Unit 5 — Geometry & Measurement · Math 8

This unit is about shapes in the real world — the diagonal across a screen, the cardboard it takes to box a gift, the water a tank can hold. You will learn the Pythagorean theorem for right triangles, then how to find the surface area and volume of prisms and cylinders, and how to build and picture 3D objects from flat nets. Each idea has a worked example, then a few problems to try. Answers are at the end of each section so you can check yourself right away. Use a calculator for the square roots and for π, and round only at the very end.

The toolkit words we will use

A right triangle has one 90° angle. The two sides that form the right angle are the legs; the longest side, across from the right angle, is the hypotenuse. A prism is a solid with two identical parallel ends (the bases) joined by flat rectangular sides; name it by its base — a triangular prism, a rectangular prism. A cylinder is like a prism but with two circular ends. Surface area is the total area of all the outside faces (units²). Volume is the space inside (units³).

A net is a flat pattern that folds up into a 3D object — unfold a cereal box and you see its net. A view is what the object looks like from one direction: the top, front, and side views together describe a solid.

1 · The Pythagorean theorem right triangles only

In any right triangle, the square of the hypotenuse equals the sum of the squares of the two legs. Write the legs as a and b and the hypotenuse as c: a² + b² = c². This works only for right triangles. To find a side, fill in what you know, then take a square root at the end.

Worked example — finding the hypotenuse

A right triangle has legs 3 cm and 4 cm. Find the hypotenuse c.

a² + b² = c² → 3² + 4² = c²

9 + 16 = c² → 25 = c²

c = √25 = 5

c = 5 cm

To find a leg, the hypotenuse is already known, so subtract: rearrange to a² = c² − b².

Worked example — finding a leg

A ladder 13 m long leans against a wall. Its base is 5 m from the wall. How high up the wall does it reach?

The ladder is the hypotenuse (c = 13); the ground is a leg (b = 5); the height is the other leg (a).

a² = c² − b² → a² = 13² − 5² = 169 − 25 = 144

a = √144 = 12

The ladder reaches 12 m up the wall.

When the answer is not a perfect square, leave the square root, then round at the end. Example: √20 ≈ 4.5 (to one decimal place).

Check yourself

Legs 6 cm and 8 cm. Find the hypotenuse.
Hypotenuse 10 m, one leg 6 m. Find the other leg.
Legs 2 cm and 3 cm. Find the hypotenuse, rounded to one decimal place.

Answers

1. 6² + 8² = 36 + 64 = 100, c = √100 = 10 cm

2. a² = 10² − 6² = 100 − 36 = 64, a = √64 = 8 m

3. 2² + 3² = 4 + 9 = 13, c = √13 ≈ 3.6 cm

2 · Surface area & volume of right prisms add the faces · stack the base

Surface area of a prism is the sum of the areas of all its faces. A neat shortcut for any right prism: add the two base areas, then add the area of the sides, which is the perimeter of the base × the length of the prism. SA = 2 × base + perimeter × length.

Volume of any right prism is the area of the base × the length (how far it is pushed back): V = base area × length. For a rectangular prism this becomes V = l × w × h and SA = 2(lw + lh + wh).

Worked example — rectangular prism

A box is 5 cm long, 3 cm wide, 4 cm high. Find its surface area and volume.

SA = 2(lw + lh + wh) = 2(5×3 + 5×4 + 3×4) = 2(15 + 20 + 12) = 2(47) = 94

SA = 94 cm²

V = l × w × h = 5 × 3 × 4 = 60

V = 60 cm³

For a triangular prism, the base is a triangle. Triangle area is A = ½ × b × h (b = base of the triangle, h = its height). The volume is that triangle area × the length of the prism. The surface area is the two triangular ends plus the three rectangles that wrap around: SA = 2(½bh) + (perimeter of triangle × length).

Worked example — triangular prism

A triangular prism has a right-triangle base with legs 3 cm and 4 cm (so the slanted side is 5 cm). The prism is 10 cm long. Find the surface area and volume.

Triangle area = ½ × 3 × 4 = 6 cm². Triangle perimeter = 3 + 4 + 5 = 12 cm.

