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Question Number 53755 by ajfour last updated on 25/Jan/19

Commented by ajfour last updated on 25/Jan/19

$$thanks,howisscalene△area$$$$obtained?$$

Commented by mr W last updated on 25/Jan/19

Commented by Kunal12588 last updated on 25/Jan/19

$$s=\frac{p+q+q+r+r+p}{2}=p+q+r$$$$\sqrt{s(s-a)(s-b)(s-c)}$$$$=\sqrt{(p+q+r)(p+q+r-p-q)(p+q+r-q-r)(p+q+r-r-p)}$$$$=\sqrt{(p+q+r)pqr}\left(itseemstoberight\right)$$$$reallycoolsi\stackrel{}{r}$$

Commented by ajfour last updated on 25/Jan/19

$$incidentlyifoundjustnowthat$$$$Area(△ABC)=\sqrt{(p+q+r)pqr}$$$$butwhy?[howtofindareaofa$$$$triangle=\sqrt{s(s-a)(s-b)(s-c)}?]$$$$whatidid$$$$p+q=c$$$$q+r=a$$$$r+p=b$$$$\Rightarrow p+q+r=s$$$$\Rightarrow p=s-a,q=s-b,r=s-c$$$$Area(△ABC)=\sqrt{s(s-a)(s-b)(s-c)}$$$$hence△=\sqrt{(p+q+r)pqr}.$$$$buthow△=\sqrt{s(s-a)(s-b)(s-c)}$$$$ifiknewihaveforgotten..$$$$pleasehelp..$$

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

$$area△=\frac{1}{2}absinC$$$$cosC=\frac{a^2+b^2-c^2}{2ab}$$$$sinC=\sqrt{1-{\left(\frac{a^2+b^2-c^2}{2ab}\right)}^2}$$$$=\sqrt{\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2}{4a^2{b}^2}}$$$$=\frac{1}{2ab}\times \sqrt{(a+b+c)(a+b-c)(c+a-b)(c-a+b)}$$$$△=\frac{1}{2}absinC$$$$=\frac{1}{2}ab\times \frac{\sqrt{(a+b+c)(a+b-c)(c+a-b)(c-a+b)}}{2ab}$$$$=\sqrt{\frac{a+b+c}{2}\times \frac{a+b-c}{2}\times \frac{a+c-b}{2}\times \frac{b+c-a}{2}}$$$$s=\frac{a+b+c}{2}\left(assume\right)$$$$=\sqrt{s\times \frac{2s-c-c}{2}\times \frac{2s-b-b}{2}\times \frac{2s-a-a}{2}}$$$$=\sqrt{\frac{s(2s-2c)}{2}\times \frac{2s-2b}{2}\times \frac{2s-2a}{2}}$$$$=\sqrt{s(s-a)(s-b)(s-c)}Proof$$

Commented by mr W last updated on 25/Jan/19

$$\sqrt{c^2-h^2}+\sqrt{b^2-h^2}=a$$$$\sqrt{c^2-h^2}=a-\sqrt{b^2-h^2}$$$$c^2-h^2=a^2-2a\sqrt{b^2-h^2}+b^2-h^2$$$$a^2+b^2-c^2=2a\sqrt{b^2-h^2}$$$${(a^2+b^2-c^2)}^2=4a^2(b^2-h^2)$$$$4a^2{h}^2={\left(2ab\right)}^2-{(a^2+b^2-c^2)}^2$$$$4a^2{h}^2=(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)$$$$4a^2{h}^2=[{(a+b)}^2-c^2][c^2-{(a-b)}^2]$$$$4a^2{h}^2=(a+b+c)(a+b-c)(c+a-b)(c-a+b)$$$$4a^2{h}^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$$$4a^2{h}^2=2s(2s-2a)(2s-2b)(2s-2c)$$$$a^2{h}^2=4s(s-a)(s-b)(s-c)$$$$\Rightarrow h=\frac{2\sqrt{s(s-a)(s-b)(s-c)}}{a}$$$$\mathrm{\Delta }=\frac{ah}{2}=\sqrt{s(s-a)(s-b)(s-c)}$$

Commented by Kunal12588 last updated on 25/Jan/19

$$theformula\sqrt{s(s-a)(s-b)(s-c)}iscalled$$$${Heron}^{\prime }sformula$$

Commented by ajfour last updated on 25/Jan/19

$$Thankseveyrone!butibelieve$$$$therehastobesomemorebasic,$$$$naturalandsymmetricalwayto$$$$theformula.$$

