Question

You are Dev Mehta/ Divya Mehta of Class VI of Bal Bharti
Public School, Dehradun. You have lost your library book which
was issued to you, in the playground. Draft a notice for the school
notice board, informing the student about it.
PLS WRITE IN NOTICE WRITING FORMAT . Pls help i cant understand .If you answer i will mark you as the BRAINLIEST

Answer 1

Answer:

\large{\underline{\bold{\orange{QuesTion}}}}QuesTion

ᴀ ꜱᴇᴍɪ-ᴄɪʀᴄᴜʟᴀʀ ꜱʜᴇᴇᴛ ᴏꜰ ᴍᴇᴛᴀʟ ᴏꜰ ᴅɪᴀᴍᴇᴛᴇʀ 28ᴄᴍ ɪꜱ ʙᴇɴᴛ ᴛᴏ ꜰᴏʀᴍ ᴀɴ ᴏᴘᴇɴ ᴄᴏɴɪᴄᴀʟ ᴄᴜᴘ. ꜰɪɴᴅ ᴛʜᴇ ᴄᴀᴘᴀᴄɪᴛʏ ᴏꜰ ᴄᴜᴘ.

\large\underline{\sf{Solution-}}Solution−

As it is given that, a semi-circular sheet of metal of diameter 28 cm is bent to form an open conical cup.

So, when the semi-circular sheet of metal is bent in to an open conical cup,

1. The radius of semi-circular sheet of metal becomes the slant height of cone

2. Circumference of the semi-circular sheet of metal becomes the circumference of the base of the cone.

Now, we have

Radius of semi-circular sheet of metal, r = 14 cm

Let assume that

Radius of conical cup be R cm

Height of conical cup be h cm

Slant height of conical cup be L cm.

Now, as radius of semi-circular sheet of metal is equals to slant height of conical cup.

\begin{gathered}\rm\implies \:L = 14 \: cm - - - (1) \\ \\ \end{gathered}⟹L=14cm−−−(1)

Further, we have Circumference of the semi-circular sheet of metal becomes the circumference of the base of the cone.

\begin{gathered}\rm \: 2\pi \: R \: = \: \pi \: r \\ \\ \end{gathered}2πR=πr

\begin{gathered}\rm \: 2R = 14 \\ \\ \end{gathered}2R=14

\begin{gathered}\rm\implies \:R = 7 \: cm - - - (2) \\ \\ \end{gathered}⟹R=7cm−−−(2)

Now, we know, slant height L, height h and radius R of a conical cup are connected by the relationship

\begin{gathered}\rm \: {L}^{2} = {R}^{2} + {h}^{2} \\ \\ \end{gathered}L2=R2+h2

\begin{gathered}\rm \: {14}^{2} = {7}^{2} + {h}^{2} \\ \\ \end{gathered}142=72+h2

\begin{gathered}\rm \: 196 = 49 + {h}^{2} \\ \\ \end{gathered}196=49+h2

\begin{gathered}\rm \: {h}^{2} = 147 \\ \\ \end{gathered}h2=147

\begin{gathered}\rm \: {h}^{2} = 7 \times 7 \times 3 \\ \\ \end{gathered}h2=7×7×3

\begin{gathered}\rm \: h = 7 \sqrt{3} \\ \\ \end{gathered}h=73

\begin{gathered}\rm \: h = 7 \times 1.732 \\ \\ \end{gathered}h=7×1.732

\begin{gathered}\rm\implies \:h \: = \: 12.12 \: cm \: \: \{approx. \} \\ \\ \end{gathered}⟹h=12.12cm{approx.}

Now, we know Volume of conical cup of radius R and height h is given by

\begin{gathered}\rm \: Volume_{(Conical\:cup)} \: = \: \dfrac{1}{3} \: \pi \: {R}^{2} \: h \\ \\ \end{gathered}Volume(Conicalcup)=31πR2h

\begin{gathered}\rm \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \times 7 \times 7 \times 12.12 \\ \\ \end{gathered}=31×722×7×7×12.12

\begin{gathered}\rm \: = \: 22 \times 7 \times 4.04 \\ \\ \end{gathered}=22×7×4.04

\begin{gathered}\rm \: = \: 622.56 \: {cm}^{2} \\ \\ \end{gathered}=622.56cm2

Hence,

\begin{gathered}\bf\implies \:Volume_{(Conical\:cup)} \: = \: 622.26 \: {cm}^{2} \\ \\ \end{gathered}⟹Volume(Conicalcup)=622.26cm2

\rule{190pt}{2pt}

{ \red{ \mathfrak{Additional\:Information}}}AdditionalInformation

\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}MoreFormulaeMoreFormulae★CSA(cylinder)=2πrh★Volume(cylinder)=πr2h★TSA(cylinder)=2πr(r+h)★CSA(cone)=πrl★TSA(cone)=πr(l+r)★Volume(sphere)=34πr3★Volume(cube)=(