What are the exact solutions of x2 – 5x – 7 = 0?
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The distance between two points (x₁,y₁),(x₂,y₂) = √[(x₂-x₁)² + (y₂-y₁)²]
let the length of the square RSTV = d
So, <span>the perimeter of RSTV = 4 d
d is the distance RS or ST or TV or VR
RS distance between R,S
ST </span><span>distance between S,T
TV </span><span>distance between T,V
VR </span>distance between V,R
<span>R(-1,5), S(-3,1), T(-7,3), and V(-5,7)
RS = S - R = </span><span>√[(-3 - -1)² + (1- 5)²] = √[(-2)² +(-4)² ]= √(4+16) = √20 ≈ 4.472
∴ </span><span><span>The perimeter of RSTV = 4 d</span> = 4 * </span><span>4.472 = 17.888
The attached figure represents an explanation of the problem.
we should note that:
</span><span>RS = ST = TV = VR = d </span><span><span>≈ 4.472</span> </span>
let the length of the square RSTV = d
So, <span>the perimeter of RSTV = 4 d
d is the distance RS or ST or TV or VR
RS distance between R,S
ST </span><span>distance between S,T
TV </span><span>distance between T,V
VR </span>distance between V,R
<span>R(-1,5), S(-3,1), T(-7,3), and V(-5,7)
RS = S - R = </span><span>√[(-3 - -1)² + (1- 5)²] = √[(-2)² +(-4)² ]= √(4+16) = √20 ≈ 4.472
∴ </span><span><span>The perimeter of RSTV = 4 d</span> = 4 * </span><span>4.472 = 17.888
The attached figure represents an explanation of the problem.
we should note that:
</span><span>RS = ST = TV = VR = d </span><span><span>≈ 4.472</span> </span>