In square PQRS,overline{PQ} parallel overline | vedclass

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Question

(A) Since $\overline{PQ} \parallel \overline{RS}$,$	riangle XPQ \sim 	riangle XRS$ by $AA$ similarity criterion (alternate interior angles are equal

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Since $\overline{PQ} \parallel \overline{RS}$,$	riangle XPQ \sim 	riangle XRS$ by $AA$ similarity criterion (alternate interior angles are equal).Therefore,the ratios of their corresponding sides are equal: $\frac{XP}{XR} = \frac{XQ}{XS}$.Substituting the given values: $\frac{2x + 4}{x + 1} = \frac{4x - 2}{4}$.Cross-multiplying gives: $4(2x + 4) = (x + 1)(4x - 2)$.$8x + 16 = 4x^2 - 2x + 4x - 2$.$8x + 16 = 4x^2 + 2x - 2$.Rearranging into a quadratic equation: $4x^2 - 6x - 18 = 0$.Dividing by $2$: $2x^2 - 3x - 9 = 0$.Factoring the quadratic: $2x^2 - 6x + 3x - 9 = 0 \implies 2x(x - 3) + 3(x - 3) = 0$.$(2x + 3)(x - 3) = 0$.Thus,$x = 3$ or $x = -1.5$.Since lengths must be positive,$x = 3$.

In $\square PQRS$,$\overline{PQ} \parallel \overline{RS}$ and $\overline{PR} \cap \overline{QS} = \{X\}$. If $XP = 2x + 4$,$XQ = 4x - 2$,$XR = x + 1$,and $XS = 4$,find the value of $x$.

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