Geometrical Constructions

Construction of angle of 90°

Steps of construction:

Step 1: Draw a ray OA.

Step 2: With its initial point O as center and any radius, draw an arc, cutting OA at C.

Step 3: With center c and same radius (as in step 2) draw an arc cutting arc at D.

Step 4: With D as center and the same radius, draw an arc cutting the arc cutting at E.

Step 5: With D and E as centers and any convenient radius (more than DE). Draw to two arcs intersecting at P.

Step 6: Join OP. Then ∠AOP = 90°

Justification: -

By construction, OC = CD = OD

Therefore, ΔOCD is an equilateral triangle. So, ∠COD = 60°

Again OD = DE = OE

Therefore, ΔODE is also an equilateral triangle. So, ∠DOE = 60°

Since, OP bisects ∠DOE, so ∠POD = 30°.

Now,

∠AOP = ∠COD + ∠DOP

= 60° + 30°

= 90°

Construction of angle of 45°

Steps of construction:

Step 1: Draw a ray OA.

Step 2: With its initial point O as center and any radius, draw an arc, cutting OA at B.

Step 3: With center B and same radius (as in step 2), cut the previous drawn arc at C.

Step 4: With C as center and the same radius, draw an arc cutting the arc drawn in step 2 cutting at D.

Step 5: With D and E as centers and any convenient radius (more than DE). Draw to two arcs intersecting at E.

Step 6: Join OE. Then ∠AOE = 90°

Step 7: Draw the bisector ‘OF’ of ∠AOE. Then, ∠AOF = 45°

Justification: -

By construction, ∠AOE = 90° and OF is the bisector of ∠AOE.

Therefore,

∠AOF = ∠AOE

=

= 45°

Steps of construction:

Step 1: Draw a ray OA.

Step 2: With its initial point O as centre and any radius, draw an arc, cutting OA at C.

Step 3: With centre C and same radius (as in step 2). Draw an arc, cutting the arc of step 2 in D.

Step 4: With C and D as centres, and any convenient radius (more than CD), draw two arcs intersecting at B. join OB. Then ∠AOB = 30°

Verification:

On measuring ∠AOB, with the protractor,

we find ∠ AOB = 30° .

Steps of construction:

Step 1: Draw an angle AOB = 90°

Step 2: Draw the bisector OC of ∠AOB, then ∠AOC = 45°

Step 3:Bisect ∠AOC, such that ∠AOD = ∠COD = 22.5°

Thus ∠AOD = 22.5°

Verification:

On measuring ∠AOD, with the protractor,

we find ∠ AOD = 22.5° .

Steps of construction:

Step 5: Draw a ray OA.

Step 6: With its initial point O as centre and any radius, draw an arc, cutting OA at C.

Step 7: With centre C and same radius (as in step 2). Draw an arc, cutting the arc of step 2 in D.

Step 8: With C and D as centres, and any convenient radius (more than CD), draw two arcs intersecting at B. join OB. Then ∠AOB = 30°

Step 9: Bisect ∠AOB intersecting at D.

Thus ∠ AOD is required angle.

Verification:

On measuring ∠AOB, with the protractor,

we find ∠ AOB = 15°.

Thus ∠AOD = 15°

Step of construction:

Step 1: Draw a ray OA.

Step 2: With its initial point O as center and any radius, draw an arc, cutting OA at C.

Step 3: With center c and same radius (as in step 2) draw an arc cutting arc at D.

Step 4: With D as center and the same radius, draw an arc cutting the arc cutting at E.

Step 5: With D and E as centers and any convenient radius (more than DE). Draw to two arcs intersecting at P.

Step 6: Join OP. Then ∠AOP = 90°

Step 7:Bisect ∠BOP so that ∠BOQ = ∠BOP

= (∠AOP - ∠AOB)

= (90° -60°) = × 30° = 15°

So, we obtain

∠AOQ = ∠AOB + ∠BOQ

= 60° + 15° = 75°

Verification:

On measuring ∠AOQ, with the protractor,

we find ∠ AOQ = 75° .

Steps of construction:

Step 1: Draw a ray OA.

Step 2: With its initial point O as center and any radius, draw an arc, cutting OA at C.

Step 3: With center c and same radius (as in step 2) draw an arc cutting arc at D.

Step 4: With D as center and the same radius, draw an arc cutting the arc cutting at E.

Step 5: With D and E as centers and any convenient radius (more than DE). Draw to two arcs intersecting at P.

Step 6: Draw ∠AOB = 120° and ∠POB = 90°

Step 7: Bisect angle POB,

Then ∠AOQ is the required angle of 105°.

Step of construction:

Step 1: Draw a ray OA.

Step 2: With its initial point O as center and any radius, draw an arc, cutting OA at C.

Step 3: With center c and same radius (as in step 2) draw an arc cutting arc at D.

Step 4: With D as center and the same radius, draw an arc cutting the arc cutting at E.

Step 5: With D and E as centers and any convenient radius (more than DE). Draw to two arcs intersecting at P.

