Question

If the radius of a sphere is increased by 4 cm, its

surface area is increased by 464p cm2. What is the

volume (in cm3

) of the original sphere?​

Answer 1

$\text{Let r be the radius of the given sphere}$

$\text{As per given data,}$

$\text{Increased surface area}=464\,\pi\,cm^2$

$4\,\pi\,(r+4)^2-4\,pi\,r^2=464\,\pi$

$4\,\pi[(r+4)^2-r^2]=464\,\pi$

$(r+4)^2-r^2=\dfrac{464\,\pi}{4\,\pi}$

$(r+4)^2-r^2=116$

$r^2+16+8\,r-r^2=116$

$16+8\,r=116$

$8\,r=116-16$

$8\,r=100$

$\implies\,r=\dfrac{100}{8}$

$\implies\bf\,r=12.5\,cm$

$\text{Volume of the original sphere}$

$=\dfrac{4}{3}\pi\,r^3$

$=\dfrac{4}{3}{\times}\dfrac{22}{7}{\times}(12.5)^3$

$=\dfrac{4}{3}{\times}\dfrac{22}{7}{\times}1,953.125$

$=8,184.52\,cm^3$

$\therefore\textbf{The volume of the original sphere is \bf\,8,184.52\,cm^3 }$

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Answer 2

The volume of the original sphere is 8181.22 cm³

Step-by-step explanation:

Let the radius of the sphere is r

The surface area of the sphere

$S=4\pi r^2$

If the radius is increased by 4 cm, the surface area is increased by 464π cm²

Therefore,

$S+464\pi=4\pi (r+4)^2$

$\implies 4\pi r^2+464\pi=4\pi(r+4)^2$

$\implies 464\pi=4\pi [(r+4)^2-r^2]$

$\implies 116=(r+4+r)(r+4-r)$

$\implies (2r+4)=\frac{116}{4}$

$\implies 2(r+2)=29$

$\implies r+2=\frac{29}{2}$

$\implies r+2=14.5$

$\implies r=12.5$ cm

Therefore, the volume of the original sphere

$V=\frac{4}{3}\pi r^3$

$\implies V=\frac{4}{3}\pi (12.5)^3$

$\implies V=8181.22$ cm³

Hope this answer is helpful.

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