Question

XYZ is an equilateral triangle and PQRS is a square. The figure has been cut and rearranged in the form of a trapezium.

What is the perimeter of the given trapezium ABEF? Take $\sqrt{2}$ = 1.4 and $\sqrt{3}$ = 1.7 (The triangle given is Equilateral)

• A. 32 cm
• B. 36 cm
• C. 32.8 cm
• D. 30.8 cm

Answer 1

The correct option is C

32.8 cm

To know the perimeter of the trapezium we need to know the following Lengths :

AB , BE , EC , CD , DF and AF

AB = CD = Height of the equilateral triangle with side 8 cm

= $\sqrt{8^2-4^2}$

= $\sqrt{48}$ cm

BE = AF = Diagonal of the square

= $\sqrt{4^2+4^2}$

=$\sqrt{32}$ cm

DF = CE = 4 cm

$\therefore$ Perimeter of the Trapezium ABEF = AB + BE + EC + CD + DF+ AF

= $\sqrt{48}$ + $\sqrt{32}$ + 4 + $\sqrt{48}$ + $\sqrt{32}$ + 4

= $4\sqrt{3}$ + $4\sqrt{2}$ + 4 + $4\sqrt{3}$ + $4\sqrt{2}$ + 4

= $8\sqrt{3}$ + $8\sqrt{2}$ + 8

= 8(1+ $\sqrt{3}$ + $\sqrt{2}$)

= 8(1+1.4+1.7)

= 8(4.1)

= 32.8 cm