Question

Construct a $\Delta$ABC, in which BC$=4.5$cm, $\angle$A$={40}^o$ and the median AM from A to BC is $4.7$cm. Find the length of the altitude from A to BC.

Answer 1

Given: In $\Delta$ABC, BC$=4.5$cm, $\angle$A$={40}^o$ and the median AM from A to BC is $4.7$cm.
Construction:
(i) Draw a line segment BC$=4.5$cm.
(ii) Draw BX such that $\angle$CBX$={40}^o$.
(iii) Draw BY$\perp$ BX.
(iv) Draw the perpendicular bisector of BC intersecting BY at O and BC at M.
(v) With O as centre and OB as radius, draw the circle.
(vi) The major arc BKC of the circle, contains the verticle angle ${40}^o$.
(vii) With M as centre draw an arc of radius $4.7$cm meeting the circle at A and A'.
(viii) Complete $\Delta$ABC or $\Delta$A'BC, which is the required triangle.
(ix) Produce CB to CZ.
(x) Draw AE$\perp$CZ.