Question

Construct a $\Delta$ABC such that AB$=6$cm, $\angle$C$={40}^o$ and the altitude from C to AB is of length $4.2$cm.

Answer 1

In $\Delta$ABC, AB$=6$cm, $\angle$C$={40}^o$.

The length of the altitude from C to AB is $4.2$cm.

Construction:

(i) Draw a line segment AB$=6$cm.

(ii) Draw AX such that $\angle$BAX$={40}^o$.

(iii) Draw AY$\perp$ AX.

(iv) Draw the perpendicular bisector of AB intersecting AY at O and AB at M.

(v) With O as centre and OA as radius, draw the circle.

(vi) The segment AKB contains the vertical angle ${40}^o$.

(vii) On the perpendicular bisector MO, mark a point H such that MH$=4.2$cm.

(viii) Draw CHC' parallel to AB meeting the circle at C and at C'.

(ix) Complete the $\Delta$ABC, which is one of the required triangles.