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attashe74 [19]
2 years ago
8

 Jalen is painting her great room including the ceiling but not the floor. The room measures 20 ft long, 30 wide and 9 ft tall.

If each paint container covers 400 sq ft, how many cans should she buy? PLEASE HELPP MEEEE asap
Mathematics
2 answers:
Juliette [100K]2 years ago
5 0

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

= 2(hl+hb)+lb

Where,

  • H is height
  • L is length
  • B is breadth or width

In our case,

  • H = 9 ft
  • L = 20 ft
  • B = 30 ft

Put the values in the formula ~

= 2(9×20+9×30)+20×30

= 2(180+270)+600

= 2×450+600

= 900+600

= 1500 ft²

Now,

Required number of cans = 1500/400 = 3.75

  • round it off to 4

➪ <em>T</em><em>h</em><em>u</em><em>s</em><em>,</em><em> </em><em>s</em><em>h</em><em>e</em><em> </em><em>n</em><em>e</em><em>e</em><em>d</em><em>s</em><em> </em><em>t</em><em>o</em><em> </em><em>b</em><em>u</em><em>y</em><em> </em><em>4</em><em> </em><em>c</em><em>a</em><em>n</em><em>s</em><em> </em><em>o</em><em>f</em><em> </em><em>p</em><em>a</em><em>i</em><em>n</em><em>t</em><em>.</em><em>.</em><em>.</em><em>~</em>

Luden [163]2 years ago
3 0

4 cans of paint

\sf --->  \ \ {\boxed {\boxed {\frac{area  \ of \ room - area \ of \ floor}{area \ the \ paint \ covers} }}}

\hookrightarrow \sf {\dfrac{2(wl+hl+hw)- (lw)}{400}}

\hookrightarrow \sf {\dfrac{2((30)(20)+(9)(20)+(9)(30))-(20*30)}{400}

\hookrightarrow \sf {\dfrac{1500}{400}

\hookrightarrow \rm {\sf {3.75}}

→ She should buy max 4 cans as she cannot buy 3.75 cans.

<em>Here </em><em>l</em><em> refers to length, </em><em>w</em><em> refers to width, </em><em>h</em><em> refers to height.</em>

<em />

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