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Question Number 67535 by mr W last updated on 28/Aug/19

Commented by mr W last updated on 28/Aug/19

$$findtheareaofthesquareA=?$$

Commented by Prithwish sen last updated on 28/Aug/19

$$\mathbf{Area}=\frac{{\mathbf{a}}^2+{\mathbf{b}}^2+{\mathbf{c}}^2}{2}$$

Commented by petrochengula last updated on 28/Aug/19

$$Area=\frac{a^2+b^2+c^2+2ac}{2}$$

Commented by mr W last updated on 28/Aug/19

$$thankssirs!$$$$A=\frac{a^2+b^2+c^2+2ac}{2}istherightanswer.$$

Commented by Prithwish sen last updated on 28/Aug/19

$$\mathrm{Sorry}\mathrm{it}\mathrm{will}\mathrm{be}{(\mathrm{a}+\mathrm{c})}^2.\mathrm{Thank}\mathrm{you}\mathrm{sir}$$

Answered by mr W last updated on 28/Aug/19

Commented by mr W last updated on 28/Aug/19

$$diagonalofsquare=d$$$$d^2=b^2+{(a+c)}^2$$$$areaofsquare=A$$$$A={\left(\frac{d}{\sqrt{2}}\right)}^2=\frac{d^2}{2}=\frac{b^2+{(a+c)}^2}{2}$$

Commented by Prithwish sen last updated on 28/Aug/19

$$\mathrm{Thanks}\mathrm{a}\mathrm{lot}.$$

Commented by TawaTawa last updated on 28/Aug/19

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

Commented by Rasheed.Sindhi last updated on 28/Aug/19

$$Theredlinesegments(havinglengths$$$$a,b\mathrm{\&}c)dividethesquareintotwo$$$$parts.Findouttheareaofanypart.$$

Commented by mr W last updated on 28/Aug/19

$$segmentbisdividedbythediagonal$$$$inratioa:c.$$$$assumec⩾a.$$$$\mathrm{\Delta }A=\frac{c}{2}\times \frac{c}{a+c}\times b-\frac{a}{2}\times \frac{a}{a+c}\times b$$$$=\frac{b(c^2-a^2)}{2(a+c)}=\frac{b(c-a)}{2}$$$$A_1=\frac{A}{2}+\mathrm{\Delta }A=\frac{b^2+{(a+c)}^2+2b(c-a)}{4}$$$$A_2=\frac{A}{2}-\mathrm{\Delta }A=\frac{b^2+{(a+c)}^2-2b(c-a)}{4}$$

Commented by Rasheed.Sindhi last updated on 29/Aug/19

سائين توھان جي مھرباني!Thanks s¡r

Commented by mr W last updated on 30/Aug/19

$$Tawhaanjeemehrbaani!$$

Commented by Rasheed.Sindhi last updated on 30/Aug/19

$$\mathrm{You}\mathrm{have}\mathrm{written}\mathbf{exactly}\mathrm{correct}$$$$\mathrm{in}\mathrm{roman}\mathrm{alphabet}!$$

Commented by Rasheed.Sindhi last updated on 30/Aug/19

Sir, ich fühle mich mit Ihnen freundlicher als jeder andere Forum-Freund von mir.

Commented by mr W last updated on 30/Aug/19

$$thankyouforsayingthat,sir!$$

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