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Areas of Triangles, Trapezoids, and Rhombi

Areas of Triangles, Trapezoids, and Rhombi. Objective : TLW find the areas of triangles, trapezoids, and rhombi. SOL G.10. What does the area of a triangle have in common with the area of a rectangle or a parallelogram?. Let’s look at a Triangle.

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Areas of Triangles, Trapezoids, and Rhombi

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  1. Areas of Triangles, Trapezoids, and Rhombi Objective: TLW find the areas of triangles, trapezoids, and rhombi. SOL G.10

  2. What does the area of a triangle have in common with the area of a rectangle or a parallelogram?

  3. Let’s look at a Triangle • Looking at the triangle below, how can you relate this image to a rectangle? • What if we drew a line perpendicular to AC through point A? through point C? What does this do for us?

  4. Let’s look at a Triangle How can we use this diagram and the information we now have to create the formula for area of a triangle? • Could we follow this with an additional step? • What would that step be?

  5. The Area of a Triangle: Let’s Try One!!!! A = (1/2) (7.3 cm)(3.4 cm) = (1/2) (24.82 cm2) = 12.41 cm2 • If a triangle has an area of A square units, a base of b units, and a corresponding height of h units, then: • A = (1/2) b h • Where b = base; h = height

  6. How can we take this formula for the area of a triangle and apply it to the formula for a trapezoid?

  7. Let’s look at a Trapezoid b1 Area of Trapezoid WXZY: = area of ΔWYX + area of ΔYZX = (1/2) (b1 )(h) + (1/2) (b2 )(h) = (1/2) (b1 + b2 ) h = (1/2) h (b1 + b2 ) h b2 Could we create a trapezoid from triangles? If so, how?

  8. Consider a Trapezoid on the Coordinate Plane Find b1 and b2: b1 = |-3 - 3| = 6 b2 = |-5 – 6| = 11 Find h: h = |4 – (-1)| = 5 A = (1/2) h (b1 + b2) = (1/2) 5 (6 + 11) = (1/2) 5 (17) = (1/2) 85 u2 = 42.5 u2 Find the area of trapezoid MNPO:

  9. How can we take what we have learned from our area formula for a triangle and apply it to the formula for the area of a rhombus?

  10. Let’s look at a Rhombus = area of KFH + area of JFK + area of IFJ + area of HFI = (1/2) ((1/2)d1)((1/2)d2)+ (1/2)((1/2)d1)((1/2)d2) + (1/2)((1/2)d1)((1/2)d2) + (1/2)((1/2)d1)((1/2)d2) = 4 [ (1/2) ((1/2)d1) ((1/2)d2) ] =4 [ (1/2) ((1/2)d1) ((1/2)d2) ] = (d1) ((1/2)d2) A = (1/2)(d1) (d2) Area of a Rhombus:

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