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# In a rectangle ABCD, AB = 25 cm and BC = 15. In what ratio does the bisector of ∠C divide AB?

**Solution:**

Given, ABCD is a __rectangle__.

The sides AB = 25 cm and BC = 15 cm.

We have to find the ratio such that the bisector of angle C divide AB.

Let CO be the bisector of ∠C.

So, ∠OCB = ∠OCD = 45°

In triangle OCB,

By angle sum property of a triangle,

∠CBO + ∠OCB + ∠COB = 180°

90° + 45° + ∠COB = 180°

135° + ∠COB = 180°

∠COB = 180° - 135°

∠COB = 45°

In triangle OCB,

∠OCB = ∠COB

This implies OB = BC

So, OB = 15 cm

According to the question,

CO divides AB in the ratio AO : OB

Let AO = x

AB = AO + OB

25 = x + 15

x = 25 - 15

x = 10 cm

Now, AO : OB = 10 : 15

= 2 : 3

Therefore, the required ratio is 2 : 3.

**✦ Try This:**** **In a rectangle ABCD, AB = 35 cm and BC = 25. In what ratio does the bisector of ∠C divide AB

**☛ Also Check: **NCERT Solutions for Class 8 Maths

**NCERT Exemplar Class 8 Maths Chapter 5 Problem 135**

## In a rectangle ABCD, AB = 25 cm and BC = 15. In what ratio does the bisector of ∠C divide AB

**Summary:**

In a rectangle ABCD, AB = 25 cm and BC = 15. The ratio at which the bisector of ∠C divide AB is 2 : 3.

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