In ∆ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5,........
Practice Set 1.2 | Q 9 | Page 15
In ∆ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.
In △ABC,
seg BD bisects ∠ABC. ...(Given)
∴ by the theorem of angle bisector of a triangle,
∴ by the theorem of angle bisector of a triangle,
∴ `"AB"/"BC" = "AD"/"DC"`
∴ `x/(x + 5) = (x – 2)/(x + 2)`
∴ x(x + 2)= (x – 2)(x + 5)
∴ x2 + 2x = x(x + 5) - 2(x + 5)
∴ x2 + 2x = x2 + 5x - 2x - 10
∴ x2 + 2x = x2 + 3x - 10
∴ x2 - x2 + 2x - 3x = - 10
∴ - x = - 10
∴ x = 10
Answer:-
Given, in triangle ABC, segment BD bisects angle ABC.
By the angle bisector theorem, we have:
AB/BC = AD/DC
Substituting the given values, we get:
x/(x+5) = (x-2)/(x+2)
Cross-multiplying, we get:
x(x+2) = (x-2)(x+5)
Expanding and simplifying, we get:
x^2 + 2x = x^2 + 3x – 10
Subtracting x^2 from both sides, we get:
2x = 3x – 10
Subtracting 3x from both sides, we get:
-x = -10
Therefore, x = 10.
Chapter 1. Similarity- Practice Set 1.2 – Page 15
Click Here for All Textbook Soutions of Chapter 1: Similarity