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如图，在菱形纸片 ABCDA B C D 中， ∠A=60∘\angle \mathrm{A}=60^{\circ} ，将纸片折刍，点 A、DA 、 D 分别落在 $A^{\...
Mar 9, 2024
如图，在菱形纸片 ABCDA B C D 中， ∠A=60∘\angle \mathrm{A}=60^{\circ} ，将纸片折刍，点 A、DA 、 D 分别落在 A′、DA^{\prime} 、 D 处，且 A′DA^{\prime} D 经过 B，EFB ， E F 为折痕，当 D′F⊥CDD^{\prime} F \perp C D 时， CFFD\frac{C F}{F D} 的值为
Solution by Steps
step 1
Given a rhombus ABCD ABCD with ∠A=60∘ \angle A = 60^\circ , when the paper is folded such that point A A and D D fall on A′ A' and D′ D' respectively, and A′D A'D passes through B B , with EF EF being the crease, and D′F⊥CD D'F \perp CD , we need to find the value of CFFD \frac{CF}{FD}
step 2
Since ABCD ABCD is a rhombus, all sides are equal. Let the side of the rhombus be s s
step 3
Triangle ABD ABD is an equilateral triangle because ∠A=60∘ \angle A = 60^\circ and AB=AD=BD=s AB = AD = BD = s
step 4
When the paper is folded, D′F⊥CD D'F \perp CD and D′F D'F bisects ∠D \angle D , creating two 30∘−60∘−90∘ 30^\circ-60^\circ-90^\circ triangles, D′FC D'FC and D′FD D'FD
step 5
In a 30∘−60∘−90∘ 30^\circ-60^\circ-90^\circ triangle, the sides are in the ratio 1:3:2 1:\sqrt{3}:2 . Therefore, FD=s2 FD = \frac{s}{2} and CD=s3 CD = s\sqrt{3}
step 6
Since CD=CF+FD CD = CF + FD , we have s3=CF+s2 s\sqrt{3} = CF + \frac{s}{2}
step 7
Solving for CF CF , we get CF=s3−s2 CF = s\sqrt{3} - \frac{s}{2}
step 8
To find CFFD \frac{CF}{FD} , we substitute the values of CF CF and FD FD into the ratio
step 9
Thus, CFFD=s3−s2s2 \frac{CF}{FD} = \frac{s\sqrt{3} - \frac{s}{2}}{\frac{s}{2}}
step 10
Simplifying the ratio, we get CFFD=23−1 \frac{CF}{FD} = 2\sqrt{3} - 1
Answer
CFFD=23−1 \frac{CF}{FD} = 2\sqrt{3} - 1
Key Concept
Properties of rhombus and special right triangles
Explanation
In a rhombus, all sides are equal, and the diagonals bisect the angles. The 30∘−60∘−90∘ 30^\circ-60^\circ-90^\circ triangle has sides in a fixed ratio, which allows us to find the lengths of the segments and the ratio CFFD \frac{CF}{FD} .
若一个四边形的一条对角线把四边形分成两个等腰三角形, 我们把这条对角线叫这个四边形的和谐线, 这个四边形叫做和谐四边形. 如菱形就是和谐四边形. 已知四边形 ABCD\mathrm{ABCD} 中, AB=AD=BC=2,∠BAD=90∘,ACA B=A D=B C=2, \angle B A D=90^{\circ}, A C 是四边形 ABCDA B C D 的和谐线, 且 AC≠CDA C \neq C D, 求四边形 ABCDA B C D 的面积
Solution by Steps
step 1
Given that AB=AD=BC=2 AB = AD = BC = 2 and ∠BAD=90∘ \angle BAD = 90^\circ , we recognize that triangle ABD ABD is a right isosceles triangle
step 2
The area of triangle ABD ABD can be calculated using the formula for the area of a right triangle, which is 12×base×height \frac{1}{2} \times \text{base} \times \text{height}
step 3
Since AB=AD=2 AB = AD = 2 , the area of triangle ABD ABD is 12×2×2=2 \frac{1}{2} \times 2 \times 2 = 2
step 4
Triangle ABC ABC is also an isosceles triangle with AB=BC=2 AB = BC = 2 and AC AC as its base
step 5
To find the area of ABC ABC , we need the height h h from B B to AC AC . Using Pythagoras' theorem in triangle ABD ABD , we find BD=22+22=8 BD = \sqrt{2^2 + 2^2} = \sqrt{8}
step 6
Since AC AC is the perpendicular bisector of BD BD , BD BD is also the height of ABC ABC from B B to AC AC
step 7
The area of triangle ABC ABC is 12×AC×BD \frac{1}{2} \times AC \times BD . We know BD=8 BD = \sqrt{8} , but we need AC AC
step 8
Using the fact that ABD ABD is a right isosceles triangle, AC AC as the hypotenuse is AC=22+22=8 AC = \sqrt{2^2 + 2^2} = \sqrt{8}
step 9
Now we can calculate the area of ABC ABC as 12×8×8=12×8=4 \frac{1}{2} \times \sqrt{8} \times \sqrt{8} = \frac{1}{2} \times 8 = 4
step 10
The total area of quadrilateral ABCD ABCD is the sum of the areas of ABD ABD and ABC ABC , which is 2+4=6 2 + 4 = 6
Answer
The area of quadrilateral ABCD ABCD is 6.
Key Concept
Area of triangles and quadrilaterals
Explanation
The area of the quadrilateral is found by dividing it into two triangles and calculating the area of each triangle separately. The properties of isosceles and right triangles are used to determine the necessary lengths for the area calculations.
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