Geometric Constructions

First we draw a triangle ABC, with AB = 5.5 cm, BC = 6 cm and CA = 4.5 cm

Now, as ΔABC is similar to ΔLMN

∴ corresponding sides will have same ratio

Now, as

⇒ LM = 4.4 cm

⇒ MN = 4.8 cm

⇒ LN = 3.6 cm

Now, make a ΔLMN, with LM = 4.4 cm, MN = 4.8 cm and LN = 3.6 cm

_{Steps of construction:}

i. Draw a triangle PQR, with PQ = 4.2 cm, QR = 5.4 cm and PR = 4.8 cm, choosing QR = 5.4 cm as base.

ii. Below QR, draw an acute angle ∠QRX.

iii. Mark four points R₁, R₂, R₃ and R₄ on RX, such that RR₁ = R₁R₂ = R₂R₃ = R₃R₄. [As ratio is 4:3, we choose 4 points]

iv. Join QR₄ and Draw TR₃ || QR₄

v. Draw LT || PQ.

First we draw a triangle RST, with RS = 4.5 cm, ∠RST = 40° cm and ST = 5.7 cm

Now, as ΔRST is similar to ΔXYZ,

∴ corresponding sides will have same ratio

Now, as

⇒ XY = 7.5 cm

⇒ YZ = 9.5 cm

Also, Corresponding angles of similar triangles are equal

⇒ ∠RST = ∠XYZ = 40°

Now, draw a triangle XYZ, with XY = 7.5 cm, ∠XYZ = 40° cm and YZ = 9.5 cm.

_{Steps of construction:}

i. Draw a triangle AMT, with AM = 6.3 cm, ∠TAM = 50° cm and AT = 5.6 cm, choosing AM as base.

ii. Below AM, draw an acute angle ∠MAX.

iii. Mark four points A₁, A₂, A₃, A₄, A₅, A₆ and A₇ on AX, such that AA₁ = A₁A₂ =…= A₆A₇ [As ratio is 7:5, we choose 7 points]

iv. Join MA₇ and Draw HA₅ || MA₇

v. Draw HE || MT

Steps of construction:

i. Draw a circle with center P and radius 3.2 cm

ii. Take a point M on the circle, Join PM.

iii. Draw AB ⊥ PM such that AB passes through M, AB is required tangent.

Steps of construction:

i. Draw a circle with center P and radius 2.7 cm

ii. Take a point M on the circle, Join PM.

iii. Draw AB ⊥ PM such that AB passes through M, AB is required tangent.

i. Draw a circle of radius 3.6 cm. Take any point C on it.

ii. Draw chord CB and an inscribed ∠CAB .

iii. With the centre A and any convenient radius draw an arc intersecting the sides of ∠BAC in points M and N.

iv. Using the same radius and centre C, draw an arc intersecting the chord CB at point R.

v. Taking the radius equal to d(MN) and centre R, draw an arc intersecting the arc drawn in the previous step. Let D be the point of intersection of these arcs. Draw line CD. Line CD is the required tangent to the circle.

Here chord = 6.6 = 2 × 3.3 cm = 2 × radius, hence PQ is diameter of the circle.

Steps of construction:

i. Draw a circle with center O, and radius 3.3 cm

ii. Draw a diameter PQ passing through center.

iii. Draw AB ⊥ OQ and CD ⊥ OP , such that AB and CD pass through Q and P respectively.

iv. AB and CD are required tangents.

Observation: AB || CD, i.e. tangents at opposite ends of diameter are parallel.

Steps of construction:

i. Draw a circle with center P and radius 3.4 cm.

ii. Take a point Q outside the circle such that PQ = 5.5 cm

iii. Draw the perpendicular bisector of PQ, which bisects PQ at O.

iv. Taking O as center and OP = OQ as radius, draw another circle, which intersects the previous circle at A and B.

v. Join AQ and BQ, which are required tangents.

Steps of construction:

i. Draw a circle with center P and radius 4.1 cm.

ii. Take a point Q outside the circle such that PQ = 7.3 cm

iii. Draw the perpendicular bisector of PQ, which bisects PQ at O.

iv. Taking O as center and OP = OQ as radius, draw another circle, which intersects the previous circle at A and B.

v. Join AQ and BQ, which are required tangents.