Question

The figure given besides shows five squares inside one another by joining the midpoints of the outer square. The side of square I is a cm. The total area of the five squares is:

(a) \(\frac{(4\sqrt2-1)a^2}{(4\sqrt2+1)}\)

(b) \(\frac{(4\sqrt2+1)a^2}{(4\sqrt2-1)}\)

(c) \(\frac{31}{16}a^2\)

(d) (7 + 3√2)a²

Answer 1

(c) \(\frac{31}{16}a^2.\)

Side of the square I = a 

∴ Area of square I = a²

= \(\sqrt{\frac{a^2}{16}+\frac{a^2}{16}}\) = \(\sqrt{\frac{2a^2}{16}}\) = \(\frac{a}{2\sqrt2}\)

∴ Sum of the areas of the five squares

= a² + \(\frac{a^2}{2}\) + \(\frac{a^2}{4}\) + \(\frac{a^2}{8}\) + \(\frac{a^2}{16}\)

= a² \(\bigg[1+ \frac{1}{2} + \frac{1}{4}+\frac{1}{8}+\frac{1}{16}\bigg]\) = \(\frac{31}{16}a^2.\)