$In the given fig ure, DE ∥ BC . (i) If DE = 4 cm, BC = 8 cm and A (△ ADE) = 25 {cm}^{2}, find A (△ ABC) . (ii) If DE : BC = 3 : 5, then find A (△ ADE) : A (□ DBCE) .$

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Solution

(i)
$In △ ABC and △ ADE, ∠ BAC = ∠ DAE (C ommon angles) . . . (1) ∠ ABC = ∠ ADE (Corresponding angles, as DE ∥ BC) . . . (2) From equations (1) and (2), we get : △ ABC ~ △ ADE [By t h e AA test of similarity] Thus, \frac{A (△ ABC)}{A (△ ADE)} = \frac{{AB}^{2}}{{AD}^{2}} = \frac{{BC}^{2}}{{DE}^{2}} = \frac{{AC}^{2}}{{AE}^{2}} \Rightarrow \frac{A (△ ABC)}{A (△ ADE)} = \frac{{BC}^{2}}{{DE}^{2}} On substituting A (△ ADE) = 25 {cm}^{2}, DE = 4 cm and BC = 8 cm, we get : \frac{A (△ ABC)}{25} = \frac{(8)^{2}}{(4)^{2}} \Rightarrow A (△ ABC) = 25 \times \frac{64}{16} = 100 {cm}^{2}$
(ii)
$In (i) we have proved that △ ABC ~ △ ADE . Thus, \frac{A (△ ABC)}{A (△ ADE)} = \frac{{AB}^{2}}{{AD}^{2}} = \frac{{BC}^{2}}{{DE}^{2}} = \frac{{AC}^{2}}{{AE}^{2}} \Rightarrow \frac{A (△ ABC)}{A (△ ADE)} = \frac{{BC}^{2}}{{DE}^{2}} \Rightarrow \frac{A (△ ADE)}{A (△ ABC)} = \frac{{DE}^{2}}{{BC}^{2}} On substituting DE : BC = 3 : 5, we get : \frac{A (△ ADE)}{A (△ ABC)} = \frac{(3)^{2}}{(5)^{2}} = \frac{9}{25} On subtracting 1 from both the sides, we get : \frac{A (△ ABC)}{A (△ ADE)} - 1 = \frac{25}{9} - 1 \Rightarrow \frac{A (△ ABC) - A (△ ADE)}{A (△ ADE)} = \frac{25 - 9}{9} \Rightarrow \frac{A (DBCE)}{A (△ ADE)} = \frac{16}{9} \Rightarrow \frac{A (△ ADE)}{A (DBCE)} = \frac{9}{16} Thus, A (△ ADE) : A (DBCE) = 9 : 16$

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