The points $A(2, 9)$,$B(a, 5)$,and $C(5, 5)$ are the vertices of a triangle $ABC$ right-angled at $B$. Find the value of $a$ and the area of $\triangle ABC$.

(A) Given that the points $A(2, 9)$,$B(a, 5)$,and $C(5, 5)$ are the vertices of a $\triangle ABC$ right-angled at $B$.
By the Pythagoras theorem,$AC^2 = AB^2 + BC^2$ $(i)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(a - 2)^2 + (5 - 9)^2} = \sqrt{a^2 - 4a + 4 + 16} = \sqrt{a^2 - 4a + 20}$.
$BC = \sqrt{(5 - a)^2 + (5 - 5)^2} = \sqrt{(5 - a)^2} = |5 - a|$.
$AC = \sqrt{(5 - 2)^2 + (5 - 9)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.
Substituting these into $(i)$:
$5^2 = (\sqrt{a^2 - 4a + 20})^2 + (5 - a)^2$.
$25 = a^2 - 4a + 20 + 25 - 10a + a^2$.
$2a^2 - 14a + 20 = 0$.
$a^2 - 7a + 10 = 0$.
$(a - 2)(a - 5) = 0$.
So,$a = 2$ or $a = 5$.
If $a = 5$,then $B$ coincides with $C$,which is not possible for a triangle. Thus,$a = 2$.
With $a = 2$,the vertices are $A(2, 9)$,$B(2, 5)$,and $C(5, 5)$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AB$.
$BC = |5 - 2| = 3$ units.
$AB = |9 - 5| = 4$ units.
Area $= \frac{1}{2} \times 3 \times 4 = 6$ sq units.