Need help ASAP test on 4/14 please help fast!!

[Guest]

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...

 

[soroban]

Hello, AirForceOne!

a) 1:1 because both triangles have the same base . . . and the same height
So 16:16 = 1:1, right? . . . right!


b) same method as ^ . . . yes!

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? . . . right!

c) The answer is 1:1.

In (a) we established that $\displaystyle \,\Delta WYZ\,=\,\Delta XYZ$
Subtract the common area: $\displaystyle \,\Delta PYZ$
Therefore: $\displaystyle \Delta WPZ\,=\,\Delta XPY$

e) The answer is 3:4.

I agree with their answer, but my method is long and messy.

Let $\displaystyle h$ = height of the trapezoid.

The area of the entire trapezoid is: $\displaystyle \,A\:=\:\frac{h}{2}(12\,+\,16)\:=\:14h$

Now consider the top and bottom triangles: $\displaystyle \,\Delta WPX$ and $\displaystyle \Delta ZPY$
They are similar and have their sides in the ratio 3:4.
$\displaystyle \;\;$Hence, their altitudes are in the ratio 3:4.

The altitude of $\displaystyle \Delta WPX$ is $\displaystyle \frac{3}{7}h$.
The altitude of $\displaystyle \Delta ZPY$ is $\displaystyle \frac{4}{7}h$

The area of $\displaystyle \Delta WPX\:=\:\frac{1}{2}(12)\left(\frac{3}{7}h\right)\:=\:\frac{18}{7}h$
The area of $\displaystyle \Delta ZPY\:=\:\frac{1}{2}(16)\left(\frac{4}{7}h\right)\:=\:\frac{32}{7}h$

This leaves: $\displaystyle \,14h\,-\,\frac{18}{7}h\,-\,\frac{32}{7}h\:=\:\frac{48}{7}h\,$ to be shared by $\displaystyle \Delta WPZ$ and $\displaystyle \Delta XPY.$
$\displaystyle \;\;$Since those triangles are equal, each has an area of $\displaystyle \frac{24}{7}h$

Therefore: \(\displaystyle \L\,\frac{\Delta WPX}{\Delta XPY}\:=\:\frac{\frac{18}{7}h}{\frac{24}{7}h}\:=\:\frac{3}{4}\)

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It just occured to me . . .

The method I used for part (e) is quite clumsy and primitive.

But if we did this at the very beginning of the problem,
$\displaystyle \;\;$the five answers wouldn't have taken so much thought, right?

So you might keep this approach in the second drawer of your arsenal.