Question

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

Answer 1

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)

Answer 2

\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)