Last reviewed: 11 November 2026
Updated November 2026. These Grade 9 area notes show how to find the area of a regular pentagon and a regular hexagon, then how to work out the surface area of triangular and rectangular based prisms, pyramids, cones and spheres, and the area of a sector or segment of a circle. Area is the measure of a surface. For a pentagonal floor tile or a hexagonal bolt head you add plane shapes; for a closed carton or a football you add every face, including curved ones.
The examples below are original practice written for Kenyan junior school learners, not an official paper. They stay inside the Grade 9 Mathematics measurements sub-strand on area in the design published under the KICD Grade Nine curriculum designs. For the full topic list start with the Grade 9 maths revision guide and the Grade 9 notes hub.
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Where Grade 9 Area Notes Sit in the Syllabus
In Grade 9 you are no longer finding only rectangles and triangles. The design asks you to calculate the area of a pentagon and a hexagon, the surface area of triangular and rectangular based prisms, the surface area of triangular, rectangular and square based pyramids, the area of a sector and a segment of a circle, and the surface area of a cone and a sphere in real-life situations. Earlier work on rectangles and circles, including the Grade 8 area of circles and sectors notes, is assumed. These notes do not cover volume; that is the next sub-strand, with its own Grade 9 volume of solids notes.
Area of a Regular Pentagon and a Regular Hexagon
A regular pentagon has five equal sides and five equal interior angles. A regular hexagon has six. A reliable method, which matches how you use cut-outs in class, is to join the centre to every vertex and add the congruent isosceles triangles. The same idea appears in OpenStax Prealgebra’s section on rectangles, triangles and trapezoids, where composite shapes are broken into familiar pieces.
For a regular polygon with n sides of length s, the area is A = (n/2) × s2 × sin(360°/n).
- Regular pentagon (n = 5): A = (5/2) s2 sin 72°. Using sin 72° ≈ 0.9511, A ≈ 2.378 s2.
- Regular hexagon (n = 6): A = (6/2) s2 sin 60° = 3 s2 (√3/2) = (3√3/2) s2. A hexagon also splits into six equilateral triangles, which is a quicker classroom check.
Worked example: pentagonal flower bed
A regular pentagonal flower bed at a school has side 8 cm on a scale drawing (the real bed will be enlarged later). Find the area of the drawing.
A = (5/2) × 82 × sin 72° = 2.5 × 64 × 0.9511 = 160 × 0.9511 = 152.18 cm2 (to 2 decimal places).
Worked example: hexagonal bolt head
A regular hexagonal bolt head has side 6 mm. Find its area.
A = (3√3/2) × 62 = (3√3/2) × 36 = 54√3. Taking √3 ≈ 1.732, A ≈ 54 × 1.732 = 93.53 mm2.
If you forget the formula, draw the six equilateral triangles: each has area (√3/4) × 36 = 9√3, and six of them give 54√3, the same result.
Surface Area of Prisms and Pyramids
Surface area is the total area of every face. The design asks you to sketch the net, then add the areas of the faces, using the plane-figure formulas you already know.
| Solid | Faces to add | Useful formula |
|---|---|---|
| Rectangular based prism (cuboid) | 3 pairs of rectangles | 2(lw + lh + wh) |
| Triangular based prism | 2 triangular ends + 3 rectangular sides | 2 × (area of triangle) + (sum of the three side rectangles) |
| Pyramid | base + triangular lateral faces | base area + (1/2) × perimeter of base × slant height, when the slant height is the same on every face |
Worked example: triangular prism
A camping tent frame is modelled as a triangular prism. The two ends are right-angled triangles with sides 3 cm, 4 cm and 5 cm, and the length of the prism is 10 cm. Find the total surface area.
- Area of one triangular end = (1/2) × 3 × 4 = 6 cm2. Two ends: 12 cm2.
- The three rectangular sides measure 3 × 10, 4 × 10 and 5 × 10, so 30 + 40 + 50 = 120 cm2.
- Total surface area = 12 + 120 = 132 cm2.
Worked example: rectangular prism and square pyramid
A closed chalk box measures 8 cm by 5 cm by 4 cm. Surface area = 2(8×5 + 8×4 + 5×4) = 2(40 + 32 + 20) = 2 × 92 = 184 cm2.
