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Mensuration Test - 25

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Mensuration Test - 25
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  • Question 1
    1 / -0
    Find the total surface area of the given cuboid.

    Solution
    Given, length $$(l)=8\ m$$
    breadth $$(b)=5\ m$$
    height $$(h)=2\ m$$

    Surface area of the cuboid $$\displaystyle =2\times \left( lb+lh+bh \right) $$

                                                  $$\displaystyle =2\times \left( 8\times 5+8\times 2+5\times 2 \right) \ m^2$$

                                                  $$\displaystyle =2\times \left( 40+16+10 \right) \ m^2$$

                                                  $$\displaystyle =2\times 66\ m^2$$

                                                  $$\displaystyle =132\ { m }^{ 2 }$$


    Hence, total surface area is $$132\ { m }^{ 2 }$$
  • Question 2
    1 / -0
    The edge of the cube measures $$9$$ cm. Find its surface area.
    Solution
    Given, edge $$= 9 \text{ cm}$$
    Total surface area of a cube $$=6\times \text{side}^2$$
    $$= 6\times 9\times 9$$
    $$= 486 \text{ cm}^2$$
  • Question 3
    1 / -0
    Kyle bought 2 snack boxes. Box A has a measure of $$2\  in\times 3\  in\times 5 \ in.$$ Box B has a measure of $$11 \ in\times 4\  in \times 3\  in.$$ Find the surface area of the boxes and their difference.
    Solution
    Surface area of rectangle is  2 x (Length x width + Length x height + Width x height)

    Here Box A has length $$= 2$$ in, height $$= 3$$ in and width $$= 5$$ in

    Therefore, the surface area of Box A is:

    $$A=2((2\times 5)+(2\times 3)+(3\times 5))=2(10+6+15)=2\times 31=62$$ in$$^2$$

    Now, Box B has length $$= 11$$ in, height $$= 4$$ in and width $$= 3$$ in

    Therefore, the surface area of Box B is:

    $$A=2((11\times 4)+(11\times 3)+(4\times 3))=2(44+33+12)=2\times 89=178$$ in$$^2$$

    Hence, the difference between the surface area of box A and box B is $$178-62=116$$ in$$^2$$.
  • Question 4
    1 / -0
    Find the surface area of the cuboid:

    Solution
    Surface area of the cuboid $$=$$ $$\displaystyle 2\times \left( lw+lh+hw \right) $$
    $$\displaystyle =2\times \left( 8\times 2.1+8\times 14+2.1\times 14 \right) $$
    $$\displaystyle =2\times \left( 16.8+112+29.4 \right) $$
    $$\displaystyle =2\times 158.2$$
    $$\displaystyle =316.4$$ $${ mm }^{ 2 }$$
  • Question 5
    1 / -0
    Calculate the total surface area of the cuboid:

    Solution
    Total surface area of the cuboid $$\displaystyle 2\times \left( lw+lh+hw \right) $$
    $$\displaystyle =2\times \left( 10\times 1+10\times 7+7\times 1 \right) $$
    $$\displaystyle =2\times \left( 10+70+7 \right) $$
    $$\displaystyle =2\times 87$$
    $$=\displaystyle 174{ cm }^{ 2 }$$

    So, option D is correct.
  • Question 6
    1 / -0
    The side of the base of a dice is $$5.5$$ mm. Find the surface area of a dice whose edges are squares.
  • Question 7
    1 / -0
    Find the surface area of a cube with side $$\displaystyle 12$$ m
    Solution
    Total surface area of a cube $$=$$ $$\displaystyle 6\times $$ side$$^{ 2 }$$
    $$\displaystyle =6\times 12\times 12$$
    $$\displaystyle =864{ m }^{ 2 }$$
  • Question 8
    1 / -0
    A cubical water tank measures $$3$$ feet sides. Find its surface area.
    Solution
    $$Surface\ area\ of\ a\ cube=6\times Side^2$$
    $$side=3ft$$
    $$Surface\ Area=6\times 3^2$$
    $$Surface\ Area=6\times 9$$
    $$Surface\ Area=54ft^2$$
  • Question 9
    1 / -0
    The length of the side is $$3.9$$ ft. Find the surface area of a cube .
  • Question 10
    1 / -0
    The diameter of a cylinder is $$12$$ m and its height is $$10$$ m. Find the curved surface area of a cylinder.
    Solution
    Curved surface area of cylinder is $$A=2πrh$$

    Here, the diameter is $$12$$ m and therefore, the radius is half of diameter that is $$r=6$$ m and height is $$10$$ m.

    Thus,

    $$A=2πrh=2\times π\times 6\times 10=120π$$ m$$^2$$

    Hence, the curved surface area of the cylinder is $$120π$$ m$$^2$$.

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