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Question Number 89412 by I want to learn more last updated on 17/Apr/20

Commented by I want to learn more last updated on 17/Apr/20

$$\mathrm{Please}\mathrm{is}\mathrm{the}\mathrm{answer}:\frac{39}{8}\mathrm{square}\mathrm{unit}.$$

Commented by jagoll last updated on 17/Apr/20

$$x=\frac{1}{4}(4+3y)$$$$y^2=4x\Rightarrow x=\frac{1}{4}{y}^2$$$$\frac{1}{4}{y}^2=\frac{3}{4}y+1$$$$\frac{1}{4}{y}^2-\frac{3}{4}y-1=0$$$$\mathrm{\Delta }=\frac{9}{16}-4\left(\frac{1}{4}\right)(-1)=\frac{25}{16}$$$$area=\frac{\mathrm{\Delta }\sqrt{\mathrm{\Delta }}}{6.a^2}=\frac{\frac{25}{16}\times \frac{5}{4}}{6\times \frac{1}{16}}$$$$=\frac{125}{64}\times \frac{16}{6}=\frac{125}{4\times 6}=\frac{125}{24}$$

Commented by jagoll last updated on 17/Apr/20

Commented by I want to learn more last updated on 17/Apr/20

Commented by I want to learn more last updated on 17/Apr/20

$$\mathrm{Sir},\mathrm{wolfram}\mathrm{says}\frac{125}{24},\mathrm{where}\mathrm{is}\mathrm{the}\mathrm{error}\mathrm{sir}\mathrm{or}\mathrm{wolfram}\mathrm{wrong}.\mathrm{Thanks}$$

Commented by I want to learn more last updated on 17/Apr/20

$$\mathrm{And}\mathrm{how}\mathrm{can}\mathrm{i}\mathrm{get}\mathrm{a}\mathrm{sir}?$$

Commented by jagoll last updated on 17/Apr/20

$$sameresultsir.itcorrect$$

Commented by I want to learn more last updated on 17/Apr/20

$$\mathrm{Thanks}\mathrm{sir}.$$

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