Question

find dimensions #plz don't spam here​

Answer 1

$\rule{200}4$

${ \frak { \underline \pink {\qquad Given\: : \qquad }}} \:$

A hexagon park maded with joining square and trapezium.

Side of square is 16 cm , parallel sides of trapezium are 8 cm and 16 cm and height of trapezium is 7 cm.

${ \frak { \underline \pink {\qquad To\:Find\: : \qquad }}} \:$

Area of the park.

${ \frak { \underline \pink {\qquad Solution\: : \qquad }}} \:$

${\red\bigstar\:\underline{\boxed {\bold \green {Area_{(Square)}\:=\:(Side)^2}}}}$

${\red\bigstar\:\underline{\boxed {\bold \green {Area_{(Trapezium)}\:=\:\dfrac{1}{2}\:\times\:(Sum\:of\:parallel\:sides)\:\times\:height(h)}}}}$

${\pink\dag\large\tt\underline{\purple{Area\: of \:park \:= \:Area \:of \:square\: +\: Area \:of \:trapezium}}}\pink\dag$

$\dag\:\underline{\frak{Putting\:all\:values\:-}}$

$\longmapsto\:\sf{[(16)^2]\:+\:\bigg(\dfrac{1}{2}\:\times\:(8\:+\:16)\:\times\:7}\bigg)$

$\longmapsto\:\sf{(16\:\times\:16)\:+\:\bigg(\dfrac{1}{2}\:\times\:24\:\times\:7}\bigg)$

$\longmapsto\:\sf{(256)\:+\:\bigg(\dfrac{24}{2}\:\times\:7}\bigg)$

$\longmapsto\:\sf{(256)\:+\:\bigg(\dfrac{24\:\times\:7}{2}}\bigg)$

$\longmapsto\:\sf{(256)\:+\:\bigg(\dfrac{168}{2}}\bigg)$

$\longmapsto\:\sf{(256)\:+\:\bigg(\cancel{\dfrac{168}{2}}}\bigg)$

$\longmapsto\:\sf{(256)\:+\:(84)}$

$\longmapsto\:\sf{(256\:+\:84)}$

$\longmapsto\:\boxed{\boxed{\sf{340}}}\:\orange\bigstar$

$\underline{\boxed {\frak {\therefore \blue {Area_{(Hexagon\:Park)}\:\leadsto\:340\:cm^2}}}}\:\red\bigstar$

$\rule{200}4$

Answer 2

$\large{\underline{\underline{\sf{ \maltese \: {★Given:-}}}}}$

Weight = 100 N

Height = 10 m

$\large{\underline{\underline{\sf{ \maltese \: {★To \: find:-}}}}}$

Potential energy = ?

$\large{\underline{\underline{\sf{ \maltese \: {Solution:-}}}}}$

➊ We know that:–

$\qquad \bull \bf \: {Weight = Mass \times Acceleration }$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\qquad \quad {:} \longrightarrow\sf \: {100 = mass \times 10 } \\\end{gathered}\end{gathered} \end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\qquad \quad {:} \longrightarrow\sf \: { \dfrac{100}{10} = mass } \\\end{gathered}\end{gathered} \end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\qquad \quad {:} \longrightarrow\sf \: { mass = \cancel\dfrac{100}{10} } \\\end{gathered}\end{gathered} \end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\qquad \quad {:} \longrightarrow\sf \: \underline{ \boxed {\sf{mass = 10 \: kg }}} \\\end{gathered}\end{gathered} \end{gathered} \end{gathered}$

➋ We know that:–

$\qquad \bull \bf \: { Potential \: energy=mass \times acceleration \times height }$