Question

The mid points of triangle ABC are $(1,-4);(6,-1);(-1,-2)$ and its centroid is G then the area of the triangle BGC is?

• A. $\frac{32}{3}$ sq. units
• B. $8$ sq. units
• C. $\frac{23}{2}$ sq. units
• D. $4$ sq. units

Answer 1

Answer: (A)

$\frac{32}{3}$ sq. units

Let mid points be $P(1,-4)$; $Q(6,-1)$; $R(-1,-2)$

$PQ=\sqrt{9+25}=\sqrt{34}$

$QR=\sqrt{1+49}=\sqrt{50}$

$PR=\sqrt{4+4}=\sqrt{8}$

Area of $PQR=\sqrt{\left(\frac{\sqrt{8}+\sqrt{50}+\sqrt{34}}{2}\right)\left(\frac{\sqrt{8}+\sqrt{50}-\sqrt{34}}{2}\right)\left(\frac{\sqrt{8}-\sqrt{50}+\sqrt{34}}{2}\right)\left(\frac{\sqrt{50}+\sqrt{34}-\sqrt{8}}{2}\right)}=8$

$\therefore$ area of ${\Delta }^{u}ABC=8\times 4=32$

$\therefore$ area of ${\Delta }^{n}BCG=\frac{32}{3}$ $\left[A\right]$