Geometry

vampirewitchreine

Junior Member
Joined
Aug 2, 2011
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82
(Teal indicates the measurements given by my book, blue is measurements that I figured out)
14-20.jpg
In the diagram, BDGF is a rectangle and FDHJ is a kite.



(I apologize for posting a list, but it makes more sense with my measurements if I post all of the questions)

14. FD=10 (FE+ED=FD)
15. FG=??? ​(This is where I need help)
16. BE=5 (BE is congruent to BA)
17. ED=5 (ED is congruent to BE)
18. GH=8 (GH2=DH2-DG2; GH2=102-62; GH2=100-36; \(\displaystyle \sqrt{GH^2}=\sqrt{64}\))
19. JH=\(\displaystyle \sqrt{73}~\approx 8.5\) (JH2=JG2+GH2; JH2=32+82;GH2=9+64; \(\displaystyle \sqrt{GH^2}=\sqrt{73}\))
 
Hello, vampirewitchreine!

Teal indicates the measurements given by my book; blue is measurements that I figured out.
View attachment 1555
In the diagram, BDGF is a rectangle and FDHJ is a kite.


\(\displaystyle 14.\;FD \,=\,10\)

\(\displaystyle 15.\;FG \,=\,? \quad\text{(This is where I need help)}\)

\(\displaystyle 16.\;BE \,=\, 5\)

\(\displaystyle 17.\;ED \,=\, 5\)

\(\displaystyle 18.\;GH \,=\,8\quad (GH^2\:=\:DH^2-DG^2 \:=\:10^2-6^2 \:=\:64 \quad\Rightarrow\quad GH = 8)\)

\(\displaystyle 19.\;JH \,=\,\sqrt{73} \quad (JH^2 \:=\:JG^2 + GH^2 \:=\:3^2+8^2 \:=\:73)\)

Your answers are all correct . . . Good work!
. . And a big thank you for showing your work!


I disagree with some of your reasons.

\(\displaystyle 14.\;FD = 10\) . . . How do you know that \(\displaystyle FE = ED = 5\,?\)
Since \(\displaystyle FDHJ\) is a kite, \(\displaystyle FD = DH = 10.\)

\(\displaystyle 16.\;BE = 5\) . . . How do you know that \(\displaystyle BE = AB\,?\)
We know that:.\(\displaystyle BG = FD = 10\) . . . The diagonals of a rectangle are equal.
And that:.\(\displaystyle BE = ED = FE = EG = 5\) . . . The diagonals of a rectangle bisect each other.

\(\displaystyle 17.\;ED = 5\) . . . explained in 16.


In 18, you found that:.\(\displaystyle GH = 8\)

Since \(\displaystyle FDHJ\) is a kite, \(\displaystyle FG = GH=8.\)
 
Good work, Hazel.

FG = sqrt(FD^2 - DG^2) ; how did you miss that ;)

Because it was completely obvious and I overlooked it

Hello, vampirewitchreine!


Your answers are all correct . . . Good work!
. . And a big thank you for showing your work!


I disagree with some of your reasons.

\(\displaystyle 14.\;FD = 10\) . . . How do you know that \(\displaystyle FE = ED = 5\,?\)
Since \(\displaystyle FDHJ\) is a kite, \(\displaystyle FD = DH = 10.\)

\(\displaystyle 16.\;BE = 5\) . . . How do you know that \(\displaystyle BE = AB\,?\)
We know that:.\(\displaystyle BG = FD = 10\) . . . The diagonals of a rectangle are equal.
And that:.\(\displaystyle BE = ED = FE = EG = 5\) . . . The diagonals of a rectangle bisect each other.

\(\displaystyle 17.\;ED = 5\) . . . explained in 16.


In 18, you found that:.\(\displaystyle GH = 8\)

Since \(\displaystyle FDHJ\) is a kite, \(\displaystyle FG = GH=8.\)

Thank you for the corrections of my reasons. I think what made me use the reasons that I had used was because of question 20 (which I didn't post). It was: What type of parallelogram is ADEB? Explain.

I answered: ADEB is a square. All sides are congruent and the diagonals are congruent to themselves, meaning that the angles are also congruent.
 
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