Question-64557

Question Number 64557 by LPM last updated on 19/Jul/19

Commented by LPM last updated on 19/Jul/19

area of △ DBC

Commented by Tony Lin last updated on 19/Jul/19

∠ A E B = 120 ° \Rightarrow ∠ E B A = 30 °

∠ E B C = 60 ° \Rightarrow ∠ A B C = 90 °

∴ △ A B C i s 30 ° - 60 ° - 90 ° t r i a n g l e

\Rightarrow B C = \frac{4}{\sqrt{3}}

∵ △ E B C i s e q u i l a t e r a l t r i a n g l e a n d

△ A E B i s i s o s c e l e s t r i a n g l e

∴ A E = E B = E C = \frac{4}{\sqrt{3}}

∠ A E D = 60 ° \Rightarrow ∠ A D E = 75 °

\frac{A E}{s i n 75 °} = \frac{D E}{s i n 45 °}

\frac{\frac{4}{\sqrt{3}}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{D E}{\frac{\sqrt{2}}{2}}

\Rightarrow D E = \frac{4 (3 - \sqrt{3})}{3}

a r e a o f △ D B C

= △ D E C + △ E B C

= \frac{1}{2} \times D E \times E C \times s i n ∠ D E C + \frac{\sqrt{3}}{4} \times {E C}^{2}

= \frac{1}{2} \times \frac{4 (3 - \sqrt{3})}{3} \times \frac{4}{\sqrt{3}} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{4} \times {(\frac{4}{\sqrt{3}})}^{2}

= 4

Commented by LPM last updated on 19/Jul/19

good

Answered by Kunal12588 last updated on 19/Jul/19

s e e A B C

t a n 60 ° = \frac{4}{B C}

\Rightarrow B C = \frac{4}{3} \sqrt{3}

s i n 60 ° = \frac{4}{A C}

A C = \frac{8}{3} \sqrt{3}

B E = B C = E C = \frac{4}{3} \sqrt{3}

A E = A C - E C = \frac{8}{3} \sqrt{3} - \frac{4}{3} \sqrt{3} = \frac{4}{3} \sqrt{3}

∵ A E = E C = \frac{4}{3} \sqrt{3}

∴ ∠ E B A = 30 °

∠ B D A = 180 ° - 75 ° - 30 ° = 75 °

∵ ∠ B D A = ∠ D A B

A B = B D = 4

D E = B D - B E = 4 - \frac{4}{3} \sqrt{3} = \frac{4 (3 - \sqrt{3})}{3} = \frac{4 \sqrt{3} (\sqrt{3} - 1)}{3}

a r (B C E) = \frac{\sqrt{3}}{4} {(B C)}^{2} = \frac{\sqrt{3}}{4} {(\frac{4 \sqrt{3}}{3})}^{2} = \frac{4 \sqrt{3}}{3}

a r (C D E) = \frac{1}{2} \times E C \times D E s i n (∠ D E C)

= \frac{1}{2} \times \frac{4 \sqrt{3}}{3} \times \frac{4 \sqrt{3} (\sqrt{3} - 1)}{3} \times s i n (180 - 60)

= \frac{1}{2} \times \frac{4 \sqrt{3}}{3} \times \frac{4 \sqrt{3} (\sqrt{3} - 1)}{3} \times \frac{\sqrt{3}}{2}

= \frac{4 \sqrt{3} (\sqrt{3} - 1)}{3}

a r (D B C) = a r (C D E) + a r (B C E)

= \frac{4 \sqrt{3} (\sqrt{3} - 1)}{3} + \frac{4 \sqrt{3}}{3}

= \frac{4 \sqrt{3}}{3} (\sqrt{3})

a r (D B C) = 4

Commented by LPM last updated on 19/Jul/19

good

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