In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE||BC. If AD=(4x-3), DB=(2x-1), AE=7cm and EC= 3cm, Find the value of x.
Answers
Answer:
The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.
It is given that AD=4x−38, BD=3x−1, AE=8x−7 and CE=5x−3. Let AC=x
Using the basic proportionality theorem, we have
BDAD=ACAE⇒3x−14x−3=x8x−5⇒x(4x−3)=(3x−1)(8x−5)⇒4x2−3x=3x(8x−5)−1(8x−5)⇒4x2−3x=24x2−15x−8x+5⇒4x2−3x=24x2−23x+5⇒24x2−23x+5−4x2+3x=0⇒20x2−20x+5=0⇒5(4x2−4x+1)=0⇒4x
The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.
It is given that AD=4x−38, BD=3x−1, AE=8x−7 and CE=5x−3. Let AC=x
Using the basic proportionality theorem, we have
BD
AD
=
AC
AE
⇒
3x−1
4x−3
=
x
8x−5
⇒x(4x−3)=(3x−1)(8x−5)
⇒4x
2
−3x=3x(8x−5)−1(8x−5)
⇒4x
2
−3x=24x
2
−15x−8x+5
⇒4x
2
−3x=24x
2
−23x+5
⇒24x
2
−23x+5−4x
2
+3x=0
⇒20x
2
−20x+5=0
⇒5(4x
2
−4x+1)=0
⇒4x
2
−4x+1=0
⇒(2x)
2
−(2×2x×1)x+1
2
=0(∵(a−b)
2
=a
2
+b
2
−2ab)
⇒(2x−1)
2
=0
⇒(2x−1)=0
⇒2x=1
⇒x=
2
1
Hence, x=
2
1
.
this is your answer