Question

In fig . seg AB || seg DC. Find the value of x.


plz help solve this problem!!!!!!!

Answer 1

AB parallel to DC

<DCA = <ACB( alternate)

<CDB = <ABD ( alternate)

DOC = AOB( vertically opp.)

So AAA

DOC~BOA

DO/OB= OC/OA

3/(x-3) = x-5)/(3x-19)

3( 3x-19) = (x-3)( x-5)

9x - 57 = x^2 - 5x - 3x + 15

x^2 - 8 x +15 - 9x +57= 0

x^2 - 17x + 72 = 0

x^2 -8x -9 x +72= 0

x( x-8) -9( x -8)= 0

( x-9)(x-8)= 0

x=9,8

Answer 2

Answer:

⇢ AB || DC

⇢ Using (AAA guarantees congruence).

⇢ AAA = Angle Angle Angle.

⇢ As you know, DOC ~ BOA

• $\rm{\dfrac{DO}{OB} ≈ \dfrac{OC}{OA}}$

Hence,

⇢ $\rm{\dfrac{3}{x - 3} = \dfrac{x - 5}{3x - 19}}$

⇢ $\rm{3 (3x - 19) = (x - 3) (x - 5)}$

⇢ $\rm{9x - 57 = x^{2} - 5x - 3x + 15}$

⇢ $\rm{x^{2} - 8x + 15 - 9x + 57 = 0}$

⇢ $\rm{x^{2} - 17x + 72 = 0}$

⇢ $\rm{x^{2} - 8x - 9x + 72 = 0}$

⇢ $\rm{x (x - 8) - 9 (x - 8) = 0}$

⇢ ${\boxed{\rm{(x - 9) (x - 8) = 0}}}$

Therefore the value of x here is (9 , 8).