Question

A polygon with an area of (4x2 + 2x - 16) square units is combined with a rectangle that has a width of (6x + 2) units and a length of (x - 5) units. Then, a polygon with an area of (2x2 - 30x - 10) square units is removed. What is the area of the final polygon?

Answer 1

Answer:

The area of new polygon is $22 x^2 + 4 x - 16$.

Step-by-step explanation:

The area of square = $4x^2 + 2x - 16$

Width of the rectangle = 6x+2

Length of the rectangle = x-5

Area of the rectangle = $(6x+2)(x-5)=6 x^2 - 28 x - 10$

Then we combined square and rectangle. The area of combined figure is the sum of area of square and rectangle.

$\text{Combined area}=(4x^2 + 2x - 16)+(6 x^2 - 28 x - 10)$

On combining like terms we get

$\text{Combined area}=(4x^2+6x^2)+ (2x-28x)+ (- 16- 10)$

$\text{Combined area}=10x^2-26x-26$

Then, a polygon with an area of $2x^2 - 30x - 10$ square units is removed. So, new area of the polygon is

$Area=(10x^2-26x-26)-(2x^2 - 30x - 10)$

$Area=(10x^202x^2)+(-26x+30x)+(-26+10)$

$Area=8 x^2 + 4 x - 16$

Therefore the area of new polygon is $22 x^2 + 4 x - 16$.

Answer 2

the Answer should be :

328