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Question Number 113092 by bobhans last updated on 11/Sep/20

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{tringle}\mathrm{where}\mathrm{the}$$$$\mathrm{sides}\mathrm{of}\mathrm{triangle}\mathrm{are}91\mathrm{cm},98\mathrm{cm},\mathrm{and}105\mathrm{cm}$$

Answered by bemath last updated on 11/Sep/20

$$\mathrm{question}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}$$$$\mathrm{triangle}\mathrm{where}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{triangle}$$$$\mathrm{are}91\mathrm{cm},98\mathrm{cm}\mathrm{and}105\mathrm{cm}?$$$$\mathrm{area}=\frac{\sqrt{{4\mathrm{a}}^2{\mathrm{b}}^2-{({\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2)}^2}}{4}$$$$\mathrm{where}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{are}\mathrm{the}\mathrm{length}\mathrm{sides}\mathrm{of}△.$$$$\mathrm{hence}\mathrm{area}=\frac{\sqrt{4\times {91}^2\times {98}^2-{({91}^2+{98}^2-{105}^2)}^2}}{4}$$$$=4116{\mathrm{cm}}^2$$

Commented by bobhans last updated on 11/Sep/20

$$\mathrm{waw}....\mathrm{the}\mathrm{new}\mathrm{formula}$$

Commented by MJS_new last updated on 11/Sep/20

$$\mathrm{the}\mathrm{formula}\mathrm{is}\mathrm{not}\mathrm{new},{\mathrm{it}}^{\prime }\mathrm{s}{\mathrm{Heron}}^{\prime }\mathrm{s}\mathrm{Formula}$$$$\mathrm{usually}\mathrm{we}\mathrm{write}\mathrm{it}$$$$Area=\frac{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}}{4}$$

Commented by bobhans last updated on 11/Sep/20

$$\mathrm{Archimedes}\mathrm{Theorem}$$$$16{\left(\mathrm{area}\right)}^2={4\mathrm{a}}^2{\mathrm{b}}^2-{({\mathrm{c}}^2-{\mathrm{a}}^2-{\mathrm{b}}^2)}^2$$$$\mathrm{area}=\sqrt{\frac{{4\mathrm{a}}^2{\mathrm{b}}^2-{({\mathrm{c}}^2-{\mathrm{a}}^2-{\mathrm{b}}^2)}^2}{16}}$$

Commented by bemath last updated on 11/Sep/20

$$\mathrm{not}\mathrm{same}\mathrm{sir}.\mathrm{i}\mathrm{think}\mathrm{it}\mathrm{from}$$$$\mathrm{Archimedes}\mathrm{theorem}.$$$$\mathrm{sorry}\mathrm{if}\mathrm{i}\mathrm{wrong}.\mathrm{hahaha}$$

Commented by MJS_new last updated on 11/Sep/20

$$\left(a+b+c)(a+b-c)\right)a+c-b)(b+c-a)=$$$$=-(a^4+b^4+c^4)+2(a^2{b}^2+a^2{c}^2+b^2{c}^2)=$$$$4a^2{b}^2-{(a^2+b^2-c^2)}^2=$$$$=4a^2{b}^2-(a^4+b^4+c^4+2(a^2{b}^2-a^2{c}^2-b^2{c}^2))=$$$$=-(a^4+b^4+c^4)+2(a^2{b}^2+a^2{c}^2+b^2{c}^2)$$

Answered by malwaan last updated on 11/Sep/20

$$\frac{91+98+105}{2}=\frac{294}{2}=147$$$$sothearea=$$$$\sqrt{147(147-91)(147-98)(147-105)}$$$$=\sqrt{147\times 56\times 49\times 42}$$$$=\sqrt{16941456}=4116$$

Commented by bemath last updated on 11/Sep/20

$$\mathrm{same}\mathrm{sir}\mathrm{with}\mathrm{my}\mathrm{answer}.$$$$\mathrm{please}\mathrm{check}$$

Commented by malwaan last updated on 11/Sep/20

$$yes$$$$andthanksforthisnewformula$$

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