Rectangle ABCD, side AB is 12 cm. The distance from the intersection of the diagonals to this side is 8 cm.

| January 5, 2021 | education

Rectangle ABCD, side AB is 12 cm. The distance from the intersection of the diagonals to this side is 8 cm. Find the area of triangle ABC.

In the rectangle ABCD, the side AB is 12 cm. The diagonals of the rectangle are equal to AC = BD and the intersection point O is divided in half: AO = OC; BО = ОD (property of rectangle diagonals). Then ΔAOB is isosceles with base AB. Let’s draw the median OK, according to the property of the median drawn to the base, it is a bisector and height, then OK = 8 cm, since according to the condition, the distance from the point of intersection of the diagonals to the side AB is 8 cm. Point K is the middle of the segment AB, point O is the middle of the segment AC, it turns out that the segment OK is the midline of the triangle ABC. Then BC = 2 • OK (by the property of the middle line of the triangle). The area of ​​the triangle ABC (angle B – straight line) is found by the formula: S = (a • h): 2; substitute the parameters of the triangle: S (Δ ABC) = (BC • AB): 2 = (OK • AB). Substitute the values: S (Δ ABC) = 8 • 12 = 96 (sq. Cm).
Answer: S (Δ ABC) = 96 sq. cm.



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