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Maths Chapter 13:- Surface Area and Volume..

☛A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.​

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Answer 1

Given :-

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube.

To Find :-

Surface Area of remaining solid

Solution :-

Given,

Diameter is l

So,

Radius = Diameter/2

Radius = l/2

                                                                                 

Surface Area of cube = 6(edge)²

SA of cube = 6(l)²

SA of cube = 6l²

                                                                                 

CSA of hemisphere = 2πr²

CSA of hemisphere = 2 × π × (l/2)²

CSA of hemisphere = 2 × π × l²/4

CSA of hemisphere = πl²/2

                                                                                 

Area of base = πr²

Area of base = π × (l/2)²

Area of base = π × l²/4

Area of base = πl²/4

                                                                                 

Now

Surface Area of soild = Surface area of cube + CSA of hemisphere - Area of base of hemisphere

Surface Area of soild = 6l² + πl²/2 - πl²/4

Surface Area of soild = 6l² + 2πl² - πl²/4

Surface Area of soild = 6l² + πl²/4

• Taking l² as common

Surface Area of soild = l²(6 + π/4)

Surface Area of soild = l²(24 + π/4)

Surface Area of soild = 1/4 × l² (π + 24)

Surface Area of soild = l²/4(π + 24)

[tex][/tex]

Answer 2

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/<❤>✨ANSWER☃️✨

ʟᴇᴛ, ʟᴇɴɢᴛʜ ᴏғ ᴇᴀᴄʜ ᴇᴅɢᴇ ᴏғ ᴄᴜʙɪᴄᴀʟ ʙʟᴏᴄᴋ

= ᴅɪᴀᴍᴇᴛᴇʀ ᴏғ ʜᴇᴍɪsᴘʜᴇʀᴇ = ʟ

∴ ʀᴀᴅɪᴜs ᴏғ ʜᴇᴍɪsᴘʜᴇʀᴇ,

$r = \frac{1}{2} unit$

ᴛᴏᴛᴀʟ sᴜʀғᴀᴄᴇ ᴀʀᴇᴀ ᴏғ ɴᴇᴡʟʏ ғᴏʀᴍᴇᴅ ᴄᴜʙᴇ

= ᴄᴜʀᴠᴇᴅ sᴜʀғᴀᴄᴇ ᴀʀᴇᴀ ᴏғ ᴄᴜʙᴇ – ᴜᴘᴘᴇʀ ᴘᴀʀᴛ ᴏғ ᴄᴜʙᴇ + ᴀʀᴇᴀ ᴏғ ʜᴇᴍɪsᴘʜᴇʀᴇ ᴡʜɪᴄʜ ɪs ᴅᴇᴘʀᴇssᴇᴅ

$➠ \: 6 {l}^{2} - \pi ({ \frac{l}{2} })^{2} + 2\pi({ \frac{l}{2} })^{2}$

$➠ \: 6 {l}^{2} + \pi {l}^{2}$

$➠ \: \frac{1}{4} (24 \: {l}^{2} + \pi {l}^{2}$

$➠ \: \frac{1}{4} {l}^{2} ( \: 24 \: + \pi \: )$

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