$In △ ABC, seg DE ∥ side BC . If 2 A (△ ADE) = A (□ DBCE), f ind AB : AD and show that BC = \sqrt{3} \times DE .$

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Solution

$First we will draw the figure of △ ABC in which DE is a line segment, such that DE ∥ BC and 2 A (△ ADE) = A (□ DBCE) .$

$From 2 A (△ ADE) = A (□ DBCE), we get : 2 A (△ ADE) = A (△ ABC) - A (△ ADE) \Rightarrow 3 A (△ ADE) = A (△ ABC) \Rightarrow \frac{A (△ ABC)}{A (△ ADE)} = \frac{3}{1} . . . . . . . . (1) Now, in △ ABC and △ ADE, ∠ BAC = ∠ DAE [C ommon angles] ∠ ABC = ∠ ADE [C orresponding angles] Therefore, by the AA similarity test, △ ABC ~ △ ADE Thus, \frac{A (△ ABC)}{A (△ ADE)} = \frac{{AB}^{2}}{{AD}^{2}} = \frac{{BC}^{2}}{{DE}^{2}} = \frac{{AC}^{2}}{{AE}^{2}} \Rightarrow \frac{A (△ ABC)}{A (△ ADE)} = \frac{{AB}^{2}}{{AD}^{2}} \Rightarrow \frac{3}{1} = \frac{{AB}^{2}}{{AD}^{2}} [From equation (1)] On taking square root of both sides, we get : \frac{AB}{AD} = \frac{\sqrt{3}}{1} Thus, we AB : AD = \sqrt{3} : 1 . Since, \frac{{AB}^{2}}{{AD}^{2}} = \frac{{BC}^{2}}{{DE}^{2}}, we ge t : \frac{AB}{AD} = \frac{BC}{DE} \Rightarrow \frac{\sqrt{3}}{1} = \frac{BC}{DE} \Rightarrow BC = \sqrt{3} \times DE$

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Similar questions

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) .

B D ⊥ s i d e A C

s e g D E ⊥ s i d e B C

D E \times B D = D C \times B E

D E ∥ B C

△ A D E

△ A B C

A D = \frac{1}{2} B D

D E

B C = 4.5 c m

△ A B C = 18 c m^{2}

D B C E

△ A B C

3 A (△ A D E) = A (D E C B),

B C : D E

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