Question

A square PQRS and an equilateral triangle XYZ are rearranged to form a figure as shown.

What is the perimeter of the ABDF ? (Take $\sqrt{2}$ = 1.4 and $\sqrt{3}$ = 1.7)

• A. 24.4

Answer 1

The correct option is A 24.4

To find the perimeter of ABDF, we need to know the following sides :

AB, BD, DF and AF

AB = XO (height of the equilateral $△$XYZ)

= $\sqrt{X{Y}^2-Y{O}^2}$

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

= $\sqrt{64-16}$

= $\sqrt{48}$ = $4\sqrt{3}$ cm

BD = 8 cm (side of equilateral $△$XYZ)

DF = 4 cm (side of square PQRS)

AF = Diagonal of square

= $\sqrt{P{Q}^2+Q{R}^2}$

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

= $\sqrt{32}$ = $4\sqrt{2}$ cm

$\therefore$ Perimeter of ABDF = AB + BD + DF + AF

= $4\sqrt{3}$ + 8 + 4 + $4\sqrt{2}$

= 6.8 + 12 + 5.6

= 24.4 cm