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10th Maths Chapter 7 Mensuration

10th Maths Chapter 7 Mensuration solutions, with the chapter's formulas and every exercise worked step by step, as per the 2026-27 Samacheer Kalvi textbook.

This chapter deals with the surface areas and volumes of solid shapes: cuboid, cube, cylinder, cone, sphere, hemisphere, their hollow forms, and the frustum of a cone. It also looks at combined solids and at objects that are melted and recast into a new shape. In the exercises, students find curved, lateral and total surface areas, work out volumes, count how many small solids fill a bigger one, and solve cost and thickness problems.

Exercises in this chapter

Key formulas

  • Cuboid — Lateral surface area

    2h(l+b)2h(l+b)

  • Cuboid — Total surface area

    2(lb+bh+lh)2(lb+bh+lh)

  • Cuboid — Volume

    l×b×hl \times b \times h

  • Cube — Lateral surface area

    4a24a^{2}

  • Cube — Total surface area

    6a26a^{2}

  • Cube — Volume

    a3a^{3}

  • Right circular cylinder — Curved surface area

    2πrh2\pi r h

  • Right circular cylinder — Total surface area

    2πr(h+r)2\pi r(h+r)

  • Right circular cylinder — Volume

    πr2h\pi r^{2} h

  • Right circular cone — Curved surface area

    πrl\pi r l

  • Right circular cone — Slant height

    l=r2+h2l=\sqrt{r^{2}+h^{2}}

  • Right circular cone — Total surface area

    πrl+πr2=πr(l+r)\pi r l+\pi r^{2}=\pi r(l+r)

  • Right circular cone — Volume

    13πr2h\dfrac{1}{3}\pi r^{2} h

  • Sphere — Curved surface area

    4πr24\pi r^{2}

  • Sphere — Total surface area

    4πr24\pi r^{2}

  • Sphere — Volume

    43πr3\dfrac{4}{3}\pi r^{3}

  • Hemisphere — Curved surface area

    2πr22\pi r^{2}

  • Hemisphere — Total surface area

    3πr23\pi r^{2}

  • Hemisphere — Volume

    23πr3\dfrac{2}{3}\pi r^{3}

  • Hollow cylinder — Curved surface area

    2π(R+r)h2\pi(R+r)h

  • Hollow cylinder — Total surface area

    2π(R+r)(R−r+h)2\pi(R+r)(R-r+h)

  • Hollow cylinder — Volume

    π(R2−r2)h\pi\left(R^{2}-r^{2}\right)h

  • Hollow sphere — Outer surface area

    4πR24\pi R^{2}

  • Hollow sphere — Total surface area

    4π(R2+r2)4\pi\left(R^{2}+r^{2}\right)

  • Hollow sphere — Volume

    43π(R3−r3)\dfrac{4}{3}\pi\left(R^{3}-r^{3}\right)

  • Hollow hemisphere — Curved surface area

    2π(R2+r2)2\pi\left(R^{2}+r^{2}\right)

  • Hollow hemisphere — Total surface area

    π(3R2+r2)\pi\left(3R^{2}+r^{2}\right)

  • Hollow hemisphere — Volume

    23π(R3−r3)\dfrac{2}{3}\pi\left(R^{3}-r^{3}\right)

  • Frustum of a right circular cone — Curved surface area

    π(R+r)l\pi(R+r)l

  • Frustum of a right circular cone — Slant height

    l=h2+(R−r)2l=\sqrt{h^{2}+(R-r)^{2}}

  • Frustum of a right circular cone — Total surface area

    π(R+r)l+πR2+πr2\pi(R+r)l+\pi R^{2}+\pi r^{2}

  • Frustum of a right circular cone — Volume

    13πh[R2+r2+Rr]\dfrac{1}{3}\pi h\left[R^{2}+r^{2}+Rr\right]

Chapter 7 one-mark practice

  1. The curved surface area of a right circular cone of height 1515 cm and base diameter 1616 cm is

    1. (1) 60π60\pi cm2^2
    2. (2) 68π68\pi cm2^2
    3. (3) 120π120\pi cm2^2
    4. (4) 136π136\pi cm2^2
    Show answer

    Answer: (4) 136π136\pi cm2^2

    1. r=162=8r=\dfrac{16}{2}=8 cm, h=15h=15 cm.

    2. l=r2+h2=64+225=289=17l=\sqrt{r^2+h^2}=\sqrt{64+225}=\sqrt{289}=17 cm.

    3. CSA =πrl=π×8×17=136π=\pi r l=\pi\times 8\times 17=136\pi cm2^2.

    Question 1 of 10

  2. If two solid hemispheres of same base radius rr units are joined together along their bases, then curved surface area of this new solid is

    1. (1) 4πr24\pi r^2 sq. units
    2. (2) 6πr26\pi r^2 sq. units
    3. (3) 3πr23\pi r^2 sq. units
    4. (4) 8πr28\pi r^2 sq. units
    Show answer

    Answer: (1) 4πr24\pi r^2 sq. units

    1. Two hemispheres joined along their bases form a sphere of radius rr.

    2. CSA of sphere =4πr2=4\pi r^2 sq. units.

    Question 2 of 10

  3. The height of a right circular cone whose radius is 55 cm and slant height is 1313 cm will be

    1. (1) 1212 cm
    2. (2) 1010 cm
    3. (3) 1313 cm
    4. (4) 55 cm
    Show answer

    Answer: (1) 1212 cm

    1. l2=r2+h2l^2=r^2+h^2

    2. 132=52+h2⇒h2=169−25=14413^2=5^2+h^2\Rightarrow h^2=169-25=144

    3. h=12h=12 cm.

