Question

Find the area of a triangle whose vertices are $(0,6\sqrt{3}),(\sqrt{35},7)$, and $(0,3)$.

• A. $15.37$
• B. $17.75$
• C. $21.87$
• D. $25.61$
• E. $39.61$

Answer 1

Answer: (C)

$21.87$

Area of triangle having vertices $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ is given by
Area $=\frac{1}{2}\times \left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

Area of triangle whose vertices are $(0,6\sqrt{3}),(\sqrt{35},7),(0,3)$

$=\frac{1}{2}\left[0\right(7-3)+\sqrt{35}(3-6\sqrt{3})+0(6\sqrt{3}-7\left)\right)]$

$=\frac{1}{2}\left[\sqrt{35}\right(3-6\sqrt{3}\left)\right]$

$=\frac{1}{2}\left[5.91\right(3-6\times 1.73\left)\right]$

$=\frac{1}{2}\times 5.91\times 7.40$

$=\frac{1}{2}\times 43.74$

$=21.87$