Class 12th

First, express area as a function of t. Suppose there is a triangle whose vertices are A (0,0), B (t,0) and C (0, t). Here, we can use the determinant formula for the area of a triangle. $\text{Area}=\frac{1}{2}{x}_1(y_2$
$\left|{\mathrm{-\; y}}_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$ Let us substitute the coordinates in the above equation: $\begin{array}{l}\text{Area}=\frac{1}{2}\left|0(0-t)+t(t-0)+0(0-0)\right|\\ =\frac{1}{2}\left|t*t\right|\end{array}$

$=\frac{1}{2}{t}^2$

$\text{So, the area as a function of}t\text{is:}$

$\mathrm{Area}\left(t\right)=\frac{1}{2}{t}^2$ Now, let us calculate area when t = 4 and substitute t = 4 into function: $\mathrm{Area}\left(4\right)=\frac{1}{2}*4^2=\frac{1}{2}*16=8$