Question

In rectangle ABCD, point E lies half way between sides AB and CD and halfway between sides AD and BC. If AB=10 and BC=2, what is the area of the shaded region?

Answer 1

Answer:

Area of shaded region is $10 \ units^2$.

Step-by-step explanation:

We are attaching the diagram for your reference.

Given:

AB = 10

BC = 2

We need to find the Area of the shaded region.

Area of the shaded region is equal to sum of  Area of triangle ABE and Area of triangle CDE.

Now first we will find area of both the triangles.

Now we know that Area of triangle is half times base times height.

In Δ ABE,

base AB = 10 (given)

Since given point E lies half way between sides AB and CD and halfway between sides AD and BC.

So height h is,

height h = $\frac{BC}{2} = \frac{2}{2} = 1$

Area of Δ ABE = $\frac{1}{2} \times base \times height = \frac{1}{2}\times 10\times 1 = 5\ units^2$

In Δ CDE,

base CD = AB = 10 (given it is a rectangle and opposite side of rectangle are equal)

Since given point E lies half way between sides AB and CD and halfway between sides AD and BC.

So height h is,

height h = $\frac{BC}{2} = \frac{2}{2} = 1$

Area of Δ CDE = $\frac{1}{2} \times base \times height = \frac{1}{2}\times 10\times 1 = 5\ units^2$

Now Area of shaded region = Area of Δ ABE + Area of Δ CDE = $5\ units^2+ 5\ units^2 = 10\ units^2$

Hence Area of shaded region is $10 \ units^2$.