Three squares of side lengths 8 , 1 2 , and 1 0 are placed side-by-side as shown. Find the area of the orange region.
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Excellent, Sir!
Though i calculated each area of trapezium using analytic geometry of slopes and all, your's is much more clean, Nice!
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I did not know how to solve it even though I posted it! :)
Define the points as shown and use the cross product formula for the area of a triangle. Calculate the areas of the two sub-triangles as shown, and combine them. The area comes out to 7 0 .
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Nice solution but I was thinking that if it is possible to find a solution only by elementary geometry?
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Yes, you can also do:
A = 2 1 ( 1 0 ) ( 1 0 ) + 2 1 ( 2 ) ( 2 0 ) = 7 0
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Thanks a lot! This problem was confusing me too much.
Area of figure = a r ( A B C ) + a r ( C B D ) - a r ( g r e e n r e c t a n g l e )
Area of figure = 2 1 ∗ 1 0 ∗ 1 2 + 2 1 ∗ 2 ∗ 3 0 − 1 0 ∗ 2
Area of figure = 7 0
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Nice solution!
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@Vinayak Srivastava ,the previous solution was wrong.
Thanks.How you have done it ?.
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I did not do it :) I did not know I how to do it, actually.
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Redraw the figure above without affecting the orange region. We find the orange region is given by:
A = 3 0 × 1 2 − 2 3 0 × 1 0 − 2 2 0 × 1 2 − 1 0 × 2 = 3 6 0 − 1 5 0 − 1 2 0 − 2 0 = 7 0