Test on 4/14. Geometry. Picture included Please help ASAP please

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April 13th 2006, 04:49 PM
AirForceOne

Test on 4/14. Geometry. Picture included Please help ASAP please

http://img.photobucket.com/albums/v2...e/441fd7bf.gif

The ones I know are know are correct but no well understood:

a) 1:1 because both triangles have the same base? So 16:16 = 1:1 right?
b) same method as ^
d) 9:16 because using the similar triangles area ratio formula. The method of similar trianges i used was AA because of the parallel lines right?

The ones I don't understand at all but I know the answers. I'm all confused:

c) The answer is 1:1. I think you can assume that the height of both triangles are the same? How can it be 1:1 if there are no other measurements of the triangles?
e) The answer is 3:4. I figured maybe you do simplfy 12/16 to get 3:4 but I have no clue why because the the triangles that the question is asking for doesn't have 12 or 16?

Im so confused. please help ASAP. Thanks... :cry:
April 13th 2006, 09:21 PM
earboth

1 Attachment(s)

Quote:

Originally Posted by AirForceOne

...
The ones I don't understand at all but I know the answers. I'm all confused:
c) The answer is 1:1. I think you can assume that the height of both triangles are the same? How can it be 1:1 if there are no other measurements of the triangles?
e) The answer is 3:4. I figured maybe you do simplfy 12/16 to get 3:4 but I have no clue why because the the triangles that the question is asking for doesn't have 12 or 16?
Im so confused. please help ASAP. Thanks... :cry:

Hello,

let H be the height of the trapezoid and h the height of triangle(ZYP) (I#ve attached a diagram to demonstrate, what I'll calculate).
Then you get the proportion:
$\frac{h}{16}=\frac{H-h}{12}$ . Solve for h and you'llget h = 4/7*H.

to e.: You get the area of triangle(XPY) by:
$A_{\Delta XPY}=A_{\Delta WXY}-A_{\Delta WXP}$

$\frac{1}{2} \cdot 12 \cdot H-\frac{1}{2} \cdot 12 \cdot \frac{3}{7} \cdot H=\frac{1}{2} \cdot 12 \cdot H \cdot \frac{4}{7}$
That means:
$\frac{A_{\Delta XPY}}{A_{\Delta WXP}}=\frac{\frac{1}{2} \cdot 12 \cdot H \cdot \frac{4}{7}}{\frac{1}{2} \cdot 12 \cdot \frac{3}{7} \cdot H}=\frac{4}{3}$

to d.: As I've shown above you can calculate the areas of the triangles in question by calculating the differences of two triangles.
Unfortunately I'm a little bit in a hurry to complete the problem, but I'm certain that you now know how to handle the problem.

Greetings and Happy Easter to you.

EB
April 13th 2006, 10:19 PM
earboth

Quote:

Originally Posted by AirForceOne

...
The ones I don't understand at all but I know the answers. I'm all confused:
c) The answer is 1:1. I think you can assume that the height of both triangles are the same? How can it be 1:1 if there are no other measurements of the triangles?
...

Hello,

to c.: as you've demonstrated:
$A_{\Delta ZYW}=A_{\Delta ZYX}$ (same base, same height). Thus

$A_{\Delta ZYW}-A_{\Delta ZYP}=A_{\Delta ZYX}-A_{\Delta ZYP}$

$A_{\Delta WZP}=A_{\Delta XYP}$ . Thus the ratio is 1:1.

Greetings

EB