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Triangles and areas in 3D

Use vector displacements to find triangle side lengths, angles and areas in three dimensions. Choose the correct vertex, test right angles and split a planar quadrilateral into triangles.

Before you startThree-dimensional distances, Pythagoras, sine rule, cosine rule and triangle area.

01 / Three noncollinear points form a triangle in a plane

The triangle may be tilted in space, but its internal geometry is two-dimensional.

cos A = (AB² + AC² − BC²)/(2 AB AC)

AB and AC meet at A; BC is the opposite side.

First use the coordinates to calculate the true side lengths. Then use ordinary triangle rules within the triangle’s plane. A flat sketch can distort both lengths and angles, so it is a guide to relationships rather than a measuring instrument.

Choose the angle vertexExplore
A triangle from three points in spaceChoose a vertex. Amber shows the two sides meeting there, blue the opposite side. The picture is a projection, so use calculated lengths and angles.ABC

A = (1, 0, 1), B = (4, 0, 5), C = (1, 4, 1).

AB = 5; AC = 4; BC = √41.

From A: AB = (3, 0, 4), AC = (0, 4, 0).

cos A = (25 + 16 − 41)/(2 × 5 × 4) = 0; A = 90°.

Both sides in the cosine rule meet at the chosen vertex. The subtracted square belongs to the opposite side. The diagram is not a scale drawing of the angles.

02 / Form the three displacement vectors

Every side length comes from the magnitude of its displacement.

A = (1, 0, 1), B = (4, 0, 5), C = (1, 4, 1). Find the side lengths.Worked example

AB = (3, 0, 4), AC = (0, 4, 0)

Subtract A from B and C.

BC = (−3, 4, −4)

Subtract B from C.

AB = 5, AC = 4, BC = √41

Take each magnitude, keeping the surd exact.

01 · Three sides

O = (0, 0, 0), P = (2, 0, 0), Q = (0, 3, 0). Find OP, OQ and PQ.

Hint

Find the displacements and their magnitudes.

Worked solution

OP = 2, OQ = 3, PQ = √13.

04 / Choose two directed sides from the same vertex

At A, use AB and AC, not BA and AC.

Identify the sides surrounding angle ABC.Worked example

The middle letter B names the vertex

Angle ABC is at B.

Use BA and BC

Both arrows start at B.

AC is the opposite side

Its square is subtracted in the cosine-rule numerator.

04 · Name the angle

Which angle is formed by CA and CB?

Hint

Both arrows start at C.

Worked solution

Angle ACB, at C.

05 · Opposite side

Which side is opposite angle BAC?

Hint

Angle BAC is at A.

Worked solution

BC.

05 / Use the cosine rule for a general angle

Keep exact squared lengths until the inverse cosine.

For the main triangle, find angle ABC.Worked example

cos B = (BA² + BC² − AC²)/(2 BA BC)

Sides adjacent to B are 5 and √41.

cos B = (25 + 41 − 16)/(10√41) = 5/√41

Simplify before rounding.

B ≈ 38.660°

C is approximately 51.340°, so the three angles sum to 180°.

06 · Another cosine-rule angle

A triangle has AB = AC = √5 and BC = √8. Find angle A.

Hint

Use (5 + 5 − 8)/(2√5√5).

Worked solution

cos A = 1/5, so A ≈ 78.463°.

06 / Compare squared side lengths to classify the triangle

Equal squared lengths identify equal lengths.

A = (0, 0, 0), B = (1, 1, 1), C = (2, 0, 0).Worked example

AB² = 3, AC² = 4, BC² = 3

Calculate from coordinate differences.

AB = BC, so the triangle is isosceles

The equal sides meet at B.

cos B = (3 + 3 − 4)/(2 × 3) = 1/3

The apex angle is about 70.529°.

07 · Equilateral in space

Show that (1, 0, 0), (0, 1, 0), (0, 0, 1) form an equilateral triangle.

Hint

Compare all three squared lengths.

Worked solution

Every squared side length is 2, so all sides have length √2. Each angle is 60°.

07 / Use two sides and their included angle

Area = ½ab sin C uses the angle between those two sides.

Find the area of the main right triangle.Worked example

AB = 5, AC = 4, angle A = 90°

These sides meet at the known angle.

