Question

4. Atable cover of dimensions 3 m 25 cm x 2 m 30 cm is spread on a table. If 30 cm of the table cover is hanging all around the table, find the area of the table cover which is hanging outside the top of the table. Also, find the cost of polishing the table top at 16 per square metre.​

Answer 1

Answer :-

$\:\\\large{\underbrace{\underline{\sf{Question's\;Analysis}}}}$

Here the concept of Areas of Rectangle has been used. We are given the dimensions of the Table Cloth. Now we can find its area. Then we can subtract the twice of the length of extra piece of table top to find the area of used Table Cloth. This area of Used table cloth will be the Area of Table Top also. To find the area of hanging table top, we need to subtract the area of table cloth and area of table top.

Let's do it !!

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★ Equations Used :-

$\\\;\large{\boxed{\sf{Area\;of\;Rectangle\;=\;\bf{Length\;\times\;Breadth}}}}$

$\\\;\large{\boxed{\sf{Area\;of\;Hanging\;Table\;Cloth\;=\;\bf{Area\;Table\;of\;Cloth\;-\;Area\;of\;Table\;Top}}}}$

$\\\;\large{\boxed{\sf{Cost\;of\;Polishing\;the\;Table\;Top\;=\;\bf{Rate_{(in\:per\:m^2)}\;\times\;Area\;of\;Table\;Top\:}}}}$

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★ Solution :-

Given,

» Dimensions of Table Cloth = 3 m 25 cm × 2 m 30 cm

= 3.25 m × 2.3 m

» Length of table cloth hanging around = 30 cm

= 0.3 m

» Rate of Polishing the table top = Rs. 16

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~ For the Area of the table cloth :-

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Area\;of\;Rectangle_{(Table\:Cloth)}\;=\;\bf{Length\;\times\;Breadth}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Area\;of\;Rectangle_{(Table\:Cloth)}\;=\;\bf{3.25\;\times\;2.3}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Area\;of\;Rectangle_{(Table\:Cloth)}\;=\;\underline{\underline{\bf{7.475\;\;m^{2}}}}}$

$\\\:\underline{\boxed{\tt{Area\;\:of\;\:Table\;\:Cloth\;=\;\bf{7.475\;\;m^{2}}}}}$

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~ For the Area of Table top :-

$\\\:\sf{:\Rightarrow\;\;\;Length\;of\;Table\;Top\;=\;\bf{Length\;of\;Table\;Cloth\;-\;2(Length\;of\;hanging\;cloth)}}$

$\\\:\sf{:\Rightarrow\;\;\;Length\;of\;Table\;Top\;=\;\bf{3.25\;-\;2(0.3)}}$

$\\\:\sf{:\Rightarrow\;\;\;Length\;of\;Table\;Top\;=\;\bf{3.25\;-\;0.6}}$

$\\\:\sf{:\Rightarrow\;\;\;Length\;of\;Table\;Top\;=\;\underline{\underline{\bf{2.65\;\;m}}}}$

Similarly,

$\\\:\sf{:\Rightarrow\;\;\;Breadth\;of\;Table\;Top\;=\;\bf{Breadth\;of\;Table\;Cloth\;-\;2(Length\;of\;hanging\;cloth)}}$

$\\\:\sf{:\Rightarrow\;\;\;Breadth\;of\;Table\;Top\;=\;\bf{2.3\;-\;2(0.3)}}$

$\\\:\sf{:\Rightarrow\;\;\;Breadth\;of\;Table\;Top\;=\;\bf{2.3\;-\;0.6}}$

$\\\:\sf{:\Rightarrow\;\;\;Breadth\;of\;Table\;Top\;=\;\underline{\underline{\bf{1.7\;\;m}}}}$

Now let's find the area ::

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Area\;of\;Rectangle_{(Table\:Top)}\;=\;\bf{Length\;\times\;Breadth}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Area\;of\;Rectangle_{(Table\:Top)}\;=\;\bf{2.65\;\times\;1.7}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Area\;of\;Rectangle_{(Table\:Top)}\;=\;\underline{\underline{\bf{4.505\;\;m^{2}}}}}$

$\\\:\underline{\boxed{\tt{Area\;\:of\;\:Table\;\:Top\;=\;\bf{4.505\;\;m^{2}}}}}$

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~ For Area of Hanging Table cloth :-

$\\\;\;\;\;\;\sf{:\Longrightarrow\;\;\;Area\;of\;Hanging\;Table\;Cloth\;=\;\bf{Area\;Table\;of\;Cloth\;-\;Area\;of\;Table\;Top}}$

$\\\;\;\;\;\;\sf{:\Longrightarrow\;\;\;Area\;of\;Hanging\;Table\;Cloth\;=\;\bf{7.475\;-\;4.505}}$

$\\\;\;\;\;\;\sf{:\Longrightarrow\;\;\;Area\;of\;Hanging\;Table\;Cloth\;=\;\bf{2.97\;\;m^{2}}}$

$\\\;\large{\underline{\underline{\rm{Thus,\;area\;of\;hanging\;cloth\;all\;around\;is\;\;\boxed{\bf{2.97\;\;m^{2}}}}}}}$

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~ For the Total Cost of Polishing the Table Top :-

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Cost\;of\;Polishing\;the\;Table\;Top\;=\;\bf{Rate_{(in\:per\:m^2)}\;\times\;Area\;of\;Table\;Top\:}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Cost\;of\;Polishing\;the\;Table\;Top\;=\;\bf{16\;\times\;4.505}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Cost\;of\;Polishing\;the\;Table\;Top\;=\;\underline{\underline{\bf{Rs.\;\;72.08}}}}$

$\\\;\large{\underline{\underline{\rm{Thus,\;total\;cost\;of\;polishing\;the\;table\;is\;\;\boxed{\bf{Rs.\;\;72.08}}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;\;Area\;of\;Square\;=\;(Side)^{2}}$

$\\\;\sf{\leadsto\;\;\;Area\;of\;Circle\;=\;\pi r^{2}}$

$\\\;\sf{\leadsto\;\;\;Area\;of\;Parallelogram\;=\;Base\:\times\:Height}$

$\\\;\sf{\leadsto\;\;\;Area\;of\;Triangle\;=\;\dfrac{1}{3}\:\times\:Base\:\times\:Height}$

$\\\:\sf{\leadsto\;\;\;Area\;of\;Semi\:-\:Circle\;=\;\dfrac{\pi r^{2}}{2}}$

$\\\;\sf{\leadsto\;\;\;Perimeter\;of\;Square\;=\;4\:\times\:Side}$

$\\\;\sf{\leadsto\;\;\;Perimeter\;of\;Circle\;=\;2\pi r}$

$\\\;\sf{\leadsto\;\;\;Perimeter\;of\;Rectangle\;=\;2\:\times\:(Length\;+\;Breadth)}$

Answer 2

Answer:

★ Equations Used :-

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★ Solution :-

Given,

» Dimensions of Table Cloth = 3 m 25 cm × 2 m 30 cm

= 3.25 m × 2.3 m

» Length of table cloth hanging around = 30 cm

= 0.3 m

» Rate of Polishing the table top = Rs. 16

_____________________________________________

~ For the Area of the table cloth :-

_____________________________________________

~ For the Area of Table top :-

Similarly,

Now let's find the area ::

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~ For Area of Hanging Table cloth :-

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~ For the Total Cost of Polishing the Table Top :-

Step-by-step explanation:

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