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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?

User Frederickf
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4 votes

the Answer should be :

328

User Ernestocattaneo
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2 votes

Answer:

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

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.

User William Holroyd
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