IN RECTANGLE ABCD ,AB=25cm BC=15cm .in what ratio does bisectorof angle C divides AB
in rectangle ABCD,AB = 25cm and BC = 15cm. In what ratio does the bisector of angle C divide AB
To determine the ratio in which the angle bisector of C divides AB in rectangle ABCD, we can use the angle bisector theorem.
The angle bisector theorem states that in a triangle, the angle bisector divides the opposite side into segments that are proportional to the lengths of the other two sides.
In this case, let's call the point where the angle bisector of C intersects AB as point E. We can denote the length of AE as 'x' and BE as 'y'.
Using the angle bisector theorem, we have the following proportion:
AE/EB = AC/CB
Plugging in the given values, we have:
x/y = (AC)/(BC)
Since AC is the diagonal of the rectangle, we can use the Pythagorean theorem to find its length. In rectangle ABCD, AD and AB are the two sides. So:
AD^2 + AB^2 = AC^2
Substituting the given values, we have:
AD^2 + 25^2 = AC^2
AD^2 + 625 = AC^2
Since ABCD is a rectangle, AD = BC = 15cm. Substituting this value:
15^2 + 625 = AC^2
225 + 625 = AC^2
850 = AC^2
Taking the square root of both sides:
√850 ≈ 29.155
So, AC ≈ 29.155 cm
Now, we can go back to our original proportion:
x/y = (AC)/(BC)
Plugging in the values we know:
x/y = (29.155)/(15)
Now we can find the ratio. We can choose any value for x or y and then find the other value using the equation. For example, let's assume x = 5.
5/y = 29.155/15
Cross-multiplying:
5 * 15 = 29.155 * y
75 = 29.155 * y
Dividing both sides by 29.155:
y ≈ 2.57
Therefore, the ratio in which the angle bisector of C divides AB is approximately 5 : 2.57.
C is a right angle (corner of a rectangle)
the bisector makes a 45º angle
so CBX is an isosceles right triangle
... BX = BC