Question

Two rectangles, ABCD and DBEF are shown in the diagram. The length of AD is 3 cm, and the length of AB is 4 cm. What is the area of rectangle DBEF?​

Answer 1

The area of rectangle DBEF would be 12 cm²

Step-by-step explanation:

Given,

In rectangle ABCD,

AB = CD = 4 cm,

AD = BC = 3 cm,

Note: opposite sides of a rectangle are equal and all interior angle are right angles.

By the Pythagoras theorem,

$BD=\sqrt{AB^2+AD^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\text{ units}$

Now, in the given diagram,

E and F are points outside the rectangle ABCD,

Such that, FD⊥BD, EB ⊥ BD and C∈ EF,

In triangles FCD and CDB,

$\angle DFC\cong \angle DCB$ ( right angles ),

$\angle DCF\cong \angle CDB$  ( Alternative interior angle theorem, EF ║ BD )

By AA similarity postulate,

$\triangle FCD\sim \triangle CDB$

∵ Corresponding sides of similar triangles are in same proportion,

$\implies \frac{FD}{CB}=\frac{CD}{BD}$

$\frac{FD}{3}=\frac{4}{5}$

$\implies FD = \frac{12}{5}\text{ units}$

Hence, the area of the rectangle, DBEF = FD × BD

$=\frac{12}{5}\times 5$

$=12\text{ square units}$

#Learn more :

Find the area and perimeter of rectangle ABCD​

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