Download Honors Geometry Section 5.2 Areas of Triangles and Quadrilaterals

Honors Geometry Section 5.2
Areas of Triangles and
Quadrilaterals
WHY?
RHL
E
F
You can see that the area of parallelogram
ABCD is equal to the area of rectangle EBCF.
For a parallelogram with base b
and height h, the area is given by
the formula:
A parallelogram = ______
b h
Note that the height is the length of the
segment perpendicular to the base from a
point on the opposite side which is called
the altitude of the parallelogram.
2s
s 3 4 3
s
+
A  15  4 3  60 3 u
2
A  10  6  60
8  x  60
x  60 / 8  7.5 u
Any triangle is half of a parallelogram. For a
triangle with base b and height h, the area
is given by the formula:
1 bh
A triangle = ________
2
The height is the length of the ____________
altitude to
the base
Example: Find the area of to the nearest 1000th.
AC
sin 25 
10
AC  4.226
BC
cos 25 
10
BC  9.063
A  .5  4.226  9.063  19.150 u
2
Example: A triangle has an area of 56
and a base of 10. Find its height.
A  1 bh
2
56  1 10h
2
h  11.2
A  .5(10)(3)  15 u
2
Trigonometry and the
Area of a Triangle
Using your knowledge of trigonometry, express h in terms of
sinC.
h
sin C 
b
b sin C  h
1
Substituting this into the formula A  bh , and using a as the
2
base we get
1
A  ab sin C
2
We have just discovered that the
area of a triangle can be expressed
using the lengths of two sides and
the sine of the included angle.
Example: Use what you have learned above to find
the area of parallelogram ABCD to the nearest 1000th.
A// gram  2( Atriangle )
A// gram
1

 2 15  25  sin 50 
2

A// gram  15  25  sin 50
A// gram  287.267cm
2
An altitude of a trapezoid is a
segment perpendicular to the two
bases with an endpoint in each of
the bases.
The length of an altitude will be the height
of the trapezoid.
1 b2 h
2
1 b1h
2
1 h(b2  b1 )
2
1 b2h  1 b1h
2
2
For a trapezoid with bases b1 and b2 and
height h, the area of a trapezoid is given by
the formula:
1
Atrap. 
2
hb1  b2 
Recall that the diagonals of both
rhombuses and kites are
perpendicular.
1
1
Akite  1

BC  AE 
2

BC DE 
2
E

BC  AE   1 BC DE 
2
2
Akite  1

BC  AE  DE 
2
1
Akite  Ar hombus 
d1d 2
2