Triangle JKL has an area of in2 and a perimeter of in.
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Answer:
√2:1
Step-by-step explanation:
First we need to know that the length of the side of the square is equal to the diameter of the inscribed circle i.e
L = di
Given the area of the square to be 9in², we can get the length of the square.
Area of a square = L²
L is the length of the square.
9 = L²
L = √9
L = 3in
Hence the length of one side of the square is 3in
This means that the diameter of the inscribed circle di is also 3in.
Circumference of a circle = π×diameter of the circle(di)
Circumference of inscribed circle = π×3
= 3π in
For the circumscribed circumscribed circle, diameter of the outer circle will be equivalent to the diagonal of the square.
To get the diagonal d0, we will apply the Pythagoras theorem.
d0² = L²+L²
d0² = 3²+3²
d0² = 9+9
d0² = 18
d0 = √18
d0 = √9×√2
d0 = 3√2 in
Hence the diameter of the circumscribed circle (d0) is 3√2 in
Circumference of the circumscribed circle = πd0
= π(3√2)
= 3√2 π in
Hence, ratio of the circumference of the circumscribed circle to the one of the inscribed will be 3√2 π/3π = √2:1
