Given that
PQ || AB
CP = x+3
PA = 3x+19
DC = x+3
CA = x
QB = 3x+4
We know that
By Basic Proportionality Theorem,
CP / PA = CQ / QB
On substituting these values in the above formula
⇛ (3x+19) / (x+3) = (3x+4) / x
On applying cross multiplication then
⇛ x(3x+19) = (3x+4)(x+3)
⇛ 3x²+19x = 3x(x+3)+4(x+3)
⇛ 3x²+19x = 3x²+9x+4x+12
⇛ 3x²+19x = 3x²+13x+12
⇛ 3x²+19x-3x²-13x = 12
⇛ (3x²-3x²)+(19x-13x) = 12
⇛ 0+6x = 12
⇛ 6x = 12
⇛ x = 12/6
⇛ x = 2
∴ , x = 2
<u>Answer:</u> The value of x for the given problem is 2
<em>Additional</em><em> comment</em><em>:</em>
<u>Basic Proportionality Theorem</u>
" A line drawn parallel to the one side of a triangle intersecting other two sides at two different points, then the line divides the other two sides in the same ratio".
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