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Question Number 68110 by TawaTawa last updated on 05/Sep/19

Commented by kaivan.ahmadi last updated on 07/Sep/19
$((y/2))^2 =1−((x/3))^2 ⇒y=±2(√(1−((x/3))^2 )) equation of BC: m=tg30=((√3)/2) y−0=((√3)/2)(x+3)⇒y=((√3)/2)x+((3(√3))/2) equation of OD: m=tg60=(√3) y=(√3)x we find point C ((x/3))^2 +(((√3)/4)x+((3(√3))/4))^2 =1⇒(1/9)x^2 +(3/(16))x^2 +(9/8)x+((27)/(16))=1⇒ 16x^2 +27x^2 +162x+243=144⇒43x^2 +162x+99=0 x_(1,2) =((−162±96)/(86))= { ((((−162+96)/(86))=((−66)/(86))=((−33)/(43)))),((((−162−96)/(86))=−3)) :} so C(((−33)/(43)),((−33(√3))/(86))+((3(√3))/2))=(((−33)/(43)),((96(√3))/(86)))=(((−33)/(43)),((48(√3))/(43))) so this diagram is not true. and now we find point D: ((x/3))^2 +(((√3)/2)x)^2 =1⇒(1/9)x^2 +(3/4)x^2 =1⇒13x^2 =36⇒ x^2 =((36)/(13))⇒x=±(√((36)/(13)))⇒D((√((36)/(13))),(√((108)/(13)))) so we have S_(OBCD) =∫_(−3) ^((−33)/(43)) (((√3)/2)x+((3(√3))/2))dx+∫_((−33)/(43)) ^0 2(√(1−((x/3))^2 ))dx +∫_0 ^(√((36)/(13))) (2(√(1−((x/3))^2 ))−(√3)x)dx$
${(\frac{y}{2})}^{2} = 1 - {(\frac{x}{3})}^{2} \Rightarrow y = \pm 2 \sqrt{1 - {(\frac{x}{3})}^{2}}$ $e q u a t i o n o f B C :$ $m = t g 30 = \frac{\sqrt{3}}{2}$ $y - 0 = \frac{\sqrt{3}}{2} (x + 3) \Rightarrow y = \frac{\sqrt{3}}{2} x + \frac{3 \sqrt{3}}{2}$ $e q u a t i o n o f O D :$ $m = t g 60 = \sqrt{3}$ $y = \sqrt{3} x$ $w e f i n d p o i n t C$ ${(\frac{x}{3})}^{2} + {(\frac{\sqrt{3}}{4} x + \frac{3 \sqrt{3}}{4})}^{2} = 1 \Rightarrow \frac{1}{9} x^{2} + \frac{3}{16} x^{2} + \frac{9}{8} x + \frac{27}{16} = 1 \Rightarrow$ $16 x^{2} + 27 x^{2} + 162 x + 243 = 144 \Rightarrow 43 x^{2} + 162 x + 99 = 0$ $x_{1, 2} = \frac{- 162 \pm 96}{86} = {\begin{cases} \frac{- 162 + 96}{86} = \frac{- 66}{86} = \frac{- 33}{43} \\ \frac{- 162 - 96}{86} = - 3 \end{cases}$ $s o C (\frac{- 33}{43}, \frac{- 33 \sqrt{3}}{86} + \frac{3 \sqrt{3}}{2}) = (\frac{- 33}{43}, \frac{96 \sqrt{3}}{86}) = (\frac{- 33}{43}, \frac{48 \sqrt{3}}{43})$ $s o t h i s d i a g r a m i s n o t t r u e .$ $a n d n o w w e f i n d p o i n t D :$ ${(\frac{x}{3})}^{2} + {(\frac{\sqrt{3}}{2} x)}^{2} = 1 \Rightarrow \frac{1}{9} x^{2} + \frac{3}{4} x^{2} = 1 \Rightarrow 13 x^{2} = 36 \Rightarrow$ $x^{2} = \frac{36}{13} \Rightarrow x = \pm \sqrt{\frac{36}{13}} \Rightarrow D (\sqrt{\frac{36}{13}}, \sqrt{\frac{108}{13}})$ $s o w e h a v e$ $S_{O B C D} = \int_{- 3}^{\frac{- 33}{43}} (\frac{\sqrt{3}}{2} x + \frac{3 \sqrt{3}}{2}) d x + \int_{\frac{- 33}{43}}^{0} 2 \sqrt{1 - {(\frac{x}{3})}^{2}} d x$ $+ \int_{0}^{\sqrt{\frac{36}{13}}} (2 \sqrt{1 - {(\frac{x}{3})}^{2}} - \sqrt{3} x) d x$
Commented by TawaTawa last updated on 08/Sep/19

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