What do you think is the shape of the field?

In the triangle given above, $\angle \mathrm{ADB}={90}^{\mathrm{o}}, \angle \mathrm{ABC}={45}^{\mathrm{o}}, \mathrm{AD}=10 \mathrm{cm}, \mathrm{AC}=20 \mathrm{cm}.$ The length of $\mathrm{BC}$ is:

In $∆\mathrm{PQR}, \angle \mathrm{Q}>\angle \mathrm{R}, \mathrm{PS}$ is the bisector of $\angle \mathrm{P}$, and $\mathrm{PT}\perp \mathrm{QR}$. If $\angle \mathrm{SPT}={28}^{°}$ and $\angle \mathrm{R}={23}^{°}$, then the measure of $\angle \mathrm{Q}$ is:

If the interior angle of a regularpolygon is $108°$, then it is a-

1. Octagon

2. Hexagon

3. Pentagon

4. Tetragon