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標題:
quadratic equation
發問:
There are a rectangle ABCD. The length of BC is 5 cm less than 3 times that of AB. Two right-angled isoceles triangles are cut away from rectangle ABCD. If the area of parallelogram BEDF obtained is 42 cm^2, find the dimensions of rectangle ABCD. THANK U!
最佳解答:
First, by the formula for the area of rectangle LxW, the area of ABCD = AB x BC =AB x (3AB - 5) =3(AB)^2 - 5AB..........(I) Second, by the sum of two triangles and the parallelogram, the area of ABCD = Parallelogram + 2 (triangles) = 42 + 2 x (AB x AB )/2 = 42 + (AB)^2..........(II) Therefore, by (I) And (II) 3(AB)^2 - 5AB = 42 + (AB)^2 2(AB)^2 -5AB-42 = 0..........(III) Solve for (III), we get AB = 6 or AB = -7/2 (rejected) So AB = 6 and BC = 3(AB) -5 = 3(6) - 5 = 13 and the dimensions are 6 x 13 = 78cm^2
quadratic equation
發問:
There are a rectangle ABCD. The length of BC is 5 cm less than 3 times that of AB. Two right-angled isoceles triangles are cut away from rectangle ABCD. If the area of parallelogram BEDF obtained is 42 cm^2, find the dimensions of rectangle ABCD. THANK U!
最佳解答:
First, by the formula for the area of rectangle LxW, the area of ABCD = AB x BC =AB x (3AB - 5) =3(AB)^2 - 5AB..........(I) Second, by the sum of two triangles and the parallelogram, the area of ABCD = Parallelogram + 2 (triangles) = 42 + 2 x (AB x AB )/2 = 42 + (AB)^2..........(II) Therefore, by (I) And (II) 3(AB)^2 - 5AB = 42 + (AB)^2 2(AB)^2 -5AB-42 = 0..........(III) Solve for (III), we get AB = 6 or AB = -7/2 (rejected) So AB = 6 and BC = 3(AB) -5 = 3(6) - 5 = 13 and the dimensions are 6 x 13 = 78cm^2
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