A boy of height 1.5m with his eye level at 1.4 m stands before a plane mirror of length 0.75 m fixed on the wall. The height of the lower edge of the mirror above the floor is 0.8 m. Then

From similar triangles we have

$\frac{\text{AB}}{\text{BE}}=\frac{\text{CD}}{\text{DE}}$

$\frac{1.4}{2x+x_0}=\frac{0.8}{x+x_0}$

$x_0=x\text{/3}$

Now$\frac{0.8}{x+x_0}=\frac{y}{x_0}$

$\left(\text{or}\right)\frac{0.8}{3x_0+x_0}=\frac{y}{x_0}$

$y=0.02\text{m}$

Thus the length of the image he can see in the mirror

$=\text{1}\text{.5}-\text{0}\text{.02}$

$=\text{1}\text{.48 m}$

So the boy cannot sit is feet