Question

Construct $△$KLM such that KM = 7.2 cm, $\angle$KLM = 72$°$, LA$\perp$KM and LA = 4.8 cm.

Answer 1

Steps of construction:
(i) Draw a line segment KM = 7.2 cm.
(ii) Through K, draw KY such that $\angle$MKY = 72$°$.
(iii) Draw KX$\perp$KY.
(iv) Draw the perpendicular bisector of KM intersecting KX at O and KM at H.
(v) With O as the centre and OK as the radius, draw a circle.
(vi) The major arc KWM of the circle contains the vertical 72$°$.
(vii) With H as the centre, cut off HP = 4.8 cm.
(viii) Draw a line LL' parallel to KM passing through P.
(ix) With L as the centre, draw a line parallel to PH intersecting KM at A.
(x) Join KL and KM

Thus, $△$KLM or $△$KL'M is the required triangle.