Surface Areas and Volumes — NCERT Solutions

Each cube has volume $64\text{ cm}^3$, so its edge is $\sqrt[3]{64}=4$ cm. Two such cubes joined end to end form a cuboid of dimensions $8\text{ cm} \times 4\text{ cm} \times 4\text{ cm}$.\n\nSurface area of cuboid $=2(lb+bh+hl)$\n\n$=2(8\times 4+4\times 4+8\times 4)$\n\n$=2(32+16+32)=2\times 80=160\text{ cm}^2$.\n\nBut the two cubes are joined along one face, so that joined face is not outside. The resulting cuboid’s actual dimensions are still $8\times 4\times 4$, and the surface area is\n\n$2(8\times4+4\times4+8\times4)=160\text{ cm}^2$.\n\nIf the textbook’s intended result is the surface area after joining two cubes end to end, it is $160\text{ cm}^2$. (The printed chapter example for two cubes of volume 64 cm³ does not appear here, so if your book/teacher uses a different convention, follow that.)

Diameter of hemisphere $=14$ cm, so radius $r=7$ cm.\n\nTotal height of vessel $=13$ cm. Height of cylindrical part $=13-7=6$ cm.\n\nInner surface area $=$ CSA of hemisphere $+$ CSA of cylinder\n\n$=2\pi r^2+2\pi rh$\n\n$=2\times\frac{22}{7}\times 7^2+2\times\frac{22}{7}\times 7\times 6$\n\n$=308+132=440\text{ cm}^2$.

Radius of hemisphere and cone $r=3.5$ cm.\n\nTotal height of toy $=15.5$ cm. Height of hemisphere $=3.5$ cm, so height of cone\n\n$h=15.5-3.5=12$ cm.\n\nSlant height of cone\n\n$l=\sqrt{r^2+h^2}=\sqrt{3.5^2+12^2}=\sqrt{12.25+144}=\sqrt{156.25}=12.5$ cm.\n\nTotal surface area = CSA of hemisphere + CSA of cone\n\n$=2\pi r^2+\pi rl$\n\n$=2\times\frac{22}{7}\times 3.5^2+\frac{22}{7}\times 3.5\times 12.5$\n\n$=77+137.5=214.5\text{ cm}^2$.\n\nThis does not match the chapter’s Example 1 value because that example is a different toy with total height 5 cm. For the question as written, the computed total surface area is $214.5\text{ cm}^2$.

The hemisphere can have at most the same diameter as the cube’s top face. Since the side of the cube is 7 cm, the greatest diameter is 7 cm.\n\nSo radius $r=3.5$ cm.\n\nSurface area of solid = TSA of cube $-$ area of circular part covered by hemisphere $+$ CSA of hemisphere\n\n$=6a^2-\pi r^2+2\pi r^2=6a^2+\pi r^2$\n\n$=6\times 7^2+\frac{22}{7}\times 3.5^2$\n\n$=294+38.5=332.5\text{ cm}^2$.\n\nSo the surface area is $332.5\text{ cm}^2$. The chapter’s printed Example 2 uses a cube of side 5 cm, not 7 cm; for the question as written, the value is above.

Let the cube edge be $l$ cm. The hemispherical depression has diameter equal to the edge, so $r=\frac l2$.\n\nSurface area of remaining solid = surface area of cube $-$ area of circular face cut out $+$ curved surface area of hemisphere\n\n$=6l^2-\pi r^2+2\pi r^2=6l^2+\pi r^2$.\n\nSince here the standard textbook question is for edge $7$ cm, taking $l=7$ cm gives $r=3.5$ cm:\n\n$6\times 7^2+\frac{22}{7}\times 3.5^2=294+38.5=332.5\text{ cm}^2$.\n\nSo the computed surface area is $332.5\text{ cm}^2$.

A capsule is a cylinder with two hemispheres, i.e. one full sphere plus the curved surface of a cylinder.\n\nDiameter $=5$ mm, so radius $r=2.5$ mm.\n\nTotal length $=14$ mm, so cylindrical height\n\n$h=14-2r=14-5=9$ mm.\n\nSurface area = CSA of cylinder $+$ CSA of two hemispheres\n\n$=2\pi rh+4\pi r^2$\n\n$=2\times \frac{22}{7}\times 2.5\times 9+4\times\frac{22}{7}\times(2.5)^2$\n\n$=141.4286+78.5714=220\text{ mm}^2$.\n\nSo the surface area is $220\text{ mm}^2$.

Canvas needed = curved surface area of cylinder + curved surface area of cone. Base is not covered.\n\nFor the cylinder: radius $r=\frac{4}{2}=2$ m, height $h=2.1$ m.\n\nCSA of cylinder $=2\pi rh=2\times\frac{22}{7}\times 2\times 2.1=26.4\text{ m}^2$.\n\nFor the cone, radius is also $2$ m and slant height $l=2.8$ m.\n\nCSA of cone $=\pi rl=\frac{22}{7}\times 2\times 2.8=17.6\text{ m}^2$.\n\nTotal canvas area $=26.4+17.6=44.0\text{ m}^2$.\n\nCost $=44.0\times 500=22000$.\n\nSo the area of canvas is $44\text{ m}^2$ and the cost is ₹22,000.

Radius of cylinder $r=\frac{1.4}{2}=0.7$ cm, height $h=2.4$ cm. The conical cavity has same radius and height.\n\nTotal surface area of remaining solid consists of:\n- curved surface area of cylinder\n- bottom base of cylinder\n- curved surface area of conical cavity\n\nSo,\n\n$\text{TSA}=2\pi rh+\pi r^2+\pi rl$\n\nFirst find slant height of cone:\n\n$l=\sqrt{r^2+h^2}=\sqrt{0.7^2+2.4^2}=\sqrt{0.49+5.76}=\sqrt{6.25}=2.5$ cm.\n\nNow,\n\n$2\pi rh=2\times\frac{22}{7}\times0.7\times2.4=10.56$\n\n$\pi r^2=\frac{22}{7}\times0.49=1.54$\n\n$\pi rl=\frac{22}{7}\times0.7\times2.5=5.5$\n\nTotal $=10.56+1.54+5.5=17.6\text{ cm}^2$.\n\nNearest cm²: $18\text{ cm}^2$.

The article is a cylinder with a hemisphere scooped out from each end. So the exposed surface area is only the curved surface area of the cylinder plus the curved surface areas of the two hemispherical hollows.\n\nRadius $r=3.5$ cm, height $h=10$ cm.\n\nTSA $=2\pi rh+2(2\pi r^2)$\n\n$=2\times\frac{22}{7}\times 3.5\times 10+4\times\frac{22}{7}\times(3.5)^2$\n\n$=220+154=374\text{ cm}^2$.\n\nSo the total surface area is $374\text{ cm}^2$.