{"id":140164,"date":"2022-04-13T14:30:33","date_gmt":"2022-04-13T09:00:33","guid":{"rendered":"https:\/\/www.aakash.ac.in\/blog\/?p=140164"},"modified":"2023-04-02T23:35:13","modified_gmt":"2023-04-02T18:05:13","slug":"constructions-class-10-maths-ncert-solutions-for-cbse-board-exam","status":"publish","type":"post","link":"https:\/\/www.aakash.ac.in\/blog\/constructions-class-10-maths-ncert-solutions-for-cbse-board-exam\/","title":{"rendered":"Constructions Class 10 Maths: NCERT Solutions For CBSE Board Exam"},"content":{"rendered":"

The artistry of Mathematics is that it is perhaps the only subject in which students can obtain a complete 100 score. However, for most students, it is a hassle for various reasons. While some people find arithmetic to be boring, others find it to be fascinating. Mathematics is not a discipline that can be muddled up into a tale. But there is one chapter in particular that every student enjoys: “Construction” <\/span>NCERT Solutions For Class 10<\/span><\/a><\/p>\n\n\n\n
Did You Know? <\/b>What is the significance of learning constructions?\u00a0<\/span>Assume one wants to become an architect in future, then they regularly need to build a perfect layout for the building. For this, they must have the proper knowledge of the map. Construction entails precisely drawing lines and angles. Learning the fundamentals of Construction is also essential for drawing road layouts. Let’s look at the fundamental geometric constructions.<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Chapter 11 Construction NCERT Solutions Exercise 11.1<\/h3>\n

Answer 1:\u00a0<\/b><\/p>\n

    \n
  • Draw a line segment AB = 7.6 cm.<\/span><\/li>\n
  • Draw an acute angle BAX on the base AB. Mark the ray as AX.<\/span><\/li>\n
  • Locate 13 points A1, A2, A3,\u2026, A13 on the ray AX so that AA1 = A1A2 = \u2026 = A12A13.<\/span><\/li>\n
  • Join A13 with B, and at A5, draw a line II to BA13, i.e. A5C. The line intersects AB at C.<\/span><\/li>\n
  • Measure AC = 2.9 cm and BC = 4.7 cm.<\/span><\/li>\n<\/ul>\n

    Answer 2:\u00a0<\/b><\/p>\n

      \n
    • Draw AC = 6 cm, with A and C as centres and radii 4 cm.<\/span><\/li>\n
    • Draw two arcs of 5 cm, each intersecting at B. Join BA and BC.<\/span><\/li>\n
    • Draw a ray AY making an acute angle with AC.<\/span><\/li>\n
    • Locate three points and R on AY, such that AP – POQR.<\/span><\/li>\n
    • Join CR.<\/span><\/li>\n
    • Through O draw a line QC’ \u0965 RC (by making an angle equal to angle ARC) meeting the line segment AC at C’.<\/span><\/li>\n
    • Similarly, through C, draw a line B’C \u0965 CB.\u00a0<\/span><\/li>\n
    • Thus, ABC is the required triangle, similar to ABC with a scale factor of 2\/3.<\/span><\/li>\n<\/ul>\n

      Answer 3:\u00a0<\/b><\/p>\n

        \n
      • Draw a triangle ABC with AB = 5 cm, BC = 7 cm and AC = 6 cm.<\/span><\/li>\n
      • Draw an acute angle CBX below BC at point B.<\/span><\/li>\n
      • Mark the ray BX as B1, B2, By, B4, B5, B6 and B7 such that BB1= B1B2 = B2B3 = B3B4 = B4B5 = B5B6, = B6B7.<\/span><\/li>\n
      • Join B5 to C.<\/span><\/li>\n
      • Draw B7C’ \u0965 B5C, where C’ is a point on extended line BC.<\/span><\/li>\n
      • Draw AC’ \u0965 AC, where A’ is a point on extended line BA.<\/span><\/li>\n
      • A’B’C’ is the required triangle <\/span>NCERT Solutions For Class 10 – All Subject – Download Free PDFs | AESL<\/span>.\u00a0<\/span><\/li>\n<\/ul>\n

        Answer 4:\u00a0<\/b><\/p>\n

          \n
        • Construct an isosceles triangle ABC in which BC is 8 cm, and altitude AD is 4 cm.<\/span><\/li>\n
        • Draw a ray BX, making an acute angle with BC.<\/span><\/li>\n
        • Locate 3 points on BX, such that BP – PQ = QR.<\/span><\/li>\n
        • Join QC.<\/span><\/li>\n
        • Through R, draw a line RC parallel to QC, meeting produced line BC at C’.<\/span><\/li>\n
        • Through C, draw a line CA \u0965 CA, meeting the produced line BA at A’.<\/span><\/li>\n
        • Thus, triangle A’BC’ is the required isosceles triangle.\u00a0<\/span><\/li>\n<\/ul>\n

