Respuesta :

GE=42  divide that by 2 to get GH=21, HE=21

DH=16  that is half of line DF, so HF=16 too

so GF would be 21+16=37



EF=13   DF=18 divide 18 by 2 and get DH=9 and HF=9

since HF=9  do 13-9=4 so EH=4

DeanR

6. That's a rhombus, so the meet of the diagonals is their midpoint, the diagonals are perpendicular and the sides are congruent.  So GH=GE/2 and DH are the legs of a right triangle with hypotenuse GD=GF.  So

[tex]GF = \sqrt{GH^2 + DH^2} = \sqrt{(42/2)^2 + (16)^2} = \sqrt{697} [/tex]

Answer: √697

7. Same story, this time we have hypotenuse EF=13, one leg FH=DF/2=9 so other leg

[tex]EH = \sqrt{EF^2-FH^2} = \sqrt{13^2 - 9^2} = \sqrt{88} = 2 \sqrt{22}[/tex]

Answer: 2√22


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