Question

ABCD is a square. Another square EFGH with the same area is placed on the square ABCD such that the point of intersection of diagonals of square ABCD and square EFGH coincide and the sides of the square EFGH are parallel to the diagonals of square ABCD. Thus a new figure is formed as shown in the figure. What is the area enclosed by the given figure, if each side of the square is 4 cm?

(a) 32(2 – √2) 

(b) \(16\big(\frac{3+\sqrt2}{2+\sqrt2}\big)\)

(c) \(32\big(\frac{2+\sqrt2}{3+\sqrt2}\big)\)

(d) None of these

Answer 1

(a) 32 (2 - √2)

As is seen in the given figure, the sides of one square are parallel to the diagonals of another square. Also, square ABCD and EFGH have same area.

⇒ Sides of square ABCD and square EFGH are 4 cm each. 

Let DP = a units 

As DP = PG = GQ = QC = a units and ∠G = 90º

∴ PQ = a√2 units

∴ DC = DP + PQ + QC = (a + a√2 + a) units = a ( 2 + √2) units

Area of ΔPGQ = \(\frac{1}{2}\) x a x a = \(\frac{a^2}{2}\) sq units.

Area of all triangles outside square ABCD = 4 x \(\frac{a^2}{2}\) = 2a² sq units

Also, side of square = 4 ⇒ a ( 2 + √2) = 4 ⇒ a = \(\frac{4}{2+\sqrt2}\)

∴ Total area of the four Δs outside square ABCD

= 2 x \(\bigg(\frac{4}{2+\sqrt2}\bigg)^2\) = \(\frac{2\times16}{(4+4\sqrt2+2)}\)

= \(\frac{16}{3+2\sqrt2}\) x \(\frac{3-2\sqrt2}{3-2\sqrt2}\) = 16 (3 - 2√2)

∴ Total area of the figure = Area of square ABCD + Area of the four Δs outside ABCD

= 16 +16 (3 -2√2)

= 16 + (4 - 2√2) = 32 (2 - √2) cm².