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Question Number 75955 by benjo last updated on 21/Dec/19

h e l p m e

Answered by benjo last updated on 21/Dec/19

Answered by mr W last updated on 21/Dec/19

Commented by mr W last updated on 21/Dec/19

\cos θ = \frac{7^{2} + 9^{2} - 8^{2}}{2 \times 7 \times 9} = \frac{11}{21}

r_{1} = \frac{7}{2 \times \sin θ} = \frac{7}{2 \times \frac{8 \sqrt{5}}{21}} = \frac{147}{16 \sqrt{5}}

r_{2} = \frac{9}{2 \times \sin θ} = \frac{9}{2 \times \frac{8 \sqrt{5}}{21}} = \frac{189}{16 \sqrt{5}}

∠ A C B = 2 (\frac{π}{2} - θ) + θ = π - θ

{A B}^{2} = r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} \cos ∠ A C B

= {(\frac{147}{16 \sqrt{5}})}^{2} + {(\frac{189}{16 \sqrt{5}})}^{2} + 2 \times \frac{147}{16 \sqrt{5}} \times \frac{189}{16 \sqrt{5}} \times \frac{11}{21}

= \frac{147^{2} + 189^{2} + 2 \times 147 \times 189 \times \frac{11}{21}}{16^{2} \times 5}

= \frac{21609}{320}

\Rightarrow A B = \frac{147 \sqrt{5}}{40}

\frac{1}{2} \times A B \times \frac{C D}{2} = \frac{1}{2} \times r_{1} \times r_{2} \times \sin ∠ A C B

\frac{147 \sqrt{5}}{40} \times \frac{C D}{2} = \frac{147}{16 \sqrt{5}} \times \frac{189}{16 \sqrt{5}} \times \frac{8 \sqrt{5}}{21}

\Rightarrow C D = \frac{9}{2} = 4.5

Commented by benjo last updated on 21/Dec/19

thanks sir .

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