Silindir, Koni ve Küre Soru Çözümü

When rotated around the leg of length 4 cm, this leg becomes the height $h = 4$ cm, and the other leg becomes the radius $r = 3$ cm. The volume of a cone is given by $$V = \frac{1}{3} \pi r^2 h.$$ Substituting, $$V = \frac{1}{3} \pi (3)^2 (4) = \frac{1}{3} \pi \cdot 9 \cdot 4 = 12\pi \text{ cm}^3.$$

Kürenin yarıçapı $R=3$ cm. En büyük hacimli koni için yükseklik $h = \frac{4R}{3} = \frac{4 \times 3}{3} = 4$ cm. Taban yarıçapı $r$ için: $r^2 = 2Rh - h^2 = 2 \times 3 \times 4 - 4^2 = 24 - 16 = 8$, so $r = \sqrt{8} = 2\sqrt{2}$ cm. Koninin hacmi: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3 \times 8 \times 4 = 32$ cm³. (π=3 alındı)

First, find the radius of the sphere from the surface area formula $4\pi R^2 = 100\pi$, so $R^2 = 25$ and $R = 5$ cm. Then, the distance from the center to the plane is $d = 3$ cm. The radius of the intersection circle is $r = \sqrt{R^2 - d^2} = \sqrt{25 - 9} = \sqrt{16} = 4$ cm.