IN RECTANGLEABCD, AB=25CM ,BC=15CM.IN WHAT RATIO DOES BISECTOR OF ANGLE CDIVIDES AB?
IN RECTANGLE ABCD ,AB=25CM ,BC=15CM.IN WHAT RATIO DOES BISECTOR OF ANGLE C DIVIDES AB?
Since the bisector meets BC at a 45° angle,
CB=PB = 15
so AP=25-15=10
To find the ratio in which the angle bisector of angle C divides AB in the rectangle ABCD, we can use the angle bisector theorem. This theorem states that in a triangle, the angle bisector of an angle divides the opposite side in the ratio of the lengths of the other two sides.
In our case, the angle bisector of angle C divides AB. Let's call the point where the angle bisector intersects AB as E. Now we need to find the ratio AE:EB.
According to the angle bisector theorem:
AE/EB = AC/CB
To use this theorem, we need to find the lengths of AC and CB.
In rectangle ABCD, AB is the length of the rectangle, and BC is the width. Given that AB = 25 cm and BC = 15 cm, we have the lengths of the other two sides.
Now let's substitute these values into the angle bisector theorem equation:
AE/EB = AC/CB
AE/EB = (AB + BC) / BC
AE/EB = (25 + 15) / 15
AE/EB = 40/15
Simplifying this ratio, we get:
AE/EB = 8/3
Therefore, the angle bisector of angle C divides AB in the ratio of 8:3.