In the given figure, ABCD is a rectangle with AD = 4 units and AE = EB. EF is perpendicular to DB and is half of DF. If the area of the triangle DEF is 5 sq. units, then what is the area of ABCD?

### Concept & Pythagorean Theorem in Area Calculations By utilizing given area relations and right-angled triangles, we can deduce unknown side lengths step-by-step to find the dimensions of the larger rectangle. ### Step-by-Step Solution * Given $\triangle DEF$ is a right-angled triangle (since $EF \perp DB$) with area 5 sq. units. * Given $EF = \frac{1}{2}DF$. Let $EF = x$, then $DF = 2x$. * Area of $\triangle DEF = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot DF \cdot EF = \frac{1}{2} \cdot (2x) \cdot x = x^2$. * Since Area = 5, $x^2 = 5$, meaning $EF = \sqrt{5}$ and $DF = 2\sqrt{5}$. * In right $\triangle DEF$, applying Pythagoras theorem to find hypotenuse $DE$: $$DE^2 = EF^2 + DF^2 = (\sqrt{5})^2 + (2\sqrt{5})^2 = 5 + 20 = 25$$ * Therefore, $DE = 5$ units. * In right $\triangle DAE$ (corner of rectangle ABCD), $AD = 4$ units and hypotenuse $DE = 5$ units. * Apply Pythagoras theorem to find $AE$: $AE = \sqrt{DE^2 - AD^2} = \sqrt{25 - 16} = \sqrt{9} = 3$ units. * Given $AE = EB$, so $AB = 2 \cdot AE = 2 \cdot 3 = 6$ units. * The area of rectangle $ABCD = AB \cdot AD = 6 \cdot 4 = 24$ sq. units. ### Exam Strategy & Shortcut Recognize the classic 3-4-5 right triangle. Once you find $DE = 5$ and know $AD = 4$, $AE$ must be 3. Since $E$ is the midpoint, the full length $AB$ is 6. Area is simply $6 \times 4 = 24$. ### Common Pitfall Misinterpreting the relation between $EF$ and $DF$, or failing to identify $\triangle DAE$ as a right-angled triangle with sides 3-4-5. ### Final Answer Therefore, the correct answer is **24 sq. units**.