The bisector of angle B of rectangle ABCD intersects side AD at point K, AK = 5 cm

The bisector of angle B of rectangle ABCD intersects side AD at point K, AK = 5 cm and KD = 7 cm. Find the area of this rectangle.

Since BK is the bisector of the right angle B, it divides the angle B in half:

ABK = KBC = 90 ° / 2 = 45 °.

Let’s find the angle AKB in the triangle ABK:

AKB = 180 ° – A – ABK = 180 ° – 90 ° – 45 ° = 45 °;

In triangle ABK, angle AKB = angle ABK = 45 °, from which it follows that triangle ABK is isosceles. In isosceles triangles, the sides are equal:

AB = AK = 5 cm;

Find the side of the rectangle AD:

AD = AK + KD = 5 + 7 = 12 cm;

The area of a rectangle is equal to the product of length and width:

S = AD * AB = 12 * 5 = 60 cm²

Answer: 60 cm²