Another Area ratio problem...arggg!!!

[Guest]

Can seem to do these...
I am completely frustrated! :x
Please help ASAP, test on friday.

Thanks!

 

[soroban]

Hello, AirForceOne!

Here are a few of them . . .


29) Triangle

[I suspect that the "8" is incorrect . . . or maybe some other part.]

Triangles $\displaystyle ADB$ and $\displaystyle ABC$ have two corresponding angles equal,
$\displaystyle \;\;$hence they are similar and their corresponding sides are proportional.

We have: $\displaystyle \,\frac{AD}{AB}\,=\,\frac{4}{7}$ . . . The ratio of their sides is $\displaystyle \,4:7$

The ratio of their areas is: \(\displaystyle \L\,\frac{A_1}{\Delta ABC}\:=\:\frac{4^2}{7^2}\:=\:\frac{16}{49}\)

Then: \(\displaystyle \L\,\frac{A_2}{\Delta ABC}\:=\:\frac{33}{49}\)

Therefore: \(\displaystyle \L\,\frac{A_1}{A_2}\:=\:\frac{16}{33}\)


29. Trapezoid

Triangles 1 and 2 are similar.

On diagonal $\displaystyle TR$, the two segments are 5 and 6.
These are a pair of corresponding sides of the two triangles.
Hence, the ratio of the sides of the two triangles is $\displaystyle \,5:6$

Therefore: \(\displaystyle \L\,\frac{A_1}{A_2}\:=\:\frac{5^2}{6^2}\:=\:\frac{25}{36}\)


30, 31. $\displaystyle \;$ There are no measurements?

 

[Guest]

Thanks a lot!

My teacher announced today that she forgot to put measurements on problems 30 and 31. Sorry the inconvience!!

Thanks again!!