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Find the perimeter of quadrilateral WXYZ with vertices W (2, 4), X (–3, 3), Y (–1, 0), and Z (3, –1). A. 17.9 B. 22 C. 13.3 D. 9.1. Unit 1-Lesson 4. Area of two dimensional figures. Objective. I can recall area formulas for parallelograms, trapezoids, and triangles.
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Find the perimeter of quadrilateral WXYZ with verticesW(2, 4), X(–3, 3), Y(–1, 0), and Z(3, –1). A. 17.9 B. 22 C. 13.3 D. 9.1
Unit 1-Lesson 4 Area of two dimensional figures
Objective • I can recall area formulas for parallelograms, trapezoids, and triangles
Before Area… • We need to recall some important mathematical tools • Pythagorean Theorem • Right Angles / Triangles
Right Angles • Right angles measure 90 degrees • Two segments / lines that form a right angle are perpendicular This symbol indicates a right angle
Right Triangle • A right triangle has exactly one right angle Can you find the right triangles?
A Right Triangle’s Best Friend • The Pythagorean Theorem • A2 + B2 = C2 • The sum of the squares of the legs of a right triangle is equal to the squareof its hypotenuse
ExamplePythagorean Theorem • Find the value of x A = 6 B = 15 C = x 62 + 152 = x2 36 + 225 = x2 261 = x2 x =
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)
Area of a Parallelogram • Be careful! NO! Base and height do not intersect at a right angle Yes! Notice the height intersects the base
Find the area of Area of a Parallelogram
Area of a Parallelogram AreaFind the height of the parallelogram. The height forms a right triangle with points Sand T with base 12 in. and hypotenuse 20 in. c2 = a2 + b2Pythagorean Theorem 202 = 122 + b2c = 20 and a = 12 400 = 144 + b2 Simplify. 256 = b2 Subtract 144 from each side. 16 = b Take the square root of each side.
The height is 16 in. UT is the base, which measures 32 in. Area of a Parallelogram Continued A = bh Area of parallelogram = (32)(16) or 512 in2b = 32 and h = 16 Answer: The area is 512 in2.
A. Find the perimeter and area of • A • B • C • D A. 88 m; 255 m2 B. 88 m; 405 m2 C. 88 m; 459 m2 D. 96 m; 459 m2
Application Perimeter and Area of a Triangle SANDBOXYou 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?
Application Perimeter and Area of a Triangle Step 1 Find the perimeter of the sandbox. Perimeter = 16 + 12 + 7.5 or 35.5 ft Step 2 Find the area of the sandbox. Area of a triangle b = 12 and h = 9
boards Application Perimeter and Area of a Triangle Step 3 Use unit analysis to determine how many of each item are needed. Boards Bags of Sand Answer: You will need 12 boards and 6 bags of sand.
Area of a Trapezoid SHAVINGFind the area of steel used to make the side of the trapezoid shown below. Area of a trapezoid h = 1, b1 = 3, b2 = 2.5 Simplify. Answer:A = 2.75 cm2
Area of a Trapezoid OPEN ENDEDMiguel 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.
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.
Use the Pythagorean Theorem to find ℓ. a2 + b2 = c2 Pythagorean Theorem 42 + ℓ2 = 52 Substitution 16 + ℓ2 = 25 Simplify. ℓ2 = 9 Subtract 16 from each side. ℓ = 3 Take 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 Answer: So, the area of the deck is 30 square feet. Simplify.
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, d1 and d2 • A= d1d2
8 10 4 14 8 12 Area of Regions 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
Other Types of Polygons • Regular Polygon – a polygon with sides that are all the same length and angles that are all the same measure
Areas of Regular Polygons Perimeter = (6)(8) = 48 apothem = Area = ½ (48)( ) = sq. units 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). 8