The height of the cone is 30 cm A small cone is cut off at the top by a - Maths - Surface Areas and Volumes

Question

The height of the cone is 30 cm A small cone is cut off at the
top by a plane parallel to the base If its volume be 1/27 of the
volume of the given cone, at what height abive the base is the
section made ?
A cone of radius 20 cm is divided into 2 parts by drawing a
plane through the mid point of its axis parallel to its base
Compare the volumes of the 2 parts
A semicircular region and a square region have equal perimeters
The area of the square region exceeds that of the semicircular
region by 4 cm2 Find the perimeters and areas of the 2
regions

Chetna Khanna · Accepted Answer

Perimeter of semi circular region = 2r + πr, where 'r' is the radius of the semi circular region.

Perimeter of square region = 4a, where 'a' is the side of square.

Given perimeter of semi circular region = Perimeter of square region

⇒ 2r + πr = 4a

⇒ r (2 + π) = 4a

$\Rightarrow a = \frac{r (2 + π)}{4} ..... (1)$

Area of square region = a² [Area of square = Side × side]

Area of semi circular region = $\frac{π r^{2}}{2}$

According to the given condition,

$a^{2} = \frac{π r^{2}}{2} + 4 \Rightarrow [\frac{r (2 + π)}{4}]^{2} = \frac{π r^{2}}{2} + 4 \Rightarrow \frac{r^{2} (2 + π)^{2}}{16} = \frac{π r^{2}}{2} + 4 \Rightarrow \frac{r^{2} (4 + π^{2} + 4 π)}{16} = \frac{π r^{2}}{2} + 4 \Rightarrow [\frac{(4 + π^{2} + 4 π)}{16} - \frac{π}{2}] r^{2} = 4$

$\Rightarrow [\frac{4 + π^{2} + 4 π - 8 π}{16}] r^{2} = 4 \Rightarrow (\frac{4 + π^{2} - 4 π}{16}) r^{2} = 4 \Rightarrow r^{2} = \frac{4 \times 16}{(4 + π^{2} - 4 π)} \Rightarrow r^{2} = \frac{64}{(π - 2)^{2}} \Rightarrow r = \sqrt{\frac{64}{(π - 2)^{2}}} \Rightarrow r = \frac{8}{(π - 2)} \Rightarrow a = \frac{8}{(π - 2)} \times \frac{(2 + π)}{4} \Rightarrow a = \frac{2 (π + 2)}{(π - 2)}$

∴ Perimeter of semi circular region = $(2 + π) \times \frac{8}{(π - 2)}$

$= \frac{8 (π - 2)}{(π - 2)}$

Perimeter of square region = $4 \times \frac{2 (π + 2)}{(π - 2)}$

$= \frac{8 (π + 2)}{(π - 2)}$

Area of semi circular region = $\frac{1}{2} π [\frac{8}{(π - 2)}]^{2}$

$= \frac{1}{2} π \times \frac{64}{(π - 2)^{2}} = \frac{32 π}{(π - 2)^{2}}$

Area of square region = $[\frac{2 (π + 2)}{(π - 2)}]^{2}$

$= \frac{4 (π + 2)^{2}}{(π - 2)^{2}}$

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