Unit notes and practice questions

Grade 10 Mathematics unit 6: Solid Figures

Everything here is for unit 6 of the Grade 10 Mathematics textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

4
Cards: 10
Practice questions on unit 6: 271–326
The textbook pages for unit 6

What unit 6 covers

1.Revision of Cylinders and Prisms
2.Pyramids, cones and Spheres
3.Frustum of pyramids and cones
4.Surface areas and volumes of composed solids
5.Applications

Short notes for unit 6

The cards filed under this chapter on the Grade 10 Mathematics page. Open one at a time.

Cross-Section Area RatioFormula

\frac{A'}{A} = \frac{k^2}{h^2}

\frac{A'}{A} = \frac{k^2}{h^2}

$A'$: Area of the horizontal cross-section $m^2$
$A$: Area of the base of the pyramid or cone $m^2$
$k$: Distance from the vertex to the cross-section plane $m$
$h$: Altitude of the pyramid or cone $m$

When you use it

Use when finding the area of a parallel slice or cross-section at a known distance from the vertex of a pyramid or cone.

Watch out

Distance k is measured from the vertex, not from the base. If distance from the base is given, subtract it from h first.

Drafted from Grade 10 Mathematics, pages 272-327, then checked twice before it went up

Lateral Surface Area of a Frustum of a ConeFormula

LSA = \pi l(r + r')

LSA = \pi l(r + r')

$LSA$: Lateral surface area of the cone frustum $m^2$
$l$: Slant height of the frustum $m$
$r$: Radius of the larger base $m$
$r'$: Radius of the smaller base $m$

When you use it

Use this to compute the curved surface area of a frustum of a right circular cone.

Watch out

This formula gives only the lateral area. To find the total surface area, add both circular base areas.

Drafted from Grade 10 Mathematics, pages 272-327, then checked twice before it went up

Surface Area of Composed SolidsCommon mistake

When you use it

Reach for this reminder whenever finding the total surface area of an object made by joining two solids together.

Watch out

Do not simply add the total surface areas of the individual solids. The interior touching faces are covered and must not be included.

Drafted from Grade 10 Mathematics, pages 272-327, then checked twice before it went up

Volume of a Frustum of a PyramidFormula

V = \frac{1}{3} h(A + A' + \sqrt{A A'})

V = \frac{1}{3} h(A + A' + \sqrt{A A'})

$V$: Volume of the frustum of a pyramid $m^3$
$h$: Altitude of the frustum $m$
$A$: Area of the lower base $m^2$
$A'$: Area of the upper base $m^2$

When you use it

Use to calculate the volume of a pyramid frustum when the upper and lower base areas and altitude are known.

Watch out

Remember to include the geometric mean term sqrt(A A') inside the parentheses.

Drafted from Grade 10 Mathematics, pages 272-327, then checked twice before it went up

Practice questions on unit 6

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

A right circular cylinder has a base radius of 3 cm and an altitude of 5 cm. What is its volume?

A.45π cm³
B.48π cm³
C.30π cm³
D.15π cm³

Show the answer and why

Answer: A 45π cm³

Why

The volume of a right circular cylinder is V = πr²h. Here r = 3 and h = 5, so V = π × 3² × 5 = π × 9 × 5 = 45π cm³.

Question 2

A right prism has a base that is a right triangle with legs 3 cm and 4 cm. The altitude of the prism is 6 cm. What is its volume?

A.6 cm³
B.36 cm³
C.72 cm³
D.84 cm³

Show the answer and why

Answer: B 36 cm³

Why

First find the base area: BA = ½ × 3 × 4 = 6 cm². The volume of a prism is V = (BA)h, so V = 6 × 6 = 36 cm³.

Question 3

A sphere has a radius of 10 cm. What is its surface area?

A.100π cm²
B.400π cm²
C.4000π/3 cm²
D.1000π cm²

Show the answer and why

Answer: B 400π cm²

Why

The surface area of a sphere is SA = 4πr². With r = 10, SA = 4π × 10² = 4π × 100 = 400π cm².

