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Mathematics · Surface Areas and Volumes · NCERT Exercises

Exercise 12.1

Complete, independently verified solutions for NCERT Exercise 12.1.

Mathematics · Chapter 12 · NCERT Exercises

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Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.

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Question

Question 1

joined cubes · cuboid surface area

Two cubes, each of volume 64 cm³, are joined end to end. Find the surface area of the resulting cuboid.

Solution

Board-exam working:

Let a be the edge of each cube.

a3=64⇒a=4 cma^3=64\Rightarrow a=4\text{ cm}

Joining the cubes end to end gives a cuboid of dimensions 8 cm × 4 cm × 4 cm.

TSA=2(lb+bh+hl)TSA=2(lb+bh+hl)
=2[(8)(4)+(4)(4)+(4)(8)]=2[(8)(4)+(4)(4)+(4)(8)]
=2(32+16+32)=160 cm2=2(32+16+32)=160\text{ cm}^2

Final answer

160 cm2\boxed{160\text{ cm}^2}

Question

Question 2

hollow vessel · inner surface area

A vessel is a hollow hemisphere mounted by a hollow cylinder. The hemisphere's diameter is 14 cm and the vessel's total height is 13 cm. Find its inner surface area.

Solution

Board-exam working:

r=142=7 cmr=\frac{14}{2}=7\text{ cm}

Cylinder height:

h=13−7=6 cmh=13-7=6\text{ cm}

The open vessel's inner area is the curved area of the cylinder plus that of the hemisphere.

A=2πrh+2πr2A=2\pi rh+2\pi r^2
=2×227×7×(6+7)=2\times\frac{22}{7}\times7\times(6+7)
=572 cm2=572\text{ cm}^2

Final answer

572 cm2\boxed{572\text{ cm}^2}

Question

Question 3

cone on hemisphere · surface area

A toy is a cone of radius 3.5 cm mounted on a hemisphere of the same radius. Its total height is 15.5 cm. Find its total surface area.

Solution

Board-exam working:

r=3.5 cmr=3.5\text{ cm}

Cone height:

h=15.5−3.5=12 cmh=15.5-3.5=12\text{ cm}

Slant height:

l=r2+h2=3.52+122=12.5 cml=\sqrt{r^2+h^2}=\sqrt{3.5^2+12^2}=12.5\text{ cm}

Only the curved surfaces are exposed.

A=πrl+2πr2=πr(l+2r)A=\pi rl+2\pi r^2=\pi r(l+2r)
=227×3.5×(12.5+7)=214.5 cm2=\frac{22}{7}\times3.5\times(12.5+7)=214.5\text{ cm}^2

Final answer

214.5 cm2\boxed{214.5\text{ cm}^2}

Question

Question 4

cube with hemisphere · greatest diameter

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution

Board-exam working:

The hemisphere's base must fit within the square top face, so its greatest diameter equals the cube's side.

d=7 cm,r=3.5 cmd=7\text{ cm},\qquad r=3.5\text{ cm}

Surface area = cube TSA − covered circular area + hemisphere CSA.

A=6a2−πr2+2πr2=6a2+πr2A=6a^2-\pi r^2+2\pi r^2=6a^2+\pi r^2
=6(7)2+227(3.5)2=6(7)^2+\frac{22}{7}(3.5)^2
=294+38.5=332.5 cm2=294+38.5=332.5\text{ cm}^2

Final answer

d=7 cm,A=332.5 cm2d=7\text{ cm},\qquad A=332.5\text{ cm}^2

Question

Question 5

cubical block · hemispherical depression

A hemispherical depression is cut from one face of a cubical block so that the hemisphere's diameter l equals the cube's edge. Determine the surface area of the remaining solid.

Solution

Board-exam working:

r=l2r=\frac l2

Start with the surface area of the cube, remove the circular opening and add the curved area of the depression.

A=6l2−πr2+2πr2A=6l^2-\pi r^2+2\pi r^2
=6l2+π(l2)2=6l^2+\pi\left(\frac l2\right)^2
=6l2+πl24=l24(24+π)=6l^2+\frac{\pi l^2}{4}=\frac{l^2}{4}(24+\pi)

Final answer

l24(24+π)\boxed{\frac{l^2}{4}(24+\pi)}

Question

Question 6

capsule · surface area

A medicine capsule is a cylinder with two hemispheres at its ends. Its total length is 14 mm and its diameter is 5 mm. Find its surface area.