Area = ½ × 5 × 4 × sin 90° = 10

The area lies in the triangle’s plane.

The projected area on a page need not be 10

Projection can shrink an area.

08 · Equilateral area

Find the area of an equilateral triangle of side √2.

Hint

Use two sides and included angle 60°.

Worked solution

½ × √2 × √2 × sin 60° = √3/2.

09 · Isosceles area

AB = AC = √5 and cos A = 1/5. Find the exact area.

Hint

sin A is positive for an internal triangle angle.

Worked solution

sin A = √(1 − 1/25) = 2√6/5. Area = ½ × 5 × 2√6/5 = √6.

08 / A perpendicular height must be measured in the triangle plane

Use the side chosen as base.

The main triangle has area 10. Find its perpendicular height to BC.Worked example

Base BC = √41

This is not one of its perpendicular legs.

10 = ½ × √41 × h

Area = half base times height.

h = 20/√41

A different base has a different corresponding height.

10 · Height from area

A triangle has area 12 cm² and a side of length 5 cm. Find its perpendicular height to that side.

Hint

12 = ½ × 5 × h.

Worked solution

h = 24/5 = 4.8 cm.

09 / Match every sine-rule angle to its opposite side

Use a known angle and its opposite side as a pair.

In the main triangle, B ≈ 38.660° is opposite AC = 4, and A = 90° is opposite BC = √41.Worked example

sin B / 4 = sin 90° / √41

Keep the opposite pairings consistent.

sin B = 4/√41

This agrees with the cosine-rule result.

B is acute because the triangle already has a right angle

Do not choose the supplementary sine solution.

11 · Avoid mismatched pairs

For triangle ABC, complete AB/sin(?) = AC/sin(?) = BC/sin(?).

Hint

Use the vertex not on each side.

Worked solution

AB/sin C = AC/sin B = BC/sin A.

12 · Sine ambiguity

If sin B = 0.6, why is B ≈ 36.870° not always the only possible triangle angle?

Hint

Sine is positive at an angle and its supplement.

Worked solution

B could also be about 143.130° if the remaining geometry allows it. Check side lengths and the other angles.

10 / Split a planar shape along a diagonal

Check the vertex order and the shape before adding areas.

A = (0, 0, 0), B = (3, 0, 4), D = (0, 2, 0), C = (3, 2, 4). Find the area of ABCD.Worked example

AB = (3, 0, 4), AD = (0, 2, 0)

The adjacent side lengths are 5 and 2.

BC = AD and DC = AB

The four points form a parallelogram in this order.

AC² = 29 = 25 + 4

Triangle ABC is right-angled at B; the other half is congruent.

Area ABCD = 2 × (½ × 5 × 2) = 10

The parallelogram is a rectangle in its tilted plane.

13 · Triangle half

What is the area of triangle ACD in that rectangle?

Hint

A diagonal divides the rectangle into equal areas.

Worked solution

5 square units.

14 · Nonplanar points

Can you automatically call four points in 3D a planar quadrilateral?

Hint

Three noncollinear points determine a plane, but the fourth might be outside it.

Worked solution

No. Check the geometry. For a nonplanar four-point boundary, an area needs a specified surface or triangulation.

11 / Use coordinates for lengths, then triangle geometry

Do not measure a projected sketch.

  • Find displacement vectors and exact squared side lengths.
  • Identify the angle vertex and opposite side.
  • Use Pythagoras only when its condition is met.
  • Use the cosine rule for an angle from three sides.
  • Use ½ab sin C with the included angle.
  • Keep base and perpendicular height paired.
  • Check planarity and vertex order for a quadrilateral.

15 · Degenerate triangle

A = (0, 0, 0), B = (1, 2, 2), C = (2, 4, 4). Do these form a nondegenerate triangle?

Hint

Compare AC with AB.

Worked solution

No. AC = 2AB, so the points are collinear and the area is zero.

16 · Units

Coordinates are measured in metres. What units do side lengths and areas have?

Hint

Area involves a product of two lengths.

Worked solution

Side lengths are in metres; areas are in square metres.

Section 1 of 11 · Three noncollinear points form a triangle in a plane