          Answer 5:\u00a0<\/b><\/p>\n

            \n
          • Draw a line segment BC = 6 cm, and at point B, draw an angle ABC = 60\u00b0.<\/span><\/li>\n
          • Cut AB = 5 cm. Join AC. We obtain a triangle ABC.<\/span><\/li>\n
          • Draw a ray BX making an acute angle with BC on the side opposite the vertex A.<\/span><\/li>\n
          • Locate 4 points A1, A2, A3 and A4 on the ray BX so that BA, = A1A2 = A2A3 = A3A4.<\/span><\/li>\n
          • Join A4 to C.<\/span><\/li>\n
          • \u00a0At A3, draw A3C’ \u0965 A4C, where C’ is a point on the line segment BC.<\/span><\/li>\n
          • At C’, draw C’A’ \u0965 CA, where A’ is a point on the line segment BA.<\/span><\/li>\n
          • Therefore triangle A’BC’ is the required triangle <\/span>NCERT Solutions For Class 10 – All Subject – Download Free PDFs | AESL<\/span>.<\/span><\/li>\n<\/ul>\n

            Answer 6:\u00a0<\/b><\/p>\n

              \n
            • Draw a line segment BC = 7 cm.<\/span><\/li>\n
            • Draw an angle ABC = 45\u00b0 and an angle ACB = 30\u00b0, i.e., angle BAC = 105\u00b0.<\/span><\/li>\n
            • We obtain triangle ABC.<\/span><\/li>\n
            • Draw a ray BX making an acute angle with BC.<\/span><\/li>\n
            • Mark four points B1, B2, B3 and B4 on BX, such that BB1 = B1B2 = B2B3 = B3B4.<\/span><\/li>\n
            • Join B3C.<\/span><\/li>\n
            • Through B4, draw a line B4C’ \u0965 B3C, intersecting the extended line segment BC at C’.<\/span><\/li>\n
            • Through C’, draw a line A’C’ \u0965 CA, intersecting the extended line segment BA at A.<\/span><\/li>\n
            • Thus, A’BC’ is the required triangle.<\/span><\/li>\n<\/ul>\n

              Answer 7:\u00a0<\/b><\/p>\n

                \n
              • Construct a triangle ABC, such that BC = 4 cm, CA = 3 cm and angle BCA = 90\u00b0.<\/span><\/li>\n
              • Draw a ray BX making an acute angle with BC.<\/span><\/li>\n
              • Mark five points B1, B2, B3, B4 and B5 on BX, such that BB1 = B1B2 = B2B3 = B3B4 = B4B5.<\/span><\/li>\n
              • Join B3C.<\/span><\/li>\n
              • Through B5, draw B5C’ \u0965 B3C intersecting BC produced at C’.<\/span><\/li>\n
              • Through C’, draw C’A’ \u0965 CA intersecting AB produced at A’.<\/span><\/li>\n
              • Thus, triangle A’BC’ is the required right triangle <\/span>NCERT Solutions For Class 10 – All Subject – Download Free PDFs | AESL<\/span>.<\/span><\/li>\n<\/ul>\n

                Chapter 11 Construction NCERT Solutions Exercise 11.2<\/h3>\n

                Answer 1:\u00a0<\/b><\/p>\n

                  \n
                • Draw a circle with a radius of 6 cm.<\/span><\/li>\n
                • Take a point P such that OP = 10 cm.<\/span><\/li>\n
                • Draw the perpendicular bisector of OP. Let M is the midpoint of OP.<\/span><\/li>\n
                • With centre M and radius PM = MO, draw a circle that cuts the given circle at S and T.<\/span><\/li>\n
                • Join PS and PT.<\/span><\/li>\n
                • Thus, PS and PT are the required tangents<\/span><\/li>\n
                • The length of tangents PS = PT = 8 cm <\/span>NCERT Solutions For Class 10 – All Subject – Download Free PDFs | AESL<\/span>.<\/span><\/li>\n<\/ul>\n

                  Answer 2:<\/b>\u00a0<\/span><\/p>\n

                    \n
                  • Draw concentric circles of radius DA = 4 cm and OP = 6 cm having the same centre O.<\/span><\/li>\n
                  • Mark these circles as C and C’.<\/span><\/li>\n
                  • Points O, A and P lie on the same line.<\/span><\/li>\n
                  • Draw a perpendicular bisector of OP, which intersects OP at O’.<\/span><\/li>\n
                  • Take O’ as the centre and draw a circle of radius OO’ which intersects circle C at points T and Q.<\/span><\/li>\n
                  • Join PT and PQ. These are the required tangents.<\/span><\/li>\n
                  • The length of these tangents is approx. 4.5 cm.<\/span><\/li>\n<\/ul>\n

