Question

in fogure of ab||cs and oa=3x-1,oc =5x-3,ob=2x+1,od=6x-5,find the value of x​

Answer 1

Answer:

GIVEN:

OA= 3x-1 , OC= 5x-3, OD= 6x-5 , BO = 2x+1

AO/OC = BO/OD

[The diagonals of a Trapezium divide each other proportionally]

(3x-1)/(5x-3) =( 2x+1)/(6x-5)

(3x-1) (6x-5) = (5x-3) ( 2x+1)

3x(6x-5) -1 (6x-5) = 2x(5x-3)+1(5x-3)

18x² -15x -6x +5 = 10x² -6x +5x -3

18x² -21x +5 = 10x² - x -3

18x² -10x² -21x + x +5+3 =0

8x² - 20x +8= 0

4(2x² -5x +2) = 0

2x² -5x +2 = 0

2x² -4x -x +2= 0

[By factorization]

2x(x-2) -1(x-2)= 0

(2x-1) (x-2) = 0

(2x-1) = 0 or (x-2) = 0

x = ½ or x = 2

If we put x= ½ in OD , The value of OD is negative.

Hence, the value of is x = 2. please mark as brilliant

Answer 2

OA=3x−1OC=5x−3OB=2x+1OD=6x−5ThediagonalsoftrapeziumdivideeachotherproportionallyOCOA=ODOB⇒(5x−3)(3x−1)=(6x−5)(2x+1)⇒(3x−1)(6x−5)=(2x+1)(5x−3)⇒18x2−21x+5=10x2−x−3⇒8x2−20x+8=0⇒2x2−5x+2=0⇒(2x−1)(x−2)=0⇒x=21or2ifweputx=21inODthevalueofODisnegativeHence,x=2

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