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 -Maths 9th

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 -Maths 9th
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(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) unitsArea of ΔPGQ = \(rac{1}{2}\) x a x a = \(rac{a^2}{2}\) sq units.Area of all triangles outside square ABCD = 4 x \(rac{a^2}{2}\) = 2a2 sq unitsAlso, side of square = 4 ⇒ a ( 2 + √2) = 4 ⇒ a = \(rac{4}{2+\sqrt2}\)∴ Total area of the four Δs outside square ABCD= 2 x \(\bigg(rac{4}{2+\sqrt2}\bigg)^2\) = \(rac{2 imes16}{(4+4\sqrt2+2)}\)= \(rac{16}{3+2\sqrt2}\) x \(rac{3-2\sqrt2}{3-2\sqrt2}\) = 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.
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