Question

Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of

15m, 10m and 7m, respectively. From each can paint of 100m2 of area is painted. How many

cans of paint will she need to paint the room?​

Answer 1

HEYA

Area to be painted = area of the walls + area of ceiling

Area to be painted = area of the walls + area of ceiling =2(hl+hb)+lb

Area to be painted = area of the walls + area of ceiling =2(hl+hb)+lb=[2×(7×15+7×10)+(15×10)]=500 sq m

Area to be painted = area of the walls + area of ceiling =2(hl+hb)+lb=[2×(7×15+7×10)+(15×10)]=500 sq mNo. of cans required =

Area to be painted = area of the walls + area of ceiling =2(hl+hb)+lb=[2×(7×15+7×10)+(15×10)]=500 sq mNo. of cans required = 100

Area to be painted = area of the walls + area of ceiling =2(hl+hb)+lb=[2×(7×15+7×10)+(15×10)]=500 sq mNo. of cans required = 100500

Area to be painted = area of the walls + area of ceiling =2(hl+hb)+lb=[2×(7×15+7×10)+(15×10)]=500 sq mNo. of cans required = 100500

Area to be painted = area of the walls + area of ceiling =2(hl+hb)+lb=[2×(7×15+7×10)+(15×10)]=500 sq mNo. of cans required = 100500 =5

Area to be painted = area of the walls + area of ceiling =2(hl+hb)+lb=[2×(7×15+7×10)+(15×10)]=500 sq mNo. of cans required = 100500 =5Hence, 5 cans are required to paint the walls and the ceiling of the cuboidal hall.

HOPE IT HELPED U.

AGAR WRONG HAI TOH JUST REPORT IT. NP☺️

Answer 2

★ Concept :-

Here the concept of LSA of Cuboid and Area of Rectangle has been used. We see that we are given the dimensions of the room. Even we know that for painting the room, we don't paint the floor. So firstly we will calculate the LSA that is area of four walls of the cuboidal room. Then we will add the area of ceiling, calculated using the formula of area of rectangle, to that area of four room. This will give us the total area to be painted. Then we will divide this by 100 m² which is area painted by each can, to get number of cans required.

Let's do it !!

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★ Formula Used ::

$\;\boxed{\sf{\pink{LSA\;of\;Cuboid\;=\;\bf{2(L\;+\;B)\:\times\:H}}}}$

$\;\boxed{\sf{\pink{Area\;of\;Rectangle\;=\;\bf{L\:\times\:B}}}}$

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★ Solution :-

» Length of the room = L = 15 m

» Breadth of the room = B = 10 m

» Height of the room = H = 7 m

» Area painted by each can = 100 m²

» Shape of room = Cuboidal

» Shape of ceiling = Rectangular

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~ For the LSA of Room ::

We know that,

$\;\sf{\rightarrow\;\;LSA\;of\;Cuboid\;=\;\bf{2(L\;+\;B)\:\times\:H}}$

By applying values, we get

$\;\sf{\rightarrow\;\;LSA\;of\;Room\;=\;\bf{2(15\;+\;10)\:\times\:7}}$

$\;\sf{\rightarrow\;\;LSA\;of\;Room\;=\;\bf{2(25)\:\times\:7}}$

$\;\sf{\rightarrow\;\;LSA\;of\;Room\;=\;\bf{50\:\times\:7}}$

$\;\bf{\rightarrow\;\;LSA\;of\;Room\;=\;\bf{\red{350\;\;m^{2}}}}$

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~ For the Area of Ceiling of Room :-

We know that,

$\;\sf{\rightarrow\;\;Area\;of\;Rectangle\;=\;\bf{L\:\times\:B}}$

By applying the values, we get

$\;\sf{\rightarrow\;\;Area\;of\;Rectangle\;=\;\bf{15\:\times\:10}}$

$\;\bf{\rightarrow\;\;Area\;of\;Rectangle\;=\;\bf{\blue{150\;\;m^{2}}}}$

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~ For Area to be painted ::

We know that,

✒ Area to be painted = LSA of Room + Area of Ceiling

✒ Area to be painted = 350 cm² + 150 cm²

✒ Area to be painted = 500 cm²

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~ For the Number of cans required ::

This is given as,

$\;\sf{\Longrightarrow\;\;\orange{Number\;of\;Cans\;=\;\bf{\dfrac{Area\;to\;be\;Painted}{Area\;painted\;by\;one\;Can}}}}$

By applying values, we get

$\;\sf{\Longrightarrow\;\;Number\;of\;Cans\;=\;\bf{\dfrac{500}{100}}}$

$\;\bf{\Longrightarrow\;\;Number\;of\;Cans\;=\;\bf{\green{5\;\;cans}}}$

$\;\underline{\boxed{\tt{Number\;\:of\;\:cans\;\:required\;=\;\bf{\purple{5\;\;cans}}}}}$

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★ More to know :-

$\;\sf{\leadsto\;\;TSA\;of\;Cuboid\;=\;2(LB\:+\:BH\:+\:LH)}$

$\;\sf{\leadsto\;\;Volume\;of\;Cuboid\;=\;L\:\times\;B\:\times\:H}$

$\;\sf{\leadsto\;\;Diagonal\;of\;Cuboid\;=\;\sqrt{L^{2}\:+\:B^{2}\:+\:H^{2}}}$

$\;\sf{\leadsto\;\;Perimeter\;of\;Rectangle\;=\;2(L\:+\:B)}$

$\;\sf{\leadsto\;\;Diagonal\;of\;Rectangle\;=\;\sqrt{L^{2}\:+\:B^{2}}}$