5. Construction of a triangle when its base, vertical angle and median are given

Construct a triangle $ABC$ in which $AB = 9.6 \ cm$, $\angle C = 48^{\circ}$ and the median $CQ$ from $C$ to $AB$ is $9.6 \ cm$. Find the length of the altitude from $C$ to $AB$.

 

Solution:

 

First, let us draw a rough figure.

 

 

Construction:

 

 

Step 1: Draw a line segment $AB$ of length $9.6 \ cm$.

 

Step 2: At $A$, draw $AD$ such that $\angle DAB = 48^{\circ}$.

 

Step 3: At $A$, draw $AE$ such that $\angle DAE = 90^{\circ}$.

 

Step 4: Draw the perpendicular bisector of $AB$, which intersects $AE$ at $P$ and $AB$ at $Q$.

 

Step 5: Draw a circle with $P$ as centre and $AP$ as radius.

 

Step 6: From $Q$, mark arcs of radius $9.6 \ cm$ on the circle. Mark them as $C$ and $R$.

 

Step 7: Join $AC$ and $BC$. Thus, $\triangle ABC$ is the required triangle.

 

Step 8: From $C$, draw a line $CW$ perpendicular to $BV$. $BV$ meets $CW$ at $Z$.

 

Step 9: The length of the altitude is $CZ = 8 \ cm$.