Find the value of x for which DE||AB in Fig. 6.8
Solution:
Given, DE||AB
We have to find the value of x.
From the figure,
CD = x+ 3
AD = 3x + 19
CE = x
BE = 3x + 4
Basic Proportionality Theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
So, CD/AC = CE/BC
AC = CD + AD = x + 3 + 3x + 19 = 4x + 22
BC = CE + BE = x + 3x + 4 = 4x + 4
Now, (x+3)/(4x+22) = x/(4x+4)
On cross multiplication,
(x+3)(4x+4) = x(4x+22)
By multiplicative and distributive property,
4x² + 4x + 12x + 12 = 4x² + 22x
Cancelling out common terms,
4x + 12x + 12 = 22x
By grouping,
16x - 22x = -12
-6x = -12
x = 12/6
x = 2
Therefore, the value of x is 2.
✦ Try This: In the adjoining figure, line AB || line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6
NCERT Exemplar Class 10 Maths Exercise 6.3 Problem 2
Find the value of x for which DE||AB in Fig. 6.8
Summary:
The value of x for which DE||AB in Fig. 6.8. is 2.
☛ Related Questions: