In the given figure, DE || BC

Solution:

In the given figure, we have DE || BC.

In ΔADE and ΔABC

∠ADE=∠B              Corresponding angles

∠DAE=∠BAC       Common

So, ∆ADE~∆ABC      (AA Similarity)

(i) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Hence 

$\frac{A r(\Delta \mathrm{ADE})}{A r(\Delta \mathrm{ABC})}=\frac{\mathrm{DE}^{2}}{\mathrm{BC}^{2}}$

$\frac{16}{\operatorname{Ar}(\Delta \mathrm{ABC})}=\frac{4^{2}}{6^{2}}$

$A r(\Delta \mathrm{ABC})=\frac{6^{2} \times 16}{4^{2}}$

$A r(\triangle \mathrm{ABC})=36 \mathrm{~cm}^{2}$

(ii) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Hence,

$\frac{A r(\Delta \mathrm{ADE})}{A r(\Delta \mathrm{ABC})}=\frac{\mathrm{DE}^{2}}{\mathrm{BC}^{2}}$

$\frac{25}{A r(\Delta \mathrm{ABC})}=\frac{4^{2}}{8^{2}}$

$\frac{25}{\operatorname{Ar}(\Delta \mathrm{ABC})}=\frac{4^{2}}{8^{2}}$

$\operatorname{Ar}(\Delta \mathrm{ABC})=\frac{8^{2} \times 25}{4^{2}}$

$A r(\triangle \mathrm{ABC})=\frac{8^{2} \times 25}{4^{2}}$

$A r(\triangle \mathrm{ABC})=100 \mathrm{~cm}^{2}$

(iii) We know that 

$\frac{A r(\Delta \mathrm{ADE})}{A r(\Delta \mathrm{ABC})}=\frac{\mathrm{DE}^{2}}{\mathrm{BC}^{2}}$

$\frac{A r(\Delta \mathrm{ADE})}{A r(\Delta \mathrm{ABC})}=\frac{3^{2}}{5^{2}}$

$\frac{A r(\triangle \mathrm{ADE})}{A r(\triangle \mathrm{ABC})}=\frac{9}{25}$

Let Area of ΔADE = 9x sq. units and Area of ΔABC = 25x sq. units

$A r[\operatorname{trapBCED}]=A r(\triangle \mathrm{ABC})-A r(\triangle \mathrm{ADE})$

$=25 x-9 x$

$=16 x$ sq units

Now,

$\frac{A r(\triangle \mathrm{ADE})}{A r(\operatorname{trapBCED})}=\frac{9 x}{16 x}$

$\frac{A r(\Delta \mathrm{ADE})}{A r(\operatorname{trapBCED})}=\frac{9}{16}$