Calculate Rectangle Perimeter: Triangle Perimeter = 20 with Side X+2

To solve this problem, we need to find the perimeter of rectangle ABCD given information about triangle BCD.

Step 1: Identify the given information
From the diagram, we can see:

• Rectangle ABCD with diagonal BD drawn
• The diagonal BD = 10
• The bottom side DC = 6
• The right side BC = x + 2
• The perimeter of triangle BCD = 20

Step 2: Find the value of x using the perimeter of triangle BCD
Triangle BCD has three sides: BD, DC, and BC.
The perimeter of triangle BCD is the sum of these three sides:
$BD + DC + BC = 20$
$10 + 6 + (x + 2) = 20$
$18 + x = 20$
$x = 2$

Step 3: Verify using Pythagoras' theorem
Since ABCD is a rectangle, angle BCD is a right angle. Triangle BCD is a right triangle with the right angle at C.
By Pythagoras' theorem: $DC^2 + BC^2 = BD^2$
With $x = 2$, we have $BC = x + 2 = 4$
Check: $6^2 + 4^2 = 36 + 16 = 52$
And $10^2 = 100$
Wait, this doesn't match. Let me reconsider the labeling.

Actually, looking more carefully: if BC = x + 2 and DC = 6, and BD = 10, then:
$6^2 + (x+2)^2 = 10^2$
$36 + (x+2)^2 = 100$
$(x+2)^2 = 64$
$x + 2 = 8$
$x = 6$

But let's verify with the perimeter condition:
Perimeter of triangle BCD = $10 + 6 + 8 = 24$

This doesn't equal 20. Let me reconsider: if the perimeter is 20, then:
$10 + 6 + BC = 20$
$BC = 4$

So if BC = 4, then $x + 2 = 4$, giving $x = 2$.

Step 4: Calculate the perimeter of rectangle ABCD
In rectangle ABCD:

• Length AB = DC = 6
• Width BC = AD = 4

Therefore, the perimeter of rectangle ABCD is 20.