Cho ∆ABC biết độ dài 3 đường trung tuyến lần lượt bằng 15; 18; 27.

a) Tính diện tích ∆ABC.

a) Gọi M, N, P lần lượt là trung điểm của BC, AC, AB.

Giả sử AM = 15; BN = 18; CP = 27.

Khi đó ta có $\left\{\begin{array}{l}\frac{2\left(A{C}^2+A{B}^2\right)-B{C}^2}{4}=A{M}^2=225\\ \frac{2\left(B{C}^2+A{B}^2\right)-A{C}^2}{4}=B{N}^2=324\\ \frac{2\left(B{C}^2+A{C}^2\right)-A{B}^2}{4}=C{P}^2=729\end{array}\right.$

$\iff \left\{\begin{array}{l}2\left(A{C}^2+A{B}^2\right)-B{C}^2=900\\ 2\left(B{C}^2+A{B}^2\right)-A{C}^2=1296\\ 2\left(B{C}^2+A{C}^2\right)-A{B}^2=2916\end{array}\right.$

$\iff \left\{\begin{array}{l}2A{C}^2+2A{B}^2-B{C}^2=900\\ 2B{C}^2+2A{B}^2-A{C}^2=1296\\ 2B{C}^2+2A{C}^2-A{B}^2=2916\end{array}\right.$

$\iff \left\{\begin{array}{l}A{B}^2=164\\ A{C}^2=704\\ B{C}^2=836\end{array}\right.$

$\iff \left\{\begin{array}{l}AB=2\sqrt{41}\\ AC=8\sqrt{11}\\ BC=2\sqrt{209}\end{array}\right.$

Nửa chu vi ∆ABC là: $p=\frac{AB+AC+BC}{2}=\frac{2\sqrt{41}+8\sqrt{11}+2\sqrt{209}}{2}$ .

Vậy diện tích ∆ABC là: $S_{\Delta ABC}=\sqrt{p\left(p-AB\right)\left(p-AC\right)\left(p-BC\right)}=120\sqrt{2}$ .