Question

Rukshar painted the outside of the cabinet of measure 1 m ×2 m ×1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

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Answer 1

Answer:

Rukshar painted the outside of the cabinet of measure 1 m ×2 m ×1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

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Answer 2

${\large{\pmb{\sf{\underline{Understanding \; the \; Question...}}}}}$

☀️ This question says that Rukshar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 metres. Means the length is 1 m, breadth is 2 m and the height is 1.5 metres. We have to find that how much surface area did she cover if she painted all except the bottom of the cabinet? This cabinet is in the shape of cuboid according to the question. And the bottom area of thia cabinet is in the shape of rectangle. Let's solve this question!

Diagram -

$\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf 1\:m}\put(7.7,6.3){\sf 2\:m}\put(11.3,7.45){\sf 1.5\:cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}$

${\large{\pmb{\sf{\underline{Given \; that...}}}}}$

● The length of cabinet = 1 m

● The breadth of cabinet = 2 m

● The height of cabinet = 1.5 m

${\large{\pmb{\sf{\underline{To \; find...}}}}}$

● She cover if she painted all except the bottom of the cabinet.

${\large{\pmb{\sf{\underline{Solution...}}}}}$

● How much surface area did she cover if she painted all except the bottom of the cabinet = 11 m sq.

${\large{\pmb{\sf{\underline{Using \; concepts...}}}}}$

● Formula to find total surface area of the cuboid.

● Formula to find area of rectangle.

● Formula to find area painted.

${\large{\pmb{\sf{\underline{Using \; formulas...}}}}}$

${\small{\underline{\boxed{\sf{TSA \: of \: cuboid \: = 2(lb + bh + hl)}}}}}$

${\small{\underline{\boxed{\sf{Area \: of \: rectangle \: = L \times B}}}}}$

${\small{\underline{\boxed{\sf{Area \: pained \: = TSA \: of \: cuboid - \: Area \: of \: rectangle}}}}}$

${\large{\pmb{\sf{\underline{Where...}}}}}$

★ TSA denotes total surface area

★ l or L denotes length

★ b or B denotes breadth

★ h denotes height

${\large{\pmb{\sf{\underline{Some \: procedure...}}}}}$

~ To solve this question firstly let's find the total surface area of cuboid by using formula, we have to put the values.

~ Then we have to find the area of rectangle by using formula, just have to put the values.

~ Now we have to use formula to find area painted just have to subtract and we get our final result.

${\large{\pmb{\sf{\underline{Full \; Solution...}}}}}$

${\sf{\pmb{\qquad \dag \: \: Finding \: TSA}}}$

${\small{\underline{\boxed{\sf{TSA \: of \: cuboid \: = 2(lb + bh + hl)}}}}}$

➟ TSA of cuboid = 2(lb + bh + hl)

➟ TSA of cuboid = 2(2×1 + 1×1.5 + 1.5×2)

➟ TSA of cuboid = 2(2+1.5+3)

➟ TSA of cuboid = 2(5+1.5)

➟ TSA of cuboid = 2(6.5)

➟ TSA of cuboid = 2 × 6.5

➟ TSA of cuboid = 13 m²

• Henceforth, 13 metres sq. is the total surface area of the cuboid.

Diagram

$\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf 1\:m}\put(7.7,6.3){\sf 2\:m}\put(11.3,7.45){\sf 1.5\:cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}$

${\sf{\pmb{\qquad \dag \: \: Finding \: Area}}}$

${\small{\underline{\boxed{\sf{Area \: of \: rectangle \: = L \times B}}}}}$

➟ Area of rectangle = L×B

➟ Area of rectangle = 1 × 2

➟ Area of rectangle = 2 m²

• Henceforth, 2 metres sq. is the area of rectangle.

Diagram -

$\begin{gathered} \sf 1m \: \: \: \: \: \: \: \: \: \: \: \\ \begin{gathered}\begin{gathered}\boxed{\begin{array}{}\bf { \red{}}\\{\qquad \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{}\\ { \sf{ }}\\ { \sf{ }} \\ \\ { \sf{ }}\end{array}}\end{gathered}\end{gathered} \sf 2m\end{gathered}$

${\sf{\pmb{\qquad \dag \: \: Finding \: Painted \: Area}}}$

${\small{\underline{\boxed{\sf{Area \: pained \: = TSA \: of \: cuboid - \: Area \: of \: rectangle}}}}}$

➟ Painted area = TSA of cuboid - Area of rectangle

➟ Painted area = (13-2) m²

➟ Painted area = 11 m²

• Henceforth, 11 metres sq. is the area to be painted