The parallelogram, area of a parallelogram, Trapezoid or trapezium, The area of the trapezoid given two parallel sides and angles at ends of the side, Diagonals of a trapezoid, determining the diagonals of the trapezoid given its four sides, A right pyramid and regular right pyramid, Sections of solids

Geometry and use of trigonometry

Applications of trigonometry in plane geometry

The parallelogram, area of a parallelogram

The parallelogram in the figure consists of two congruent triangles, DABD and DBCD, therefore its area

Triangles, BEC and ABD, are congruent as are ABS and DSC, thus the area of parallelogram equals the area of the triangle AEC. We use the formula for the area of a triangle given two adjacent sides and

angle between them,

Trapezoid or trapezium

The area of a trapezoid given its four sides

The trapezoid in the figure consists of a triangle and a parallelogram. According to Heron’s formula for the area of a triangle

The area of the parallelogram A_P = c · h where h is the height of the triangle too. Therefore, the area of the triangle

Thus, the area of the trapezoid given its four sides

The area of the trapezoid given two parallel sides and angles at ends of the side

The area of trapezoid given two parallel sides (bases), a and c, and angles, a and b at ends of the side a

applying the sine law

so, the area of the trapezoid

Diagonals of a trapezoid, determining the diagonals of the trapezoid given its four sides

In DABC, using the cosine law e² = a² + b² - 2abcosb,

and in DACD, e² = c² + d² - 2cdcosd,

then a² + b² - 2ab cosb = c² + d² - 2cdcosd

and since cosd = cos(180° - a) = - cosa

then 2cdcosa + 2abcosb = a² - c² + b² - d² (1)

Similarly, in DABD, f² = a² + d² - 2adcosa,

and in DBCD, f² = b² + c² - 2bccosg,

thus, a² + d² - 2adcosa = b² + c² - 2bccosg and since cosdg = cos(180° - b) = - cosb

then 2adcosa + 2bccosb = a² - b² + d² - c² (2)

To separate the term containing cosa from (1) and (2), multiply (1) by c and (2) by a, then subtract first equation from second thus, obtained is

2(a² - c²) · d cosa = (a² - c²) · (a - c) + (d² - b²) · (a + c)

To separate the term containing cosb from (1) and (2), multiply (1) by a and (2) by c, then subtract second equation from first thus, obtained is

2(a² - c²) · b cosb = (a² - c²) · (a - c) - (d² - b²) · (a + c)

Derived expressions for e² and f ² will be positive only if a < b + c + d and a - c > b - d.

From the above figure it follows that first condition must be satisfied, the trapezoid to be closed, and second condition must be satisfied, the triangle EBC to be closed.

Applications of trigonometry in solid geometry

A right pyramid and regular right pyramid

A pyramid is a polyhedron with one polygonal face, the base, (not necessarily a regular polygon) and all lateral faces triangular with a common vertex (apex).

A right pyramid is a pyramid for which the line joining the centroid of the base (the point of coincidence of the medians) and the apex is perpendicular to the base.

A regular pyramid is a right pyramid whose base is a regular polygon and the other faces are congruent isosceles triangles. Note that,

- if all lateral edges of a pyramid form equal angle with the base then,

a) all lateral edges are equal,

b) the pyramid’s altitude foot is the center of the circumcircle of the base.

- If all lateral faces of a pyramid form the same face-to-base dihedral angle then,

a) the slant heights of all faces are equal,

b) the pyramid’s altitude foot is the center of the incircle,

c) B = S_lat. · cosa, where B is the area of the base and S_lat. is the lateral surface area.

Thus, for example in the regular pentagonal pyramid shown in the right figure,

h = s · sina and r = h · cota, then the area of the DABO

Example: Known is surface area of a regular triangular pyramid S and given is the base-to-face angle a, determine the base edge a.

Solution: Using above formula for the surface area of a regular pyramid,

Sections of solids

Sections of solids examples

Example: Vertex angles of faces of a regular triangular pyramid with a side of the base a, are all equal and denoted a, as is shown in the down figure. Determine face to face dihedral angles and the area of section that the plane passing through a side of the base perpendicular to the opposite lateral edge cuts of the pyramid.

Solution: All face to face angles are the same and one of them, denoted j, is shown in the figure.

Section BCE is an isosceles triangle of the area

Example: Through the vertex of a cone laid is a plane that makes the angle j with its base and cuts the segment bounded by arc that subtends the central angle a, as is shown in the below figure. The distance of the plane from the center of the base equals d. Determine the volume of the cone.

Solution: Angle VOC equals angle ODV = j since they have mutually perpendicular sides.

Example: Lateral faces of a square pyramid are inclined to the base by angle a. Through the base’s edge laid is a plane that forms with the base of the pyramid the angle b, as shows the below figure. Find the area of the section if the side of base is a.

Solution: Through midpoints E and F of the two opposite sides of base passes the plane EFV which is perpendicular to the base and intersects the plane of the trapezoid BCC₁B₁ along the segment FG (the altitude of the trapezoid).

From DEFG by using the sine law

and since triangles, ADV and B₁C₁V are similar

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