5. In the given figure, if AB || CD, then the value of x is
A А
B
3x - 4
2X-
4x - 2
2x + 4
M
D
CM
(1) 3
(2) 2
(3 4
(4) 5
9.
Answers
Answer:
(1) For Fig.a,
AB∥CD [ Given ]
∴ Quadrilateral ABCD is a trapezium.
∴
CO
AO
=
DO
BO
[ Diagonals of trapezium divides each other proportionally ]
⇒
4x−2
4
=
2x+4
x+1
⇒ 4(2x+4)=(x+1)(4x−2)
⇒ 8x+16=4x
2
−2x+4x−2
⇒ 4x
2
−6x−18=0
⇒ 2x
2
−3x−9=0
⇒ (x−3)(2x+3)=0
∴ x=3 or x=
2
−3
(2) For Fig.(b),
AB∥CD [ Given ]
∴ Quadrilateral ABCD is a trapezium.
∴
CO
AO
=
DO
BO
[ Diagonals of trapezium divides each other proportionally ]
⇒
5x−3
3x−1
=
6x−5
2x+1
⇒ (3x−1)(6x−5)=(2x+1)(5x−3)
⇒ 18x
2
−15x−6x+5=10x
2
−6x+5x−3
⇒ 8x
2
−20x+8=0
⇒ 2x
2
−5x+2=0
⇒ (x−2)(2x−1)=0
∴ x=2 or x=
2
1
(3) For Fig.(c),
AB∥CD [ Given ]
∴ Quadrilateral ABCD is a trapezium.
∴
CO
AO
=
DO
BO
[ Diagonals of trapezium divides each other proportionally ]
⇒
x−3
3x−19
=
4
x−4
⇒ (3x−19)(4)=(x−4)(x−3)
⇒ 12x−76=x
2
−3x−4x+12
⇒ x
2
−19x+88=0
⇒ (x−8)(x−11)=0
∴ x=8 or x=11