Question

a table cover of dimensions 3m 25cm x 2m 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 meter​

Answer 1

Answer:

To find the cost of polishing the table top, we have to find out its area, 

Given, 

Length of cover  =3m25cm

=3.25m 

Breadth of cover =2m30cm

=2.30m 

Now, area of the table cover =3.25×2.30

=7.475m²

Give, 30cm width of cloth is outside the table an each side.

Hence, 

length of table =3.25−2×0.30

=3.25−0.6=2.65m 

Breadth of table =2.30−2×0.30

=2.30−0.6=1.70m 

Now, area of the top of the table 

=2.65×1.70m

=4.505m²

So, Area of hanging cover of table = Area of cover of table - area of the top of the table

=7.475−4.505

=2.97m²

It has been given that Rs. 16 per square meter is the cost of polishing the table. 

So, cost of polishing top of table =16 x area of top of table 

=16×4.505

=Rs.72.08

So, Rs.72.08 will be needed to polish the top of table.

Answer 2

$\sf\large\underbrace \red{Required \: Solution \: :}$

Let ABCD be a table cover and shaded region represents the hanging part of table cover having width 30cm in the figure given along side. obviously , the rectangular part PQRS of the table cover lie on table top.

• Length of the table cover = 3m 25m = 3 × 100 + 25 = 325 cm.

• Breadth of the table cover = 2m 30m = 2 × 100 + 30 = 230 cm.

Also,

• Length of the top of the table = 325 - 2 × 30 = 265 cm.

• Breadth of the top of the table = 230 - 2 × 30 = 170 cm

Area of table cover, which is hanging outside the top of the table.

$\begin{gathered} \rm \bf \: Area \: of \: the \: rectangular \: ABCD \: - \: Area \: of \: the \: rectangular \: PQRS .\end{gathered}$

$\begin{gathered} \sf \qquad \implies \: 325 \: cm \: \times \: 230 \: cm \: - \: 265 \: cm \: \times \: 170 \: cm \end{gathered}$

$\begin{gathered} \sf \qquad \implies \: {74750 \: cm}^{ \:2} \: - \: {45050 \: cm}^{ \:2} . \end{gathered}$

$\begin{gathered} \sf \qquad \implies \: \: {29700 \: cm}^{ \:2} . \end{gathered}$

$\begin{gathered} \rm \bf \: Now, \: Area \: of \: the \: top \: of \: the \: table = Area \: of \: the \: rectangle \: PQRS .\end{gathered}$

$\begin{gathered} \sf \qquad \implies \: {265 \: cm} \: \times \: {170 \: cm} . \end{gathered}$

$\begin{gathered} \sf \qquad \implies \: \: {45050 \: cm}^{ \:2} . \end{gathered}$

$\begin{gathered} \sf \qquad \implies \: \dfrac{45050}{10000} \: {m}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [ {1 \: m}^{2} \: = 10000 \: {cm}^{2} ] \end{gathered}$

$\begin{gathered} \sf \qquad\large \implies \: 4.505 \: {m}^{2} \end{gathered}$

$\begin{gathered} \rm \large\therefore \: Cost \: of \: polishing \: the \: table \: top \\ \\ \sf = \sf \large ₹ \: 16 × 4.505 \end{gathered} \\ =\sf\large\boxed {\red{ ₹ \: 72.08}}$

SO, your answer is ₹ 72.08 .