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: \(\displaystyle \Delta WPZ\,=\,\Delta XPY\)
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.