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Sagot :
Side of circumscribed square (bigger square) = 4 cm
Radius of the inscribed circle is equal to side of bigger square = 4 cm
Diagonal of the inscribed square in circle is equal to the radius of the circle=4 cm
To compute for the area of the inscribed square, find its side using the Pythagorean Theorem (because the the diagonal of the square divides the square into two congruent right triangles)
Let x be the length of the shorter legs which are congruent.
Hypotenuse or the diagonal = 4 cm
(4 cm)² = x² + x²
16 cm² = 2x²
[tex] \sqrt{16} = \sqrt{2x ^{2} } [/tex]
[tex]4cm = x \sqrt{2} [/tex]
[tex] \frac{4}{ \sqrt{2} } = \frac{x \sqrt{2} }{ \sqrt{2} } [/tex]
[tex]x = 2 \sqrt{2} [/tex]
Area of inscribed square (the smaller square):
[tex]A = (2 \sqrt{2} ) ^{2} [/tex]
Area = 4 (2)
Area = 8 cm²
Radius of the inscribed circle is equal to side of bigger square = 4 cm
Diagonal of the inscribed square in circle is equal to the radius of the circle=4 cm
To compute for the area of the inscribed square, find its side using the Pythagorean Theorem (because the the diagonal of the square divides the square into two congruent right triangles)
Let x be the length of the shorter legs which are congruent.
Hypotenuse or the diagonal = 4 cm
(4 cm)² = x² + x²
16 cm² = 2x²
[tex] \sqrt{16} = \sqrt{2x ^{2} } [/tex]
[tex]4cm = x \sqrt{2} [/tex]
[tex] \frac{4}{ \sqrt{2} } = \frac{x \sqrt{2} }{ \sqrt{2} } [/tex]
[tex]x = 2 \sqrt{2} [/tex]
Area of inscribed square (the smaller square):
[tex]A = (2 \sqrt{2} ) ^{2} [/tex]
Area = 4 (2)
Area = 8 cm²
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