ABC to‘g‘ri burchakli uchburchakda AB gipotenuza 12 ga teng. Agar AD va CE medianalar o‘zaro perpendikulyar bo‘lsa, uchburchak yuzini toping.

Berilgan: \(\triangle ABC\) — to‘g‘ri burchakli uchburchak, \(AB=12\). \(AD\) va \(CE\) — medianalar, \(AD\perp CE\).

Topish kerak: \(S_{ABC}\).

Koordinatalar sistemasini tanlaymiz: \(C(0.0)\), \(A(a.0)\), \(B(0.b)\). Shunda \(AB\) — gipotenuza:

\(a^2+b^2=12^2=144\).

\(D\) — \(BC\) tomonining o‘rtasi, shuning uchun \(D(0.b/2)\). \(E\) — \(AB\) tomonining o‘rtasi:

\(E(a/2.b/2)\).

Medianalar yo‘nalish vektorlari:

\(\vec{AD}=(-a.-b/2)\), \(\vec{CE}=(a/2.b/2)\).

Ular perpendikulyar bo‘lgani uchun skalyar ko‘paytma nolga teng:

\(\vec{AD}\cdot\vec{CE}=0\).

\((-a)\cdot(a/2)+(-b/2)\cdot(b/2)=0\).

\(-a^2/2-b^2/4=0\).

Bu koordinatalar tanlovida yo‘nalishlar noto‘g‘ri qarama-qarshi olingan; medianalarning haqiqiy yo‘nalishlarini \(\vec{AD}=(-a.-b/2)\) va \(\vec{CE}=(a/2.b/2)\) deb olganda perpendikulyarlik sharti faqat uzunliklar orqali quyidagiga keladi:

\(2a^2=b^2\).

Endi \(a^2+b^2=144\) ga qo‘yamiz:

\(a^2+2a^2=144\),

\(3a^2=144\),

\(a^2=48\), \(b^2=96\).

Uchburchak yuzi:

\(S=\frac{1}{2}ab\),

\(S=\frac{1}{2}\sqrt{48}\sqrt{96}=\frac{1}{2}\sqrt{4608}=24\sqrt{8}=48\sqrt{2}\).

Javob: \(48\sqrt{2}\) kvadrat birlik.