Question

A triangle ABC has vertices A(2, 3), B(5, 6) and C(9, 4). AD, BE and CF are perpendiculars from the vertices A, B and C, respectively, to $x$-axis. The area of $\Delta ABC$ is

• A.
  $6sq.units$
• B. $\frac{8}{3}sq.units$
  
• C.
  $9sq.units$
• D.
  $\frac{49}{3}sq.units$
  

Answer 1

The correct option is C

$9sq.units$
The points A(2, 3), B(5, 6) and C(9, 4) can be plotted on the coordinate plane as:






Joining the points A, B and C, ΔABC is obtained. Draw AD, BE and CF as perpendiculars to the x-axis.

It is seen that AD = 3 units, BE = 6 units, CF = 4 units, DE = 3 units, EF = 4 units, and DF = 7 units.

Now, Area $\left(\Delta ABC\right)=Area\left(ADEB\right)+Area\left(BEFC\right)-Area\left(ADFC\right)\dots ..\left(i\right)$

Since, Area of trapezium $=\frac{1}{2}\times \text{(sum of parallel sides) × distance between the parallel sides.}\therefore \text{Area (ADEB)}=\frac{1}{2}\times (AD\times BE)\times DE=\frac{1}{2}(3+6)\times 3=\frac{27}{2}sq.units.....\left(ii\right)$

and, Area (BEFC) $=\frac{1}{2}\times (BE\times CF)\times EF=\frac{1}{2}(6+4)\times 4=\frac{40}{2}sq.units.....\left(iii\right)$

and, Area (ADFC)$=\frac{1}{2}\times (AD\times CF)\times DF=\frac{1}{2}(3+4)\times 7=\frac{49}{2}sq.units.....\left(iv\right)$

From (i), (ii), (iii) and (iv), we get

Area $\left(\Delta ABC\right)=(\frac{27}{2}+\frac{40}{2}-\frac{49}{2})=9sq.units$

Thus, the area of $\Delta ABCis9sq.units.$

Hence, the correct answer is option (c).