{"id":9278,"date":"2024-02-20T10:32:01","date_gmt":"2024-02-20T05:02:01","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=9278"},"modified":"2024-02-21T17:13:56","modified_gmt":"2024-02-21T11:43:56","slug":"ts-6th-class-maths-solutions-chapter-13-ex-13-5","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-6th-class-maths-solutions-chapter-13-ex-13-5\/","title":{"rendered":"TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5"},"content":{"rendered":"

Students can practice Telangana SCERT Class 6 Maths Solutions<\/a> Chapter 13 Practical Geometry Ex 13.5 to get the best methods of solving problems.<\/p>\n

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.5<\/h2>\n

Question 1.
\nConstruct \u2220ABC = 60\u00b0 without using protractor.
\nAnswer:
\n\"TS<\/p>\n

Steps of construction:<\/p>\n

    \n
  1. Draw a line l and choose a point B on it.<\/li>\n
  2. Place the pointer of the compasses at B and draw an arc of convenient radius which cuts the line \/ at a point say C.<\/li>\n
  3. Take C as centre and with the same radius (as in step 2) in draw an arc.<\/li>\n
  4. Now take B as centre and with the same radius (as in step 2), draw another arc cutting the previous arc (drawn in step 2) at A.<\/li>\n
  5. Join AB, we get \u2220ABC whose measure is 60\u00b0.<\/li>\n<\/ol>\n

    \"TS<\/p>\n

    Question 2.
    \nConstruct an angle of 120\u00b0 with using protractor and compasses.
    \nAnswer:
    \nWith using compasses:
    \n\"TS<\/p>\n

    Steps of construction:<\/p>\n

      \n
    1. \u00a0Draw any ray OA.<\/li>\n
    2. Place the pointer of the compasses at O with O as centre and any convenient radius draw an arc cutting OA at M.<\/li>\n
    3. With M as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at P.<\/li>\n
    4. With P as centre and without altering the radius (as in step 2) draw an arc which cuts the first arc at Q.<\/li>\n
    5. Join OQ. Then \u2220AOQ is the required angle.<\/li>\n<\/ol>\n

      With Using protractor:
      \n\"TS<\/p>\n

      Steps of construction :<\/p>\n

        \n
      1. Draw a ray QR of any length.<\/li>\n
      2. Place the centre point of the protractor at
        \nQ and the line aligned with the \\(\\overline{\\mathrm{QR}}\\).<\/li>\n
      3. Mark at a point P at 120\u00b0.<\/li>\n
      4. Join QP. \u2220PQR is the required angle.<\/li>\n<\/ol>\n

        Question 3.
        \nConstruct the following angles using ruler and compasses. Write the steps of construction in each case.
        \n(i) 75\u00b0
        \nAnswer:
        \n\"TS<\/p>\n

        Steps of construction:<\/p>\n

          \n
        1. Draw any ray OA.<\/li>\n
        2. Place the pointer of the compasses at O. With ‘O’ as centre and any convenient radius draw an arc cutting OA at M.<\/li>\n
        3. With M as centre and without altering radius (as in step 2) draw an arc which, cuts the first arc at P.<\/li>\n
        4. With P as centre and without altering the radius (as in step 2) draw an arc which cuts the first arc at Q.<\/li>\n
        5. Join OQ. \u2220AOQ = 120\u00b0 (\u2235\u2220AOP + \u2220POQ = 60\u00b0)
          \n\u2220AOQ = \u2220AOP + \u2220POQ = 60\u00b0 + 60\u00b0 = 120\u00b0<\/li>\n
        6. Draw OB the perpendicular bisector of \u2220POQ.
          \nNow \u2220POB = \u2220QOB = 30\u00b0
          \n\u2234\u2220AOB = 90\u00b0<\/li>\n
        7. Draw OC the perpendicular bisector of \u2220POB.
          \nNow \u2220POC = \u2220BOC = 15\u00b0<\/li>\n
        8. \u2220AOC = \u2220AOP + \u2220OC
          \n= 60\u00b0 + 15\u00b0
          \n= 75\u00b0
          \n(or)
          \n\u2220AOC = \u2220AOB – \u2220BOC
          \n= 90\u00b0 – 15\u00b0
          \n= 75\u00b0
          \n\u2220AOC is the required angle whose measure is 75\u00b0.<\/li>\n<\/ol>\n

          \"TS<\/p>\n

          (ii) 15\u00b0
          \nAnswer:
          \n\"TS<\/p>\n

          Steps of construction:<\/p>\n

            \n
          1. Draw a line l and mark a point O on it.<\/li>\n
          2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line \/ at a point say A.<\/li>\n
          3. With the pointer at A (as centre) and the same radius as in the step – 2, now draw an arc that passes through O.<\/li>\n
          4. Let the two arcs intersect at B. Join OB. We get \u2220BOA whose measure is 60\u00b0.<\/li>\n
          5. Draw the bisector OC of \u2220AOB. Now \u2220AOC = 30\u00b0.
            \n[\u2220AOC = \u2220BOC = 30\u00b0 ]<\/li>\n
          6. Draw the bisector OD of \u2220AOC. Now \u2220AOD = 15\u00b0.
            \n[\u2220AOD = \u2220COD = 15\u00b0]<\/li>\n
          7. \u2220AOD = 15\u00b0 is the required angle.<\/li>\n<\/ol>\n

