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Unit 1.4 Surface area right pyramids & cones
Name: Date:
Shape definitions to know: memorize
Tetrahedron 4 triangles joined
Square pyramid
Square bottom, 4 attached triangle faces meet at point at top.
Rectangular pyramid
Rectangular bottom, 4 triangle faces
Pentagonal pyramid
Pentagon bottom [5 sides], 5 triangle faces
Hexagonal pyramid
Hexagon bottom [6 sides], 6 triangle faces
Triangular pyramid
Triangle bottom, 3 triangle faces … same as tetrahedron
Apex The point
where the faces meet.
Height how tall the pyramid is [centre of floor to apex]. INSIDE height … or CENTRE height.
Slant height on an angle, how tall is the triangle outside. … distance on outside – up the middle of the outside face.
RIGHT pyramid
The faces are ALL triangles … base is any polygon shape.
Net The ‘exploded’ form of a 3D shape.
Or
Also, recall … and memorize the formulae for area of
shapes. “area” means FLAT object
Square 𝑨 = 𝒍 × 𝒘 𝒐𝒓 𝑨 = 𝒔𝟐
Rectangle 𝑨 = 𝒍 × 𝒘
Triangle 𝑨 =
𝟏
𝟐𝒃𝒉 𝒐𝒓 𝑨 =
𝒃𝒉
𝟐
Circle 𝑨 = 𝝅𝒓𝟐
Remember: r = radius half way across circle
d = diameter all the way across
Surface area – means the outside area of a 3D object [NOT
flat]. Must use the ‘net’.
UNITS for area and surface area are ‘squared’: cm2, ft2, …
So, to find the surface area of a right square pyramid, you
will USE ONLY the base length … and slant height.
So … to calc SURFACE AREA of the pyramid, see the net
and see that you need area of the middle square … and
area of all 4 triangles.
Of course … you
won’t always be
given these … you
might have to
calculate them!
So if you are given …
Then surface area = area [square] + 4[area of triangle]
𝑺𝑨 = [𝟖𝒄𝒎 × 𝟖𝒄𝒎] + 𝟒 [𝟖𝒄𝒎 × 𝟏𝟎𝒄𝒎
𝟐]
= 𝟔𝟒𝒄𝒎𝟐 + 𝟒 [𝟖𝟎𝒄𝒎𝟐
𝟐]
= 𝟔𝟒𝒄𝒎𝟐 + 𝟒[𝟒𝟎𝒄𝒎𝟐]
= 𝟔𝟒𝒄𝒎𝟐 + 𝟏𝟔𝟎𝒄𝒎𝟐
= 𝟐𝟐𝟒𝒄𝒎𝟐
e.g. 1) page 28: Determine the area of this regular
tetrahedron … given its slant height.
Always write
in your units.
ALWAYS pay
attention to
them!!
Solution: (1) Think of the net … what shapes are here?
(2) Calc area of one triangle
* “Regular” tetrahedron means
EVERY side is identical.
(3) Multiply that one triangle by 4, since there
are 4 triangles in a tetrahedron.
So: 𝑨∆ =𝒃×𝒉
𝟐 =
𝟗.𝟎𝒄𝒎×𝟕.𝟖𝒄𝒎
𝟐 =
𝟕𝟎.𝟐𝒄𝒎𝟐
𝟐 = 𝟑𝟓. 𝟏 𝒄𝒎𝟐
CHECK!! Are you done?
Don’t forget: 𝑺𝑨 = 𝟒 × 𝟑𝟓. 𝟏 𝒄𝒎𝟐 = 𝟏𝟒𝟎. 𝟒 𝒄𝒎𝟐
e.g. 2 page 29) A right rectangular pyramid has base
dimensions 8 ft by 10 ft, and height of 16 ft. Calculate the
SA of the pyramid to the nearest square foot.
Solution:
[1] Sketch and label the diagram.
[2] Sketch and label the NET!!
* If you don’t do this … you’ll miss an important
fact watch
NET: Gives you area of the square.
To get area of each triangle, you need height of each
TRIANGLE … not height of the actual pyramid. BUT you
use height of the pyramid … and base lengths … to
calculate triangle heights [via PYTHAGORUS].
Notice that
you are not
told slant
height!
𝒂𝟐 + 𝒃𝟐 = 𝒄𝟐
(𝟏𝟔𝒇𝒕)𝟐 + (𝟒𝒇𝒕)𝟐 = 𝒄𝟐
𝟐𝟓𝟔𝒇𝒕𝟐 + 𝟏𝟔𝒇𝒕𝟐 = 𝒄𝟐
𝟐𝟕𝟐𝒇𝒕𝟐 = 𝒄𝟐
Thus 𝒄 = √𝟐𝟕𝟐𝒇𝒕𝟐 = 𝟏𝟔. 𝟒𝟗 𝒇𝒕 [square root of ft2 = ft]
… and that is only the SIDE of that triangle … and its
partner.
You NEED the red-
dashed line – or you
can’t calc area of that
triangle. You KNOW
the 16 ft height … and
the red line [4ft]. Use
pythag to calc.
Now look again …
𝒂𝟐 + 𝒃𝟐 = 𝒄𝟐
(𝟏𝟔𝒇𝒕)𝟐 + (𝟓𝒇𝒕)𝟐 = 𝒄𝟐
You also NEED the
green-dashed line – or
you can’t calc area of
that triangle. You
KNOW the 16 ft height
… and the green line
[5ft]. Use pythag to
calc.
𝟐𝟓𝟔𝒇𝒕𝟐 + 𝟐𝟓𝒇𝒕𝟐 = 𝒄𝟐
𝟐𝟖𝟏𝒇𝒕𝟐 = 𝒄𝟐
Thus 𝒄 = √𝟐𝟖𝟏𝒇𝒕𝟐 = 𝟏𝟔. 𝟕𝟔 𝒇𝒕 [square root of ft2 = ft]
… and that is FINALLY the SIDE of that triangle … and its
partner.
Now you can actually answer the question. The surface
area is:
𝑺𝑨 = 𝒂𝒓𝒆𝒂 𝒓𝒆𝒄𝒕𝒂𝒏𝒈𝒍𝒆 + 𝒂𝒓𝒆𝒂 𝟐 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆𝒔 + 𝒂𝒓𝒆𝒂 𝒐𝒕𝒉𝒆𝒓 𝟐
= (𝟏𝟎𝒇𝒕 × 𝟖𝒇𝒕) + 𝟐 × (𝟏𝟎𝒇𝒕 × 𝟏𝟔. 𝟒𝟗𝒇𝒕
𝟐) + 𝟐 × (
𝟖𝒇𝒕 × 𝟏𝟔. 𝟕𝟔𝒇𝒕
𝟐)
= 𝟖𝟎𝒇𝒕𝟐 + 𝟏𝟔𝟒. 𝟗𝒇𝒕𝟐 + 𝟏𝟑𝟒. 𝟎𝟖𝒇𝒕𝟐
= 𝟑𝟕𝟖. 𝟗𝟖 𝒇𝒕𝟐
** Note that your units are correct for SA … so you can be
quite sure that you did it right!
Done part 1 of section 1.4
ASSIGNMENT:
CYU 1 page 28
then pages 34-35: q’s 4, 5, 9, 13a
** p.s. LATERAL AREA means sides & top only … not
bottom.
Next do CYU 2 page 29
Then do pages 34-35: q’s 8a, 10, 13b, & 18.
THIS is the order I TAUGHT this stuff. It’s likely easiest to
do the work in this order.