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In the given figure, ABCD is a square. EFGH is a square formed by pointing the mid points of sides of ABCD. LMNO is a square formed by joining the mid points of sides of EFGH. A circle is inscribed inside LMNO. If area of circle is 38.5 cm2, then what is the area (in cm2) square ABCD?

In the given figure, ABCD is a square. EFGH is a square formed by pointing the mid points of sides of ABCD. LMNO is a square formed by joining the mid points of sides of EFGH. A circle is inscribed inside LMNO. If area of circle is 38.5 cm2, then what is the area (in cm2) square ABCD? Correct Answer 196

Given Area of circle = 38.5

⇒ 22/7r2 = 38.5

⇒ r2 = 12.25

⇒ r = 3.5 cm

⇒ Side of square LMNO = 2 × 3.5 = 7 cm

In ΔNHO, suppose NH = OH = a

Now applying Pythagoras theorem in ΔNHO,

⇒ a2 + a2 = 49

⇒ a = 7/√2

⇒ NH = OH = 7/√2

⇒ Side of square EFGH = 2 × 7/√2 = 7√2 cm

In ΔFDH, suppose FD = DH = b cm

Now applying Pythagoras theorem in ΔFDH,

⇒ b2 + b2 = 98

⇒ b = 7 cm

⇒ FD = DH = 7 cm

⇒ Side of square ABCD = 2 × 7 = 14 cm

∴ Area of square ABCD = 196 cm2

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