Aim: How do we find the area of special quadrilaterals and triangles?
Objectives: to derive and apply the area formulas for special quadrilaterals.
Do Now: P454: #17
Draw radii from endpoints of each chord. We will see that the shaded region is consisted of four triangles (two equilateral
triangles, two isosceles triangles) and two sectors.
Or we construct a rectangle, which will break the shaded region into two isosceles triangles and 2 sectors. Either way,
A4 3
4
3
Lesson Development: We will prove the area formulas for parallelograms, triangles, rhombus and trapezoids.
Whenever the unit is missing, remember you were instructed to put down square units. Area is defined as how
many square units (squares with side 1) can fit. So the area of a rectangle (A=(b)(h) is a basic fact.
We will demonstrate using a diagram that a parallelogram
can always be turned into a rectangle. We can show that
the triangle on the left that is being shifted to the right to
form the rectangle are congruent by AAS congruence
theorem (h = h, both have right angles, and A  B by
supplements of a congruent angle are congruent).
Therefore the area of a parallelogram is (b)(h)
EX: what is the area of a parallelogram with a 45 angle and sides 6 and 10.
Let AD = 6 and CD = 10 and mD  45 . We already
know the length of the base. All we know is the height,
which can be determined by drawing an altitude from A to
h
CD .
6
3 2
2
A  10(3 2)  30 2
We will use this to establish the area formula for a triangle.
When we put two congruent triangles together to form a quadrilateral, we can show that alternate interior angles
are congruent which show that opposite sides are parallel, which means the quadrilateral is a parallelogram.
1
Therefore, the area of the triangle is one-half the area of a parallelogram, A  bh .
2
EX: Find the area of a triangle whose sides measure 8, 8 and 6.
In an isosceles triangle, the altitude bisects the base. We
can prove it using HL congruence theorem. So the base is
bisected into two segments of 3 each. Using Pythagorean
Theorem, the height is
82  32  55 . The area is
1
6 55  3 55
2
Equilateral triangle:
Proof: Let h be the angle bisector to C . This splits
ABC into two 30  60  90 triangles. h 
A
3
a . So
2
1
3
3
a(
a)  a2
2
2
4
EX: Derive the area formula for the rhombus in terms of its diagonals.
The diagonals bisect each other and divide the rhombus
into four congruent triangles. Therefore the area of the
1
2
rhombus is 4( d1d 2 )  2d1d 2
Derive the area formula for a trapezoid:
The trapezoid can be partitioned into three figures, two
triangles and one rectangle.
1
1
xh  b1h  (b2  b1  x )h
2
2
1
1
1
1
A  xh  b1h  b2h  b1h  xh
2
2
2
2
2b h 1
1
1
A  1  b2h  b1h  h[b1  b2 ]
2
2
2
2
A
HW#28: P431 – 433: 5, 6, 10 – 15, 17, 34a-e, 35, 38**
**EC: If correct solution is handed in before unit test, +2 pts on your test.
HW#28 Solutions
5) 16 3
6) 48
10) 60
13) 30 2
14) 6 3
15)
34) a) 100
35) 10 and 20
b) 100 2
11) 16
25 3
17) 240
2
c) 100 3
12) 9 3
d) 200
e) 100 3