Constructing Triangles

A triangle is a three-sided polygon. It has three sides, three vertices and three angles. We know that a unique triangle can be constructed if
(i) all three sides are given
(ii) two sides and included angle are given
(iii) two angles and the included side is given
(iv) the measure of the hypotenuse and a side is given in the right triangle.

For constructing triangles from given data, we generally make use of the given congruency conditions and construct the required triangle. Let us discuss the technique of constructing different types of triangles when specific conditions are given:

Construction of Triangles – Case 1

Given the base of a triangle, its base angle and sum of other two sides

For constructing ∆ABC such that base BC, base angle ∠B and the sum of other two sides, i.e. AB + AC are given, the following steps of construction is followed:

Steps of Construction of a Triangle

1. Draw the base BC of ∆ABC as given and construct ∠XBC of the required measure at B as shown.

Constructing Triangles - Step 1

2. Keeping the compass at point B cut an arc from the ray BX such that its length equals to AB + AC at point P and join it to C as shown

Constructing Triangles - Step 23. Now measure ∠BPC and from C, draw an angle equal to ∠BPC as shown

Constructing Triangles - Step 3

∆ABC is the required triangle. This can be proved as follows:

Sl.No Statement Reason
Base BC and ∠B are drawn as given. Now in ∆ACP,
1 ∠ACP = ∠APC By Construction
2 AC = AP ∆ACP is isosceles
3 AB = BP – AP = BP – AC From Statement 2
4 AB + AC = BP Proved

Construction of Triangles – Case 2

Given the base of a triangle, its base angle and difference of the other two sides

For constructing ∆ABC such that base BC, base angle∠B and difference of the other two sides, i.e. AB – AC or  AC-AB is given, then for constructing triangles such as these two cases can arise:

  1. AB > AC
  2. AC > AB

The following steps of construction are followed for the two cases:

Steps of Construction if AB > AC:

1. Draw the base BC of ∆ABC as given and construct ∠XBC of the required measure at B as shown.

Constructing Triangles case 2 - step 12. From the ray, BX cut an arc equal to AB – AC at point P and join it to C as shown

3. Draw the perpendicular bisector of PC and let it intersect BX at point A as shown:Constructing Triangles case 2 - step 34. Join AC, ∆ABC is the required triangle.

Constructing Triangles case 2 - step 4

Steps of Construction if AC > AB:

1. Draw the base BC of ∆ABC as given and construct ∠XBC of the required measure at B as shown.

 

Constructing Triangles case 3 -12. From the ray BX cut an arc equal to AB – AC at point P and join it to C. In this case P will lie on the opposite side to the ray BX. Draw the perpendicular bisector of PC and let it intersect BX at point A as shown

Constructing Triangles case 3 - 23. Join the points A and C, and hence ∆ABC is the required triangle.

Leave a Comment

Your email address will not be published. Required fields are marked *

BOOK

Free Class