Commented by peter frank last updated on 25/Jan/19

$$perimeteroftriangle$$$$s=\frac{a+b+c}{2}$$$$2s=a+b+c$$$$from$$$$A=\frac{1}{2}ab\sin c$$$$A=\frac{1}{2}ab2\sin \frac{c}{2}\cos \frac{c}{2}$$$$A^2=a^2{b}^2\sin {}^2\frac{c}{2}\cos {}^2\frac{c}{2}$$$$\sin {}^2\frac{c}{2}=?$$$$\cos {}^2\frac{c}{2}=?$$$$\sin {}^{2}c=\frac{1-\cos c}{2}$$$$from$$$$c^2=a^2+b^2-2ab\cos c$$$$\cos c=\frac{a^2+b^2-c^2}{2ab}$$$$\sin {}^2\frac{c}{2}=\frac{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)}{2}$$$$\sin {}^2\frac{c}{2}=\frac{c^2+2ab-a^2-b^2}{4ab}$$$$\sin {}^2\frac{c}{2}=\frac{c^2-{(a-b)}^2}{4ab}$$$$\sin {}^{2}c=\frac{(a+c-b)(c-(a-b)}{4ab}$$$$recall$$$$2s=a+b+c$$$$a+c=2s-b$$$$\sin {}^2\frac{c}{2}=\frac{(a+c-b)(c-(a-b)}{4ab}$$$$\sin {}^2\frac{c}{2}=\frac{(s-b)(s-a)}{ab}.....\left(i\right)$$$$recall$$$$\cos {}^2\frac{c}{2}=\frac{1+\cos c}{2}$$$$c^2=a^2+b^2-2ab\cos c$$$$\cos c=\frac{a^2+b^2-c^2}{2ab}$$$$\cos {}^2\frac{c}{2}=\frac{1+\left(\frac{a^2+b^2-c^2}{2ab}\right)}{2}$$$$\cos {}^2\frac{c}{2}=\frac{a^2+b^2+2ab-c^2}{4ab}$$$$\cos {}^2\frac{c}{2}=\frac{s(s-c)}{ab}......\left(ii\right)$$$$A^2=a^2{b}^2\sin {}^2\frac{c}{2}\cos {}^2\frac{c}{2}$$$$A^2=a^2{b}^2\frac{(s-b)(s-a)}{ab}.\frac{s(s-c)}{ab}$$$$A^2=s(s-a)(s-b)(s-c)$$$$A=\sqrt{s(s-a)(s-b)(s-c)}$$$$$$

Answered by MJS last updated on 25/Jan/19

$$\mathrm{these}\mathrm{things}\mathrm{I}\mathrm{remember},\mathrm{but}\mathrm{some}\mathrm{further}$$$$\mathrm{steps}\mathrm{are}\mathrm{missing}$$$$\left(1\right)\mathrm{draw}\mathrm{a}\mathrm{triangle}\mathrm{and}\mathrm{its}\mathrm{incircle}(\mathrm{center}=I)$$$$\left(2\right)\mathrm{draw}\mathrm{the}\mathrm{radii}\mathrm{\perp }\mathrm{to}\mathrm{the}\mathrm{sides}\mathrm{which}\mathrm{leads}$$$$\mathrm{to}\mathrm{the}\mathrm{points}{I}_a,I_b,I_c$$$$\left(3\right)\mid {AI}_b\mid =\mid {AI}_c\mid =i$$$$\mid {BI}_a\mid =\mid {BI}_c\mid =j$$$$\mid {CI}_a\mid =\mid {CI}_b\mid =k$$$$\left(4\right)\mathrm{the}\mathrm{area}\mathrm{of}ABC\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{the}\mathrm{sum}$$$$\mathrm{of}\mathrm{the}\mathrm{areas}\mathrm{of}\mathrm{the}6\mathrm{rectangular}\mathrm{sub}-$$$$\mathrm{triangles}\mathrm{we}\mathrm{just}\mathrm{created}:$$$$△=r(i+j+k)$$$$\left(5\right)\mathrm{the}\mathrm{area}\mathrm{of}ABC\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{the}\mathrm{sum}$$$$\mathrm{of}\mathrm{the}\mathrm{three}\mathrm{triangles}ABI,BCI,CAI$$$$△=\frac{r}{2}(a+b+c)$$$$$$$$\left(6\right)\mathrm{we}\mathrm{know}$$$$i+j=c$$$$i+k=b$$$$j+k=a$$$$\mathrm{we}\mathrm{can}\mathrm{put}s=\frac{a+b+c}{2}$$$$\Rightarrow i=s-a;j=s-b;k=s-c$$$$$$$$\mathrm{there}\mathrm{was}\mathrm{something}\mathrm{with}\mathrm{the}\mathrm{radii}\mathrm{of}\mathrm{the}$$$$\mathrm{excircles}\mathrm{to}\mathrm{eliminate}r...$$

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