The logo shown is symmetrical about one of its diagonals. Enter the angle measures in the green triangle,to the nearest degree. (Hint: First find an angle in a blue triangle.) T…

Question

asked Jul 23, 2023 67.8k views

1 Answer

Tim Perry · Answer 1 · 2023-07-29T19:02:33+0000

Solution

For this case we can do the following:

We can find the lenght for AC:

$AC=\sqrt[]{2^2+4^2}=\sqrt[]{20}$

And similarly for EC we have:

$EC=\sqrt[]{2^2+2^2}=\sqrt[]{8}$

And finally Ae like this:

$AE=\sqrt[]{4^2+2^2}=\sqrt[]{20}$

then we can conclude that:

< CAE= 36.9º

< AEC= 71.6º

< ACE= 71.6º

Then for the area we can do the following:

$A=16mm^2-(4\cdot2)/(2)-(2\cdot2)/(2)-(4\cdot2)/(2)=16-4-2-4=6\operatorname{mm}^2$