Surface Areas and Volumes — CBSE Class 10 Maths Important Questions

Answer: (a) $462\ \text{cm}^2$462 cm²

TSA of a solid hemisphere $=3\pi r^2=3\times\dfrac{22}{7}\times 49=462\ \text{cm}^2.$=3π r²=3× 22/7× 49=462 cm².

Answer: (a) $19.404\ \text{cm}^3$19.404 cm³

The largest cone has radius $r=2.1\ \text{cm}$r=2.1 cm (half the edge) and height $h=4.2\ \text{cm}$h=4.2 cm (the edge).

$V=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\times\dfrac{22}{7}\times 2.1^2\times 4.2=19.404\ \text{cm}^3.$V=1/3π r²h=1/3× 22/7× 2.1²× 4.2=19.404 cm³.

Answer: (a) $550\ \text{cm}^2$550 cm²

$l=\sqrt{7^2+24^2}=\sqrt{625}=25\ \text{cm}.$l=√(7²+24²)=√625=25 cm. CSA $=\pi rl=\dfrac{22}{7}\times 7\times 25=550\ \text{cm}^2.$=π rl=22/7× 7× 25=550 cm².

Answer: (a) $332.5\ \text{cm}^2$332.5 cm²

The cut removes a flat circle of area $\pi r^2$π r² but exposes a curved hemisphere of area $2\pi r^2$2π r²:

$\text{SA}=6a^2-\pi r^2+2\pi r^2=6(49)+\pi r^2=294+\dfrac{22}{7}\times 3.5^2=294+38.5=332.5\ \text{cm}^2.$SA=6a²-π r²+2π r²=6(49)+π r²=294+22/7× 3.5²=294+38.5=332.5 cm².

Answer: (d) A is false but R is true.

The curved surface area of a hemisphere is $2\pi r^2$2π r² (R is true), but its total surface area also includes the flat circular base, giving $2\pi r^2+\pi r^2=3\pi r^2$2π r²+π r²=3π r²; hence A is false.

$\dfrac{\text{Volume of cone}}{\text{Volume of cylinder}}=\dfrac{\frac{1}{3}\pi r^2h}{\pi r^2h}=\dfrac{1}{3}$(Volume of cone)/(Volume of cylinder)=1/3π r²hπ r²h=1/3, so the ratio is $1:3.$1:3.

Volume of cone $=\dfrac{1}{3}\times 360=120\ \text{cm}^3.$=1/3× 360=120 cm³.

Each cube has side $=\sqrt[3]{64}=4\ \text{cm}.$=3√64=4 cm. The cuboid formed has dimensions $8\times 4\times 4\ \text{cm}.$8× 4× 4 cm.

Surface area $=2(lb+bh+hl)=2(8\times 4+4\times 4+4\times 8)=2(32+16+32)=2\times 80=160\ \text{cm}^2.$=2(lb+bh+hl)=2(8× 4+4× 4+4× 8)=2(32+16+32)=2× 80=160 cm².

Radius $r=3.5\ \text{cm}.$r=3.5 cm. Height of cone $=15.5-3.5=12\ \text{cm}.$=15.5-3.5=12 cm.

Slant height $l=\sqrt{r^2+h^2}=\sqrt{3.5^2+12^2}=\sqrt{12.25+144}=\sqrt{156.25}=12.5\ \text{cm}.$l=√(r²+h²)=√(3.5²+12²)=√(12.25+144)=√156.25=12.5 cm.

TSA $=\text{CSA of cone}+\text{CSA of hemisphere}=\pi rl+2\pi r^2=\pi r(l+2r)$=CSA of cone+CSA of hemisphere=π rl+2π r²=π r(l+2r)
$=\dfrac{22}{7}\times 3.5\times(12.5+7)=11\times 19.5=214.5\ \text{cm}^2.$=22/7× 3.5×(12.5+7)=11× 19.5=214.5 cm².

Radius $r=2.5\ \text{mm}$r=2.5 mm. Length of the cylindrical part $h=14-2\times 2.5=9\ \text{mm}.$h=14-2× 2.5=9 mm.

$\text{SA}=\text{CSA of cylinder}+2\times\text{CSA of hemisphere}=2\pi rh+4\pi r^2=2\pi r(h+2r)$SA=CSA of cylinder+2× CSA of hemisphere=2π rh+4π r²=2π r(h+2r)
$=2\times\dfrac{22}{7}\times 2.5\times(9+5)=220\ \text{mm}^2.$=2× 22/7× 2.5×(9+5)=220 mm².

