Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
In the given figure, $A B C D$ is a rectangle with $A D=$ 4 units and $A E=E B . E F$ is perpendicular to DB and is half of DF. If the area of the triangle DEF is 5 sq. units, then what $18 \sqrt{3}$ sq. units
(b)20 sq. units
(c)24 sq. units
(d)28 sq. units
Answer
Answer (as printed): C
Explanation
Let $E F=x$ units. Then, $D F=2 x$ units. $$\frac{1}{2} \times E F \times D F=5 \Rightarrow \frac{1}{2} \times x \times 2 x=5 \Rightarrow x^{2}=5 \Rightarrow x=\sqrt{5} .$$ $\therefore D E=\sqrt{(D F)^{2}+(E F)^{2}}=(2 \sqrt{5})^{2}+(\sqrt{5})^{2}=\sqrt{25}=5$ units $\therefore A E=\sqrt{(D E)^{2}-(A D)^{2}}=\sqrt{5^{2}-4^{2}}=\sqrt{9}=3$ units. $\mathrm{AB}=2 \mathrm{AE}=6$ units. ∴ Area of rect. ABCD $=\mathrm{AB} \times \mathrm{AD}$ $$=(6 \times 4) \text { sq. units }=24 \text { sq. units. }$$
Explanation as extracted from the printed page; notation may be imperfect.