Question

Given quadrangle $ABCD$ with right angles in $B$ and $D$. We also know that $\left|BC\right|$ = 1, $\left|CD\right|$ = 4, and $\left|DA\right|$ = 3. What is the area of $ABCD$?

• A. $8$
• B. $6+\sqrt{6}$
• C. $8.5$
• D. $17$
• E. $12+2\sqrt{6}$

Answer 1

The correct option is B $6+\sqrt{6}$

$|BC|=1,|CD|=4,|DA|=3{|AC|}^2={|AD|}^2+{|CD|}^2=3^2+4^2=25\therefore |AC|=5⟹{|AC|}^2={|AB|}^2+{|BC|}^2⟹25={|AB|}^2+1\therefore |AB|=2\sqrt{6}$

Area of triangle ABCD = Area of triangle ACD + Area of triangle ACB = $\frac{1}{2}\times 3\times 4+\frac{1}{2}\times 2\sqrt{6}\times 1=6+\sqrt{6}$