Question

From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq. ft of the remaining portion of triangle ABC is

• A. $225\sqrt{3}$
• B. $\frac{500}{\sqrt{3}}$
• C. $\frac{275}{\sqrt{3}}$
• D. $\frac{250}{\sqrt{3}}$

Answer 1

The correct option is B

$\frac{500}{\sqrt{3}}$

Area of triangle when all three sides are given is given by $\sqrt{\left\{s\right(s-a\left)\right(s-b\left)\right(s-c\left)\right\}}$
s = semi-perimeter of triangle $ABC=\frac{(40+35+25)}{2}=50$
Area of triangle $ABC=\sqrt{(50\times 10\times 15\times 25)}=250\sqrt{3}sq.ft.$
Since, the centroid is a point of intersection of medians which divide the triangle in two equal halves apart from the centroid G dividing the median in the ratio 2:1
Therefore, Area of triangle $GBC=\left(\frac{1}{3}\right)\times$ Area of triangle ABC
Hence, Required Area = $\left(\frac{2}{3}\right)\times$ Area of triangle $ABC=\left(\frac{2}{3}\right)\times 250\sqrt{3}=\frac{500}{\sqrt{3}}sq.ft.$