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!