    Question 3 of 10

  4. If the radius of the base of a right circular cylinder is halved keeping the same height, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is

    1. (1) 1:21:2
    2. (2) 1:41:4
    3. (3) 1:61:6
    4. (4) 1:81:8
    Show answer

    Answer: (2) 1:41:4

    1. New volume =π(r2)2h=πr2h4=\pi\left(\dfrac{r}{2}\right)^2h=\dfrac{\pi r^2h}{4}.

    2. Ratio =πr2h/4πr2h=14=\dfrac{\pi r^2h/4}{\pi r^2h}=\dfrac{1}{4}.

    Question 4 of 10

  5. The total surface area of a cylinder whose radius is 13\dfrac{1}{3} of its height is

    1. (1) 9πh28\dfrac{9\pi h^2}{8} sq.units
    2. (2) 24πh224\pi h^2 sq.units
    3. (3) 8πh29\dfrac{8\pi h^2}{9} sq.units
    4. (4) 56πh29\dfrac{56\pi h^2}{9} sq.units
    Show answer

    Answer: (3) 8πh29\dfrac{8\pi h^2}{9} sq.units

    1. Given r=h3r=\dfrac{h}{3}.

    2. TSA =2πr(r+h)=2πh3(h3+h)=2\pi r(r+h)=2\pi\dfrac{h}{3}\left(\dfrac{h}{3}+h\right)

    3. =2πh3×4h3=8πh29=2\pi\dfrac{h}{3}\times\dfrac{4h}{3}=\dfrac{8\pi h^2}{9}.

    Question 5 of 10

  6. In a hollow cylinder, the sum of the external and internal radii is 1414 cm and the width is 44 cm. If its height is 2020 cm, the volume of the material in it is

    1. (1) 5600π5600\pi cm3^3
    2. (2) 1120π1120\pi cm3^3
    3. (3) 56π56\pi cm3^3
    4. (4) 3600π3600\pi cm3^3
    Show answer

    Answer: (2) 1120π1120\pi cm3^3

    1. R+r=14R+r=14, width R−r=4R-r=4, h=20h=20.

    2. Volume of material =π(R2−r2)h=π(R+r)(R−r)h=\pi(R^2-r^2)h=\pi(R+r)(R-r)h

    3. =π×14×4×20=1120π=\pi\times 14\times 4\times 20=1120\pi cm3^3.

    Question 6 of 10

  7. If the radius of the base of a cone is tripled and the height is doubled then the volume is

    1. (1) made 66 times
    2. (2) made 1818 times
    3. (3) made 1212 times
    4. (4) unchanged
    Show answer

    Answer: (2) made 1818 times

    1. New volume =13π(3r)2(2h)=13π⋅9r2⋅2h=\dfrac{1}{3}\pi(3r)^2(2h)=\dfrac{1}{3}\pi\cdot 9r^2\cdot 2h

    2. =18×13πr2h=18\times\dfrac{1}{3}\pi r^2h.

    3. So the volume is made 1818 times.

    Question 7 of 10

  8. The total surface area of a hemi-sphere is how much times the square of its radius.

    1. (1) π\pi
    2. (2) 4π4\pi
    3. (3) 3π3\pi
    4. (4) 2π2\pi
    Show answer

    Answer: (3) 3π3\pi

    1. TSA of hemisphere =3πr2=3\pi r^2.

    2. So it is 3π3\pi times the square of the radius.

    Question 8 of 10

  9. A solid sphere of radius xx cm is melted and cast into a shape of a solid cone of same radius. The height of the cone is

    1. (1) 3x3x cm
    2. (2) xx cm
    3. (3) 4x4x cm
    4. (4) 2x2x cm
    Show answer

    Answer: (3) 4x4x cm

    1. Volume of sphere == volume of cone

    2. 43πx3=13πx2h\dfrac{4}{3}\pi x^3=\dfrac{1}{3}\pi x^2h

    3. h=4xh=4x cm.

    Question 9 of 10

  10. A frustum of a right circular cone is of height 1616cm with radii of its ends as 88cm and 2020cm. Then, the volume of the frustum is

    1. (1) 3328π3328\pi cm3^3
    2. (2) 3228π3228\pi cm3^3
    3. (3) 3240π3240\pi cm3^3
    4. (4) 3340π3340\pi cm3^3
    Show answer

    Answer: (1) 3328π3328\pi cm3^3

    1. V=13πh(R2+Rr+r2)V=\dfrac{1}{3}\pi h(R^2+Rr+r^2)

    2. =13π(16)(400+160+64)=\dfrac{1}{3}\pi(16)(400+160+64)

    3. =163π(624)=3328π=\dfrac{16}{3}\pi(624)=3328\pi cm3^3.

    Question 10 of 10

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