                    Answer 3:<\/b><\/p>\n

                      \n
                    • Draw a circle with a centre and radius of 3 cm.<\/span><\/li>\n
                    • Produce the circle’s diameter to both ends up to P and such that OP = OQ = 7 cm.<\/span><\/li>\n
                    • Mark the mid-points M and M’ of OP and OQ, respectively.<\/span><\/li>\n
                    • With centres M and M’ and radii MP and MO, draw two circles.<\/span><\/li>\n
                    • The circle with centre M intersects the given circle at R and S. The circle with centre M intersects the given circle at T and U.<\/span><\/li>\n
                    • Join PR, PS, QT and QU.<\/span><\/li>\n
                    • Thus, we have PR and PS as tangents from P and OT and QU as another pair of tangents from Q drawn to the given circle <\/span>NCERT Solutions For Class 10 – All Subject – Download Free PDFs | AESL<\/span>.<\/span><\/li>\n<\/ul>\n

                      Answer 4:\u00a0<\/b><\/p>\n

                        \n
                      • Draw a circle of radius 5 cm.<\/span><\/li>\n
                      • As tangents are inclined to each other at an angle of 60\u00b0 therefore, the angle between the circle’s radius is 120\u00b0. (Use quadrilateral property)<\/span><\/li>\n
                      • Draw radii OA and OB inclined to each other at an angle of 120\u00b0.<\/span><\/li>\n
                      • At points A and B, draw 90\u00b0 angles. The arms of these angles intersect at point P.<\/span><\/li>\n
                      • PA and PB are the required tangents.<\/span><\/li>\n<\/ul>\n

                        Answer 5:\u00a0<\/b><\/p>\n

                          \n
                        • Draw a line segment AB = 8 cm.<\/span><\/li>\n
                        • Draw two circles with centres A and B and radii 4 cm and 3 cm.<\/span><\/li>\n
                        • Mark the midpoint M of AB.<\/span><\/li>\n
                        • With centre M and radius AM = BM, draw a circle intersecting the two circles at P, Q, R, and S.<\/span><\/li>\n
                        • Join AP, AQ, BR and BS.<\/span><\/li>\n<\/ul>\n

                          Answer 6:\u00a0<\/b><\/p>\n

                            \n
                          • Draw a right triangle ABC with AB = 6 cm, BC = 8 cm and angle B = 90\u00b0.<\/span><\/li>\n
                          • From B, draw BD perpendicular to AC.<\/span><\/li>\n
                          • Draw the perpendicular bisector of BC, which intersects BC at point O’.<\/span><\/li>\n
                          • Take O’ as the centre and O’B as the radius. Draw a circle C’ that passes through points B, C and D.<\/span><\/li>\n
                          • Join O’A and draw a perpendicular bisector of O’A, which intersects O’A at point K.<\/span><\/li>\n
                          • Take K as the centre, draw an arc of radius KO’ to intersect the previous circle C’ at T.<\/span><\/li>\n
                          • Join AT, AT is a required tangent.<\/span><\/li>\n<\/ul>\n

                            Answer 7:\u00a0<\/b><\/p>\n

                              \n
                            • Draw a circle.<\/span><\/li>\n
                            • Take two non-parallel chords AB and CD of the circle.<\/span><\/li>\n
                            • Draw perpendicular bisectors of these chords intersect at O, which is the centre of the circle.<\/span><\/li>\n
                            • Take a point P outside the circle.<\/span><\/li>\n
                            • Join OP.<\/span><\/li>\n
                            • Mark the midpoint M of OP.<\/span><\/li>\n
                            • With M as centre and radius equal to MP = OM, draw a circle intersecting the first circle at and R.<\/span><\/li>\n
                            • Join PQ and PR.<\/span><\/li>\n<\/ul>\n

                              Conclusion<\/h3>\n

                              Students can use the mathematical principles of chapter 11 Construction in various other units such as measurements, geometry, and trigonometry. A clear understanding of these concepts will ease students’ higher class preparation. Studying effectively for CBSE class 10 Maths Construction will help them score higher grades in the geometry section. Using these notes will enable students to clear their CBSE class 10 Maths board exams <\/span>NCERT Solutions For Class 10 – All Subject – Download Free PDFs | AESL<\/span> with flying colours!\u00a0\u00a0<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

                              The artistry of Mathematics is that it is perhaps the only subject in which students can obtain a complete 100 score. However, for most students, it is a hassle for various reasons. While some people find arithmetic to be boring, others find it to be fascinating. Mathematics is not a discipline that can be muddled […]<\/p>\n","protected":false},"author":1,"featured_media":230871,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3581],"tags":[29,1568,2881,2606],"class_list":["post-140164","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-cbse","tag-board-exams","tag-cbse-class-10","tag-cbse-class-10-term-2-exams","tag-cbse-term-2-exams"],"yoast_head":"\nClass 10 Maths Constructions Chapter 11: NCERT Solutions For CBSE Board Exam<\/title>\n<meta name=\"description\" content=\"Constructions Class 10 Maths Chapter 11: If you are appearing for CBSE Class 10 Maths exam then Constructions is one of the important topics.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" 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