Line segments are drawn from the vertices of the large square to the midpoint of the opposite side to form a small square. If side of the square is 20 meter . Find the area of the smaller square?
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using Pythagoras therom you can get the length of the line segments which would be sq ft of 500. Then you can use area formula to calculate the height of every triangle.
20×10×1/2 = √500x(BJ)×1/2
similarly BJ=AI=DL=CK
since square is formed inside, triangle AIE is rt angle triangle hence using Pythagoras therom you can find length IE
(IE)= √(100-(BJ^2))
then area of quadrilateral IEBJ can be calculated as:
1/2×20×10= 1/2×10×BJ + 1/2×AI×IE + area IEBJ
then you can find the area of sq by
20×20 = 1/2×20×10×2 + 2×area IEBJ + area sq IJKL
20×10×1/2 = √500x(BJ)×1/2
similarly BJ=AI=DL=CK
since square is formed inside, triangle AIE is rt angle triangle hence using Pythagoras therom you can find length IE
(IE)= √(100-(BJ^2))
then area of quadrilateral IEBJ can be calculated as:
1/2×20×10= 1/2×10×BJ + 1/2×AI×IE + area IEBJ
then you can find the area of sq by
20×20 = 1/2×20×10×2 + 2×area IEBJ + area sq IJKL
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