Construct a traingle of sides 5 cm, 8 cm, 12 cm and then construct a parallelogram whose area is equal to area of that triangle and one of whose angles is 60∘. The dimensions of such parallelogram are:
9 cm, 4 cm
9 cm, 4 cm
Construct a triangle ABC such that AB = 5 cm, BC = 12 cm, CA = 8 cm.
Method of Construction:
(1) Draw a straight line BC of length 12 cm by the use of a ruler.
(2) Draw an arc of radius 5 cm at B of BC and an arc of radius 8 cm at C. Let the later one intersect the previous arc at A. A, B and A, C should be joined. With that, the triangle ABC is constructed, the sides of which are 5 cm, 12 cm, 8 cm.
(3) Determine D, the midpoint of BC. [Using a perpendicular bisetor]
(4) Draw PQ through A, parallel to BC.
(5) Draw (\angle CDE\) at D equal to ∠X = 60∘, the side DE of which intersect PQ at E.
(6) Cut the part EF from EQ equal to DC.
(7) Join C and F.
(8) DEFC is the required parallelogram. By measurement, the parallelogram measures ( 6 cm,2.6 cm ). [approx].