Jump to content
idhammuqoddas

Segitiga biasa tapi membingungkan

Recommended Posts

Misal ada \(E\) pada perpanjangan \(AB\) sedemikian sehingga \(AE\) tegak lurus ama \(CE\).

Misal \(BE=m\). Lalu, misal ada \(F\) pada \(CE\) sedemikian sehingga \(DF\) tegak lurus ama \(CE\). Akibatnya, didapat \(BDFE\) persegi panjang, sehingga \(DF=BE=m\)

Lalu, dengan kesebangunan, dimana kita tahu bahwa segitiga\(ABD\) sebangun ama \(CDF\), didapat \(AD=\frac{PC.AB}{DF}=\frac{100}{m}\)

Akhirnya, dengan pythagoras, didapat \(AC^2=AE^2+CE^2 => (\frac{100}{m}+10)^2=(10+m)^2+3m^2\), yang nanti didapat \(m=(500)^{(\frac{1}{3})}\) (moga bener), sehingga \(AD=\frac{100}{(500)^{(\frac{1}{3})}}\)

Edited by Rizky Maulana
Gk apa2

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now


×