Constructions

STEPS:

1. With B as centre and a suitable radius, draw an arc which cuts ray BA at point D and ray BC at point E.
2. Taking D and E as centres and with equal radii draw arcs which intersect each other at point F. In this step, each equal radius must be more than half the length DE.
3. Join B and F and produce to get the ray BF.

Result: Ray BF is the required bisector of the given angle ∠ABC.

Proof:

Join DF and EF.

In ΔBDF and ΔBEF:

• BD = BE [Radii of the same arc]
• DF = EF [Radii of the equal arcs]
• BF = BF [Common]
• ∴ ΔBDF ≅ ΔBEF [By SSS]
• ∴ ∠DBF = ∠EBF [By cpctc]

i.e., ∠ABF = ∠CBF

∴ BF bisects ∠ABC. Hence Proved.

STEPS:

1. Draw a line BC of any suitable length.
2. With B as centre and any suitable radius, draw an arc which cuts BC at point D.
3. With D as centre and radius same, as taken in step (ii), draw one more arc which cuts previous arc at point E.
4. Join BE and produce upto any point A.

Then, ∠ABC = 60°

STEPS:

1. Draw a line BC of any suitable length.
2. Taking B as centre and with any suitable radius, draw an arc which cuts BC at point D.
3. Taking D as centre, draw an arc of the same radius, as taken in step (ii), which cuts the first arc at point E.

1. Taking E as centre and radius same, as taken in step (ii), draw one more arc which cuts the first arc at point F.
2. Join BF and produce upto any suitable point A.

Then, ∠ABC = 120°

STEPS:

1. Construct angle ∠ABC = 60° by compass.
2. Draw BD, the bisector of angle ∠ABC.

Then, ∠DBC = 30°

STEPS:

1. Construct angle ∠ABC = 120° by using compass.
2. Draw PB, the bisector of angle ∠EBG.

Then, ∠PBC = 90°

Alternative Method:

1. Draw a line segment BC of any suitable length.
2. Produce CB upto an arbitrary point O.
3. Taking B as centre, draw an arc which cuts OC at points D and E.
4. Taking D and E as centres and with equal radii draw arcs which cut each other at point P. (The radii in this step must be of length more than half of DE.)
5. Join BP and produce.

Then, ∠PBC = 90°

STEPS:

1. Draw ∠PBC = 90°
2. Draw AB which bisects angle ∠PBC.

Then, ∠ABC = 45°

Alternative Method:

1. Construct ∠ABC = 60°
2. Draw BD, the bisector of angle ∠ABC.
3. Draw BE, the bisector of angle ∠ABD.

Then, ∠EBC = 45°

STEPS:

1. Construct ∠ABC = 120° and ∠PBC = 90°
2. Draw BO, the bisector of ∠ABP.

Then, ∠OBC = 105°

STEPS:

1. Draw line segment BC of any suitable length. Produce CB upto any point O.
2. With B as centre, draw an arc (with any suitable radius) which cuts OC at points D and E.
3. With D as centre, draw an arc of the same radius, as taken in step 2, which cuts the first arc at point F.
4. With F as centre, draw one more arc of the same radius, taken in step 2, which cuts the first arc at point G.
5. Draw PB, the bisector of angle ∠EBG.

Now ∠FBD = ∠GBF = ∠EBG = 60°

Then, ∠PBC = 150°

STEPS:

1. Construct ∠PBC = 150° and ∠GBC = 120°
2. Construct BQ, the bisector of angle ∠PBG.

Then, ∠QBC = 135°

Example: Draw an equilateral triangle having each side of 2.5 cm.

STEPS:

1. Draw the line segment BC = 2.5 cm.
2. Through B, construct ray BP making angle 60° with BC. i.e. ∠PBC = 60°
3. Through C, construct CQ making angle 60° with BC i.e., ∠QCB = 60°
4. Let BP and CQ intersect each other at point A.

Then, ΔABC is the required equilateral triangle.

