Finding Ratio of area of small triangle to big triangle

[Guest]

Need help ASAP. No clue how to do this problem:

Many, Many thanks.

 

[soroban]

Hello, AirForceOne!

Very funny . . . we don't need any of those numbers!

I'll refer to the whole figure as \(\displaystyle W\) and to the shaded region as \(\displaystyle S\).


(a) Since the base is bisected, \(\displaystyle S\) has half the base of \(\displaystyle W\).
They both have the same height.

Area of \(\displaystyle W:\;\;\frac{1}{2}bh\)

Area of \(\displaystyle S:\;\;\frac{1}{2}\left(\frac{1}{2}b)h\:=\:\frac{1}{4}bh\)

The ratio is: \(\displaystyle \L\,\frac{S}{W}\:=\:\frac{\frac{1}{4}bh}{\frac{1}{2}bh}\:=\:\frac{1}{2}\)


(b) An angle bisector divides the opposite sides into segments proportional to the adjacent sides.

So the base is divided in the ratio 2:5.
The base of \(\displaystyle S\) is \(\displaystyle \frac{5}{7}\) of the base of \(\displaystyle W.\)

Since they have the same height:\(\displaystyle \L\:\frac{S}{W}\,=\,\frac{5}{7}\)


(c) I assume that the base of \(\displaystyle S\) is parallel to the base of \(\displaystyle W.\)

By similar triangles, the base of \(\displaystyle S\) is half the base of \(\displaystyle W\)
\(\displaystyle \;\;\)and the height of \(\displaystyle S\) is half the height of \(\displaystyle W.\)

Hence, the area of \(\displaystyle S\) is \(\displaystyle \frac{1}{4}\) the area of \(\displaystyle W.\)

Therefore: \(\displaystyle \L\,\frac{S}{W}\:=\:\frac{1}{4}\)

 

[Guest]

Thanks a lot! I think I got it!

Thanks!