X

Available for CBSE, ICSE and State Board syllabus.

Call our LearnNext Expert on 1800 419 1234 (tollfree)

OR submit details below for a call back

Summary

Videos

References

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 Î”PQR, QR = 'a' cm, âˆ PQR = xÂ°, and PQ + PR = 'b' cm.

Step 1: Draw the base QR = 'a' cm.

Step 2: Draw âˆ XQR = xÂ°.

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, Î”PQR 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 Î”ABC, given BC = 'a' cm, âˆ B = xÂ° 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 âˆ XBC = xÂ°.

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 âˆ XBC = xÂ° 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 Î”ABC, given the perimeter (AB + BC + CA) = 'a' cm, âˆ B = xÂ° and âˆ C = yÂ°.

Steps of construction:

Step 1: Draw the line segment XY = 'a' cm.

Step 2: Draw the ray XL at X making an angle of xÂ° with XY.

Step 3: Draw the ray YM at Y making an angle of yÂ° with XY.

Step 4: Draw angle bisector of âˆ LXY.

Step 5: Draw angle bisector of âˆ MYX such that it intersects the angle bisector of âˆ 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.

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 Î”PQR, QR = 'a' cm, âˆ PQR = xÂ°, and PQ + PR = 'b' cm.

Step 1: Draw the base QR = 'a' cm.

Step 2: Draw âˆ XQR = xÂ°.

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, Î”PQR 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 Î”ABC, given BC = 'a' cm, âˆ B = xÂ° 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 âˆ XBC = xÂ°.

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 âˆ XBC = xÂ° 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 Î”ABC, given the perimeter (AB + BC + CA) = 'a' cm, âˆ B = xÂ° and âˆ C = yÂ°.

Steps of construction:

Step 1: Draw the line segment XY = 'a' cm.

Step 2: Draw the ray XL at X making an angle of xÂ° with XY.

Step 3: Draw the ray YM at Y making an angle of yÂ° with XY.

Step 4: Draw angle bisector of âˆ LXY.

Step 5: Draw angle bisector of âˆ MYX such that it intersects the angle bisector of âˆ 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.