Proof:

Since, ∠ABC = ∠ACB = 60°

∴ ∠BAC = 180° – (60° + 60°) = 60°

∴ All the angles of the ΔABC drawn are equal.

∴ All the sides of the ΔABC drawn are equal.

∴ ΔABC is the required equilateral triangle. Hence Proved.

Alternate method:

If one side is 2.5 cm, then each side of the required equilateral triangle is 2.5 cm.

1. Draw BC = 2.5 cm
2. With B as centre, draw an arc of radius 2.5 cm
3. With C as centre, draw an arc of radius 2.5 cm
4. Let the two arcs intersect each other at point A. Join AB and AC.

Then, ΔABC is the required equilateral triangle.

Example: Construct a triangle with 3 cm base and sum of other two sides is 8 cm and one base angle is 60°.

STEPS:

1. Draw BC = 3 cm
2. At point B, draw PB so that ∠PBC = 60°
3. From BP, cut BD = 8 cm.
4. Join D and C.
5. Draw perpendicular bisector of CD, which meets BD at point A.
6. Join A and C.

Thus, ΔABC is the required triangle.

Proof:

Since, OA is perpendicular bisector of CD

• ∴ OC = OD
• ∠AOC = ∠AOD = 90°
• Also, OA = OA [Common]
• ∴ ΔAOC ≅ ΔAOD [By SAS]
• ∴ AC = AD

∴ BD = BA + AD

= BA + AC

= Given sum of the other two sides

Thus, base BC and ∠B are drawn as given and BD = AC. Hence Proved.

Example: Construct a right triangle, when one side is 3.8 cm and the sum of the other side and hypotenuse is 6 cm.

STEPS:

1. Draw AB = 3.8 cm
2. Through B, draw line BP so that ∠ABP = 90°
3. From BP, cut BD = 6 cm
4. Join A and D.
5. Draw perpendicular bisector of AD, which meets BD at point C.

Thus, ΔABC is the required triangle.

Example: Construct a triangle with base of 8 cm and difference between the length of other two sides is 3 cm and one base angle is 60°.

STEPS:

1. Draw BC = 8 cm
2. Through point B, draw BP so that ∠PBC = 60°
3. From BP cut BD = 3 cm.
4. Join D and C.
5. Draw perpendicular bisector of DC; which meets BP at point A.
6. Join A and C.

Thus, ΔABC is the required triangle.

Proof:

Since OA is perpendicular bisector of CD

• ∴ OD = OC
• ∠AOD = ∠AOC = 90°
• And, OA = OA [Common]
• ∴ ΔAOD ≅ ΔAOC [By SAS]
• ∴ AD = AC [By cpctc]

Now, BD = BA – AD

= BA – AD

= BA – AC

= Given difference of other two sides.

Thus, the base BC and ∠B are drawn as given and BD = BA – AC. Hence Proved.

STEPS:

1. Draw BC = 8 cm
2. Through B, draw line BP so that angle ∠PBC = 60°.
3. Produce BP backward upto a suitable point Q.
4. From BQ, cut BD = 3 cm.
5. Join D and C.
6. Draw perpendicular bisector of DC, which meets BP at point A.
7. Join A and C.

Thus, ΔABC is the required triangle.

Proof:

Since, OA is perpendicular bisector of CD

• ∴ OD = OC
• ∠AOD = ∠AOC = 90°
• And OA = OA [Common]
• ∴ ΔAOD ≅ ΔAOC [By SAS]
• ∴ AD = AC [By cpctc]

Now, BD = AD – AB

= AC – AB [∵ AD = AC]

= Given difference of other two sides.

Thus, the base BC and ∠B are drawn as given and BD = AC – AB. Hence Proved.

Example: Construct a triangle ABC with AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°

STEPS:

1. Draw a line segment PQ = 12 cm
2. At P, construct line PR so that ∠RPO = 45° and at Q, construct a line QS so that ∠SQP = 60°.
3. Draw bisector of angles ∠RPQ and ∠SQP which meet each other at point A.
4. Draw perpendicular bisector of AP, which meets PQ at point B.
5. Draw perpendicular bisector of AQ, which meets PQ at point C.
6. Join AB and AC.

