Answer:
The other side =
units
Step-by-step explanation:
A right triangle has a side of length
units and;
the hypotenuse is 
According to Pythagorean theorem,
a² + b² = c² , where a and b are the two legs of a triangle and c is the hypotenuse.
Lets say a² =
and c² = (
b² = c² - a²
b² =
-
= 
So b =
=
units.
So the other side =
units.
Answer:
Option (B)
Step-by-step explanation:
From the figure attached,
ΔBCD and ΔACE are the similar triangles.
Therefore, by the property of similar triangles, corresponding sides of these similar triangles will be proportional.





AB + 6 = 24
AB = 18
Option (B) will be the answer.
im not sure how to explain it, but i know its b
You know you are finished dividing when you can no longer simply & when you plug it back in to check the answer
(You copied 'upper left' twice, and you left out 'lower right'.
But we know what you mean.)
If those are the corners of the wall, then they're ALL in the plane of
the wall (coplanar with it).