A square-based model pyramid has base 6 cm by 6 cm and each triangular face has slant height 5 cm (the height of the triangular face, measured down the middle of a side). Base area = 36 cm2. Four triangles: 4 × (1/2) × 6 × 5 = 60 cm2. Total surface area = 96 cm2.
Sectors, Segments, Cones and Spheres
A sector is a “slice of pie” bounded by two radii and an arc. A segment is the smaller (or larger) region between a chord and its arc, so its area is the sector minus the triangular portion. The same π that you used in Grade 8 appears in OpenStax Prealgebra’s chapter on circles and irregular figures.
- Area of a sector = (θ/360) × πr2, where θ is the angle at the centre in degrees.
- Area of a segment = area of sector − area of the triangle formed by the two radii and the chord.
- Curved surface area of a cone = πrl, where l is the slant height. A closed cone also includes the base πr2, so total surface area = πr(l + r). An open cone (a funnel without a lid) uses only πrl.
- Surface area of a sphere = 4πr2, the formula used for spherical balls in the design.
Worked example: sector and segment
A circle has radius 14 cm. A sector with angle 90° is cut out. Taking π = 22/7:
Area of sector = (90/360) × (22/7) × 14 × 14 = (1/4) × 22 × 2 × 14 = 154 cm2.
The two radii and the chord form an isosceles right-angled triangle with area (1/2) × 14 × 14 = 98 cm2. Area of the minor segment = 154 − 98 = 56 cm2.
Worked example: cone and sphere
A closed conical paper cup has radius 7 cm and slant height 25 cm. Total surface area = πr(l + r) = (22/7) × 7 × (25 + 7) = 22 × 32 = 704 cm2. If the cup is open at the top, use only the curved surface: πrl = 22 × 25 = 550 cm2.
A football modelled as a sphere of radius 7 cm has surface area 4πr2 = 4 × (22/7) × 49 = 4 × 22 × 7 = 616 cm2.
Original Practice Questions With Worked Answers
- Find the area of a regular hexagon of side 10 cm. A = (3√3/2) × 100 = 150√3 ≈ 150 × 1.732 = 259.8 cm2.
- A regular pentagon has side 4 cm. Find its area, using sin 72° = 0.9511. A = (5/2) × 16 × 0.9511 = 40 × 0.9511 = 38.04 cm2.
- A closed rectangular carton measures 12 cm by 9 cm by 5 cm. Find the surface area of card used. 2(108 + 60 + 45) = 2 × 213 = 426 cm2.
- A sector of radius 21 cm has angle 60°. Taking π = 22/7, find its area. (60/360) × (22/7) × 21 × 21 = (1/6) × 22 × 3 × 21 = 231 cm2.
- A sphere has diameter 14 cm. Find its surface area, taking π = 22/7. Radius = 7 cm; 4πr2 = 616 cm2, the same working as the football above.
- A learner adds only the four walls of a cuboid and forgets the lid and the base. What was missed? The two faces measuring length × width. Always open the net and count every face, including hidden ones.
Cover the answers, attempt each question, then check your working line by line. Convert units first if a question mixes cm and m; the metric unit converter is useful for that check. For timed practice, open the Grade 9 revision papers or test yourself in the Grade 9 CBC Challenge. These notes are free; full revision access is available with the KSh 20 Day Pass or the KSh 500 Monthly Pass.
Frequently Asked Questions
What is covered in Grade 9 area notes?
Grade 9 area covers regular pentagons and hexagons, the surface area of triangular and rectangular based prisms, the surface area of triangular, rectangular and square based pyramids, sectors and segments of a circle, and the surface area of cones and spheres.
How do you find the area of a regular hexagon?
Split it into six equilateral triangles, or use A = (3√3/2) s2, where s is the side length. For side 6 cm the area is 54√3, about 93.53 cm2.
What is the difference between a sector and a segment?
A sector is the pie-slice between two radii and an arc. A segment is the region between a chord and the arc, found by subtracting the triangular portion from the sector.
Is the surface area of a cone always πr(l + r)?
That formula is for a closed cone, which includes the circular base. An open cone or funnel uses only the curved surface πrl. Read the question to see whether a lid or base is included.