Question Number 179228 by mahdipoor last updated on 26/Oct/22
$${prove}\:{in}\:{right}\:{triangle}\::\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$−−−−−− \\ $$
Answered by a.lgnaoui last updated on 27/Oct/22
![Aire (S)du trapeze[ADEB] S=ab+(c^2 /2)=(1/2)(a+b)^2 (c^2 /2)=(1/2)(a^2 +b^2 )⇒ c^2 =a^2 +b^2](https://www.tinkutara.com/question/Q179287.png)
$${Aire}\:\left({S}\right){du}\:{trapeze}\left[\mathrm{ADEB}\right] \\ $$$${S}={ab}+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\frac{{c}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\Rightarrow\:\:\:{c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$
Commented by a.lgnaoui last updated on 27/Oct/22
Commented by a.lgnaoui last updated on 27/Oct/22
$${this}\:{is}\:{a}\:{simple}\:{methode}\:{to} \\ $$$${prove}\:{the}\:{value}\:{of}\:\left({c}\right)\:{for} \\ $$$${a}\:{right}\:{triangle}\:\left(\measuredangle{CAB}=\mathrm{90}°\right) \\ $$