Capsule formed by a cylinder and two hemispherical ends, length 14 millimetres and diameter 5 millimetres.
Figure 12-10 from the supplied NCERT chapter.

Solution

Board-exam working:

r=52=2.5 mmr=\frac52=2.5\text{ mm}

The two hemispherical ends together have length equal to the diameter, so the cylindrical length is:

h=14−5=9 mmh=14-5=9\text{ mm}

Surface area = cylinder CSA + surface area of a sphere.

A=2πrh+4πr2=2πr(h+2r)A=2\pi rh+4\pi r^2=2\pi r(h+2r)
=2×227×2.5×(9+5)=2\times\frac{22}{7}\times2.5\times(9+5)
=220 mm2=220\text{ mm}^2

Final answer

220 mm2\boxed{220\text{ mm}^2}

Question

Question 7

tent · canvas area and cost

A tent is a cylinder surmounted by a cone. The cylindrical part is 2.1 m high and 4 m in diameter; the conical top has slant height 2.8 m. Find the canvas area and its cost at ₹500 per m². The base is not covered.

Solution

Board-exam working:

r=42=2 m,h=2.1 m,l=2.8 mr=\frac42=2\text{ m},\qquad h=2.1\text{ m},\qquad l=2.8\text{ m}

Canvas area = cylinder CSA + cone CSA.

A=2πrh+πrl=πr(2h+l)A=2\pi rh+\pi rl=\pi r(2h+l)
=227×2×[2(2.1)+2.8]=\frac{22}{7}\times2\times[2(2.1)+2.8]
=447×7=44 m2=\frac{44}{7}\times7=44\text{ m}^2

Cost:

44×Rs. 500=Rs. 2200044\times\text{Rs. }500=\text{Rs. }22000

Final answer

Canvas area = 44 m²; cost = ₹22,000.

Question

Question 8

conical cavity · remaining surface area

A conical cavity of height 2.4 cm and diameter 1.4 cm is hollowed from a solid cylinder with the same height and diameter. Find the total surface area of the remaining solid to the nearest cm².

Solution

Board-exam working:

r=1.42=0.7 cm,h=2.4 cmr=\frac{1.4}{2}=0.7\text{ cm},\qquad h=2.4\text{ cm}

Slant height of the conical cavity:

l=r2+h2=0.72+2.42=2.5 cml=\sqrt{r^2+h^2}=\sqrt{0.7^2+2.4^2}=2.5\text{ cm}

Remaining surface = outer cylinder CSA + bottom base + inner cone CSA.

A=2πrh+πr2+πrlA=2\pi rh+\pi r^2+\pi rl
=πr(2h+r+l)=\pi r(2h+r+l)
=227×0.7×(4.8+0.7+2.5)=\frac{22}{7}\times0.7\times(4.8+0.7+2.5)
=17.6 cm2≈18 cm2=17.6\text{ cm}^2\approx18\text{ cm}^2

Final answer

18 cm2 (nearest cm2)\boxed{18\text{ cm}^2\text{ (nearest cm}^2\text{)}}

Question

Question 9

scooped hemispheres · surface area

A wooden article is made by scooping a hemisphere from each end of a solid cylinder. The cylinder is 10 cm high and has radius 3.5 cm. Find the article's total surface area.

Cylinder with a hemispherical cavity scooped from each circular end.
Figure 12-11 from the supplied NCERT chapter.

Solution

Board-exam working:

The exposed surface is the cylinder's curved surface plus the curved surfaces of two hemispherical cavities.

A=2πrh+2(2πr2)A=2\pi rh+2(2\pi r^2)
=2πr(h+2r)=2\pi r(h+2r)
=2×227×3.5×(10+7)=2\times\frac{22}{7}\times3.5\times(10+7)
=22×17=374 cm2=22\times17=374\text{ cm}^2

Final answer

374 cm2\boxed{374\text{ cm}^2}