Question

Construct a $\mathrm{\Delta }ABC$ in which the base BC=5cm, $\mathrm{\angle }BAC={40}^{\circ }$and the median from A to BC is 6cm. Also measure the length of the altitude from A.

Answer 1

Hint: In this question first we need to draw line segment BC of length 5cm and then draw some extra construction to get the triangle ABC. And using construction we have to find the length of the altitude from vertex A.

Complete step-by-step answer:
Construction Steps:-
Step1: Draw a line segment BC=5cm.
Step2: Draw BX such that $\mathrm{\angle }CBX={40}^{\circ }$.
Step3: Draw BY perpendicular to BX.
Step4: Draw the perpendicular bisector of BC intersecting BY at O and BC at M.
Step5: With O as centre and OB as radius, draw the circle.
Step6:The major arc BKC of the circle contains the vertex angle ${40}^{\circ }$.
Step7: With M as centre draw an arc of radius 6cm meeting the circle at A and A’.
Step8: Complete the$\mathrm{\Delta }ABC$on A’BC which is the required triangle.
Step9: Produce CB to CZ.
Step10: Draw AE perpendicular to CZ.
Step11: On measuring the length of the altitude it comes to 5.1cm.

Note:- Whenever you get this type of problem the key concept to solve this is to learn the different terms like perpendicular bisector which is a line segment which is perpendicular to a given line segment and passes through the midpoint of this line segment and altitude which is a line segment through a vertex and perpendicular to a line segment containing the base(side opposite to the vertex).