Class-IX Unit-11 Constructions

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Introduction

Sometimes we need to create precise & accurate figures. Construction helps in creating these precise & accurate figures.  Construction is used in designing Automobile, spaceships, building, roads etc.

We will need Protractor, Compass & Scale in this chapter.

Construction 1: To construct the bisector of a given angle ABC

Steps of Construction:

1. Taking B as centre and any radius, draw an arc to intersect the rays BA and BC, say at E and D respectively
2. Next, taking D and E as centers and with the radius more than 1/2 DE, draw arcs to intersect each other, say at F.
3. Draw the ray BF . This ray BF is the required bisector of the angle ABC.
   
4. Construction 2: To construct the perpendicular bisector of a given line segment.
   Steps of Construction:
   1. Taking A and B as centers and radius more than ½ AB, draw arcs on both sides of the line segment AB (to intersect each other). 2. Let these arcs intersect each other at P and Q.
   2. Join PQ. Let PQ intersect AB at the point M.
      
   3. Then line PMQ is the required perpendicular bisector of AB.
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   5. Construction 3: To construct an angle of 600 at the initial point of a given ray.
      Steps of Construction:
      1. Taking A as centre and some radius, draw an arc of a circle, which intersects AB, say at a point D.
      2. Taking D as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point E.
      3. Draw the ray AC passing through E. Then ∠ CAB is the required angle of 60°.
         
      4. Some Constructions of Triangles
         Construction 4: To construct a triangle, given its base, a base angle and sum of other two sides.
         Steps of Construction:
         1. Draw the base BC and at the point B make an angle, say XBC equal to the given angle.
         2. Cut a line segment BD equal to AB + AC from the ray BX.
         3. Join DC and make an angle DCY equal to ∠BDC.
         4. Let CY intersect BX at A. Then, ABC is the required triangle
            
         5. Construction 5: To construct a triangle given its base, a base angle and the difference of the other two sides
            Given the base BC, a base angle, say ∠B and the difference of other two sides AB – AC or AC – AB, you have to construct the triangle ABC. Clearly there are following two cases:
            Case (i) : Let AB > AC that is AB – AC is given.
            Steps of Construction:
            1. Draw the base BC and at point B make an angle say XBC equal to the given angle.
            2. Cut the line segment BD equal to AB – AC from ray BX.
            3. Join DC and draw the perpendicular bisector, say PQ of DC.
            4. Let it intersect BX at a point A. Join AC. Then ABC is the required triangle
               
            5. Case (ii) : Let AB < AC that is AC – AB is given.
               Steps of Construction:
               1. Draw the base BC and at point B make an angle say XBC equal to the given angle.
               2. Cut the line segment BD equal to AC – AB from line BX extended on opposite side of line segment BC.
               3. Join DC and draw the perpendicular bisector, say PQ of DC.
               4. Let PQ intersect BX at A. Join AC. Then, ABC is the required triangle.
                  
               5. Construction 6: To construct a triangle, given its perimeter and its two base angles.
                   Steps of Construction:
                  1. Draw a line segment, say XY equal to BC + CA + AB.
                  2. Make angles LXY equal to ∠B and MYX equal to ∠C.
                  3. Bisect ∠LXY and ∠ MYX. Let these bisectors intersect at a point A .
                  4. Draw perpendicular bisectors PQ of AX and RS of AY.
                  5. Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC . Then ABC is the required triangle.
                     
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