1. Area Topic 13: Lesson 14 Perimeter and Area in a Coordinate Plane Holt Geometry Texas ©2007

2. Objectives and Student Expectations
TEKS: G7A, G7B, G8A
The student will use two-dimensional coordinate systems to represent figures
The student will use slopes to investigate geometric relationships
The student will fine areas of circles and composite figures

3. Example: 1 Step 1 Draw the polygon.
Draw and classify the polygon with vertices E(–1, –1), F(2, –2), G(–1, –4), and H(–4, –3). Find the perimeter and area of the polygon.

4. Example: 1continued
Step 2 EFGH appears to be a parallelogram. To verify this, use slopes to show that opposite sides are parallel.
slope of EF =
slope of GH =
slope of FG =
slope of HE =
The opposite sides are parallel, so EFGH is a parallelogram.

5. Example: 1continued
Step 3 Since EFGH is a parallelogram, EF = GH, and FG = HE.
Use the Distance Formula to find each side length.
perimeter of EFGH:

6. Example: 1continued
To find the area of EFGH, draw a line to divide EFGH into two triangles. The base and height of each triangle is 3. The area of each triangle is
Area of EFGH = ½bh + ½bh
A = ½(3)(3) + ½(3)(3)
A =
A = 9 units2.

7. Example: 2
Draw and classify the polygon with vertices L(–2, 1), M(-2, 3), N(0, 3), and P(1, 0). Find the perimeter and area of the polygon.
Step 1 Draw the polygon.

8. Example: 2 continued
Step 2 LMNP appears to be a kite. To verify this, find the lengths of the sides to show that two pairs of consecutive sides
are congruent.
Two pairs of consecutive sides are congruent, so LMNP is a kite.

9. Example: 2 continued
Step 3 Find the perimeter of EFGH:

10. Example: 2 continued Step 4 To find the area of LMNP, the formula
A = ½ d1d2 could be used because the diagonals of a kite are perpendicular. And the lengths of both diagonals could be found using the distance formula. Let’s see if we can come up with a more “user friendly” way.

11. Example: 2 continued Step 4 Enclose the kite with a rectangle.
Area = rectangle – triangle – triangle
Area = bh – ½bh – ½bh
Area = (3)(3) – ½(3)(1) – ½(3)(1)
Area = 9 – 1.5 – 1.5
Area = 6 units2

12. Example: 3 Find the area of the polygon with vertices
A(–4, 0), B(2, 3),
C(4, 0), and D(–2, –3).
Draw the polygon and enclose it in a rectangle.
Area = rectangle - ∆a - ∆b - ∆c - ∆d
Area = bh – ½ bh – ½ bh – ½ bh – ½ bh
Area = (8)(6) – ½(6)(3) – ½(2)(3) – ½(3)(6) – ½(2)(3)
Area = 48 – 9 – 3 – 9 – 3
Area = 24 units2.

13. Example: 4 Find the area of the polygon with verticesb d c
Find the area of the
polygon with vertices
K(–2, 4), L(6, –2),
M(4, –4), and N(–6, –2).
Draw the polygon and enclose it in a rectangle.
Area = rectangle - ∆a - ∆b - ∆c - ∆d
Area = bh – ½ bh – ½ bh – ½ bh – ½ bh
Area = (12)(8) – ½(4)(6) – ½(8)(6) – ½(2)(2) – ½(10)(2)
Area = 96 – 12 – 24 – 2 – 10
Area = 48 units2.

14. Example: 5 Area = rectangle – triangle – circle
Find the area of the rectangle outside of the triangle and the circle.
Answer in both
algebraic form
and rounded
to the tenth.
Area = rectangle – triangle – circle
Area = bh – ½ bh -  r2
Area = (12)(4) – ½ (6)(4) –  22
Area = 48 – 
Area =  in2
Area ≈ 23.4 in2

15. H.O.T.S. EXAMPLE:
SEE IF YOU CAN FIGURE THIS OUT!
This example is NOT included
in the notes.

16. Example: 6 Show that the area does not change when the pieces are
rearranged.

17. Understand the Problem
Example: 6 cont. 1
Understand the Problem
The parts of the puzzle appear to form two trapezoids with the same bases and height that contain the same shapes, but one appears to have an area that is larger by one square unit. 2
Make a Plan
Find the areas of the shapes that make up each figure. If the corresponding areas are the same, then both figures have the same area by the Area Addition Postulate. To explain why the area appears to increase, consider the assumptions being made about the figure. Each figure is assumed to be a trapezoid with bases of 2 and 4 units and a height of 9 units. Both figures are divided into several smaller shapes.

18. Example: 6 continued Solve 3 Find the area of each shape. Left figure
Right figure
top triangle:
top triangle:
top rectangle:
top rectangle:
A = bh = 2(5) = 10 units2
A = bh = 2(5) = 10 units2

19. Example: 6 continued Solve 3 Find the area of each shape. Left figure
Right figure
bottom triangle:
bottom triangle:
bottom rectangle:
bottom rectangle:
A = bh = 3(4) = 12 units2
A = bh = 3(4) = 12 units2

20. Example: 6 continued 3 Solve
The areas are the same. Both figures have an area of
= 26.5 units2.
If the figures were trapezoids, their areas would be
A = (2 + 4)(9) = 27 units2. By the Area Addition
Postulate, the area is only 26.5 units2, so the figures must not be trapezoids. Each figure is a pentagon whose shape is very close to a trapezoid.

21. Example: 6 continued Look Back 4
The slope of the hypotenuse of the smaller triangle is 4. The slope of the hypotenuse of the larger triangle is 5. Since the slopes are unequal, the hypotenuses do not form a straight line. This means the overall shapes are not trapezoids.