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virtus

  • 3 years ago

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  1. virtus
    • 3 years ago
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    |dw:1342145485490:dw| In the figure above, if the area of triangle AQD is 20 square cm and AQ: QB = 2:3 find the area of the parallelogram ABCD

  2. SmoothMath
    • 3 years ago
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    Okay, so that triangle and the parallelogram have the same height, which is what will let us solve this. Let me define the variable h=height, both of the triangle and parallelogram Area of the triangle = (1/2)h*(AQ) = 20 Area of the parallelogram = base*height = (AQ + QB)*h QB = (3/2)*AQ

  3. SmoothMath
    • 3 years ago
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    (AQ + QB) = (AQ + (3/2)AQ) = (5/2)AQ So the area of the parallelogram = (5/2)AQ*h And we know that (1/2)AQ*h = 20 from the triangle.

  4. virtus
    • 3 years ago
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    thanks smoothmath

  5. JohnHanShanghai
    • 3 years ago
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    100

  6. virtus
    • 3 years ago
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    thankyou @JohnHanShanghai

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