A) 8 cm and 12 cm
B) 6 cm and 10 cm
C) 12 cm and 16 cm
D) None of these
Correct Answer: C
Solution :
Let side of a square = x cm |
Side of another square = (x + 4) cm |
\[\Rightarrow \] \[{{x}^{2}}+{{(x+4)}^{2}}=400\] (by condition) |
\[{{x}^{2}}+{{x}^{2}}+16+8x=400\] |
\[\Rightarrow \] \[2{{x}^{2}}+8x-384=0\] |
\[{{x}^{2}}+4x-192=0\]\[\Rightarrow \]\[(x-12)(x+16)=0\] |
\[\Rightarrow \]\[x-12=0\]or\[x+16=0\] |
\[x=12\] or \[x=-16\](not possible) |
So, side of one square =12 can |
Side of another square \[=\text{ }12+4=16\text{ cm}\] |
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