What is the area, in square feet, of the shaded part of the rectangle below?

i hope this is what you were looking for.

see attached graph

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.

step-by-step explanation:

at each of the parallel lines adjacent angles are supplementary. the angles have special names identifying their positions with respect to the parallel lines and transversal. they are corresponding angles, alternate interior angles, or alternate exterior angles. an angles is congruent to its matched angle.