Geometry Chapter 11/12 Review
Shape: Rectangle
Formula A= Base x Height
Special case: Square. The same formula works, but you can also use A= Side x Side or
A= (Side)2
Height = 6
Base = 8
Square with side of 7
Area = 7x7=49
Area = 8 x 6 = 48
Shape: Parallelogram
Formula: A= Base x Height
side is not the height.
side= 7
Major difficulty: If one side is the base, then the other
height=6
base= 10 Area = 10x6=60. Note: The side=7 is not used.
8
8
4√3
60°
60°
20
Draw an altitude from the upper left vertex
4
20
This forms a 30°-60°-90° triangle with the short leg 4
and the long leg = height = 4√3.
This makes the area =20 x 4√3 = 80√3
Shape : Triangle
First formula : A=Base x Height / 2
Major difficulty: Must have a right angle or
perpendicular relationship between the height and base.
A= (B x H) / 2
Height
Base
8
8
8
8
Then the area = 6√55 / 2 = 3√55
h
3
3
6
6
Find the area by
first drawing an altitude.
Then solve for h:
h 2+9= 64
h=√55
Shape : Trapezoid
A=
First formula :
b2=8
h=6
Ê b1 + b 2 ˆ
h
Ë 2 ¯
Second formula :
Area = (8+10)(6)=54
2
M=8
h=5
Area = 8 x 5 = 40
b1=10
14
14
10
60°
10
20
5
60°
Draw a height from the upper
left vertex, making a 30°-60°-90°
triangle.
5√3
20
This makes the small leg 5, the big
leg 5√3. A=(1/2)(14+20)(5√3)= 85√3
A = Mh where M = median
Shape : Kite Formula: A= (d1 x d2) / 2
considered as a kite for this formula.
Note: Both a square and a rhombus can be
d1
d2
For the problem below, realize that the diagonals are perpendicular. That creates 45°45°-90° and 30°-60°-90° triangles. That gives the parts of the diagonals as shown.
With that one diagonal = 4+4=8 and the other =4 +4√3 which does not reduce.
The area = 8(4 + 4√3)/2 = 4(4 + 4√3) = 16 + 16√3
4
45° 4
60°
45° 4
60°
4
4√3
Shape: Equilateral triangle. Formula A= s2√3/4
Example: The perimeter of an equilateral triangle = 30. Find the area. Since the
perimeter = 30, each side = 30/3 = 10. The area = 102√3/4=100√3/4=25√3.
Shape : Regular polygon. Formula A= (a x p)/2 where a is the apothem and p the
perimeter. Note: A regular polygon has all sides congruent and all angles congruent.
a
Example : Find the area of a regular hexagon which has each side being 10.
a
a=5√3
5
10
10
Draw the segment as shown in the 2nd hexagon. This makes a 30°-60°-90° triangle.
The 10 for the side of the hexagon is split to 5. This makes the apothem 5√3.
The perimeter = 6 x 10 = 60. The area = (1/2)(5√3)(60) = (30)(5√3) = 150√3.
Shape: Circle. Formula: A= π r2.
Example diameter = 10. The radius = 10/2 = 5. The area = π x 52=25π.
Shape: Sector of a circle.
r=10
80°
The formula is a modification of the circle area formula. Find the area of the whole circle
and then find the fraction that the sector makes up.
The area of the whole circle = π x 102= 100π.
The fraction is 80°/360° = 2/9.
2/9 x 100π = 200π/9.
Shape : Segment of a circle.
r=10
60°
To find the area first find the area of the circle, then the sector, and then the triangle.
The segment's area = the sector area - the triangle area.
Area of the circle = π x 102= 100π.
Area of the sector = ((1/6) x 100π = 50π/3.
Area of the triangle = 102√3/4=25√3.
Area of the segment = 50π/3 - 25√3.
Shape: Triangle Formula: Hero's formula. This will find the area of a triangle even if
you know none of the angles.
A = s(s - a)(s - b)(s - c)
The sides of a triangle can be represented as a,b, and c. The s stands for the semiperimeter of the triangle. Add up all the sides and divide by 2 and you will have the
semiperimeter.
Example problem: What is the area of a triangle with sides 11,13 and 8?
Find the semiperimeter s = (11+13+8)/2 = 16.
A
= 16(16 - 11)(16 - 13)(16 - 8) = 16(5)(3)(8) = 4 5x3x4x2
= 4x2 5x3x2 = 8 30
Sometimes, you will need to break the numbers int o smaller pieces.
8x15x21 = 4x2
x 5x3 x
3x7 = 4 x 3x3 x 2x5x7 = 2x3 70 = 6 70
Shape: Inscribed quadrilateral. Formula: Brahmaguptas formula.
