In the diagram all measurements are in centimetres.
The perimeter of the quadrilateral is twice the perimeter of the triangle.
Work out the perimeter of the quadrilateral.

Perimeter of the quadrilateral = 48 cm

Solution:

The dimensions of the quadrilateral are 2x, 2x – 1, x – 3 and 3x.

The dimensions of the triangle are x + 3, 5 and x + 3.

Perimeter of the polygon = Sum of the sides of the polygon

The perimeter of the quadrilateral  = 2 × Perimeter of the triangle

2x + 2x – 1 + x – 3 + 3x = 2 × (x + 3 + 5 + x + 3)

8x – 4 = 2 × (2x + 11)

8x – 4 = 4x + 22

Add 4 on both sides of the equation.

8x = 4x + 26

Subtract 4x on both sides of the equation.

4x = 26

Divide by 4 on both sides of the equation.

x = 6.5

Perimeter of the quadrilateral  = 2 × (x + 3 + 5 + x + 3)

Perimeter of the quadrilateral = 2 × (6.5 + 3 + 5 + 6.5 + 3)

                                                  = 2 × (9.5 + 5 + 9.5)

                                                  = 2 × 24

                                                  = 48

Perimeter of the quadrilateral = 48 cm

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

we now create space between the x and y terms with parenthesis.

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

we add the squares to both sides.

simplify.

and write the quadratics in factored form.

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.