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This is the second of our many quizzes on Surface area. Click on this to access more such competitive quizzes on Mensuration.

Surface area Quiz 2

mathematics

 A cubical room, opened from the top having a volume of $\text{729}{{\text{m}}^{3}}$ was painted from inside and outside excluding the base of room and it cost $\text{Rs 1620}$ to paint the room. What was the cost of painting per ${{\text{m}}^{2}}$ area?

 Recall formula for volume of cube and evaluate side of cube. Later, evaluate the lateral surface area of the cube and then the cost of painting per ${{\text{m}}^{2}}$ by dividing cost of painting room by lateral surface area of room.

$\text{Volume of cubical room is equal to 729 }{{\text{m}}^{3}}. \\ \text{Also, Volume of cube is }{{a}^{3}}. \\ \text{Then, for cubical room } \\ \text{=}{{a}^{3}}=\text{729 }{{\text{m}}^{3}} \\ \Rightarrow a=9\text{ m} \\ \text{Paint was applied on only side faces of the room i}\text{.e}\text{. only four faces}\text{.} \\ \text{Lateral surface area of the cube is }4{{a}^{2}}. \\ \text{Then, lateral surface for cubical room whose side is 9 m,} \\ \text{=4(9}{{\text{)}}^{2}}\text{ }{{\text{m}}^{2}} \\ \Rightarrow 324\text{ }{{\text{m}}^{2}} \\ \text{Cost of painting the cubical room was Rs}\text{. 1620}\text{.} \\ \text{so, cost of painting per }{{\text{m}}^{2}}\text{ will be} \\ \text{=}\dfrac{\text{Cost of painting the cubical room}}{\text{Lateral Surface Area of the room}} \\ \Rightarrow \dfrac{1620}{324}/{{\text{m}}^{2}} \\ \Rightarrow \text{Rs }5/{{\text{m}}^{2}} $ 

 From a cube of side $4c\text{m}$, a small cubical block with side $1c\text{m}$ was scooped out from the middle of the big cube. What is the surface area of the $3-D$ figure formed? <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/24d09148-9e7d-4716-a634-7f87673ba5145929762158326330850.png"/>

 Determine area of smaller cube for only 5 of its faces. Evaluate the area of the bigger cube according to the figure. Add both areas for the required answer.

$\text{From the figure, top face of smaller cube is not present}\text{.} \\ \text{So, area of smaller cube will be equal to 5 times the area of face of square}\text{.} \\ \text{i}\text{.e}\text{. 5}{{a}^{2}} \\ =5{{\left( 1 \right)}^{2}}\text{ c}{{\text{m}}^{2}} \\ \Rightarrow 5\text{ c}{{\text{m}}^{2}} \\ \text{Now, in bigger cube, there is small face of side 1 cm is missing,} \\ \text{then, surface area of larger cube will be equal to,} \\ \text{=}\left( \text{6(4}{{\text{)}}^{2}}-{{1}^{2}} \right)\text{ c}{{\text{m}}^{2}} \\ \Rightarrow 63\text{ c}{{\text{m}}^{2}} \\ \text{Thus, the required area is} \\ \text{= (63 + 5) c}{{\text{m}}^{2}} \\ \Rightarrow 68\text{ c}{{\text{m}}^{2}} $ 

 For the wooden block buoyancy test, the large cubic block was sliced evenly and was divided among $8$ students. The block which was given to the students, had a volume of $\text{64c}{{\text{m}}^{3}}$. Then what was the surface area of the big block which was sliced into smaller cubes? <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d85bb747-3a27-462d-a35a-b7e59ac6fb962252617276059375800.png"/>

 Recall formula for volume of cube and evaluate side of cube. Evaluate the side of the larger cube according to the figure and then determine the surface area of the large wooden block.

$\text{Volume of cube is }{{a}^{3}}. \\ \text{so, for blocks which students had,} \\ \text{= }{{a}^{3}}=6\text{4 c}{{\text{m}}^{3}} \\ \Rightarrow a=4\text{ cm} \\ \text{Now, from the figure, the side of larger cube is twice the side of smaller cubes}\text{.} \\ \text{thus, side of larger cube is,} \\ \text{=2}\times 4\text{ cm} \\ \Rightarrow 8\text{ cm} \\ \text{Surface area of cube is 6}{{a}^{2}}. \\ \text{So, surface area of larger wooden block is} \\ \text{=6(8}{{\text{)}}^{2}}\text{ c}{{\text{m}}^{2}} \\ \Rightarrow 384\text{ c}{{\text{m}}^{2}} $ 

 Every face of the cube when detached and arranged on the floor to make a rectangle, perimeter of rectangle with length of $9$ cm , was $30$cm as seen in figure. Then what is the surface area of the cube? <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/19f19536-14f9-4984-bb3e-599a134840993099494612650968251.png"/>

 Find the breadth of the rectangle using the formula of perimeter of rectangle. Then, evaluate the side of the cube and then calculate the surface area of the cube.

$\text{Perimeter of rectangle is }2\left( l+b \right). \\ \text{Then,} \\ \Rightarrow 2(l+b)=30\text{ cm} \\ \Rightarrow b=6\text{ cm} \\ \text{From figure, side of cube is half the breadth of the rectangle}\text{.} \\ \text{so, side of cube will be} \\ \text{=}\dfrac{6}{2}\text{ cm} \\ \Rightarrow 3\text{ cm} \\ \text{Surface area of cube is equal to }6{{a}^{2}}. \\ \text{thus, surface area of cube whose side is 3 cm is,} \\ \text{=6(3}{{\text{)}}^{2}}\text{ c}{{\text{m}}^{2}} \\ \Rightarrow 54\text{ c}{{\text{m}}^{2}} $ 

 Ram, the painter, was asked to paint some objects which were losing paints. He is capable of painting $1\text{ c}{{\text{m}}^{2}}$  area in $1$ second. The first object was, cube with side $10\text{ cm}$, second object was hemisphere of radius $\text{7 cm}$ and third object was cuboid of dimension$10\text{ cm, 5 cm},3\text{ cm}$. Then, arrange these objects in increasing order of time taken to paint them by Ram.

 Recall the formula for total surface area of cube, hemisphere and cuboid and evaluate their surface area. Arrange them in increasing order of area to get correct order.

$\text{Surface area of cube is }6{{a}^{2}}. \\ \text{then, area of cube with side }10\text{ }cm\text{,} \\ =6({{10}^{2}})\text{ }c{{m}^{2}} \\ \Rightarrow 600c{{m}^{2}} \\ \text{The total surface area of the hemisphere is }3\pi {{r}^{2}}. \\ \text{then, surface area of hemisphere whose radius is }7\text{ }cm\text{ is} \\ =3\pi {{7}^{2}}\text{ }c{{m}^{2}} \\ \Rightarrow 462\text{ }c{{m}^{2}} \\ \text{The total surface area of the cuboid is }2(lb+bh+hl). \\ \text{then, total surface area of cuboid whose length is }10\text{ }cm\text{, breadth is }5\text{ }cm\text{ and height is }2\text{ }cm\text{, is } \\ =2[(10\times 5)+(5\times 3)+(3\times 10)]\text{ }c{{m}^{2}} \\ \Rightarrow 190\text{ }c{{m}^{2}} \\ \text{Clearly, Surface Area of Cuboid, Surface Area of Hemisphere, Surface Area of Cube} \\ \text{thus, the order is }3,\text{ }2,\text{ }1.$