The diagonals of a square and that of a rectangle with length 12cm and breadth 8cm are equal. Find the side of the square correct to two decimal places. 30 points
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Answer:
Step-by-step explanation:
diagonal of the rectangle
\sqrt{12 {}^{2} + ( \frac{0.8}{2}) {}^{2} } \\ = \sqrt{144 + \frac{16}{100} } \\ = \sqrt{ \frac{14416}{100} } \\ = \frac{120.06}{10} \\ = 12 \: (almost)
now the diagonal of a square of side a
√2 a
now
√2 a=12
a=12/√2=8.49 cm
therefore the side of the square is 8.49cm or 8.5 cm (approx)
Answered by
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\sqrt{12 {}^{2} + ( \frac{0.8}{2}) {}^{2} } \\ = \sqrt{144 + \frac{16}{100} } \\ = \sqrt{ \frac{14416}{100} } \\ = \frac{120.06}{10} \\ = 12 \: (almost)
now the diagonal of a square of side a
√2 a
now
√2 a=12
a=12/√2=8.49 cm
therefore the side of the square is 8.49cm or 8.5 cm (approx)
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