ΔABC is similar to ΔPQR, AB = 4 cm, BC = 8 cm and AC = 10 cm. If QR =…

For \(\Delta \)ABC and \(\Delta \)PQR, since \(\Delta \)ABC is similar to \(\Delta \)PQR, the ratio of corresponding sides \(\frac{QR}{BC}\) = \(\frac{16}{8}\) = 2

The ratio of the areas of \(\Delta \)ABC and \(\Delta \)PQR will be the square of the ratio of their corresponding sides.

= \(\frac{Area\;of\;PQR}{Area\;of\;ABC}\) = \( {\frac{QR}{BC} }^{2 } \) = \( {2 }^{2 } \) = 4

Now the area of \(\Delta \)ABC

= s = \(\frac{4\;+\;8\;+\;10}{2}\) = 11 cm

= Area of \(\Delta \)ABC = \(\sqrt {11(11\;-\;4)\;(11\;-\;8)\;(11\;-\;10) }\)

= \(\sqrt {11 \;×\; 7\;×\;3\;×\;1 }\) = \(\sqrt { 231}\) \( {cm }^{2 } \)

= Area of \(\Delta \)PQR = 4 x Area of \(\Delta \)ABC = 4\(\sqrt { 231}\) \( {cm }^{2 } \)

Therefore, Area of \(\Delta \)PQR is 4\(\sqrt { 231}\) \( {cm }^{2 } \).