Notes On Constructions of Triangles - CBSE Class 9 Maths
Measurements of at least three parts of a triangle are required for the construction of a triangle. But all the combinations of three parts are not sufficient for the purpose. For example, it is not possible to construct a unique triangle when the measurements of two sides and an angle which is not included in between the given sides are given. A triangle can be constructed when (i) the base, one base angle and the sum of the other two sides are given (ii) the base, a base angle and the difference between the other two sides are given (iii) perimeter and two base angles are given. Construction of a triangle when the base, one base angle and the sum of the other two sides of the triangle are given.                               Construction of ΔPQR, QR = 'a' cm, ∠PQR = x°, and PQ + PR = 'b' cm. Step 1: Draw the base QR = 'a' cm. Step 2: Draw ∠XQR = x°. Step 3: Mark an arc S on QX such that QS = 'b' cm. Step 4: Join RS. Step 5: Draw the perpendicular bisector of RS such that it intersects QS at P. Step 6: Join PR. Thus, ΔPQR is the required triangle. Construction of a triangle when the base, a base angle and the difference between the other two sides of the triangle are given.                                                   In ΔABC, given BC = 'a' cm, ∠B = x° and difference of two sides AB and AC is equal to 'b' cm. Case I: AB > AC Step 1: Draw the base BC = 'a' cm. Step 2: Make ∠XBC = x°. Step 3: Mark a point D on ray BX such that BD = 'b' cm. Step 4: Join DC. Step 5: Draw the perpendicular bisector of DC such that, it intersects the ray BX at a point A. Step 6: Join AC. Thus, ABC is the required triangle. Case II: AB < AC Step 1: Draw the base BC = 'a' cm. Step 2: Make ∠XBC = x° and extend ray BX in the opposite direction. Step 3: Mark a point D on the extended ray BX such that BD = 'b' cm. Step 4: Join DC. Step 5: Draw the perpendicular bisector of DC such that, it intersects the ray BX at a point A. Step 6: Join AC. Thus, ABC is the required triangle. Construction of a triangle when the perimeter and two base angles of the triangle are given. Construction of ΔABC, given the perimeter (AB + BC + CA) = 'a' cm, ∠B = x° and ∠C = y°. Steps of construction: Step 1: Draw the line segment XY = 'a' cm. Step 2: Draw the ray XL at X making an angle of x° with XY. Step 3: Draw the ray YM at Y making an angle of y° with XY. Step 4: Draw angle bisector of ∠LXY. Step 5: Draw angle bisector of ∠MYX such that it intersects the angle bisector of ∠LXY at a point A. Step 6: Draw the perpendicular bisector of  AX such that it intersects XY at a point B. Step 7: Draw the perpendicular bisector of  AY such that it intersects XY at a point C. Step 8: Join AB and AC. Thus, ABC is the required triangle.

#### Summary

Measurements of at least three parts of a triangle are required for the construction of a triangle. But all the combinations of three parts are not sufficient for the purpose. For example, it is not possible to construct a unique triangle when the measurements of two sides and an angle which is not included in between the given sides are given. A triangle can be constructed when (i) the base, one base angle and the sum of the other two sides are given (ii) the base, a base angle and the difference between the other two sides are given (iii) perimeter and two base angles are given. Construction of a triangle when the base, one base angle and the sum of the other two sides of the triangle are given.                               Construction of ΔPQR, QR = 'a' cm, ∠PQR = x°, and PQ + PR = 'b' cm. Step 1: Draw the base QR = 'a' cm. Step 2: Draw ∠XQR = x°. Step 3: Mark an arc S on QX such that QS = 'b' cm. Step 4: Join RS. Step 5: Draw the perpendicular bisector of RS such that it intersects QS at P. Step 6: Join PR. Thus, ΔPQR is the required triangle. Construction of a triangle when the base, a base angle and the difference between the other two sides of the triangle are given.                                                   In ΔABC, given BC = 'a' cm, ∠B = x° and difference of two sides AB and AC is equal to 'b' cm. Case I: AB > AC Step 1: Draw the base BC = 'a' cm. Step 2: Make ∠XBC = x°. Step 3: Mark a point D on ray BX such that BD = 'b' cm. Step 4: Join DC. Step 5: Draw the perpendicular bisector of DC such that, it intersects the ray BX at a point A. Step 6: Join AC. Thus, ABC is the required triangle. Case II: AB < AC Step 1: Draw the base BC = 'a' cm. Step 2: Make ∠XBC = x° and extend ray BX in the opposite direction. Step 3: Mark a point D on the extended ray BX such that BD = 'b' cm. Step 4: Join DC. Step 5: Draw the perpendicular bisector of DC such that, it intersects the ray BX at a point A. Step 6: Join AC. Thus, ABC is the required triangle. Construction of a triangle when the perimeter and two base angles of the triangle are given. Construction of ΔABC, given the perimeter (AB + BC + CA) = 'a' cm, ∠B = x° and ∠C = y°. Steps of construction: Step 1: Draw the line segment XY = 'a' cm. Step 2: Draw the ray XL at X making an angle of x° with XY. Step 3: Draw the ray YM at Y making an angle of y° with XY. Step 4: Draw angle bisector of ∠LXY. Step 5: Draw angle bisector of ∠MYX such that it intersects the angle bisector of ∠LXY at a point A. Step 6: Draw the perpendicular bisector of  AX such that it intersects XY at a point B. Step 7: Draw the perpendicular bisector of  AY such that it intersects XY at a point C. Step 8: Join AB and AC. Thus, ABC is the required triangle.

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