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Question Number 67116 by TawaTawa last updated on 23/Aug/19

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by Kunal12588 last updated on 23/Aug/19

$$10\times 10-\frac{1}{2}\times 3\times 3-2\times \frac{1}{2}\times 10\times 7$$$$100-4.5-70$$$$=30-4.5=25.5squareunits$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{I}\mathrm{dont}\mathrm{get}\mathrm{the}\mathrm{break}\mathrm{down}\mathrm{here}\mathrm{sir}.\mathrm{Help}\mathrm{me}\mathrm{to}\mathrm{understand}.$$$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}.$$

Answered by Kunal12588 last updated on 23/Aug/19

$$area=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 10& 7& 1\\ 7& 10& 1\end{array}\right|$$$$=\frac{1}{2}\mid (100-49)\mid =25.5squaredunits$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

Answered by mr W last updated on 23/Aug/19

$$\frac{1}{2}\times 3\sqrt{2}\times (10\sqrt{2}-\frac{3}{\sqrt{2}})$$$$=\frac{1}{2}\times 3(20-3)$$$$=\frac{51}{2}$$$$=25.5$$

Commented by mr W last updated on 23/Aug/19

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{Please}\mathrm{let}\mathrm{me}\mathrm{see}\mathrm{how}\mathrm{you}\mathrm{get}3\sqrt{2}(10\sqrt{2}-\frac{3}{\sqrt{2}})\mathrm{in}\mathrm{the}$$$$\mathrm{diagram}.\mathrm{I}\mathrm{am}\mathrm{learning}\mathrm{your}\mathrm{approaches}.\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}.$$

Commented by mr W last updated on 23/Aug/19

$$AD=BD=3$$$$\Rightarrow AB=3\sqrt{2}$$$$\Rightarrow CD=\frac{3}{\sqrt{2}}$$$$OD=10\sqrt{2}$$$$\Rightarrow OC=OD-CD=10\sqrt{2}-\frac{3}{\sqrt{2}}$$$$A_{shade}=\frac{AB\times OC}{2}=\frac{3\sqrt{2}(10\sqrt{2}-\frac{3}{\sqrt{2}})}{2}$$$$=\frac{3(20-3)}{2}=\frac{3\times 17}{2}=\frac{51}{2}$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{Wow},\mathrm{great},\mathrm{i}\mathrm{understand}\mathrm{sir}.$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{Sir},\mathrm{help}\mathrm{me}\mathrm{check}.\mathrm{I}\mathrm{have}\mathrm{solved}\mathrm{the}\mathrm{one}\mathrm{you}\mathrm{said}\mathrm{i}\mathrm{should}\mathrm{try}$$

Answered by Kunal12588 last updated on 23/Aug/19

$$questionasksforeasywaybutletstryahardway$$$${eq}^{n}oflinepassingthroughOand(7,10)$$$$isy-0=\frac{y-10}{x-7}(x-0)$$$$\Rightarrow y=\frac{10}{7}x$$$${eq}^{n}oflinepassingthroughOand(10,7)$$$$isy-0=\frac{y-7}{x-10}(x-0)$$$$\Rightarrow y=\frac{7}{10}x$$$${eq}^{n}oflinepassingthrough(7,10)and(10,7)$$$$y-10=\frac{y-7}{x-10}(x-7)$$$$\Rightarrow xy-10x-10y+100=xy-7x-7y+49$$$$\Rightarrow x+y=17$$$$\Rightarrow y=17-x$$$$area={\int }_0^7\frac{10}{7}xdx+{\int }_7^{10}(17-x)dx-{\int }_0^{10}\frac{7}{10}xdx$$$$=\frac{10}{7}\times \frac{49}{2}+17(10-7)-(\frac{100}{2}-\frac{49}{2})-\frac{7}{10}\times \frac{100}{2}$$$$=35+51-\frac{51}{2}-35$$$$=\frac{51}{2}=25.5$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{Wow},\mathrm{great}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by TawaTawa last updated on 23/Aug/19

$$\mathrm{But}\mathrm{sir},\mathrm{how}\mathrm{do}\mathrm{i}\mathrm{choose}\mathrm{area}\mathrm{using}\mathrm{integration}$$

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