QUADRILATERALS
Quadrilateral: polygon with 4 sides
Fill in the blanks below with an appropriate term, definition or diagram:
Trapezoid: __________________________________________________
____________________: quadrilateral in which opposite sides are parallel
Rectangle: __________________________________________________
_____________________: parallelogram in which all sides are equal
Square:
(1)
rhombus in which the angles are _______________________
(2)
Rectangle in which all the sides are _____________________
Examples:
Ex 1) Calculate the area.
8 cm
4 cm
6 cm
* include a diagram*
12 cm
1
bh
2
1
= (12)(6)
2
= 36
A =
∴ The area is 36 cm2
b = 12 cm, h = 6 cm * use an appropriate formula & state givens WITH units*
*substitute with brackets but WITHOUT units*
* include a concluding statement WITH units*
Ex 2) Find the value of the variable if A = 69 cm².
8 cm
h ( a + b)
2
h((15) + (8))
(69) =
2
138 = h (23)
138
=h
23
6=h
A =
h
15 cm
a = 15 cm, b = 8 cm, A = 69 cm2
∴The height is 6 cm.
Ex. 3) Find the area of the shaded region.
14 mm
1 mm
A = Asquare – Atriangle
1
= s2 - bh
2
1
= (14)² - (1)(14)
2
= 196 - 7
= 189
s = 14 mm, b = 1 mm, h = 14 mm
∴The area of the shaded region is 189 mm2
Ex 4) Find the perimeter of the following.
8 cm
4 cm
1
π d + d + 2b
2
1
= π (4 ) + (4 ) + 2 (8)
2
= 2π + 20
P=
d = 4 cm, b = 8 cm
≈ 26.3
∴ The perimeter is 26.3 cm
HOMEWORK:
Grade 9 Textbook:
p. 432 # 1abce, 2bcde, 3, 4, 10, 12, 15
(Include diagrams with your answers!)
3-2 3-Dimensional Geometry Terminology
Solid: a 3-dimensional figure with a closed surface and a measurable volume
Cube
Cylinder
Polyhedron: A 3-dimensional figure with faces that are polygons
There are two types of polyhedrons: Prisms & Pyramids
Prism: - polyhedron with two parallel, identical, polygonal faces (bases)
- they are distinguished from each other by the shape of their base
Triangular Prism
Cube
( Square Prism)
Rectangular Prism
Pentagonal Prism
Pyramid: a polyhedron whose base is a polygon and other faces are triangles that meet at a common point
Square-based Pyramid
Triangular-based Pyramid
Hexagonal Pyramid
ROUND BODIES
Round Body: solid that has at least one curved surface
There are three types of round bodies: Cylinders, Cones, & Spheres
Cylinder:
solid in which the two opposite faces are equal circles and the straight line joining their centres
is perpendicular to the base
Cone:
solid with a circular base and a curved lateral surface that extends from the base to a point
called the vertex
Sphere:
solid in which all the points are the same distance from a fixed, interior point called the centre
Examples:
Ex. 1) Find the volume of a cylinder if d = 12.6 cm and h = 15.8 cm.
V = πr 2 h
r = 6.3 cm, h = 15.8 cm
= π (6.3) 2 (15.8)
≈ 1969
15.8 cm
∴ The volume is approximately 1969 cm³
12.6 cm
Ex. 2) A silo has a radius of 2.35 m and a height of 10.24 m. Find the area of the exposed
surface of the silo to the nearest m².
2.35 m
Atotal = A1 + A2
= 2πrh + 2πr 2
(2)
10.24 m
(1)
r = 2.35 m, h = 7.89 m
= 2π (2.35)(7.89) + 2π (2.35) 2
= 48 128π
≈ 151
∴ The area of the exposed surface is approximately 151 m².
*height of the cylinder is:
10.24 – 2.35 = 7.89*
HOMEWORK:
(Include diagrams with your answers!)
Grade 9 Textbook:
pg. 441 # 1a, 2a, 3b, 4a, 5, 6, 7, 9
pg. 447 # 1b, 2, 3, 5, 6
pg 454 # 1b, 2a, 4, 5
pg. 459 # 1a, 3, 4, 5
pg 465 # 1a, 3, 4
3-3 Applications
Ex. 1) A cone fits just inside a cylinder. The volume of the cylinder is 9 425 cm³. What is
the total surface area of the cone to the nearest square centimetre?
V = 9425 cm3
Vcylinder = π r 2 h
(9 425) = π (10)2 h
9425
= h
π (10) 2
30 ≈ h
r = 10 cm
h2 + r2 = s2
(30)² + (10)² = s2
1000 = s2
± 31.6 ≈ s
h = 30 cm r = 10 cm
20 cm
Since s is a side length s > 0 ∴ s ≈ 31.6
A cone = π r 2 + π rs
h = 30 cm
s = 31.6 cm
= π (10) 2 + π (10)(31.6)
= 416π
≈ 1307
∴ The area of the cone is 1307 cm²
Ex2)
A tennis ball fits just inside a plastic cube whose sides measure 40 mm. Determine
the volume of the empty space inside the cube.
Vempty space = Vcube − Vball
c = 40 mm
4
= c 3 − πr 3
3
4
= (40) 3 − π (20) 3
3
32 000
= 64 000 −
π
3
≈ 30 489.7
r = 20 mm
40 mm
40 mm
40 mm
∴ The volume of the empty space is 30 489.7 mm³
HOMEWORK:
(Include diagrams with your answers!)
Grade 9 Textbook:
pg. 441 # 11, 16
pg. 447 # 8, 10
pg. 454 # 14
pg. 466 # 7, 8
Worksheet: Application Questions #1-11