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NCERT Exemplar · Class 10 Mathematics Surface Areas and Volumes

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EXERCISE 12.2 1–8 (part 3 of 7)

  1. Write 'True' or 'False' and justify your answer in the following:

    Exercise 1

    Two identical solid hemispheres of equal base radius r cm\displaystyle r \mathrm{~cm} are stuck together along their bases. The total surface area of the combination is 6πr2\displaystyle 6 \pi r^2.

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    NCERT’s answer
    False
    False. Joining hides the two flat circular faces; only the two curved surfaces remain exposed, giving a sphere of radius \(\displaystyle r\).\[\text{TSA of one hemisphere (curved + flat)} = 3\pi r^2 \] \[3\pi r^2+3\pi r^2=6\pi r^2 \quad \text{(counts the hidden flat faces)} \] \[\text{Actual TSA} = 2(2\pi r^2)=4\pi r^2 \]

  2. Exercise 2

    A solid cylinder of radius r\displaystyle r and height h\displaystyle h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is 4πrh+4πr2\displaystyle 4 \pi r h+4 \pi r^2.

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    NCERT’s answer
    False
    False. The two faces where the cylinders meet are hidden; only the top and bottom circles are exposed.\[\text{CSA} = 2\pi r(2h)=4\pi rh \] \[\text{Exposed bases} = \pi r^2+\pi r^2=2\pi r^2 \] \[\text{TSA} = 4\pi rh+2\pi r^2 \ne 4\pi rh+4\pi r^2 \]

  3. Exercise 3

    A solid cone of radius r\displaystyle r and height h\displaystyle h is placed over a solid cylinder having same base radius and height as that of a cone. The total surface area of the combined solid is πr[r2+h2+3r+2h]\displaystyle \pi r\left[\sqrt{r^2+h^2}+3 r+2 h\right].

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    NCERT’s answer
    False
    False. Only the cone's curved surface, the cylinder's curved surface, and the cylinder's own base are exposed; the shared circle in between is hidden.\[l=\sqrt{r^2+h^2},\quad \text{CSA(cone)}=\pi r l,\quad \text{CSA(cyl)}=2\pi rh,\quad \text{base}=\pi r^2 \] \[\text{TSA} = \pi rl+2\pi rh+\pi r^2=\pi r\left[\sqrt{r^2+h^2}+2h+r\right] \] the radius term carries coefficient \(\displaystyle r\), not \(\displaystyle 3r\).

  4. Exercise 4

    A solid ball is exactly fitted inside the cubical box of side a\displaystyle a. The volume of the ball is 43πa3\displaystyle \frac{4}{3} \pi a^3.

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    NCERT’s answer
    False
    False. A ball that exactly fits inside a cube of side \(\displaystyle a\) touches the centre of every face, so its diameter equals \(\displaystyle a\), not its radius.\[r=\frac{a}{2} \] \[V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi\left(\frac{a}{2}\right)^3=\frac{\pi a^3}{6} \]

  5. Exercise 5

    The volume of the frustum of a cone is 13πh[r12+r22−r1r2]\displaystyle \frac{1}{3} \pi h\left[r_1^2+r_2^2-r_1 r_2\right], where h\displaystyle h is vertical height of the frustum and r1,r2\displaystyle r_1, r_2 are the radii of the ends.

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    NCERT’s answer
    False
    False. The standard frustum-volume identity adds the \(\displaystyle r_1r_2\) cross term; it does not subtract it.\[V=\frac{1}{3}\pi h\left[r_1^2+r_2^2+r_1r_2\right] \] which differs from the stated \(\displaystyle r_1^2+r_2^2-r_1r_2\) in the sign of the middle term.

  6. Exercise 6

    The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig. 12.7\displaystyle 12.7 is πr23[3h−2r]\displaystyle \frac{\pi r^2}{3}[3 h-2 r]. NCERT_Question_Class10_Maths_Exemplar_Ch12_Ex12-2_Q6

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    NCERT’s answer
    True
    True. The vessel's capacity is the cylinder's volume minus the volume of the hemisphere that bulges up from the base. \[V=\pi r^2h-\frac{2}{3}\pi r^3 \] \[=\frac{\pi r^2}{3}\left[3h-2r\right] \]

  7. Exercise 7

    The curved surface area of a frustum of a cone is πl(r1+r2)\displaystyle \pi l\left(r_1+r_2\right), where l=h2+(r1+r2)2,r1\displaystyle l=\sqrt{h^2+\left(r_1+r_2\right)^2}, r_1 and r2\displaystyle r_2 are the radii of the two ends of the frustum and h\displaystyle h is the vertical height.

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    NCERT’s answer
    False
    False. Taking \(\displaystyle r_1>r_2\), the right triangle has legs \(\displaystyle h\) and \(\displaystyle r_1-r_2\); the slant height uses the difference of the radii.\[l^2=h^2+(r_1-r_2)^2 \quad \text{(Pythagoras)} \] \[l=\sqrt{h^2+(r_1-r_2)^2} \ne \sqrt{h^2+(r_1+r_2)^2} \] \[\text{CSA}=\pi l(r_1+r_2) \quad \text{(this part holds)} \]

  8. Exercise 8

    An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to curved surface area of frustum of a cone + area of circular base + curved surface area of cylinder

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    NCERT’s answer
    True
    True. The sheet forms exactly three surfaces; the stand is hollow, so it adds no disc of its own.\[\text{Sheet area} = \text{CSA(frustum)}+\text{area of circular base}+\text{CSA(cylinder)} \]