Question

Draw a triangle ABC with side BC = 6 cm, AB = 5cm, and ∠ABC = 60°, then construct a triangle whose sides are 3/4 of the corresponding sides of ΔABC.

Answer 1

Step1: Construct segment BC of 6 cm

Step2: construct a ray at angle 60° above BC from point B

Step3: take 5 cm in compass because of AB = 5 cm, keep the needle of the compass on point B and mark an arc intersecting the ray drawn in step 2. Mark the intersection point as A and join A and C hence ΔABC is ready

Step4: draw a ray at any angle from point B below BC

Step5: take any distance in compass and keeping the needle of the compass on point A cut an arc on ray constructed in step4 and name that point X₁. Keeping the distance in compass same keep the needle of the compass on point X₁ and cut an arc on the same ray and mark that point as X₂. Draw 4 such parts (greater of 3 and 4 in 3/4), i.e. by repeating this process mark points upto X₄

Step6: join X₄ and C and from X₃ (because X₃ is the third point 3 being smaller in 3/4) construct line parallel to X₃C and mark the intersection point with BC as D

BD is three forth of BC

Step7: construct line parallel to AC from point D and mark the intersection point with AB as E thus ΔBDE which have sides three forth of ΔABC is ready.