Question

In ΔABC, D, E and F are the mid-points of sides AB, BC and CA respectively. If the coordinates of D, E and F are (–2, 5), (3, –4) and (7, a) respectively and area of ΔABC is 212 sq. units, then the possible value of a can be

• A. -7
• B. -10
• C. 10
• D. 24

Answer 1

The correct option is C 10

Given that D, E, F are mid points of AB, BC and CA of Δ ABC.



$Area\left(\Delta DEF\right)=\frac{1}{2}|-2(-4-a)+3(a-5)+7(5+4)|$

$=\frac{1}{2}|8+2a+3a-15+63|$

$=\frac{1}{2}|5a+56|\text{sq.units}$

Now, we know that if D, E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ABC then,

$Ar\left(\Delta DEF\right)=\frac{1}{4}\times Ar\left(\Delta ABC\right)$

Thus, $Ar\left(\Delta ABC\right)=4Ar\left(\Delta DEF\right)$

$\Rightarrow 4\times \frac{1}{2}|5a+56|=212$

$\Rightarrow 10a+112=212or,-212=10a+112$

$\Rightarrow 10a=100or,10a=-324$

$\Rightarrow a=10unitsor,a=-32.4units$

Hence, the correct answer is option (c).