            (iii) 105\u00b0
            \nAnswer:
            \n\"TS<\/p>\n

            Steps of construction:<\/p>\n

              \n
            1. Draw any ray OM.<\/li>\n
            2. Place the pointer of the compasses at O. With O as centre and any convenient radius draw an arc cutting OM at A.<\/li>\n
            3. With A as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at B.<\/li>\n
            4. With B as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at B.<\/li>\n
            5. Join QC. \u2220AOC = 120\u00b0.<\/li>\n
            6. We know that \u2220AOB = \u2220BOC = \\(\\frac{1}{2}\\) \u2220AOC = \\(\\frac{1}{2}\\) \u00d7 120\u00b0 = 60\u00b0<\/li>\n
            7. Draw OD the bisector of \u2220BOC.
              \n\u2234 \u2220BOD = \u2220DOC = \\(\\frac{1}{2}\\) \u2220BOC = \\(\\frac{1}{2}\\) \u00d7 60\u00b0 = 30\u00b0<\/li>\n
            8. Draw OE the bisector of \u2220DOC.
              \n\u2234 \u2220DOE = \u2220EOC = \\(\\frac{1}{2}\\) \u2220DOC = \\(\\frac{1}{2}\\) \u00d7 30\u00b0 = 15\u00b0<\/li>\n
            9. Now \u2220AOE = \u2220AOD + \u2220DOE = 90\u00b0 + 15\u00b0 = 105\u00b0<\/li>\n
            10. \u2220AOE = 105\u00b0 is the required angle.<\/li>\n<\/ol>\n

              \"TS<\/p>\n

              Question 4.
              \nDraw the angles given in Q. 3 using a protractor.
              \nAnswer:
              \nThe angles given in Q. 3 are 75\u00b0, 105\u00b0 and 15\u00b0.<\/p>\n

              (i) 75\u00b0
              \n\"TS<\/p>\n

              Steps of construction:<\/p>\n

                \n
              1. Draw a ray QR of any length.<\/li>\n
              2. Place the centre point of the protractor at Q and the line aligned with the \\(\\overline{\\mathrm{QR}}\\).<\/li>\n
              3. Mark a point P at 75\u00b0.<\/li>\n
              4. Join QP. \u2220RQP is the required angle.<\/li>\n<\/ol>\n

                (ii) 105\u00b0
                \n\"TS<\/p>\n

                Steps of construction:<\/p>\n

                  \n
                1. Draw a ray QR of any length.<\/li>\n
                2. Place the centre point of the protractor at Q and the line aligned with the \\(\\overline{\\mathrm{QR}}\\).<\/li>\n
                3. Mark a point P at 105\u00b0<\/li>\n
                4. Join QP. \u2220RQP is the required angle.<\/li>\n<\/ol>\n

                  (iii) 15\u00b0
                  \n\"TS<\/p>\n

                  Steps of construction:<\/p>\n

                    \n
                  1. Draw a ray QR of any length.<\/li>\n
                  2. Place the centre point of the protractor at Q and the line aligned with the \\(\\overline{\\mathrm{QR}}\\).<\/li>\n
                  3. Mark a point P at 15\u00b0<\/li>\n
                  4. Join QP. \u2220RQP is the required angle.<\/li>\n<\/ol>\n

                    Question 5.
                    \nConstruct \u2220ABC = 50\u00b0 and then draw another angle \u2220XYZ equal to \u2220ABC without using a protactor.
                    \nAnswer:
                    \n\"TS<\/p>\n

                    Steps of construction:<\/p>\n

                      \n
                    1. Make an angle \u2220ABC = 50\u00b0 v<\/li>\n
                    2. Draw a line l and choose a point Y on it.<\/li>\n
                    3. Use the same compasses setting to draw an arc with Y as centre, cutting l in Z.<\/li>\n
                    4. Set your compasses to the length MN.<\/li>\n
                    5. With the same radius place the compasses pointer at Z and draw an arc to cut the arc drawn earlier at X.<\/li>\n
                    6. Join XY. This gives us \u2220XYZ. It has the same measure as \u2220ABC i.e,, 50\u00b0.<\/li>\n<\/ol>\n

                      \"TS<\/p>\n

                      Question 6.
                      \nConstruct \u2220DEF = 60\u00b0. Bisect it, measure each half by using a protractor.
                      \nAnswer:
                      \n\"TS<\/p>\n

                      Steps of construction :<\/p>\n

                        \n
                      1. Draw a line P and mark a point ‘E’ on it.<\/li>\n
                      2. Place the pointer of the compasses at E and draw an arc of convenient radius which cuts the line \\(\\overleftrightarrow{\\mathrm{PQ}}\\) at a point say F.<\/li>\n
                      3. With the pointer at F (as centre), now draw an arc that passes through E.<\/li>\n
                      4. Let the two arcs intersect at D. Join ED. We get \u2220DEF whose measure is 60\u00b0.<\/li>\n
                      5. Draw OC the bisector of \u2220DEF.<\/li>\n
                      6. Now \u2220FEC = \u2220CED = 30\u00b0.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                        Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.5 to get the best methods of solving problems. TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.5 Question 1. Construct \u2220ABC = 60\u00b0 without using protractor. Answer: Steps of construction: Draw a line l and choose a point … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9278"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=9278"}],"version-history":[{"count":4,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9278\/revisions"}],"predecessor-version":[{"id":9296,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9278\/revisions\/9296"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=9278"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=9278"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=9278"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}