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標題:
math!畢氏定理.
發問:
我想問一條數 有一塊長方形的金屬片abcd,其AC=24cm,DC=12cm,將這塊金屬片熔化後,再鑄成面積相等的正方形pqrs,金屬片的周界改變了多少? 更新: help!
AD^2 = AC^2 - DC^2 = 24^2 - 12^2 = 432 AD = sqrt 432. So area = AD x DC = 12 x sqrt 432 = 12 sqrt 432. perimeter = 2(AD + DC) = 2(sqrt 432 + 12) = 65.57. Area of square PQRS = 12 sqrt 432 therefore , side of the square = sqrt ( 12 sqrt 432) = 15.79. Perimeter of square = 4 x 15 79 = 63.17. So change in perimeter = 65.57 - 63.17 = decreased 2.4 cm.
其他解答:
Let x be the length of the square x^2 = 24 * 12 x = 12 * sqrt(2) x = 16.97 cm (2 d.p.) Change of Parameter = 4 * 16.97 - (24+12)*2 = 67.88 - 72 = - 4.12 cm Change of Parameter in % = (- 4.12 / 72) * 100% = - 5.72 %
math!畢氏定理.
發問:
我想問一條數 有一塊長方形的金屬片abcd,其AC=24cm,DC=12cm,將這塊金屬片熔化後,再鑄成面積相等的正方形pqrs,金屬片的周界改變了多少? 更新: help!
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最佳解答:AD^2 = AC^2 - DC^2 = 24^2 - 12^2 = 432 AD = sqrt 432. So area = AD x DC = 12 x sqrt 432 = 12 sqrt 432. perimeter = 2(AD + DC) = 2(sqrt 432 + 12) = 65.57. Area of square PQRS = 12 sqrt 432 therefore , side of the square = sqrt ( 12 sqrt 432) = 15.79. Perimeter of square = 4 x 15 79 = 63.17. So change in perimeter = 65.57 - 63.17 = decreased 2.4 cm.
其他解答:
Let x be the length of the square x^2 = 24 * 12 x = 12 * sqrt(2) x = 16.97 cm (2 d.p.) Change of Parameter = 4 * 16.97 - (24+12)*2 = 67.88 - 72 = - 4.12 cm Change of Parameter in % = (- 4.12 / 72) * 100% = - 5.72 %
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