# 1. A kite has two diagonals. We’ll label them. 2. The diagonals in a kite make a right angle. 3. Let’s put a rectangle around the kite because we know.

## Presentation on theme: "1. A kite has two diagonals. We’ll label them. 2. The diagonals in a kite make a right angle. 3. Let’s put a rectangle around the kite because we know."— Presentation transcript:

1. A kite has two diagonals. We’ll label them. 2. The diagonals in a kite make a right angle. 3. Let’s put a rectangle around the kite because we know the formula for the area of a rectangle. 4. The diagonals have the same lengths as the sides of the rectangle, so… 5. Fill in the empty space around the kite with a different color. 6. The kite takes up half the space of the entire rectangle, so… Diagonal #1 Diagonal #2 90 Diagonal #2 Diagonal #1 1 2 1 11 2 22 Area of a Kite = ½ (d 1 )(d 2 ) Area of the Rectangle = (d 1 )(d 2 )

Area = ½ (d 1 )(d 2 ) = ½ (9+35)(12+12) = ½ (44)(24) = 528 square units 37 12 15 37 35 9 12 Be sure to identify the diagonals right away.

Area = ½ (d 1 )(d 2 ) = ½ (6+15)(8+8) = ½ (21)(16) = 168 square units 17 6 10 17 6 10 17 x 88 8 y 15 Be sure to identify the diagonals right away. Use the Pythagorean Theorem to find the missing pieces of the diagonals

1. A Rhombus has two diagonals. Label them. 2. The diagonals make a right angle. 3. Let’s put a rectangle around the rhombus because we know the formula for the area of a rectangle. 4. The diagonals have the same lengths as the sides of the rhombus, so… 5. Fill in the empty space around the rhombus with a different color. 6. The rhombus takes up half the space of the entire rectangle, so… Diagonal #1 Diagonal #2 90 Diagonal #2 Diagonal #1 1 2 1 11 2 22 Area of a Rhombus = ½ (d 1 )(d 2 ) Area of the Rectangle = (d 1 )(d 2 ) Special Note: …because a rhombus is a parallelogram, the formula for parallelograms will sometimes work as well.

6010 Area = ½ (d 1 )(d 2 ) = ½ (5+5)( + ) = ½ (10)( ) = square units Be sure to identify the diagonals right away. Use the 30-60-90 triangle rules to find the missing pieces of the diagonals 60 30 10 hypotenuse Long leg Short leg 5 5 5

1. A trapezoid has two bases and a height. Let’s label them. 2. Let’s add a third horizontal line through the middle of the figure. 3. Now let’s put a rectangle over the trapezoid. 4. Finally, lets cut off the ends of the trapezoid and move them. 5. This means the trapezoid and the rectangle have the same area. Area = (Avg of Bases) (Height) Base 1 Base 2 Avg of Bases Height The Area of a Trapezoid = (Avg of Bases) (Height) OR ½ (B 1 + B 2 )(Height)

Area=(Avg of Bases)(Height) =½ (B 1 + B 2 )(Height) =½ (20 + 10)(8) =½ (30)(8) = 120 square units 10 8 20 Be sure to identify the bases and the height right away. Base 1 Base 2

Area=(Avg of Bases)(Height) =½ (B 1 + B 2 )(Height) =½ (16 + 4)(Height) =½ (20)(6) = 60 square units 4 16 Be sure to identify the bases and the height right away. Base 1 Base 2 45 Use the 45-45-90 triangle rules to find the height 4 16 - 4 = 12 12 / 2 = 6 6 6 6 Same 6 We need to find a side on the triangle so we need to break Base 1 into three pieces. h=6

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