Maths Question
2008-02-19 7:20 am
The length of the diagonal of a rectangle is 25 cm and the difference of the lengths of the sides is 3 cm. Find the length of each side to one decimal place.
回答 (5)
2008-02-19 8:04 am
✔ 最佳答案
Let x be the length of one side of the rectanglex-3 be the adjacent side of x of the rectangle
x^2 + (x - 3)^2 = 25^2
x^2 + x^2 - 6x + 9 = 625
2x^2 - 6x -616 = 0
x^2 - 3x - 308 = 0
x = 19.1 cm or -16.1 cm (rej)
So, the length of the rectangle is 19.1cm
and the width of the rectangle is 16.1 cm
2008-02-20 7:31 pm
x² + (x+3)² = 25²
x² + x² + 6x +9 = 625
2x² + 6x - 616 = 0
x² + 3x - 308 = 0
x = 16.1 or -19.1(rejected)
lengths of rectangle are 16.1cm and 19.1cm
x² + x² + 6x +9 = 625
2x² + 6x - 616 = 0
x² + 3x - 308 = 0
x = 16.1 or -19.1(rejected)
lengths of rectangle are 16.1cm and 19.1cm
參考: me
2008-02-19 6:20 pm
Let x cm be the length of the rectangle,
(x-3) cm be the width of the rectangle:
x^2 + (x - 3)^2 = 25^2
x^2 + x^2 - 6x + 9 = 625
2x^2 - 6x -616 = 0
x^2 - 3x - 308 = 0
x = 19.1 cm or -16.1 cm (rej)
So, the length of the rectangle is 19.1cm
and the width of the rectangle is 16.1 cm
(x-3) cm be the width of the rectangle:
x^2 + (x - 3)^2 = 25^2
x^2 + x^2 - 6x + 9 = 625
2x^2 - 6x -616 = 0
x^2 - 3x - 308 = 0
x = 19.1 cm or -16.1 cm (rej)
So, the length of the rectangle is 19.1cm
and the width of the rectangle is 16.1 cm
2008-02-19 8:29 am
let x be the sides.
(x+3)^2 + x^2 = 25^2
x^2+6x+9+x^2=625
2x^2+6x-616=0
x^2+3x-308=0
(x+3)^2 + x^2 = 25^2
x^2+6x+9+x^2=625
2x^2+6x-616=0
x^2+3x-308=0
2008-02-19 8:10 am
Let x be the length, therefore width be (x-3) cm
x2+(x-3)2=252
2x2-6x-616=0
x2-3x-308=0
x=19.1 (Correct to one decimal place) or x=-16.1 (Rejected)
x2+(x-3)2=252
2x2-6x-616=0
x2-3x-308=0
x=19.1 (Correct to one decimal place) or x=-16.1 (Rejected)
參考: me
收錄日期: 2021-04-16 22:56:22
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