Transcript 5.5 Area of a Triangle From Geometry, we know the area formula for a triangle is A = ½bh But there are other.

5.5 Area of a Triangle
From Geometry, we know the area formula for a triangle is
A = ½bh
But there are other ways too!
Area of a triangle K 
K  12 bc sin A OR
K  s(s  a)(s  b)(s  c)
where semiperimeter
s = ½(a + b + c)
Ex 1) Find area of △DEF if m∠D = 56.9°, m∠E = 71.4°, d = 46.7 cm
E
71.4°
D
56.9°
Need 2 sides & an included angle…
let’s find stuff
180 – 56.9 – 71.4 = 51.7
46.7
51.7°
e
F
52.835
e
46.7

sin 71.4 sin 56.9
e = 52.835
K  12 (46.7)(52.835)(sin51.7)
K = 968 cm2
Ex 2) A triangular sign with side measures of 11, 13, and
15 in. requires a brace perpendicular to the longest side from
the opposite vertex. Determine the length of the brace.
(We will use two separate formulas for area!)
13
11
15
s  12 (11  13  15)  19.5
K  19.5(19.5  11)(19.5  13)(19.5  15)
 69.629
want this altitude … use K = ½bh
69.629  12 (15)(h)
9.3 = h
9.3 in
We can also add or subtract areas of various shapes.
Reminders for Area:
• square  s2 or 12 d1d2
s2 3
• equilateral △ 
4
• circle   r 2
• sector  r  (θ in rads)
1
2
2
OR
x
 r 2
360
(x is central angle in degrees)
(from Geometry)
*Hint: Draw a “plan” for what you want to add or subtract!
Ex 3) Determine area of the shaded region.
118°
44
44
Sector – Triangle
118
2
  (44) 
360
1
2
(44)(44)(sin118)
1993.585 – 854.693
1138.892
Ex 4) Determine area of the polygon.
17.4
13.9
x
11.578
41°
11.2
8.7
Use law of cosines to get x
x  17.42  11.22  2(17.4)(11.2)(cos 41)
x = 11.578
Ex 4) Determine area of the polygon.
17.4
x
11.578
13.9
II
41°
I
11.2
8.7
△I 
s  12 (17.4  11.2  11.578)  20.089
K  (20.089)(20.089  17.4)(20.089  11.2)(20.089  11.578)
= 63.928
△II  s  12 (13.9  8.7  11.578)  17.089
K  (17.089)(17.089  13.9)(17.089  8.7)(17.089  11.578)
= 50.194
Total Area = 63.928 + 50.194 = 114.122
Homework
#505 Pg 276 #1, 3, 5, 11, 15, 22, 24, 29, 31, 35,
39, 41, 47, 51
Answers to Evens:
22) 439.8 cm2
24) 18 times larger