Step 6: Join OP. Then ∠AOP = 90°

Step 7: Bisect ∠AOP towards 180°

Let us draw an equilateral triangle of side 4.5 cm.

Step of construction:

Step 1: Draw BC = 4.5 cm

Step 2: With B and C as centres and radii equal to BC = 4.5 cm, draw two arcs on the same side of BC, intersecting each other at A.

Step 3: Join AB and AC.

ΔABC is the required equilateral triangle.

Justification:

Since by construction:

AB = BC = CA = 4.5 cm

Therefore ∆ ABC is an equilateral triangle.

Let us assume the base to be 5.5cm and base angle to be 50°

∴, AB = 5.5 cm and ∠B = 50°

We know that,

In an isosceles triangle, opposite sides are equal and opposite angles are equal.

So, ∠B = ∠A = 50°

and AC = BC

Steps of construction:

Step 1: Draw base AB = 5.5cm.

Step 2: At vertex B, Draw a ray constructing angle of 50°.

Step 3: Now draw another ray at A constructing an angle of 50°

Step 4: Mark the point of intersection as C.

ABC is the required triangle.

Given: base BC = 7cm, AB + AC = 12 cm and

∠B = 75° of Δ ABC.

Required: To construct a ΔABC

Steps of construction:

Step 1: Draw a segment BC of length of 7 c

Step 2: At vertex B, construct ∠B = 75° and produce a ray BP.

Step 3: Mark an arc on ray BP cutting at D such that BD = 12cm.

Step 4: Draw segment CD.

Step 5: Construct the perpendicular bisector of segment CD.

Step 6: Name the point of intersection of ray BP and the perpendicular bisector of CD as A.

Step 7: Draw segment AB.

ΔABC is the required triangle.

Given: Base QR = 8 cm, PQ-PR = 3.5cm and

∠Q = 60° of Δ PQR.

Required: To construct a triangle PQR.

Steps of construction:

Step 1: Draw a segment QR of length 8cm.

Step 2: Draw ray QL such that ∠Q = 60°

Step 3: Mark an arc on opposite ray QL i.e. QS cutting at D such that QD = 3.5cm.

Step 4: Draw segment RD.

Step 5: Construct the perpendicular bisector of segment RD.

Step 6: Name the point of intersection of ray QL and the perpendicular bisector of RD as P.

Step 7: Draw segment PR.

ΔPQR is the required triangle.

Given: ∠Y = 30°, ∠Z = 60° and perimeter of ΔXYZ = 10 cm

Steps of construction:

Step 1: Step 1: raw a line segment QR of 10.5cm.

Step 2: From point Q draw a ray QD at 30° and from R draw a ray RE at 60°.

Step 3: Draw an angle bisector of Q and R, two angle bisectors intersect each other at point X.

Step 4: Draw a line bisector of QX and XR respectively these two-line bisectors intersect at point Y and Z

Step 5: Join XY AND XZ.

Step 6: Δ XYZ is required triangle.

Given base(BC) = 7.5cm and AB + AC = 15cm and ∠B = 90°

Steps of construction:

Step 1: Draw the base BC = 7.5cm

Step 2: Make an ∠XBC = 90° at the point B of base BC.

Step 3: Cut the line segment BD equals to AB + AC i.e. 15cm from the ray

XB.

Step 4: Join DC and make an angle bisector of ∠DCB.

Step 5: Let Y intersect BX at A.

Thus, ΔABC is the required triangle.

Given an angle of 90° and chord 5cm

Steps of construction:

Rough image:

Explanation:

x + x + 90° = 180°

[using sum of all angles in a triangle = 180°]

⇒ 2x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 900°

⇒ x = = 45°

Step 1: Draw a line segment AB = 5cm

Step 2: Draw an angle of 45° on point A and B to intersect at O.

Step 3: With centre ‘O’ and radius OA and OB, draw the circle.

Step 4: Mark a point ‘C’ on the arc of the circle. Join AC and BC.

we get ∠CAB = 90°.

Thus, ACB is the required circle segment.

Given an angle of 45° and chord 5cm

Steps of construction:

Step 1: Draw a line segment AB = 5cm

Step 2: Draw an angle of 45° on point A and B to intersect at O.

Step 3: With centre ‘O’ and radius OA and OB, draw the circle.

Step 4: Mark a point ‘C’ on the arc of the circle. Join AC and BC.

we get ∠ACB = 45°.

Thus, ACB is the required circle segment.

Given an angle of 120° and chord 5cm

Rough Image :

Explanation:

x + x + 120° = 180°

[using sum of all angles in a triangle = 180°]

⇒ 2x + 120° = 180°

⇒ 2x = 180° - 120°

⇒ 2x = 60°

⇒ x = = 30°

Steps of construction:

Step 1: Draw a line segment AB = 5cm

Step 2: Draw an angle of 30° on point A and B to intersect at O.

Step 3: With centre ‘O’ and radius OA and OB, draw the circle.

Step 4: Mark a point ‘C’ under the chord AB and on the arc of the

circle . Join AC and BC.

we get M∠A0B = 240°.

Thus, ACB is the required circle segment.