1. A.17.9 B.22 C.13.3 D.9.1 Find the perimeter of quadrilateral WXYZ with vertices W(2, 4), X(–3, 3), Y(–1, 0), and Z(3, –1).

2. Unit 1-Lesson 4 Area of two dimensional figures

3. Objective I can recall area formulas for parallelograms, trapezoids, and triangles

4. Before Area… We need to recall some important mathematical tools Pythagorean Theorem Right Angles / Triangles

5. Right Angles Right angles measure 90 degrees Two segments / lines that form a right angle are perpendicular This symbol indicates a right angle

6. Right Triangle A right triangle has exactly one right angle Can you find the right triangles?

7. A Right Triangle’s Best Friend The Pythagorean Theorem A 2 + B 2 = C 2 The sum of the squares of the legs of a right triangle is equal to the square of its hypotenuse

8. Example Pythagorean Theorem Find the value of x A = 6 B = 15 C = x 6 2 + 15 2 = x 2 36 + 225 = x 2 261 = x 2 x =

9. Why? In order to find surface area, you need to use know the height of the figure Height – perpendicular distance (you will see a right angle)

10. Area of a Parallelogram

11. Be careful! Yes! Notice the height intersects the base NO! Base and height do not intersect at a right angle

12. Area of a Parallelogram Find the area of

13. Area of a Parallelogram Area Find the height of the parallelogram. The height forms a right triangle with points S and T with base 12 in. and hypotenuse 20 in. c 2 = a 2 + b 2 Pythagorean Theorem 20 2 = 12 2 + b 2 c = 20 and a = 12 400= 144 + b 2 Simplify. 256= b 2 Subtract 144 from each side. 16= bTake the square root of each side.

14. Area of a Parallelogram Continued A= bhArea of parallelogram = (32)(16) or 512 in 2 b = 32 and h = 16 The height is 16 in. UT is the base, which measures 32 in. Answer: The area is 512 in 2.

15. A.A B.B C.C D.D A.88 m; 255 m 2 B.88 m; 405 m 2 C.88 m; 459 m 2 D.96 m; 459 m 2 A. Find the perimeter and area of

16. Area of a Triangle

17. Application Perimeter and Area of a Triangle SANDBOX You need to buy enough boards to make the frame of the triangular sandbox shown and enough sand to fill it. If one board is 3 feet long and one bag of sand fills 9 square feet of the sandbox, how many boards and bags do you need to buy? What is it asking for?

18. Application Perimeter and Area of a Triangle Step 1Find the perimeter of the sandbox. Perimeter = 16 + 12 + 7.5 or 35.5 ft Step 2Find the area of the sandbox. Area of a triangle b = 12 and h = 9

19. Application Perimeter and Area of a Triangle Step 3Use unit analysis to determine how many of each item are needed. Boards Bags of Sand boards Answer: You will need 12 boards and 6 bags of sand.

20. Area of a Trapezoid

21. SHAVING Find the area of steel used to make the side of the trapezoid shown below. Area of a trapezoid h = 1, b 1 = 3, b 2 = 2.5 Simplify. Answer: A = 2.75 cm 2

22. Area of a Trapezoid OPEN ENDED Miguel designed a deck shaped like the trapezoid shown below. Find the area of the deck. Read the Problem You are given a trapezoid with one base measuring 4 feet, a height of 9 feet, and a third side measuring 5 feet. To find the area of the trapezoid, first find the measure of the other base.

23. Solve the Test Item Draw a segment to form a right triangle and a rectangle. The triangle has a hypotenuse of 5 feet and legs of ℓ and 4 feet. The rectangle has a length of 4 feet and a width of x feet.

24. Use the Pythagorean Theorem to find ℓ. a 2 + b 2 =c 2 Pythagorean Theorem 4 2 + ℓ 2 =5 2 Substitution 16 + ℓ 2 =25Simplify. ℓ 2 =9Subtract 16 from each side. ℓ=3Take the positive square root of each side. By Segment Addition, ℓ + x = 9. So, 3 + x = 9 and x = 6. The width of the rectangle is also the measure of the second base of the trapezoid. Area of a trapezoid Substitution Simplify. Answer: So, the area of the deck is 30 square feet.

25. Area of a Rhombus or Kite The area A of a rhombus or a kite is one half the product of the length of its diagonals, d 1 and d 2 A= d 1 d 2 25

26. 26 Area of Regions 8 10 12 414 8 The area of a region is the sum of all of its non-overlapping parts. A = ½(8)(10) A= 40 A = (12)(10) A= 120 A = (4)(8) A=32 A = (14)(8) A=112 Area = 40 + 120 + 32 + 112 = 304 sq. units

27. Other Types of Polygons Regular Polygon – a polygon with sides that are all the same length and angles that are all the same measure

28. 28 Areas of Regular Polygons Perimeter = (6)(8) = 48 apothem = Area = ½ (48)( ) = sq. units 8 If a regular polygon has an area of A square units, a perimeter of P units, and an apothem of a units, then A = ½ (a)(p).