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