Question

Two rectangles $ABCD$ and $DBEF$ are as shown in the figure. The area of rectangle $DBEF$ (in square units) is _______.

• A. $10$
• B. $12$
• C. $14$
• D. $15$

Answer 1

Answer: (B)

$12$

From the figure given, $△ADB$ forms a right angled triangle.

We know that square of hypotenuse of a triangle is equal to sum of squares of adjacent sides.

$\Rightarrow (BD{)}^2=(AB{)}^2+(AD{)}^2$

$\Rightarrow (BD{)}^2=(4{)}^2+(3{)}^2=25$

$\Rightarrow BD=\sqrt{25}=5$

Consider $△BCD$,

Area of $△BCD$ $=\frac{1}{2}\times BC\times CD=\frac{1}{2}\times BD\times CG$

$\Rightarrow BD\ast CG=BC\ast CD$ ............(i)

Area of rectangle $BDFE=BD\times FE=BD\times CG=BC\times CD$ ($\because FE=CG$ and eqn (i))

Area of rectangle $BDFE=BC\times CD=3\times 4=12$

Option (B) is correct.