To learn this concept, first we should be aware of operation in polynomials.
Adding and subtracting polynomials is nothing but combining the like terms.
When we multiply two polynomials, we will follow the order given below.
(1) Signs (2) Number (3) Variable
Let us see how it works,
Multiply (5x2) and (-2x3)
= (5x2) x (-2x3)
= -10x2+3
= -10x5
Find the shaded area A in terms of x for :
Example 1 :
Solution :
Area of shaded region
= Area of large rectangle - Area of small rectangle
Area of rectangle = length ⋅ width
Large rectangle :
Length = 4x and width = 3x
Area of large rectangle = (4x ⋅ 3x)
= 12x2
Small rectangle :
Length = 2x and width = x
Area of small rectangle = (2x ⋅ x)
= 2x2 -------(2)
(1) - (2)
= 12x2-2x2
= 10x2
So, area of shaded region is 10x2 square meter.
Example 2 :
Solution :
Shaded region
= Area of rectangle - Area of triangle
Area of rectangle = length ⋅ width and
Area of triangle = (1/2) ⋅ base ⋅ height
Length = 3x and width = 2x
Area of rectangle = 3x(2x) ==> 6x2 ---(1)
Area of triangle = (1/2)⋅ x ⋅ x
= x2/2 ---(2)
(1) - (2)
Area of shaded region = 6x2 - (x2/2)
= (12x2 - x2)/2
= 11x2/2
So, area of shaded region is 11x2/2 square meter.
Example 3 :
Solution :
Area of shaded region
= Area of large rectangle - Area of small rectangle
Large rectangle :
length = 2x+5, width = 3x
Area of large rectangle = 3x(2x+5)
= 6x2 + 15x ----(1)
Small rectangle :
length = 2x, width = x
Area of small rectangle = 2x(x)
= 2x2 ----(2)
(1) - (2)
Area of shaded region = 6x2 + 15x - 2x2
Combining like terms, we get
= 4x2 + 15x
So, area of shaded region is 4x2 + 15x square meter.
Example 4 :
Solution :
Area of shaded region
= Area of rectangle - Area of square
Rectangle :
Length = x+6, width = x+2
Area of rectangle = (x+6)(x+2)
= x2+2x+6x+12
= x2+8x+12 ----(1)
Square :
Side length = x
Area of square = x2 ----(2)
(1) - (2)
Area of shaded region = x2+8x+12-x2
By combining like term, we get
= 8x+12
So, area of shaded region is 8x+12 square meter.
Example 5 :
Solution :
Area of rectangle = length ⋅ width
= (2x+3)(x+7)
= 2x(x) + 2x(7) + 3(x) + 3(7)
= 2x2 + 14x + 3x + 21
= 2x2 + 17x+ 21
So, area of rectangle is 2x2 + 17x+ 21 square meter.
Example 6 :
Solution :
Area of shaded region
= Area of rectangle + Area of square
Rectangle :
Length = 2x and width = x
Area of rectangle = 2x(x)
= 2x2
Square :
Length of square = x
Area of square = x(x)
= x2
Area of shaded region = 2x2 + x2
= 3x2
So, area of shaded region is 3x2 square meter.
Example 7 :
Solution :
Area of shaded region = Area of circle - Area of triangle
Area of circle = πr2
Area of triangle = (1/2) ⋅ (2r) ⋅ r ==> r2
Area of shaded region = πr2 - r2
= r2(π - 1)
So, area of shaded region is r2(π - 1) square meter.
Example 8 :
Solution :
Area of shaded region = Area of rectangle - Area of triangle
Area of circle = bh
Area of triangle = (1/2) ⋅ b ⋅ h ==> bh/2
Area of shaded region = bh - (bh/2)
= (bh/2)
So, area of shaded region is (bh/2) square meter.
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