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Compulsory Education Curriculum Standard Experimental Textbook Mathematics Grade 8 Volume II P 1 10 Answer to Questions 10
1. It is an isosceles trapezoid. It is proved that AE = AB, CE = CB because triangle AEC is equal to triangle ABC, AD = BC, CD = AB because quadrilateral ABCD is rectangular, AE = CD, CE = AD, so triangle ADC is equal to triangle CEA. After a point d, e is DG, vertical AC is g, g, EH is vertical AC is H.

AC times DG, s triangle AEC= 1/2.

AC times EH, so DG=EH. Because DG is parallel to EH, the quadrilateral DGHE is a parallelogram, because DE is parallel to AC and DE is not equal to AC, so the quadrilateral ADEC is a trapezoid, and because AD=CE, the quadrilateral AC and DE are isosceles trapezoid.

2. In RT triangle ACD, AC=5, because S triangle ADC= 1/2.

Advertising time CD= 1/2

AC times DG, that is, 1/2 times 3 times 4= 1/2 times 5 times DG, so DG= 12/5. In the RT triangle ADG, AG=9/5, so CH=AG=9/5, so GH=AC-AG-CH=5.

-

9/5

-

9/5

=

7/5, because DEHG is a parallelogram, so DE=GH=7/5, so the perimeter of trapezoidal ACED is AD+AC+EC+DE=3+5+3+.

7/5

=

62/5, so S trapezoid ACED= 1/2(DE+AC) times DG= 1/2.

Multiply (7/5+5) by 12/5.

= 192/25

I am exhausted.