Download Unit three Geometry and measurement Contents: Lesson (1): The

Unit three
Geometry and measurement
Contents:
Lesson (1): The distance between two points in the coordinate plane
Lesson (2): Geometric transformation: Translation
Lesson (3): Area of the circle
Lesson (4): The lateral area and the total area of: (The cube – The cuboid)
Lesson (1): The distance between two points in the coordinate plane
a) The distance between two points on a ray:
The distance between the two points A and B is:
Length of AB = coordinate of the ending point – coordinate of the starting point. = 9 – 2 = 7
cm.
AC = ……. - …….. = …….. cm.
BC = ……. - …….. = …….. cm.
b) The distance between two points in the coordinate plane:
From the opposite figure: A (2 , 2) , B (6 , 2) , and C (2 , 6)
AB = ……………… units.
AC = ……………… units.
The type of the triangle Δ ABC with respect to its
sides is ………..
Note for triangles types:
All triangles have 3 sides and 3 angles which always add up to 180°.
Triangles are classified in 2 ways1) By the number of equal sides they have:
• scalene - all 3 sides have different lengths
• isosceles - 2 sides have equal lengths
• equilateral - all 3 sides are equal
2) By the types of angles they have:
• acute triangle - all 3 angles are acute (less than 90°)
• right triangle - has one right angle (a right angle = 90°)
• obtuse triangle - has one obtuse angle (an obtuse angle is greater than 90° and less
than 180°)
c) Calculate the distance between two points on a straight line.
When calculating the distance between two points on the integers number line we care about:
1- The absolute value which is
= | number of the ending point – number of the starting point |
2- The properties of addition and subtraction in Z.
AB = | B – A | = | 5 – (-2) | = | 5 + 2 | = 7 units.
DE = …………. = …………. = …………… units.
d) Calculate the distance between two points in the coordinate plane Z.
From opposite figure :
A (-2 , 1) , B (3 , 1) and D (-2 , 5)
AB // X- X+
AB = | B – A | = | 3 – (-2) | = 5 cm.
AD = ……….. = …………… = ………… cm.
Determine the position of the point C (3 , 5) , then satisfy that the shape ABCD is a
parallelogram, Calculate its perimeter and its area.
Lesson (2): Geometric transformation: Translation
The geometric transformation transforms each point in the plane in to a point A′ in the same
plane.
In the opposite figure A′ B′ is the image of AB by reflection in L.
A′ B′ = AB
A′ B′ // AB
Translation transformation: Two things must be known for the translation to happen,
* Magnitude of the translation.
* Direction of the translation.
First: Translation of a point in the plane.
(a) In the page plane.
From opposite figure: Required: translate the point A by
distance 4cm in the direction of MN.
Solution:
Draw a ray from A parallel to MN to take its direction as AB in
the opposite figure.
Determine the point A′ on AB such that AA′ = 4cm.
A′ is the image of A by translation its magnitude is 4cm in direction of MN.
(b) In the coordinate plane for integer numbers.
In the opposite figure, find the image of the two points
A (2 , 3) , B (-3 , 1) by translation (x + 3 , y + 2)
Solution:
the magnitude and direction of translation:
displacement 3cm in the direction of x+ followed by
displacement 2cm in the direction of y+.
A′ = (2 + 3 , 3 + 2) = (5 , 5)
B′ = (-3 + 3 , 1 + 2) = (0 , 3)
Second: translation of a line segment in the plane.
In the opposite figure, Find the image of the line segment AB
where : A (2 , 3) , B (-2 , 0) by translation (x + 3 , y – 2).
Solution:
First: Determine the magnitude and direction of translation
which are:
displacement 3 cm in the direction of X+ followed by
displacement 2cm in the direction of Y- .
Second:
A′ = (2 + 3 , 3 – 2) = (5 , 1)
B′ = (-2 + 3 , 0 – 2) = (1 , -2 )
Third: The translation of a geometric shape in the coordinate plane:
In the opposite figure, Find the image of the Δ ABC where:
A (0 , 1) , B (2 , 3) and C (-1.4) by translation
(X + 2 , y + 3).
Solution :
First: Determine the magnitude and direction of translation
which are : displacement 2 cm in the direction of X+
followed by displacement 3 cm in the direction of y+.
Second: Find the image of each point one by one as follow :
A′ = (0 + 2, 1 + 3) = (2 , 4)
B′ = (2 + 2, 3 + 3) = (4 , 6)
C′ = (-1 + 2, 4 + 3) = (1 , 7)
Third: Determine the points A′ , B′ and C′ in the plane, then join between them. We found Δ
A′B′C′ is the image of Δ ABC by translation (x + 2 , y + 3) .
Lesson (3): Area of the circle
The surface area of the circle = r²
- r= d/2
d
, d =2 * r
Circumference of the circle = 2 r =  d
-
d: diameter
- r:
radius
- d= CA = BD
- r= AM=BM=CM=DM
Notice that: is the approximately ratio between the circumference and the diameter of the
circle it is or 3.14
Lesson (4): The lateral area and the total area of: (The cube – The cuboid)
First: The cube:
The lateral area of the cube = Area of one face x 4
The lateral area of the cube = Perimeter of the base x height
The Total area of the cube = Area of one face x 6
The sum of edge lengths of a cube is 84 cm. Find its lateral area and its total area.
Solution:
The cube has 12 edge; so the length of one edge = 84/ 12 = 7 cm.
The Lateral Area of the cube= Area of one face x 6 = 7x7x6 = 294 cm²
Second: The Cuboid:
Example (4) :
A room in the form of cuboid its inner dimensions are :
5 m length, 3.5m width and 3 m height. It is wanted to paint its lateral walls. The cost price of
one square meter is LE 9. Calculate the required cost.
Solution :
The lateral area of the room’s wall = Perimeter of the base x height
= (length + width) x 2 x height
= (5 + 3.5) x 2 x 3 = 8.5 x 2 x 3 = 17 x 3 = 51 m²
The cost price = 51 x 9 = LE 459.
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A cuboid shaped box with a square base its length is 9 cm and its height is 20 cm. Calculate the
lateral area and total area.
Solution:
The lateral Area of the Cuboid = Perimeter of the base X height
= (9 x 4) x 20 = 720 cm²
The total area = Lateral Area + Area of two bases
= 720 + 2 (9 x 9) =720 + 162 = 882 cm²