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ABCD is a parallelogram of area "S". E and F are the mid points of the sides AD and BC respectively. If G is any point on the line EF. then the area of ΔAGB is equal to

Question

ABCD is a parallelogram of area "S". E and F are the mid points of the sides AD and BC respectively. If G is any point on the line EF. then the area of ΔAGB is equal to

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Solution

The area of triangle AGB can be found using the formula for the area of a triangle, which is 1/2 * base * height.

In a parallelogram, opposite sides are equal in length, so the base of triangle AGB is equal to the length of side AD or BC.

Since E and F are the midpoints of AD and BC respectively, EF is a line segment that cuts the parallelogram into two equal areas.

Therefore, the height of triangle AGB is equal to the distance from G to line segment EF, which is half the height of the parallelogram.

So, the area of triangle AGB is 1/2 * base * height = 1/2 * AD * (1/2 * height of parallelogram) = 1/4 * area of parallelogram = 1/4 * S.

Therefore, regardless of where G is on line segment EF, the area of triangle AGB is always 1/4 * S.

This problem has been solved

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