Question-216239

Question Number 216239 by mr W last updated on 01/Feb/25

Commented by mr W last updated on 01/Feb/25

f i n d a r e a o f t r i a n g l e

Commented by SonGoku last updated on 01/Feb/25

A_{△} = \frac{A B \times C A \times \sin (\cos^{- 1} (- \frac{{B C}^{2} - {C A}^{2} - {A B}^{2}}{2 \times C A \times A B}))}{2}

A_{△} = \frac{\sqrt{4} \times \sqrt{5} \times \sin (\cos^{- 1} (- \frac{{(\sqrt{3})}^{2} - {(\sqrt{5})}^{2} - {(\sqrt{4})}^{2}}{2 \times \sqrt{5} \times \sqrt{4}}))}{2}

A_{△} = \frac{2 \sqrt{5} \sin (\cos^{- 1} (- \frac{3 - 5 - 4}{2 \times \sqrt{5} \times 2}))}{2}

A_{△} = \sqrt{5} \sin (\cos^{- 1} (- \frac{- 6}{4 \sqrt{5}}))

A_{△} = \sqrt{5} \sin (\cos^{- 1} (\frac{3}{2 \sqrt{5}}))

A_{△} = \sqrt{5} \sin (\cos^{- 1} (\frac{3 \sqrt{5}}{2 \times 5}))

A_{△} = \sqrt{5} \sin (\cos^{- 1} (\frac{3 \sqrt{5}}{10}))

A_{△} = \sqrt{5} \sin (47.867585 °)

A_{△} = \sqrt{5} \times 0.741596

A_{△} \approx 1.658259

\begin{matrix} A_{△} \approx 1.66 u^{2} \end{matrix}

Commented by Rasheed.Sindhi last updated on 01/Feb/25

p l e a s e s t a n d i n t h e q u e u e o f “ {a n s w e r s}^{″} .

Answered by ajfour last updated on 01/Feb/25

a + b = \sqrt{5}

\sqrt{4 - c^{2}} + \sqrt{3 - c^{2}} = \sqrt{5}

7 - 2 c^{2} + 2 \sqrt{4 - c^{2}} \sqrt{3 - c^{2}} = 5

(4 - c^{2}) (3 - c^{2}) = {(c^{2} - 1)}^{2}

- 7 c^{2} + 12 = 1 - 2 c^{2}

5 c^{2} = 11

△ = \frac{\sqrt{5}}{2} c = \frac{\sqrt{5}}{2} \times \frac{\sqrt{11}}{5} = \frac{\sqrt{11}}{2}

Answered by efronzo1 last updated on 01/Feb/25

\cos θ = \frac{9 - 3}{4 \sqrt{5}} = \frac{3}{2 \sqrt{5}}

\sin θ = \sqrt{1 - \frac{9}{20}} = \sqrt{\frac{11}{20}} = \frac{\sqrt{55}}{10}

Area of △ = \frac{1}{2} . 2 . \sqrt{5} . \frac{\sqrt{55}}{10}

= \frac{\sqrt{11}}{2}

Answered by Frix last updated on 01/Feb/25

\frac{\sqrt{(a + b + c) (a + b - c) (a + c - b) (b + c - a)}}{4} =

= \frac{\sqrt{2 (a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2}) - (a^{4} + b^{4} + c^{4})}}{4} =

= \frac{\sqrt{2 (12 + 15 + 20) - (9 + 16 + 25)}}{4} =

= \frac{\sqrt{94 - 50}}{4} = \frac{\sqrt{44}}{4} = \frac{\sqrt{11}}{2}

Answered by MathematicalUser2357 last updated on 02/Feb/25

s = \frac{\sqrt{3} + \sqrt{4} + \sqrt{5}}{2} = 1 + \frac{\sqrt{3} + \sqrt{5}}{2} \approx 2.984

A = \sqrt{s (s - \sqrt{3}) (s - 2) (s - \sqrt{5})} = \sqrt{(1 + \frac{\sqrt{3} + \sqrt{5}}{2}) (1 + \frac{\sqrt{5} - \sqrt{3}}{2}) (\frac{\sqrt{3} + \sqrt{5}}{2} - 1) (1 + \frac{\sqrt{3} - \sqrt{5}}{2})}

Using this,

(1 + \frac{\sqrt{A} - \sqrt{B}}{2}) (1 + \frac{\sqrt{B} - \sqrt{A}}{2}) = 1 - \frac{{(\sqrt{A} - \sqrt{B})}^{2}}{4}

Continuing,

= \sqrt{{\frac{{(\sqrt{3} + \sqrt{5})}^{2}}{4} - 1} {1 - \frac{{(\sqrt{3} - \sqrt{5})}^{2}}{4}}} = \sqrt{- {1 - \frac{{(\sqrt{3} + \sqrt{5})}^{2}}{4}} {1 - \frac{{(\sqrt{3} - \sqrt{5})}^{2}}{4}}}

= \sqrt{- [1 - {\frac{{(\sqrt{3} + \sqrt{5})}^{2}}{4} + \frac{{(\sqrt{3} - \sqrt{5})}^{2}}{4}} + \frac{{(\sqrt{3} + \sqrt{5}) (\sqrt{3} - \sqrt{5})}^{2}}{16}]}

= \sqrt{- [1 - \frac{{(\sqrt{3} + \sqrt{5})}^{2} + {(\sqrt{3} - \sqrt{5})}^{2}}{4} + \frac{{{(\sqrt{3})}^{2} - {(\sqrt{5})}^{2}}^{2}}{16}]}

= \sqrt{- {1 - \frac{16}{4} + \frac{{(- 2)}^{2}}{16}}} = \sqrt{- (1 - 4 + \frac{1}{4})} = \sqrt{- (- \frac{11}{4})} = \sqrt{\frac{11}{4}} = \frac{\sqrt{11}}{2} \approx 1.658 ✓