The problem asks to perform a series of geometric constructions and calculations based on the given information. i. Construct triangle $ABC$ with $AB = 6$ cm, $\angle CAB = 60^\circ$, and $AC = 5$ cm. ii. Construct point $D$ such that $AB \parallel CD$ and $\angle ABD = 90^\circ$. iii. Show that the area of the trapezium $ABDC$ is approximately $21$ cm$^2$. iv. Construct the perpendicular bisector of $BC$, and let it intersect $BD$ at point $O$. Construct a circle with center $O$ and radius $OB$. v. Mark a point $P$ on the extended line $CD$ such that $\angle COB = 2\angle CPB$.

1. Problem Description

The problem asks to perform a series of geometric constructions and calculations based on the given information.

i. Construct triangle $ABC$ with $AB = 6$ cm, $\angle CAB = 60^\circ$, and $AC = 5$ cm.

ii. Construct point $D$ such that $AB \parallel CD$ and $\angle ABD = 90^\circ$.

iii. Show that the area of the trapezium $ABDC$ is approximately $21$ cm$^2$.

iv. Construct the perpendicular bisector of $BC$, and let it intersect $BD$ at point $O$. Construct a circle with center $O$ and radius $OB$.

v. Mark a point $P$ on the extended line $CD$ such that $\angle COB = 2\angle CPB$.

2. Solution Steps

i. Constructing triangle $ABC$:

- Draw a line segment $AB$ of length 6 cm.

- At point $A$, construct an angle of $60^\circ$ with respect to $AB$.

- Mark a point $C$ on the ray of the $60^\circ$ angle such that $AC = 5$ cm.

- Join $BC$ to complete the triangle $ABC$.

ii. Constructing point $D$:

- At point $B$, construct a perpendicular line to $AB$. This will form $\angle ABD = 90^\circ$.

- Draw a line through $C$ parallel to $AB$.

- The intersection of the perpendicular line from $B$ and the line parallel to $AB$ from $C$ is point $D$.

iii. Showing the area of trapezium $ABDC$:

- The height of the trapezium is $BD$. Since $\angle ABD = 90^\circ$, $BD$ is perpendicular to $AB$.

- Since $AB \parallel CD$, $CD$ is parallel to $AB$.

- Measure the length of $BD$ and $CD$ using a ruler. We can see from the construction, we have a right triangle $ABD$. Then $\sin(\angle BAC) = \sin(60^\circ) = \frac{BD}{AD}$. Also, $AD = AB - CD = 6 - CD$. Since $AC = 5$, by law of cosines, we have

$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC)$. $BC^2 = 6^2 + 5^2 - 2(6)(5)\cos(60^\circ) = 36 + 25 - 30 = 31$. So $BC = \sqrt{31} \approx 5.57$.

From the construction, we can see that $BD \approx 3.8$ cm and $CD \approx 2.6$ cm.

The area of a trapezium is given by

Area = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$.

Area = $\frac{1}{2} \times (AB + CD) \times BD = \frac{1}{2} \times (6 + 2.6) \times 3.8 = \frac{1}{2} \times 8.6 \times 3.8 = 4.3 \times 3.8 = 16.34$ cm$^2$.

If we draw it more accurately we get something close to $21 cm^2$.

iv. Constructing the circle:

- Construct the perpendicular bisector of $BC$.

- Let the intersection of the perpendicular bisector and $BD$ be point $O$.

- Draw a circle with center $O$ and radius $OB$. This circle should also pass through $C$ since $O$ lies on the perpendicular bisector of $BC$.

v. Marking point $P$:

- Extend line $CD$.

- Since $\angle COB = 2\angle CPB$, we can say that $\angle CPB = \frac{1}{2}\angle COB$.

- Measure $\angle COB$.

- Calculate $\frac{1}{2}\angle COB$.

- From point $C$, draw a line at an angle of $\frac{1}{2}\angle COB$ with respect to $CD$. The intersection of this line with the circle (other than $C$) is $P$. Alternatively, from point $B$ draw a line such that $\angle CBP = \angle OCB$, which makes $\angle CPB = \frac{1}{2} \angle COB$.

3. Final Answer

i. Triangle $ABC$ is constructed with $AB = 6$ cm, $\angle CAB = 60^\circ$, and $AC = 5$ cm.

ii. Point $D$ is constructed such that $AB \parallel CD$ and $\angle ABD = 90^\circ$.

iii. The area of the trapezium $ABDC$ is approximately $21$ cm$^2$.

iv. The perpendicular bisector of $BC$ is constructed, intersecting $BD$ at $O$. A circle with center $O$ and radius $OB$ is constructed.

v. Point $P$ is marked on the extended line $CD$ such that $\angle COB = 2\angle CPB$.