Some Constructions of Triangles

By using the properties of the triangle and the above basic constructions we can construct triangles.

Construction 1: How to construct a triangle if its base, a base angle and sum of other two sides are given?

Given the base BC, a base angle  \angleB and the sum of other two sides AB + AC of a triangle ABC, now we need to construct it.

Steps of Construction:

Step 1: First of all, draw the base BC and at the point B make an  \angleXBC equal to the given angle.

Step 2: From the ray BX, cut the line segment BD = AB + AC.

Step 3: Join DC now which make  \angleDCY =  \angleBDC.

Step 4: When CY intersect BX at A then it form the required triangle i.e. ABC.

Construction 2: How to construct a triangle if its base, a base angle and the difference of the other two sides are given?

Given the base BC, a base angle i.e.  \angleB and the difference of the other two sides AB – AC or AC – AB, we need to construct the triangle ABC. There could be two cases:

Case (i): If AB > AC that is AB – AC is given.

Steps of Construction:

Step 1: Draw the base BC which is equal to a cm and at point B make an \angle XBC = x°.

Step 2: From ray, BX cut the line segment BD which is equal to AB – AC.

Step 3: Join DC and draw the perpendicular bisector of DC.

Step 4: This perpendicular bisector intersects BX at a point A. By joining A to C we get the required triangle i.e. ∆ABC.

Case (ii): If AB < AC that is AC – AB is given.

Steps of Construction:

Step 1: Draw the base BC and at point B make an \angle XBC.

Step 2: From ray, BX cut the line segment BD which is equal to AC – AB from the line BX by extending it on opposite side of line segment BC.

Step 3: Join DC and draw the perpendicular bisector of DC.

Step 4: Let PQ intersect BX at A and by joining A to C, we get the required triangle ∆ABC.

Construction 3: How to construct a triangle if its perimeter and two base angles are given?

Given the base angles, say \angle B and \angle C and BC + CA + AB, you have to construct the triangle ABC.

Steps of Construction:

Step 1: Draw a line segment XY = BC + CA + AB.

Step 2: Make \angle LXY = \angle B and \angle MYX = \angle C.

Step 3: Now bisect \angle LXY and \angle MYX. These bisectors will intersect at a point A.

Step 4: Draw perpendicular bisectors PQ of AX and RS of AY.

Step 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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