Question

A hemisphere 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. plz give explanation step by step ​

Answer 1

$\red{\textbf{Answer}}$ ➧

Step-by-step explanation:

$\blue{\texttt{Given \: that \: :-}}$

$diameter = side \: of \: cube \: = l$

$\blue{\textbf{surface \: area \: of \: solid}}$➧

= Area of cube +curved surface area of hemisphere - base area of hemisphere .

•$\blue{\texttt{Area \: of \: cube}}$➧

$here, \: side = l$

$area \: of \: cube \: = 6(side {)}^{2}$

•$\blue{\textbf{curved \: surface \: area \: of \: hemisphere}}$

$diameter \: of \: hemisphere = l$

$hence \: radius \: r \: = \frac{diameter}{1} = \large\frac{l}{2}$

$curved \: surface \: area \: of \: hemisphere$➧$2\pi \: {r}^{2}$

$=2 \pi( \large\frac{l}{2} {)}^{2}$

$= 2\pi \large\frac{ {l}^{2} }{4}$

$= \large\frac{\pi {l}^{2} }{2}$

•$\blue{\textbf{Base \: area \: of \: hemisphere}}$➧$\pi \: {r}^{2}$

$\pi( \large\frac{l}{2} {)}^{2} = \pi \large\frac{ {l}^{2} }{4}$

$= \large\frac{\pi {l}^{2} }{4}$

•$\red{\textbf{surface \: area \: of \: solid}}$➧

$6 {l}^{2} + \large\frac{\pi {l}^{2} }{2} - \frac{\pi {l}^{2} }{4}$

$6 {l}^{2} + \large\frac{2\pi {l}^{2} - \pi {l}^{2} }{4}$

${6l}^{2} + \large\frac{\pi {l}^{2} }{4} = {l}^{2}(6 + \large\frac{\pi}{4} )$

${l}^{2} ( \large\frac{6(4) + \pi}{4} ) = {l}^{2} ( \large\frac{24 + \pi}{4} )$

$= \large\frac{1}{4} {l}^{2} (\pi + 24)$

$\underline\blue{\bold{so,\: surface \: of \: solid }}$➧$\large\frac{1}{4} {l}^{2} (\pi + 24)$

Answer 2

Answer:

Given :-

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

To Find :-

SA of remaining solid

Solution :-

At first we will half the diameter to make radius

R = L/2

Now,

Finding CSA of hemisphere

CSA = 2πr²

CSA = 2π(l/2)²

CSA = π(l²/4)

• Cancelling 4 and 2

CSA = π(l²/2)

Base area = πr² + πr

Base area = π(l²/2)² + π l²/2

Base area = πl²/4

SA = CSA - Base area

6l² = πl²/2 - πl²/4

• LCM of 2 and 4 as 4

6l² = 2πl² - πl²/4

6l² + πl²/4 = l² (6 + π/4)

¼ l² (π + 24)