A solid sphere of radius 12 inches is melted and cast into a right circular cone whose base diameter is $\sqrt[]{2}$  times its slant height. If the radius of the sphere and the cone are the same, how many such cones can be made and how much material is left out?

<p>The solid sphere is melted and recast into cones. The volume of material is the same before and after casting.</p><p>Volume of the sphere =&nbsp;<math><mrow><mrow><mo> (</mo><mrow><mfrac><mrow><mn>4</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow><mo>)</mo></mrow><mi>π</mi><msup><mrow><mi>r</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow></math>&nbsp;. (1)</p><p>Volume of the right circular cone =&nbsp;<math><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mi>π</mi><msup><mrow><mi>R</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>H</mi></mrow></math>&nbsp;. (2)</p><p>Diameter of the base of the cone</p><p>=&nbsp;<math><mrow><mroot><mrow><mn>2</mn></mrow><mrow></mrow></mroot></mrow></math>&nbsp;slant height</p><p>=&nbsp;<math><mrow><mroot><mrow><mn>2</mn></mrow><mrow></mrow></mroot></mrow></math>&nbsp;S (say). (3)</p><p>4R = 2S = 2 (H² + R²)</p><p>2R² = 2H</p><p>R = H. (4)</p><p>Volume of the cone =&nbsp;<math><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mi>π</mi><msup><mrow><mi>R</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>H</mi></mrow></math></p><p><math><mrow><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mi>π</mi><msup><mrow><mi>r</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>.</mo><mo stretchy="false"> (</mo><mi>r</mi><mo stretchy="false">)</mo></mrow></math>&nbsp; (since, R = r)</p><p><math><mrow><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mi>π</mi><msup><mrow><mi>r</mi></mrow><mrow><mn>3</mn></mrow></msup></mrow></math></p><p>From equations (1) and (4), it can be seen that the melted material creates exactly 4 cones of the specified dimensions. No material is left over.</p>

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