The bisector of angle A of rectangle ABCD intersects side BC at point K. Find the ratio of the area of triangle ABK

| March 10, 2021 | education

The bisector of angle A of rectangle ABCD intersects side BC at point K. Find the ratio of the area of triangle ABK to the area of trapezoid AKCD if AB: AD = 3: 5.

Since ABCD is a rectangle, and AK is a bisector of a right angle, then the angle BAK = BAD / 2 = 90/2 = 45.

The AВK triangle is rectangular with an acute angle of 45, then it is also isosceles, which means BK = AB.

Let the length of the side AB = 3 * X cm, then the side AD = 5 * X cm, BK = 3 * X cm.

The area of the rectangle ABCD is equal to: Savsd = AB * BC = 15 * X ^ 2 cm.

The area of the triangle is ABK = AB ^ 2/2 = 9 * X ^ 2/2 = 4.5 * X ^ 2.

Then the area of the AKСD trapezoid is equal to:

Saxd = Svsd – Sawk = 15 * X ^ 2 = 4.5 * X ^ 2 = 10.5 * X ^ 2.

Sawk / Saxd = 4.5 * X ^ 2 / 10.5 * X ^ 2 = 45/105 = 3/7.

Answer: Squares are referred to as 3/7.



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