Question-127547

Question Number 127547 by Mathgreat last updated on 30/Dec/20

Answered by mr W last updated on 31/Dec/20

Commented by mr W last updated on 31/Dec/20

Δ {A D}^{'} B \equiv Δ C D B

A B = B C = \sqrt{1^{2} + 2^{2}} = \sqrt{5}

A_{Δ A B C} = \frac{\sqrt{5} \times \sqrt{5}}{2} = \frac{5}{2}

A_{Δ A D C} = A_{Δ A B C} - (A_{Δ C D B} + A_{Δ A B D})

= A_{Δ A B C} - A_{{A D}^{'} B D}

= A_{Δ A B C} - (A_{Δ {A D}^{'} D} + A_{Δ {D D}^{'} B})

= \frac{5}{2} - (\frac{\sqrt{2} \times \sqrt{2}}{2} + \frac{1 \times 1}{2})

= \frac{5}{2} - \frac{3}{2}

= 1

Commented by mr W last updated on 31/Dec/20

o r

∠ B D C = ∠ {B D}^{'} A = 90 °

∠ A D C = 360 ° - 90 ° - 45 ° - 90 ° = 135 °

A_{Δ A D C} = \frac{\sqrt{2} \times 2 \times \sin 135 °}{2} = 1

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