7. In the adjoining figure, ABCD is a trapezium
in which CD || AB and its diagonals intersect
at O. If AO = (2x + 1) cm, OC = (5x - 7) cm,
DO = (7x-5) cm and OB = (7x+1) cm, find
the value of x.
Answers
Answer:
Answer is:2
Step-by-step explanation:
Here we have been given,
ABCD Is a trapezium in which CD parallel to AB and its diagonals intersect at 0. AO=5x-7cm,OC=2x+1cm,DO= 7x-5cm and OB=7x+1cm.
We know that the diagonals of a Trapezium Divide each other proportionally.
So, \frac{AO}{OC} = \frac{DO}{OB}
OC
AO
=
OB
DO
⇒ \frac{5x - 7}{2x + 1} = \frac{7x - 5}{7x + 1}
2x+1
5x−7
=
7x+1
7x−5
⇒ 35x^{2} - 49x + 5x - 7 = 14x^{2} + 7x - 10x -535x
2
−49x+5x−7=14x
2
+7x−10x−5
⇒ 21x^{2} - 41x - 2 = 021x
2
−41x−2=0
⇒ 21x^{2} - 42x + x -2 = 021x
2
−42x+x−2=0
⇒ 21x(x-2) + 1(x-2) = 021x(x−2)+1(x−2)=0
⇒ (21x + 1) (x-2) = 0(21x+1)(x−2)=0
⇒ x - 2 = 0 or 21x + 1 = 0x−2=0 or21x+1=0
⇒ x = 2 or x = \frac{-1}{21}x=2 orx=
21
−1
x = \frac{-1}{21} is Rejectedx=
21
−1
isRejected
Therefore x = 2x=2 .