SA = 2(6) + (12 × 10) = 12 + 120 = 132

SA = 132 cm²

V = base area × length = 6 × 10 = 60

V = 60 cm³

Check yourself

A rectangular prism is 6 cm × 2 cm × 5 cm. Find its volume and surface area.
A triangular prism has a triangular base with base 6 cm and height 4 cm; the prism is 9 cm long. Find its volume.
The same triangle (sides 6 cm, 5 cm, 5 cm; height 4 cm to the 6 cm base) on a 9 cm prism — find its surface area.

Answers

4. V = 6×2×5 = 60 cm³. SA = 2(6×2 + 6×5 + 2×5) = 2(12 + 30 + 10) = 2(52) = 104 cm²

5. Triangle area = ½ × 6 × 4 = 12 cm². V = 12 × 9 = 108 cm³

6. Perimeter = 6 + 5 + 5 = 16 cm. SA = 2(12) + (16 × 9) = 24 + 144 = 168 cm²

3 · Surface area & volume of cylinders round the base

A cylinder has two circular ends and a curved side. Unroll the curved side and it is a rectangle: its height is the cylinder's height h, and its width is the circle's circumference 2πr. So the surface area is the two circles plus that rectangle: SA = 2πr² + 2πrh. The volume is the circle area × the height: V = πr²h. Here r is the radius (half the diameter) and π ≈ 3.14.

Worked example — cylinder

A can has radius 3 cm and height 10 cm. Find its surface area and volume. Use π ≈ 3.14 and round to one decimal place.

SA = 2πr² + 2πrh = 2π(3²) + 2π(3)(10) = 2π(9) + 2π(30) = 18π + 60π = 78π

78 × 3.14 = 244.92

SA ≈ 244.9 cm²

V = πr²h = π(3²)(10) = π(9)(10) = 90π

90 × 3.14 = 282.6

V ≈ 282.6 cm³

Watch the diameter. If a problem gives the diameter, halve it to get the radius before you start. A 14 cm diameter means r = 7 cm.

Check yourself

A cylinder has radius 5 cm and height 8 cm. Find its volume (π ≈ 3.14, one decimal place).
The same cylinder — find its surface area (one decimal place).
A pipe has diameter 10 cm and height 20 cm. Find its volume (one decimal place).

Answers

7. V = π(5²)(8) = π(25)(8) = 200π ≈ 200 × 3.14 = 628.0 cm³

8. SA = 2π(5²) + 2π(5)(8) = 50π + 80π = 130π ≈ 130 × 3.14 = 408.2 cm²

9. Diameter 10 → r = 5. V = π(5²)(20) = 500π ≈ 500 × 3.14 = 1570.0 cm³

4 · Building, viewing & netting 3D objects flat ↔ solid

A net is the unfolded skin of a solid. A rectangular prism nets into 6 rectangles (3 matching pairs). A triangular prism nets into 2 triangles and 3 rectangles. A cylinder nets into 2 circles and 1 rectangle (the unrolled curved side). Counting the faces in the net is a good check before you add up surface area.

Views. We describe a solid by its top, front, and side views — each is a flat 2D drawing of what you see looking straight at that face. From three views you can rebuild the object, often from linking cubes.

Worked example — match net to solid

A net shows 2 circles and 1 rectangle. What solid is it?

Two circular ends and one curved side → a cylinder.

A cylinder.

Worked example — views of a stack

Three cubes sit in a row, with one extra cube on top of the leftmost cube.

Front view: an L-shape — a row of three squares with one square above the left end.

Top view: a row of three squares.

Different directions give different outlines — that is why we need all three views.

Check yourself

A net has 6 rectangles in 3 matching pairs. What solid is it?
How many rectangles are in the net of a triangular prism?

Answers

10. A rectangular prism.

11. Three rectangles (plus the two triangular ends).

If you want to go further

The Pythagorean theorem is really about areas: build a literal square on each side of a right triangle, and the two smaller squares (on the legs) have areas that add up to the big square (on the hypotenuse). The "3-4-5" and "5-12-13" triangles are famous because all three sides come out as whole numbers — try to find another set that works.