Example 1 :
Find, in simplest form, an expression for the perimeter P of:
Solution :
The given triangle is a isosceles triangle. Side lengths of the given triangle are x + 3, x + 3 and x.
Perimeter of the triangle given = x + 3 + x + 3 + x
= 3x + 6
So, perimeter of the triangle is 3x + 6.
Example 2 :
Solution :
The given figure is a rectangle.
Perimeter of rectangle = 2(length + width)
Length = 3x - 4 and width = x + 1
= 2(3x - 4 + x + 1)
= 2(4x - 3)
= 8x - 6
So, the required perimeter is 8x - 6.
Example 3 :
Solution :
Perimeter of the shape = 2(2x + 5) + 3(x - 2)
= 4x + 10 + 3x - 6
= 7x + 4
Example 4 :
Find, in simplest form, an expression for the area A of :
Solution :
The given shape is rectangle.
Area of rectangle = length x width
length = x + 1 and width = 3x
= (x + 1) ⋅ 3x
= 3x^{2 }+ 3x
Example 5 :
Solution :
Area of triangle = (1/2) ⋅ base ⋅ height
= 1/2 ⋅ (x + 4) ⋅ 3
= (3/2)(x + 4)
Example 6 :
Solution :
Area of given shape = Area of square + Area of triangle
Side length of square = x + 2, base of triangle = x + 2 and height of the triangle = x.
Area of square = (x + 2)(x + 2)
= x^{2 }+ 4x + 4
Area of triangle = (1/2) ⋅ base ⋅ height
= (1/2) ⋅ (x + 2) ⋅ x
= x(x + 2)/2
Area of the shape = (x^{2 }+ 4x + 4) + x(x + 2)/2
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