{"id":31885,"date":"2026-09-21T15:18:22","date_gmt":"2026-09-21T09:48:22","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31885"},"modified":"2026-09-21T16:07:58","modified_gmt":"2026-09-21T10:37:58","slug":"constructions-and-tilings-ncert-solutions-class-7-maths-ganita-prakash","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/constructions-and-tilings-ncert-solutions-class-7-maths-ganita-prakash\/","title":{"rendered":"Constructions and Tilings – NCERT Solutions Class 7 Maths (Ganita Prakash)"},"content":{"rendered":"\n

Constructions and Tilings<\/strong><\/strong> – NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 7 Maths (Ganita Prakash).<\/p>\n\n\n\n

NCERT Solutions Class 7<\/h2>\n\n\nEnglish Poorvi<\/a>\n\n\nHindi Malhar<\/a>\n\n\nMaths Ganita Prakash<\/a>\n\n\nScience Curiosity<\/a>\n\n\nSocial Exploring Society<\/a>\n\n\n\n

Constructions and Tilings<\/strong><\/strong> \u2013 NCERT Solutions<\/h2>\n\n\n\n
\n\n\n\n

Q.1: Given a line segment XY, how do we draw its perpendicular bisector using only an unmarked ruler and a compass?<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Steps:<\/strong><\/p>\n\n\n\n

    \n
  1. Taking some fixed radius, from X and then Y, construct two sufficiently long arcs above XY. Name the point where the arcs meet as A.
    <\/li>\n\n\n\n
  2. Using the same radius, from X and then Y, construct two sufficiently long arcs below XY. Name the point where the arcs meet as B.
    <\/li>\n\n\n\n
  3. AB is the required perpendicular bisector.<\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

    Q.2: Construct at least 4 different angles. Draw their bisectors.<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    Steps for Angle Bisection:<\/strong><\/p>\n\n\n\n

      \n
    1. Mark points A and B such that OA = OB.
      <\/li>\n\n\n\n
    2. Choosing any sufficiently long radius, cut arcs from A and B, keeping the radius same. Mark the point of intersection as C.
      <\/li>\n\n\n\n
    3. OC bisects {tex}\\angle {\\text{AOB}}{\/tex}.<\/li>\n<\/ol>\n\n\n\n

      So, a 45o<\/sup> angle can be constructed by first constructing a 90o<\/sup> angle and then bisecting it.<\/p>\n\n\n\n

      Similarly, you can draw next 3 angles and bisect them. <\/p>\n\n\n\n

      \n\n\n\n

      Q.3: How do we construct a 45o<\/sup> angle using only a ruler and a compass?<\/p>\n\n\n\n

      Solution:<\/p>\n\n\n\n

      Steps for Angle Bisection:<\/strong><\/p>\n\n\n\n

      \"\"\/<\/figure>\n\n\n\n
        \n
      1. Mark points A and B such that OA = OB.<\/li>\n\n\n\n
      2. Choosing any sufficiently long radius, cut arcs from A and B, keeping the radius same. Mark the point of intersection as C.<\/li>\n\n\n\n
      3. OC bisects {tex}\\angle {\\text{AOB}}{\/tex}.<\/li>\n<\/ol>\n\n\n\n

        So, a 45o<\/sup> angle can be constructed by first constructing a 90o<\/sup> angle and then bisecting it.<\/p>\n\n\n\n

        \n\n\n\n

        Q.4: How do we construct a 60o<\/sup> angle?<\/p>\n\n\n\n

        Solution:<\/p>\n\n\n\n

        We get a 60o<\/sup> angle if we construct an equilateral triangle! We can use the following steps for this.
        Step 1:<\/strong>

        Construct an arc with centre A and any radius.
        Step 2:<\/strong>

        With the same radius, cut another arc from B that meets the first arc. Let C be the point at which the arcs meet.
        We have {tex}\\angle {\\text {CAX}} = 60^o{\/tex}.<\/p>\n\n\n\n

        \n\n\n\n

        Q.5: How will you construct 30o<\/sup> and 15o<\/sup> angles?<\/p>\n\n\n\n

        Solution:<\/p>\n\n\n\n

        Constructing a 30o<\/sup> Angle
        Steps:<\/strong><\/p>\n\n\n\n

          \n
        1. Draw a ray.<\/li>\n\n\n\n
        2. At its endpoint, construct a {tex}{6 0}^{\\boldsymbol{\\circ}}{\/tex} angle using an equilateral triangle construction.<\/li>\n\n\n\n
        3. Bisect this {tex}60^{\\circ}{\/tex} angle using a compass (draw arcs from both rays and join their intersection to the vertex).<\/li>\n\n\n\n
        4. The bisected angle is {tex}{3 0}^{\\circ}{\/tex}.
          So {tex} 60^{\\circ} \\div 2=30^{\\circ} {\/tex}<\/strong><\/li>\n<\/ol>\n\n\n\n

          Constructing a {tex}15^{\\circ}{\/tex} Angle
          Steps:<\/strong><\/p>\n\n\n\n

            \n
          1. First construct a {tex}{3 0}^{\\boldsymbol{\\circ}}{\/tex} angle (as above).<\/li>\n\n\n\n
          2. Bisect this {tex}30^{\\circ}{\/tex} angle again by the same arc-intersection method.<\/li>\n\n\n\n
          3. The resulting angle is {tex}{1 5}^{\\boldsymbol{\\circ}}{\/tex}.
            So {tex} 30^{\\circ} \\div 2={1 5}^{\\circ} {\/tex}<\/li>\n<\/ol>\n\n\n\n

            NCERT Solutions Class 7 Maths (Ganita Prakash)<\/h2>\n\n\n\n
              \n
            1. Large Numbers Around Us<\/a><\/li>\n\n\n\n
            2. Arithmetic Expressions<\/a><\/li>\n\n\n\n
            3. A Peek Beyond the Point<\/a><\/li>\n\n\n\n
            4. Expressions Using Letter-Numbers<\/a><\/li>\n\n\n\n
            5. Parallel And Intersection Lines<\/a><\/li>\n\n\n\n
            6. Number Play<\/a><\/li>\n\n\n\n
            7. A Tale of Three Intersecting Lines<\/a><\/li>\n\n\n\n
            8. Working With Fractions<\/a><\/li>\n\n\n\n
            9. Geometric Twins<\/a><\/li>\n\n\n\n
            10. Operations with Integers<\/a><\/li>\n\n\n\n
            11. Finding Common Ground<\/a><\/li>\n\n\n\n
            12. Another Peek Beyond the Point<\/a><\/li>\n\n\n\n
            13. Connecting the Dots<\/a><\/li>\n\n\n\n
            14. Constructions and Tilings<\/a><\/li>\n\n\n\n
            15. Finding the Unknown<\/a><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

              Constructions and Tilings – NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 7 Maths (Ganita Prakash). NCERT Solutions Class 7 Constructions and Tilings \u2013 NCERT Solutions Q.1: Given a line segment XY, how do we draw its perpendicular bisector using only an unmarked ruler and … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[281,2082,2106],"tags":[216],"class_list":["post-31885","post","type-post","status-publish","format-standard","hentry","category-ncert-solutions","category-ncert-solutions-class-7","category-ncert-solutions-class-7-maths-ganita-prakash","tag-ncert-solutions"],"yoast_head":"\nConstructions and Tilings - NCERT Solutions Class 7 Maths (Ganita Prakash) | myCBSEguide<\/title>\n<meta name=\"description\" content=\"Constructions and Tilings - NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/constructions-and-tilings-ncert-solutions-class-7-maths-ganita-prakash\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Constructions and Tilings - 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