class 9 Maths Chapter 11 Constructions Ex 11.1\",\"datePublished\":\"2023-02-16T05:33:08+00:00\",\"dateModified\":\"2023-02-16T05:34:03+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/\"},\"wordCount\":1254,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"image\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2023\\\/01\\\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1.png\",\"keywords\":[\"Class 9th Chapter 11 Constructions in English\",\"Ncert Solution Class 9\",\"NCERT Solutions\",\"NCERT Solutions Class 9 Maths Chapter 11 in english\",\"NCERT Solutions Class 9 Maths in English\",\"NCERT Solutions in English\"],\"articleSection\":[\"Class 9 Maths\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/\",\"name\":\"NCERT Solutions Class 9 Maths Chapter 11 Constructions Ex 11.1 | TheExamPillar NCERT\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/#primaryimage\"},\"image\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2023\\\/01\\\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1.png\",\"datePublished\":\"2023-02-16T05:33:08+00:00\",\"dateModified\":\"2023-02-16T05:34:03+00:00\",\"description\":\"NCERT Solutions Class 9 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter - 11 Constructions Exercise 11.1 has been provided here to help the students in solving the questions from this exercise.\u00a0\",\"breadcrumb\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/#primaryimage\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2023\\\/01\\\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1.png\",\"contentUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2023\\\/01\\\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1.png\",\"width\":374,\"height\":281,\"caption\":\"NCERT Class 9 Solutions Maths\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\\\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"NCERT Solutions Class 9 Maths Chapter 11 Constructions Ex 11.1\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#website\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/\",\"name\":\"TheExamPillar NCERT\",\"description\":\"\",\"publisher\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"alternateName\":\"NCERT Solution\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":[\"Person\",\"Organization\"],\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\",\"name\":\"Admin\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"contentUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"width\":512,\"height\":512,\"caption\":\"Admin\"},\"logo\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\"},\"sameAs\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\"],\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/author\\\/ncert_eng_vikram\\\/\"}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"NCERT Solutions Class 9 Maths Chapter 11 Constructions Ex 11.1 | TheExamPillar NCERT","description":"NCERT Solutions Class 9 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter - 11 Constructions Exercise 11.1 has been provided here to help the students in solving the questions from this exercise.\u00a0","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/","og_locale":"en_US","og_type":"article","og_title":"NCERT Solutions Class 9 Maths Chapter 11 Constructions Ex 11.1","og_description":"NCERT Solutions Class 9 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter - 11 Constructions Exercise 11.1 has been provided here to help the students in solving the questions from this exercise.\u00a0","og_url":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/","og_site_name":"TheExamPillar NCERT","article_published_time":"2023-02-16T05:33:08+00:00","article_modified_time":"2023-02-16T05:34:03+00:00","og_image":[{"url":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1.png","type":"","width":"","height":""}],"author":"Admin","twitter_card":"summary_large_image","twitter_creator":"@exampillar","twitter_site":"@exampillar","twitter_misc":{"Written by":"Admin","Est. reading time":"13 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/#article","isPartOf":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/"},"author":{"name":"Admin","@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"headline":"NCERT Solutions Class 9 Maths Chapter 11 Constructions Ex 11.1","datePublished":"2023-02-16T05:33:08+00:00","dateModified":"2023-02-16T05:34:03+00:00","mainEntityOfPage":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/"},"wordCount":1254,"commentCount":0,"publisher":{"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"image":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/#primaryimage"},"thumbnailUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1.png","keywords":["Class 9th Chapter 11 Constructions in English","Ncert Solution Class 9","NCERT Solutions","NCERT Solutions Class 9 Maths Chapter 11 in english","NCERT Solutions Class 9 Maths in English","NCERT Solutions in English"],"articleSection":["Class 9 Maths"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/","url":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/","name":"NCERT Solutions Class 9 Maths Chapter 11 Constructions Ex 11.1 | TheExamPillar NCERT","isPartOf":{"@id":"https:\/\/theexampillar.com\/ncert\/#website"},"primaryImageOfPage":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/#primaryimage"},"image":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/#primaryimage"},"thumbnailUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1.png","datePublished":"2023-02-16T05:33:08+00:00","dateModified":"2023-02-16T05:34:03+00:00","description":"NCERT Solutions Class 9 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter - 11 Constructions Exercise 11.1 has been provided here to help the students in solving the questions from this exercise.\u00a0","breadcrumb":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/#primaryimage","url":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1.png","contentUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.1-Ans1.png","width":374,"height":281,"caption":"NCERT Class 9 Solutions Maths"},{"@type":"BreadcrumbList","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/theexampillar.com\/ncert\/"},{"@type":"ListItem","position":2,"name":"NCERT Solutions Class 9 Maths Chapter 11 Constructions Ex 11.1"}]},{"@type":"WebSite","@id":"https:\/\/theexampillar.com\/ncert\/#website","url":"https:\/\/theexampillar.com\/ncert\/","name":"TheExamPillar NCERT","description":"","publisher":{"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"alternateName":"NCERT Solution","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/theexampillar.com\/ncert\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":["Person","Organization"],"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1","name":"Admin","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","url":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","contentUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","width":512,"height":512,"caption":"Admin"},"logo":{"@id":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png"},"sameAs":["https:\/\/theexampillar.com\/ncert"],"url":"https:\/\/theexampillar.com\/ncert\/author\/ncert_eng_vikram\/"}]}},"_links":{"self":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/4573","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/comments?post=4573"}],"version-history":[{"count":4,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/4573\/revisions"}],"predecessor-version":[{"id":5087,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/4573\/revisions\/5087"}],"wp:attachment":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/media?parent=4573"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/categories?post=4573"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/tags?post=4573"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}