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Exercise 2.4 · Q4

Q.A cone is surmounted on a hemisphere as shown here. The radius of the solid formed is 6 cm and its height is 21 cm. Find:

i) Curved surface area of the solid
ii) Volume of the solid. [Figure: a cone of height h sitting on top of a hemisphere of radius R, together forming one composite solid.]
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A cone (radius 6 cm) sits atop a hemisphere of the same radius; total height of the composite solid is 21 cm, giving CSA ≈530.93 cm² and volume ≈1018.29 cm³.

For radius rr, cone height hh, slant height l=r2+h2l=\sqrt{r^2+h^2}: CSAcone=πrl\text{CSA}_{\text{cone}}=\pi rl, CSAhemisphere=2πr2\text{CSA}_{\text{hemisphere}}=2\pi r^2, Vcone=13πr2hV_{\text{cone}}=\dfrac13\pi r^2h, Vhemisphere=23πr3V_{\text{hemisphere}}=\dfrac23\pi r^3.

  1. The hemisphere and cone share the same radius (surmounted): r=6r=6 cm.
  2. Total height of solid =21=21 cm =hcone+r=h_{\text{cone}}+r (hemisphere's flat face is the cone's base) ⇒hcone=21−6=15\Rightarrow h_{\text{cone}}=21-6=15 cm.
  3. Slant height l=r2+h2=62+152=36+225=261≈16.16l=\sqrt{r^2+h^2}=\sqrt{6^2+15^2}=\sqrt{36+225}=\sqrt{261}\approx16.16 cm.
  4. (i) CSA of solid =πrl+2πr2=π(6)(16.16)+2π(36)=\pi rl+2\pi r^2=\pi(6)(16.16)+2\pi(36).
  5. πrl=227×6×16.1555≈304.65 cm2\pi rl=\dfrac{22}{7}\times6\times16.1555\approx304.65\text{ cm}^2; 2πr2=227×72=15847≈226.29 cm22\pi r^2=\dfrac{22}{7}\times72=\dfrac{1584}{7}\approx226.29\text{ cm}^2.
  6. Total CSA ≈304.65+226.29=530.93 cm2\approx304.65+226.29=530.93\text{ cm}^2. …

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