Question 4

A conical tent for a wedding in Gondar is 6 m high. The radius of its base is 8 m. What is the slant height of the tent?

A.14 m
B.√28 m
C.9 m
D.10 m

Show the answer and why

Answer: D 10 m

Why

The slant height of a right circular cone is l = √(h² + r²). Here h = 6 and r = 8, so l = √(6² + 8²) = √(36 + 64) = √100 = 10 m.

Question 5

A regular square pyramid has a base edge of 4 cm and a slant height of 5 cm. What is its total surface area?

A.40 cm²
B.16 cm²
C.56 cm²
D.80 cm²

Show the answer and why

Answer: C 56 cm²

Why

The perimeter of the base is p = 4 × 4 = 16 cm. The lateral surface area is LSA = ½pl = ½ × 16 × 5 = 40 cm². The base area is 4 × 4 = 16 cm². So the total surface area is TSA = LSA + BA = 40 + 16 = 56 cm².

Question 6

A regular square pyramid has a base edge of 4 cm and an altitude of 7 cm. What is its volume?

A.112/3 cm³
B.112 cm³
C.28/3 cm³
D.56/3 cm³

Show the answer and why

Answer: A 112/3 cm³

Why

The base is a square with area BA = 4 × 4 = 16 cm². The volume of a pyramid is one third of the base area times the altitude: V = ⅓ × 16 × 7 = 112/3 cm³.

Question 7

The base area of a pyramid is 270 cm² and its altitude is 6 cm. A horizontal cross-section is taken 4 cm from the vertex. What is the area of this cross-section?

A.90 cm²
B.120 cm²
C.160 cm²
D.30 cm²

Show the answer and why

Answer: B 120 cm²

Why

For a cross-section at distance k from the vertex, the area of the cross-section divided by the base area equals k²/h². Here k = 4 and h = 6, so the ratio is 4²/6² = 16/36 = 4/9. The cross-section area is 270 × 4/9 = 120 cm².

Question 8

A frustum of a right circular cone has top and bottom diameters of 6 m and 10 m. Its height is 12 m. What is its volume? Give your answer in terms of π.

A.49π m³
B.147π m³
C.588π m³
D.196π m³

Show the answer and why

Answer: D 196π m³

Why

The radii are r = 5 m (bottom) and r′ = 3 m (top), since the radius is half the diameter. The frustum volume is V = ⅓πh(r² + r′² + rr′) = ⅓π × 12 × (25 + 9 + 15) = ⅓π × 12 × 49 = 196π m³.

Question 9

The areas of the upper and lower bases of a frustum of a pyramid are 36 cm² and 81 cm². The altitude of the frustum is 5 cm. What is the altitude of the whole pyramid?

A.5 cm
B.10 cm
C.15 cm
D.20 cm

Show the answer and why

Answer: C 15 cm

Why

Let k be the distance from the vertex to the upper base. The area ratio gives 36/81 = k²/(5 + k)², so 6/9 = k/(5 + k). Cross-multiplying: 6(5 + k) = 9k, so 30 = 3k and k = 10 cm. The altitude of the whole pyramid is 5 + k = 5 + 10 = 15 cm.

Question 10

A solid is made of a cylinder with radius 4 cm and height 7 cm, with a cone of altitude 3 cm on top. The cone has the same radius of 4 cm. The overall height of the solid is 10 cm. What is the total volume of the solid? Give your answer in terms of π.

A.128π cm³
B.112π cm³
C.92π cm³
D.176π cm³

Show the answer and why

Answer: A 128π cm³

Why

The volume of the solid is the cylinder volume plus the cone volume. Cylinder: πr²h = π × 16 × 7 = 112π cm³. Cone: ⅓πr²h = ⅓π × 16 × 3 = 16π cm³. The total volume is 112π + 16π = 128π cm³.

The app holds 12 questions on this unit, marked as you go. These are the first 10.

The textbook pages for unit 6

Unit 6 runs from page 271 to page 326 of the Ministry of Education edition.

Ministry of Education edition, English, 373 pages

Download the PDF

All Grade 10 Mathematics short notes

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