A = (s - a)(s - b)(s - c)(s - d)
The formula only works if the quadrilateral can be inscribed in a quadrilateral. Other
than that, it works very much like Hero's formula.
Example: What is the area of an inscribed quadrilateral with sides 4,5,7, and 10?
s= (4+5+7+10)/2= 13.
A
= (13 - 4)(13 - 5)(13 - 7)(13 - 10) = (9)(8)(6)(3) = 3 (4x2)(2x3)(3)
= 3x2 (2x2)(3x3) = 3x2x2x3 = 36.
Shape: Prism Surface Area Formula: Surface Area = Base Area + Lateral Area
This is abbreviated as S.A. = B.A. + L.A.
The base area consists of the two triangles. The lateral area is the 3 rectangles. The
rectangles are drawn as parallelograms to give a three dimensional effect.
To find the base area draw an altitude from the 16 side of either triangle.
10
10
10
10
h
8
h=6
8
8
8
Find the h with the Pythagorean theorem. This gives h as 6.
The triangle area = (1/2)(16 x 6) = 48.
B.A. = 2 x 48 = 96.
The lateral area consists of the three rectangles 16 by 20, 10 by 20, and 10 by 20.
L.A. = 16 x 20 + 10 x 20 + 10 x 20 = 320 + 200 + 200 = 720.
The Surface Area = S.A. = 720 + 96 = 816.
Another prism example.
The surface area has 6 faces:
2 of 6 by 10,
2 of 6 by 5, and
2 of 5 by 10.
S.A.= 2( 6 x 10 + 6 x 5 + 5 x 10) = 280.
5
10
6
Shape: Pyramid
Surface Area = L.A. + B.A.
The L.A. consists of triangle sides. The base area varies, but it is typically a square.
Example: Find the surface area of a regular square pyramid with a lateral edge of 13 and
base edge of 10.
13
13
13
SL.
12.
5
5
10
5
10
5
10
First draw a slant height. This splits the base edge to 5 and 5. Find the slant height with
the
triangle with the Pythagorean Theorem: 52+SL2=132. This gives the slant height as 12.
Each triangle has an area of (1/2)(10 x 12) = 60.
L.A. = 4 x 60 =240.
B.A. = 10 x 10 = 100
S.A. = 240 + 100 = 340.
Shape: Cylinder. Surface Area Formula: B.A. = 2πr2 (The two circles.) L.A. = 2πrh.
Example: Find the surface area of a cylinder with radius of 3 and height of 8.
r=3
h=8
B.A. = 2 π x 32= 18π
L.A. = 2 π x 3 x 8 = 48π
S.A. = 18π + 48π = 66π
Shape: Cone. Surface Area Formula: B.A. = πr2 (The circle at the base) L.A. = πrL
where L is the slant height.
Example; Find the surface area of a cone with radius 5
and a height of 12.
By the Pythagorean theorem, L = 13.
B.A. = π x 52 = 25π
L.A. = π x 5 x 13 = 65π
S.A. = 25π + 65π = 90π.
L
h
Shape: Sphere. Surface Area Formula: S.A. = 4πr2.
The sphere has no base at all.
r
Example: Find the surface area of a sphere with
radius of 6.
S.A. = 4π x 62= 4π x 36 = 144π.
radius
Shape: Prism, Box, or Cylinder
Volume: V = Bh where B is the base area.
To find the base area draw an altitude from the 16 side of either triangle.
10
10
10
10
h
8
h=6
8
8
8
Find the h with the Pythagorean theorem. This gives h as 6.
The triangle area = (1/2)(16 x 6) = 48.
V=Bh = 48 x 20 = 960
Box example.
Use any face as the base.
B=10 x 6 = 60
5
V= 60 x 5 = 300
10
6
Cylinder example
r= 3
h=10
B= π x 32=9π
V= Bh = 9π x 10 = 90π
Shape: Pyramid or Cone.
Formula V= (1/3) Bh where B is the base area.
Cone example:
Find the volume of a cone with radius 6 and slant height of 10.
Using the Pythagorean theorem, we have
62+h2=102
which solves to h = 8.
B= π x 62=36π
L
h
V= (1/3) B h = (1/3) x 36π x 8=
12π x 8 = 96π
r
Pyramid example:
Find the volume of a regular square pyramid with base edges of 10 and a lateral height of
13.
13
13
13
12
5
5
5
10
10
10
First use the square base to find the segments marked 5. Then use the Pythagorean
theorem to find the altitude of 12.
V= (1/3)(10 x 10) (12) = 1200/3 = 400.
Shape: Sphere. Formula: V = (4/3)π r3
Example: Find the volume of a sphere with radius 6.
V = (4/3)π x 63= (4/3) π x 216 = 4 π x 72 =288 π