q-3] The surface area of a sphere a cube are equal Prove that their volumes are in - Maths - Areas of Parallelograms and Triangles

Question

q-3] The surface area of a sphere & a cube
are equal Prove that their volumes are in the ratio 1:v22/7/6
[paye=22/7]

q-4] The area of 3 adjacent faces of a cuboid
are x,y,z if the volume is v , prove that [v]2=xyz

q-5] Wall paper ,312cm long and 25cm wide is
required to cover the wall of a room length of the room is 7m and
its breadth is twice its height Determine the height of the
room

Gursheen kaur. · Accepted Answer

a) Surface area of a sphere = $4 π r^{2}$

Surface area of a cube= $6 a^{2}$

∴According to the question ,

Surface area of a sphere=Surface area of a cube

⇒ $4 π r^{2}$ = $6 a^{2}$

$\Rightarrow r^{2} = \frac{6 a^{2}}{4 π} \Rightarrow r^{2} = \frac{6 a^{2} \times 7}{4 \times 22} = \frac{42 a^{2}}{88} \Rightarrow r^{2} = \frac{21 a^{2}}{44} \Rightarrow r = \sqrt{\frac{21}{44}} a$

$R a t i o = \frac{V o l u m e o f s p h e r e}{V o l u m e o f c u b e} = \frac{\frac{4}{3} π r^{3}}{a^{3}} = \frac{\frac{4}{3} \times \frac{22}{7} \times {[\sqrt{\frac{21}{44}}]}^{3} a^{3}}{a^{3}} = \frac{\frac{4}{3} \times \frac{22}{7} \times {[\sqrt{\frac{21}{44}}]}^{3}}{1} = \frac{\frac{4}{3} \times \frac{22}{7} \times {[\sqrt{\frac{21}{44}}]}^{3}}{1} = \frac{\frac{4}{3} \times \frac{22}{7} \times \sqrt{\frac{21}{44}} \times \sqrt{\frac{21}{44}} \times \sqrt{\frac{21}{44}}}{1} = \frac{\frac{4}{3} \times \frac{22}{7} \times \frac{21}{44} \times \frac{1}{2} \times \sqrt{\frac{21}{11}}}{1} = \frac{\sqrt{\frac{21}{11}}}{1}$

b) Let the edges of cuboid are a,b,c
and faces of a cuboid can be written as
x = ab ......(i)
y = bc ....(ii)
z = ac .....(iii)
As , we know that

volume of cuboid, V = abc
Squaring both sides, we get,

$V^{2} = {(a b c)}^{2} V^{2} = a^{2} b^{2} c^{2} V^{2} = a b \times b c \times c a H e r e, a^{2} b^{2} c^{2} = a a \times b b \times c c = a b \times b c \times c a V^{2} = x \times y \times z [F r o m (i), (i i) a n d (i i i)] V = \sqrt{x \times y \times z}$

c) length of wall paper=312 cm = 3.12 m

and width of wall paper =25 cm = 0.25 m

∴Area of wall paper= Length of wall paper $\times$ width of wall paper

⇒ Area of wall paper = 3.12m $\times$ 0.25 m

⇒ Area of wall paper= 0.78 m²

Now,

length of the room , l= 7m

Let the height of the room = h m

As, the breadth is twice its height, then the breadth of the room = 2h m

∴C.S.A. of room = Area of wall paper

$\Rightarrow 2 h (l + 2 h) = 0.78 \Rightarrow 2 h (7 + 2 h) = 0.78 \Rightarrow 14 h + 4 h^{2} = 0.78 \Rightarrow 2 h^{2} + 7 h - 0.39 = 0 \Rightarrow h = \frac{- 7 \pm \sqrt{7^{2} - 4 \times 2 (- 0.39)}}{2 \times 2} = \frac{- 7 \pm \sqrt{49 3.12}}{4} \Rightarrow h = \frac{- 7 \pm \sqrt{52.12}}{4} = \frac{- 7 \pm 7.22}{4} \Rightarrow h = \frac{- 7 + 7.22}{4}, \frac{- 7 - 7.22}{4} \Rightarrow h = \frac{22}{4}, \frac{- 14.22}{4} \Rightarrow h = 0.055, - 3.555$

We will neglect the negative term as the height cannot be negative.

∴ h = 0.055 m²

or h = 5.5 cm = height of the room

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