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The circumference of rectangle ABCD is 16cm. On each side of rectangle ABCD, draw a square with that side as the length. It is known that the sum of the four squares is 68cm2 Finding the area of rectangle ABCD
16/2=8,68/2=34,8*8=64,64-34=30,30/2=15
*It's easy to understand when drawing
As shown in the figure, in ladder ABCD, ad ‖ BC, point E is on BC, AE = be, point F is the midpoint of CD, and AF ⊥ ab. if ad = 2.7, AF = 4, ab = 6, then the length of CE is ()
A. 22B. 23−1C. 2.5D. 2.3
Extend the intersection of AF and BC at g. ∵ ad ∥ BC, ∵ d = ∥ FCG, ∥ DAF = ∥ g. DF = CF, ≌ AFD ≌ GFC. ∥ Ag = 2AF = 8, CG = ad = 2.7. ∵ AF ⊥ AB, ab = 6, ∥ BG = 10. ∥ BC = bg-cg = 7.3. ∥ AE = be, ∥ BAE = ∥ B. ≌ EAG = ∥ age. ∥ AE = Ge. ∥ be = 12bg = 5
As shown in the figure, the quadrilateral ABCD is inscribed in ⊙ o, BD is the diameter of ⊙ o, AE ⊥ CD, the perpendicular foot is e, Da bisects ⊙ BDE. (1) prove that AE is the tangent of ⊙ o; (2) if ⊥ DBC = 30 ° de = 1cm, find the length of BD
(1) It is proved that connecting OA, ∵ Da bisection ∵ BDE, ∵ BDA = ∵ EDA. ∵ OA = OD, ∵ ODA = ∵ oad, ∵ oad = ∵ EDA, ∵ OA ∥ CE. ∵ AE ⊥ CE, ∵ AE ⊥ OA. ∵ AE are tangent lines of ⊙ O. (2) ∵ BD is the diameter, ∵ BCD = ∵ bad = 90 °, ∵ DBC = 30 °, ∵ BDC = 60 °, ∵ BDE = 120 °. ∵ Da bisection ∵ BDE, ∵ BDA = ∵ EDA = 60 °. ∵ abd = ∵ ead = 30 °. ∵ in RT △ AED, ∵ AED = 90 °, ∵ ead = 30 °, ∵ ad = 2DE. ∵ in RT △ abd, ∵ bad = 90 °, ∵ abd = 30 °, ∵ BD = 2ad = 4; De
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