Volume of cuboid $=15\times 10\times 3.5=525\ \text{cm}^3.$=15× 10× 3.5=525 cm³.

Volume of one conical depression $=\dfrac{1}{3}\times\dfrac{22}{7}\times 0.5^2\times 1.4=0.367\ \text{cm}^3$=1/3× 22/7× 0.5²× 1.4=0.367 cm³ (to 3 d.p.), so four depressions $=1.467\ \text{cm}^3.$=1.467 cm³.

Volume of wood $=525-1.467=523.53\ \text{cm}^3.$=525-1.467=523.53 cm³.

Radius $r=0.7\ \text{cm}$r=0.7 cm, height $h=2.4\ \text{cm}$h=2.4 cm, slant height $l=\sqrt{0.7^2+2.4^2}=\sqrt{0.49+5.76}=\sqrt{6.25}=2.5\ \text{cm}.$l=√(0.7²+2.4²)=√(0.49+5.76)=√6.25=2.5 cm.

(i) TSA $=\text{CSA of cylinder}+\text{area of base}+\text{CSA of cone}=2\pi rh+\pi r^2+\pi rl=\pi r(2h+r+l)$=CSA of cylinder+area of base+CSA of cone=2π rh+π r²+π rl=π r(2h+r+l)
$=\dfrac{22}{7}\times 0.7\times(4.8+0.7+2.5)=2.2\times 8.0=17.6\ \text{cm}^2.$=22/7× 0.7×(4.8+0.7+2.5)=2.2× 8.0=17.6 cm².

(ii) Volume $=\pi r^2h-\dfrac{1}{3}\pi r^2h=\dfrac{2}{3}\pi r^2h=\dfrac{2}{3}\times\dfrac{22}{7}\times 0.49\times 2.4=2.464\ \text{cm}^3.$=π r²h-1/3π r²h=2/3π r²h=2/3× 22/7× 0.49× 2.4=2.464 cm³.

$r=3.5\ \text{cm},\ h=10\ \text{cm}.$r=3.5 cm, h=10 cm.

(i) TSA $=\text{CSA of cylinder}+2\times\text{CSA of hemisphere}=2\pi rh+4\pi r^2=2\pi r(h+2r)$=CSA of cylinder+2× CSA of hemisphere=2π rh+4π r²=2π r(h+2r)
$=2\times\dfrac{22}{7}\times 3.5\times(10+7)=22\times 17=374\ \text{cm}^2.$=2× 22/7× 3.5×(10+7)=22× 17=374 cm².

(ii) Volume $=\pi r^2h-2\times\dfrac{2}{3}\pi r^3=\dfrac{22}{7}\times 12.25\times 10-\dfrac{4}{3}\times\dfrac{22}{7}\times 42.875$=π r²h-2× 2/3π r³=22/7× 12.25× 10-4/3× 22/7× 42.875
$=385-179.67=205.33\ \text{cm}^3.$=385-179.67=205.33 cm³.

(i) $l=\sqrt{r^2+h^2}=\sqrt{1.5^2+2^2}=\sqrt{2.25+4}=\sqrt{6.25}=2.5\ \text{m}.$l=√(r²+h²)=√(1.5²+2²)=√(2.25+4)=√6.25=2.5 m.

(ii) Volume $=\pi r^2h_{\text{cyl}}+\dfrac{1}{3}\pi r^2h_{\text{cone}}=\dfrac{22}{7}(1.5)^2(7)+\dfrac{1}{3}\cdot\dfrac{22}{7}(1.5)^2(2)$=π r²h_cyl+1/3π r²h_cone=22/7(1.5)²(7)+1/3· 22/7(1.5)²(2)
$=49.5+4.71=54.21\ \text{m}^3.$=49.5+4.71=54.21 m³.

(iii) Curved area $=2\pi rh_{\text{cyl}}+\pi rl=2\times\dfrac{22}{7}\times 1.5\times 7+\dfrac{22}{7}\times 1.5\times 2.5=66+11.79=77.79\ \text{m}^2.$=2π rh_cyl+π rl=2× 22/7× 1.5× 7+22/7× 1.5× 2.5=66+11.79=77.79 m².