A square PQRS and an equilateral triangle XYZ are rearranged to form a figure as shown.
What is the perimeter of the ABDF ? (Take √2 = 1.4 and √3 = 1.7)
24.4 cm
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)
= √XY2−YO2
= √82−42
= √64−16
= √48 = 4√3 cm
BD = 8 cm (side of equilateral △XYZ)
DF = 4 cm (side of square PQRS)
AF = Diagonal of square
= √PQ2+QR2
= √42+42
= √32 = 4√2 cm
∴ Perimeter of ABDF = AB + BD + DF + AF
= 4√3 + 8 + 4 + 4√2
= 6.8 + 12 + 5.6
= 24.4 cm