In given figure, AB||DC. Find the value of x.

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Solution

Since the diagonals of trapezium divide each other proportionally.

$∴$ $\frac{A O}{O C} = \frac{B O}{O D}$

$\Rightarrow$ $\frac{3 x - 19}{x - 5} = \frac{x - 3}{3}$

$\Rightarrow$ $3 (3 x - 19) = (x - 5) (x - 3)$

$\Rightarrow$ $9 x - 57 = x^{2} - 8 x + 15$

$\Rightarrow$ $x^{2} - 17 x + 72 = 0$

$\Rightarrow$ $(x - 8) (x - 9) = 0$

$\Rightarrow$ $x - 8 = 0$ or $x - 9 = 0$

$\Rightarrow x = 8$ or $x = 9$

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In given figure, AB||DC. Find the value of x.

Since the diagonals of trapezium divide each other proportionally.∴ AOOC=BOOD⇒ 3x−19x−5=x−33⇒ 3(3x−19)=(x−5)(x−3)⇒ 9x−57=x2−8x+15⇒ x2−17x+72=0⇒ (x−8)(x−9)=0⇒ x−8=0 or x−9=0 ⇒x=8 or x=9