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Question Number 48388 by Tawa1 last updated on 23/Nov/18

Commented by Tawa1 last updated on 23/Nov/18

$Find the Area of the triangle$
	Answered by mr W last updated on 23/Nov/18

	$C B = \frac{∣ 5 \times 1 - 5 \times 3 ∣}{\sqrt{1^{2} + 3^{2}}} = \sqrt{10}$ $O C = \sqrt{5^{2} + 5^{2}} = 5 \sqrt{2}$ $O B = \sqrt{50 - 10} = 2 \sqrt{10}$ $o r$ $\tan ∠ C O B = \frac{\frac{5}{5} - \frac{1}{3}}{1 + \frac{5 \times 1}{5 \times 3}} = \frac{10}{20} = \frac{1}{2}$ $O B = \frac{C B}{\tan ∠ C O B} = 2 \sqrt{10}$ $\tan ∠ A O B = \frac{2 - \frac{1}{3}}{1 + \frac{2 \times 1}{3}} = \frac{5}{5} = 1$ $A B = O B \tan ∠ A O B = 2 \sqrt{10}$ $A r e a o f t r i a n g l e = \frac{O B \times A B}{2} = \frac{2^{2} \times 10}{2} = 20$
	Commented by Tawa1 last updated on 24/Nov/18

	$God bless you sir$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Nov/18

	$e q n A B (y - 5) = - 3 (x - 5)$ $y - 5 + 3 x - 15 = 0$ $y + 3 x = 20$ $p o i n t A i n t e r s e c t i o n b e t w e e n y + 3 x = 20 a n d y = 2 x$ $2 x + 3 x = 20 x = 4 s o y = 8 A (4, 8)$ $B i s t h e i n t e r s e c t i i n b e t w e e n y = \frac{x}{3} a n d y + 3 x = 20$ $\frac{x}{3} + 3 x = 20 x = 6 y = 2$ $s o A (4, 8) B (6, 2) O (0, 0)$ $a r e a o f O A B$ $\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$ $= \frac{1}{2} [4 (2 - 0) + 6 (0 - 8) + 0 (8 - 2)]$ $= \frac{1}{2} [- 40]$ $s o a r e a i s 20 u n i t . . .$
	Commented by Tawa1 last updated on 24/Nov/18

	$God bless you sir$
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