Question

10. A verandah of width 4.5 m is constructed around a room outside it. If the dimensions of the room are 18.5
m x 14 m, find the following:
(a) Area of the verandah
(b) The cost of paving tiles in the verandah at the rate of ₹20 per m2.​

Answer 1

Answer ::

• The area of path is 222m².
• The cost of paving tiles in the verandah at the rate of ₹20 per m² is Rs. 4230.

Step-by-step explanation ::

To Find :-

• Area of the verandah
• The cost of paving tiles in the verandah at the rate of ₹20 per m²

Solution :-

Given that,

• A verandah of width 4.5 m is constructed around a room outside it.
• The dimensions of the room are 18.5m × 14m

ACCORDING THE QUESTION,

• Area of verandah

Length of the room = 18.5m

Breadth of the room = 14m

Area of whole room = ( Length of room × Breadth of room )

= ( 18.5 × 14 )m²

= 259m²

Length of the verandah excluding the room

= [ 18.5 - ( 4.5 + 4.5 ) ]m

= [ 18.5 - 9 ]m

= 9.5m

Breadth of the verandah excluding the room

= [ 14 - ( 4.5 + 4.5 ) ]m

= [ 14 - 9 ]m

= 5m

Area of verandah excluding the room

= ( 9.5 × 5 )m²

= 47.5m²

Area of path = ( Area of room - Area of the verandah excluding the room )

= 259m² - 47.5m²

= 211.5m² ★

The area of path is 211.5m².

• The cost of paving tiles in the verandah at the rate of ₹20 per m²

= Rs. ( 211.5 × 20 )

= Rs. 4230 ★

The cost of paving tiles in the verandah at the rate of ₹20 per m² is Rs. 4230.

Answer 2

☆Given Question :-

A verandah of width 4.5 m is constructed around a room outside it. If the dimensions of the room are 18.5m x 14 m, find the following:

(a) Area of the verandah

(b) The cost of paving tiles in the verandah at the rate of ₹20 per m^2.

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$\huge \orange{AηsωeR} ✍$

☆Given :-

• A verandah of width 4.5 m is constructed around a room outside it.
• The dimensions of the room are 18.5m x 14 m

☆To Find :-

• (a) Area of the verandah
• (b) The cost of paving tiles in the verandah at the rate of₹20 per square metre.

☆Formula used:-

$\boxed{\sf \: ⟼Area\:of\:Rectangle=Length × Breadth}$

$\begin{gathered}\Large{\bold{\purple{\underline{CaLcUlAtIoN\::}}}} \end{gathered}$

_____________________________________________

☆Case :- 1

$\sf \: ⟼\:Dimension \: of \: Room$

$\sf \: Length \: of \: Room \: = 18.5 \: m$

$\sf \: Breadth \: of \: room \: = 14 \: m$

$\boxed{\sf \: ⟼Area\:of\:Rectangle=Length × Breadth}$

$\sf \: ⟼Area\:of \: room \: = 18.5 \times 14 = 259 \: {m}^{2}$

☆Case :- 2.

$\sf \: ⟼ \:Outside \: Dimensions$

$\sf \: Length = 18.5 + 2 \times 4.5 = 27.5 \: m$

$\sf \: Breadth = 14 + 2 \times 4.5 = 23 \: m$

$\sf \: ⟼Area\:of \: outside \: rectangle \: = 27.5 \times 23 = 632.5 \: {m}^{2}$

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☆Now, to calculate the area of Verandah

Area of Verandah = Area of outside rectangle - Area of room

$\sf \: ⟼Area \: of \: Verandah = 632.5 - 259 = 373.5 \: {m}^{2}$

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☆To find the cost of paving the tiles.

$\sf \: According \: to \: statement$

☆The cost of paving tiles in the verandah at the rate of ₹20 per square metre.

$\sf \: Cost \: of \: paving \: tiles\: 1 \: {m}^{2} = ₹20$

$\sf \: Cost \: of \:paving \: 373.5 {m}^{2} = 373.5 \times 20 = ₹ \: 7470$

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$\begin{gathered}\begin{gathered}\bf So = \begin{cases} &\bf{Area \: of \: Verandah=373.5 \: m^2} \\ &\bf{Cost \: of \: paving =₹7470} \end{cases}\end{gathered}\end{gathered}$

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☆ More information: -

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length ²+breadth ²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²