﻿ Three Dimensional Geometry

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## Three Dimensional Geometry

(i) Common solids and Nets of Solids All solids are three dimensional

This is because they have length and width, this makes them to occupy space.

Examples

(i) Common solids and Nets of Solids All solids are three dimensional

This is because they have length and width, this makes them to occupy space.

Examples

This can be best demonstrated by the following nets and their respective solids.

Geometric properties of common solids

Solids have length, area and volume since measurements on them are said to be three dimensional. A vertex on a solid is a point where three or more edges meet. An edge is a line where faces meet. A plane / face is a set of points in a flat surface and extends indefinitely in all directions. In many solids some face, line or line and faces are parallel while others are not.

Example

Consider the triangular prism below. a) Name the line(s) which are parallel i. AB ii. BC iii. BF b) Name a face parallel to face ABF c) Name the line where face ABCD and ABF intersect. d) Name the point of intersection of i. Line DA and face ABF ii. Line DC and face BCEF.

Solution

a) . Line(s) parallel to (i). AB is DC (ii). BC are AD and FE (iii). BF is CE b) Face parallel to ABF is DCE c) Line of intersection of ABCD and ABF is AB. d) Point of intersection of (i) Line DA and face ABF is A (ii) Line DC and face BCEF is C

1.Pythagorus theorem

This theorem states that in any right-angled triangle the sum of the squares on the two shorter sides equals to the square of the hypothenuse; i.e. AB2 + BC2 = AC2 ( Height 2 + Base 2 = Hypothenuse2)

Example: Calculate the length of the side KM shown below given that the triangle is right angled at M.

Solution:

(KM)2 + (ML)2 = (KL)2
letting KM = y cm
y
2 + 52 = 132
y
2 + 25 = 169
y
2 = 169 - 25
y
2 = 144
y = 12 cm

TRIGONOMETRIC RATIOS
There are three main Basic trigonometric rations in a right angled triangle. These are sine, cosine and the tangent.
Example 1 Calculate the size of the angle θ in the figure below.

Solution
The side 5 is on the opposite side of the angle θ and side 7 is the adjacent side: the ratio tan θ will be used.
Tan
θ = Opp/ adj = 5/ 7

Tan θ = 0.7143 (look for the angle whole tan = 0.7143
θ = 35.54o

Example 2
Find the size of the angle θ shown below.

Solution
In this case the ratio sin is 10 be used since the lines given are opposite and hypotenuse.
Sine
θ = Opp/ Hyp = 3/ 8

Sin θ = 0.375
θ = 22.02

Example 3
Find the length of the side shown with label y on the figure show below given that the images in right angled at B.
Solution
The relationship between the side 12cm and the angle 300 given is that so the side is the adjacent side and y is the hypotenuse of the of the triage therefore we use the ratio cosine
i.e. Cos θ = Adj/Hyp ; Cos 30
o = 12/y

Y = 12/Cos 30o ; y = 12/0.866

Y = 13.85 cm

SINE AND COSINE RULE

The sine and cosine rules are used to solve for the sides and angles of triangle which are not right angles.
The sine rule: consider the non-right angled triangle ABC below.
The sine Rule will given by
a/Sin A = b/ = c/sin C = 2R where R is the Radius of the circumcircle of the triangle.

The sine Rule is used to
(i) Determine the Radius of the circumcircle
(ii) Solve for a triangle completely given two
a) Angles and a side
b) Sides and the angle opposite on of the sides

Example
Solve completely the triangle XYZ in which X = 42o ; x = 8.4 cm and y = 10.2 cm

Solution :
The triangle is drawn as shown below

Using sine rule

x /Sin x = y/Sin y = z/sin z = 2R

But we do not have R therefore

8.4/Sin 42 = 10.2/Sin y = 8.4 Sin y =10.2 sin 42

Sin Y = (10.2 x sin 42)/8.4

Sin y = 0.8125 - y = 54.34
Therefore :
Angle XYZ = Angle Y = 54.34o
Angle XZY = 180 - (42 + 54.34) = 180 - 96.34
Angle XZY = Angle z = 83.66o
Using sine rule Again implies that Z/Sin Z = x / Sin x
Z = 8.4/Sin 83.66 = 8.4/sin 42 which implies that

Z = ( 8.4 sin 83.66)/ Sin 42

Z = 12.48 cm

NB: The triangle has been solved completely since all its angles and sides are now known

The Cosine rule
By use of the non-right angled triangle shown below

The Cosine rule
a2 = b2 + c2 - 2bc cos A

Or

b2 = a2 + c2 - 2ac Cos B
c2 + a2 + b2 - 2ab Cos B

All these alternative forms of cosine Rule are used to find the :
(i) The third side of a triangle given two sides and an included angle
(ii) Three angle given three sides

Example
Solve completely for the following triangle given that P = 6.9cm, Q = 12.3cm and R = 7.6cm

Solution
The triangle is drawn and labeled as shown below
By use of the cosine rule

p2 = q2 + r2 - 2qr cos P

6.92 = (12.3)2 + (7.6)2 - 2 (12.3) (7.6) Cos P

47.61 = 151.29 + 57.76 -186.96 cos P

47.61 = 209.05 - 186.96 Cos P

-161.44/-186.96 = -186.96 Cos P

Cos P = -161.44/-186.96

Cos P = 0.8635

P = 30.29o

By use of cosine Rule still

R2 = p2 + q2 - 2(p)(q) Cos R
(7.6)2 =6.92 + 12.32 - 2(6.9) 912.30 Cos R
57.76 = 47.61 + 151.29 - 169.74 Cos R
57.76 = 198.90 - 169.74 Cos R
141.14 = -169.74 Cos R
Cos R = -141.14/ -169.74 which implies that Cos R = 0.8315

R = 33.75o

From the triangle PQR then therefore Angle Q can be obtained by
Q = 180 - (30.29 + 33.75)
Q =180 - 64.04
Q = 115.96o

The triangle has been solved completely since all its angles and sides are now known.

OBJECTIVE

Three Dimensional Geometry

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Three Dimensional Geometry

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