Question : PQRS is a square with a side of 10 cm. A, B, C, and D are mid-points of PQ, QR, RS, and SP, respectively. Then, the perimeter of the square ABCD so formed is:
Option 1: $10\sqrt{2}\ \text{cm}$
Option 2: $20\sqrt{2}\ \text{cm}$
Option 3: $25\sqrt{2}\ \text{cm}$
Option 4: $15\sqrt{2}\ \text{cm}$
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Correct Answer: $20\sqrt{2}\ \text{cm}$
Solution :
Given: PQRS is a square with a side of 10 cm. A, B, C, and D are PQ, QR, RS, and SP mid-points, respectively.
The perimeter of the square is $4a$, where $a$ is its side.
PA = AQ = QB = 5 cm
Also, $\triangle AQB$ is a right-angled triangle at $Q$.
In $\triangle AQB$, from Pythagoras's theorem, we get,
⇒ $(AB)^2=(AQ)^2+(QB)^2$
⇒ $AB=\sqrt{(AQ)^2+(QB)^2}$
⇒ $AB=\sqrt{(5)^2+(5)^2}$
⇒ $AB=\sqrt{50}$
$\therefore AB=5\sqrt{2}\ \text{cm}$
So, the perimeter of the square ABCD $=4 × 5\sqrt{2}= 20\sqrt{2}\ \text{cm}$
Hence, the correct answer is $20\sqrt{2}\ \text{cm}$.
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