F.5 Maths geometry(2)

YTC

2015-04-07 11:49 pm

Question: http://postimg.org/image/j0yfwou1l/

My calculation:
http://postimg.org/image/4vsmuvl09/
http://postimg.org/image/gicqw08bd/
http://postimg.org/image/42g129ezd/
which is mainly use the parallel lines, property of isosceles triangle and tan.

I think my calculation is too time-consuming.
Is there any better methods which do not use too many steps?
Thank you !!

回答 (1)

用戶10012

2015-04-08 1:46 am

✔ 最佳答案

Triangle A1D1E1 is congruent to triangle B1C1E1 (RHS), therefore E1 is the mid - point of A1B1. A1E1 = 0.5 cm. A1D1 = 1 cm.
A1D1 is parallel to A2D2 (both are side of a square), so triangle A1D1E1 is similar to triangle A2D1D2.
That is A1E1/A1D1 = D1D2/A2D2
Let A2D2 = x = side of square A2B2C2D2.
so 0.5/1 = D1D2/x
D1D2 = 0.5x
D1C1 = D1D2 + D2D2 + C2C1 = 0.5x + x + 0.5x = 1
2x = 1
x = 1/2 = length of A2B2.

2015-04-07 17:50:46 補充：
Using the same method, A3B3 = (1/2)(1/2) = 1/4 cm.

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