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Question Number 100543 by mhmd last updated on 27/Jun/20

Commented by mr W last updated on 27/Jun/20

$$\left(1\right)$$$$y^2=4+4-2y$$$$\Rightarrow y^2+2y-8=0$$$$\mathrm{\Delta }=2^2+4\times 8=36$$$$Area=\frac{\mathrm{\Delta }\sqrt{\mathrm{\Delta }}}{6a^2}=\frac{36\sqrt{36}}{6\times 1^2}=36$$$$\left(4\right)$$$$2y^2+2x=0$$$$\Rightarrow 2y^2+y-1=0$$$$\mathrm{\Delta }=1^2+4\times 1\times 1=5$$$$Area=\frac{\mathrm{\Delta }\sqrt{\mathrm{\Delta }}}{6a^2}=\frac{5\sqrt{5}}{6\times 2^2}=\frac{5\sqrt{5}}{24}$$$$$$$$seeQ89781$$

Answered by Rio Michael last updated on 27/Jun/20

$$1\searrow \swarrow 6$$$$122\searrow \swarrow 4=4$$$$-12-3\searrow \swarrow 7=14$$$$71\searrow \swarrow 6=-18$$$$$$$$\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}[4+14-18-(12-12+7)]=\frac{7}{2}\mathrm{square}\mathrm{units}$$

Answered by Rio Michael last updated on 27/Jun/20

$$\mathrm{Curves}\mathrm{intersect}\mathrm{at}\mathrm{the}\mathrm{points}(-2,5),(2.3,0.7)\mathrm{and}(1.25,-1.50)$$$$\Rightarrow -25$$$$162.30.7-1.4$$$$0.851.25-1.53.45$$$$3-256.25$$$$A=\frac{1}{2}[-1.4+3.45+6.25-(16+0.85+3)]\approx 5.78\mathrm{square}\mathrm{units}$$

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