Question

In a quadrilateral the sides are $9,40,28,15$ units and the angle between first sides is a right angle. Find the area of quadrilateral

• A. $106sq.units$
• B. $206sq.units$
• C. $306sq.units$
• D. $406sq.units$

Answer 1

The correct option is C $306sq.units$

R.E.F image

As ${}^{\prime }AB{C}^{\prime }$ is pt. triangle

$\Rightarrow AC=\sqrt{9^2+{40}^2}=41$

$ar\left(ABCD\right)=ar\left(ABC\right)+ar\left(ACD\right)$

$ar\left(ABC\right)=\sqrt{S(S-a)(S-b)(S-c)}$

here $S=\frac{41+40+9}{2}=45$

$a=9,b=40,c=41$

$\Rightarrow ar\left(ABC\right)=\sqrt{45\times 5\times 4\times 36}=180$

$ar\left(ACD\right)=\sqrt{S^1(S^1-a^1)(S^1-b^1)(S^1-C^1)}$

$S^1=\frac{41+28+15}{2}=42$ & $a^1=15$ & $b^1=28$

$\Rightarrow ar\left(ACD\right)=\sqrt{42\times 1\times 27\times 14}=126$

$\therefore ar\left(ABCD\right)=126+180=306$