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Sagot :
Hello,
Rectangle A: 6cm (x+2)cm
Rectangle B: 3cm (2x+1)cm
The formula for the perimeter of a rectangle is: P=2(base+height)
Then:
[tex]P_A=2(b+h) \\ P_A=2*[6+(x+2)] \\ P_A=2*(x+8) \\ P_A=2x+16 \\ \\ \\ P_B=2(b+h) \\ P_B=2*[3+(2x+1)] \\ P_B=2*(2x+4) \\ P_B=4x+8 [/tex]
But we know that the perimeters are the same, so:
[tex]P_A=P_B \\ 2x+16=4x+8 \\ 8=2x \\ \boxed{x=4} \\ \\ Replacing: \\ \\ Lenght\,\,of\,\,A=(x+2)cm\\ Lenght\,\,of\,\,A=(4+2)cm\\ \boxed{Lenght\,\,of\,\,A=6cm}\\ \\ Lenght\,\,of\,\,B=(2x+1)cm\\ Lenght\,\,of\,\,B=(2*4+1)cm\\ \boxed{Lenght\,\,of\,\,B=9cm}[/tex]
With the answer of A, we realize that this figure is actually a square.
Rectangle A: 6cm (x+2)cm
Rectangle B: 3cm (2x+1)cm
The formula for the perimeter of a rectangle is: P=2(base+height)
Then:
[tex]P_A=2(b+h) \\ P_A=2*[6+(x+2)] \\ P_A=2*(x+8) \\ P_A=2x+16 \\ \\ \\ P_B=2(b+h) \\ P_B=2*[3+(2x+1)] \\ P_B=2*(2x+4) \\ P_B=4x+8 [/tex]
But we know that the perimeters are the same, so:
[tex]P_A=P_B \\ 2x+16=4x+8 \\ 8=2x \\ \boxed{x=4} \\ \\ Replacing: \\ \\ Lenght\,\,of\,\,A=(x+2)cm\\ Lenght\,\,of\,\,A=(4+2)cm\\ \boxed{Lenght\,\,of\,\,A=6cm}\\ \\ Lenght\,\,of\,\,B=(2x+1)cm\\ Lenght\,\,of\,\,B=(2*4+1)cm\\ \boxed{Lenght\,\,of\,\,B=9cm}[/tex]
With the answer of A, we realize that this figure is actually a square.
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