1 thought on “What is the area of largest square that can be inscribed in a circle of radius 6cm.”
A square that perfectly fits inside a circle is inscribed in the circle. The square’s corners will touch, but not intersect, the circle’s boundary, and the square’s diagonal will equal the circle’s diameter. Also, as is true of any square’s diagonal, it will equal the hypotenuse of a 45°-45°-90° triangle.
hence diameter of circle = diaglnal of square
2 × r =
[tex]2 \times r = \sqrt{2} \times \: side \: of \: square \\ side \: of \: square = \frac{2 \times 6}{ \sqrt{2} } \\ area \: of \: square = ({side})^{2} \\ area \: of \: square = {( \frac{2 \times 6}{ \sqrt{2} } )}^{2} \\ area \: of \: square = \frac{4 \times 36}{2} \\ area \: of \: square =64 \: {cm}^{2} [/tex]
A square that perfectly fits inside a circle is inscribed in the circle. The square’s corners will touch, but not intersect, the circle’s boundary, and the square’s diagonal will equal the circle’s diameter. Also, as is true of any square’s diagonal, it will equal the hypotenuse of a 45°-45°-90° triangle.
hence diameter of circle = diaglnal of square
2 × r =
[tex]2 \times r = \sqrt{2} \times \: side \: of \: square \\ side \: of \: square = \frac{2 \times 6}{ \sqrt{2} } \\ area \: of \: square = ({side})^{2} \\ area \: of \: square = {( \frac{2 \times 6}{ \sqrt{2} } )}^{2} \\ area \: of \: square = \frac{4 \times 36}{2} \\ area \: of \: square =64 \: {cm}^{2} [/tex]