Question

n rectangle ABCD, point E lies half way between sides AB and CD and halfway between sides AD and BC. If AB=11 and BC=2, what is the area of the shaded region? Write your answer as a decimal, if necessary.

Answer 1

Answer:

Step-by-step explanation:

Hello!

For the rectangle ABCD

AB= DC= 11

BC= AD= 2

Point E lies halfway between AB and CD

The shaded are forms two triangles, I'll refer to the upper triangle as "Triangle one" and the lower triangle will be "triangle 2"

The area of a triangle is calculated as

$a= \frac{bh}{2}$

b= base

h= height

Triangle 1

b₁= AB= 11

$h_1= \frac{BC}{2}= \frac{2}{2}= 1$

$a_1= \frac{b_1h_1}{2}= \frac{11*1}{2}= 5.5$

Triangle 2

b₂= DC= 11

$h_2= \frac{BC}{2}= \frac{2}{2} = 1$

$a_2= \frac{b_2h_2}{2}= \frac{11*1}{2}= 5.5$

Now you add the areas of both triangles to get the area of the shaded region:

a₁ + a₂= 5.5 + 5.5= 11

Since point E is halfway to all sides of the rectangle, even tough it doesn't see so, the shaded area is equal to half the area of the rectangle:

area= bh= DC*AD= 11*2= 22

area/2= 22/12= 11

I hope this helps!