Thus, ΔABC is the required triangle.

Proof:

Since, MB is perpendicular bisector of AP

• ∴ ΔQNC ≅ ΔANC [By SAS]
• PB = AC

Similarly, NC is perpendicular bisector of AQ.

• ∴ ΔQNC ≅ ΔANC [By SAS]
• ∴ CQ = AC [By cpctc]

Now, PQ = PB + BC + CQ

= AB + BC + AC

= Given perimeter of the ΔABC drawn.

Also, ∠BPA = ∠BAP [As ΔPMB ≅ ΔAMB]

∴ ∠ABC = ∠BPA + ∠BAP

[Ext. angle of a triangle = sum of two interior opposite angles]

∠ABC = ∠BPA + ∠BAP = 2∠BPA = ∠RPB = ∠ACB [Given]

∠ACB = ∠CQA + ∠CQA

= 2∠CQA [∵ ΔQNC ≅ ΔANC ∴ ∠CQA = ∠CAQ]

= ∠SQC = Given base angle ∠ACB.

Thus, given perimeter = perimeter of ΔABC.

given one base angle = angle ABC

and, given other base angle = angle ACB.

Example: Construct an equilateral triangle if its altitude is 3.2 cm.

STEPS:

1. Draw a line PQ and mark a point D on it.
2. Construct a ray DE perpendicular to PQ.
3. Cut off DA = 3.2 cm from DE.
4. Construct ∠DAR = 30° = (1/2 × 60°). The ray AR intersects PQ at B.
5. Cut off line segment DC = BD.
6. Join A and C. We get the required ΔABC.

Example: Construct a right angled triangle whose hypotenuse measures 8 cm and one side is 6 cm.

STEPS:

1. Draw a line segment AC = 8 cm.
2. Mark the midpoint O of AC.
3. With O as centre and radius OA, draw a semicircle on AC.
4. With A as centre and radius equal to 6 cm, draw an arc, cutting the semicircle at B.
5. Join A and B; B and C. We get the required right angled triangle ABC.

Example: Construct a ΔABC in which BC = 6.4 cm, altitude from A is 3.2 cm and the median bisecting BC is 4 cm.

STEPS:

1. Draw BC = 6.4 cm
2. Bisect BC at L.
3. Draw EF ∥ BC at a distance 3.2 cm from BC.

1. With L as centre and radius equal to 4 cm, draw an arc, cutting EF at A.
2. Join A and B; A and C. We get the required triangle ABC.

Example: Construct a ΔABC in which ∠B = 30° and ∠C = 60° and the perpendicular from the vertex A to the base BC is 4.8 cm.

STEPS:

1. Draw any ray line PQ.

1. Take a point B on line PQ and construct ∠QBR = 30°
2. Draw a line EF ∥ PQ at a distance of 4.8 cm from PQ, cutting BR at A.
3. Construct an angle ∠FAC = 60°, cutting PQ at C.
4. Join A and C. We get the required triangle ABC.

Example: Construct a triangle ABC, the lengths of whose medians are 6 cm, 7 cm and 6 cm.

STEPS:

1. Construct a ΔAPQ with AP = 6 cm, PQ = 7 cm and AQ = 6 cm.
2. Draw the medians AE and PF of ΔAPQ intersecting each other at G.
3. Produce AE to B such that GE = EB
4. Join B and Q and produce it to C, such that BQ = QC
5. Join A and C. We get the required triangle ABC.

• Points lying on the perpendicular bisector of the line segment joining the two points are equidistant from them.
• A triangle is unique if:
  • Two sides and the included angle is given.
  • Three sides are given.
  • Two angles and the included side is given.
  • In a right triangle, hypotenuse and one side is given.
• Perimeter of a triangle is equal to sum of its three sides.
• Altitude to a given line segment intersects the line at 90°.
• Median of a line segment bisects the given line segment.
• Perpendicular bisector of a line segment bisects the line segment at right angles.