Question

A room is 6m long, 5m broad and 4m high. It has two doors, each measuring 1m by 1m. Find the cost of distempering the walls at ₹25/m2​

Answer 1

Answer:

Cost of distempring @ of 25m^2 = Rs 2150

Explanation:

>> We are given with dimensions of the room, which are, 6m long, 5m broad, and 4m high.

>> 2 Doors are there of 1m by 1m.

>> Target is to find the cost of distempering walls at Rs 25/m^2

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We have to basically find out the area of 4 walls excluding the area of windows as no paint is to be applied there.

Our room is a cuboid, so we know that the lateral surface area of a cuboid is = $\sf 2h(l+w)$

So, let's put the known values into this formula and equate them.

$\bf \implies Lateral \ surface \ area = 2h(l+w)$

$\implies \sf Surface \ area\ of \ walls \ of \ the room = 2(4)[6m+5m]$

$\implies \sf Surface \ area \ of \ walls \ of \ the \ room = 8m[11m]$

$\implies \sf Surface \ area \ of \ walls \ of \ the \ room = 88m^2$

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Now, let's find the area of a window measuring 1m by 1m. Don't forget that there are 2 such windows!

So, the area of those windows, = 2 × [1m ×1m]

Area of windows = 2m²

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So the area we need to apply distemper is = 88m² - 2m² = 86m²

Now, what's left? We have to find the cost of destempering them.

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Given that distempering is done @ of Rs 25/m^2 so,

let's apply direct proportion:

Cost of distempring @ of 25m^2 = 25 × 86

=> Cost of distempering @ of 25m^2 = Rs 2150

Answer 2

ANSWER

T.S.A( ROOM) - WINDOW

FIND THE THE AREA LEFT

MULTIPLY WITH COST

***NOTE- T.S.A =2(LB +BH + HL)

HOPE IT HELPS

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