MCQ
In the given figure if $\text{PS}\parallel\text{QR}$ and $\text{PQ}\parallel\text{SR}$ and $AT = AQ = 6, AS = 3, TS = 4,$ then :
- ✓$x = 3, y = 4.$
- B$x = 1, y = 2.$
- C$x = 2, y = 3.$
- D$x = 4, y = 5.$
✓
Answer
Correct option: A.
$x = 3, y = 4.$
In triangles $\text{APQ }$ and $\text{ ATS}$
$\angle\text{PAQ}=\text{TAS}, \ [$Verticaly opposite angles$]\ $
$\angle\text{PQA}=\text{ATS},\ [$Alternate angles$]$
$\therefore\triangle\text{APQ}\sim\triangle\text{AST}\ [\text{AA}$ similarity$]$
$\therefore\frac{\text{AQ}}{\text{AT}}=\frac{\text{AP}}{\text{AS}}$
$\Rightarrow\frac{6}{6}=\frac{x}{3}$
$\Rightarrow\text{x}=\frac{6\times3}{6}=3$
And $\frac{\text{AQ}}{\text{AT}}=\frac{\text{PQ}}{\text{ST}} $
$\Rightarrow\frac{6}{6}=\frac{\text{x}}{4}$
$\Rightarrow\text{y}=\frac{4\times6}{6}=4$
$\Rightarrow$ Therefore $,\text{x}=3, \text{ y}=4$
$\angle\text{PAQ}=\text{TAS}, \ [$Verticaly opposite angles$]\ $
$\angle\text{PQA}=\text{ATS},\ [$Alternate angles$]$
$\therefore\triangle\text{APQ}\sim\triangle\text{AST}\ [\text{AA}$ similarity$]$
$\therefore\frac{\text{AQ}}{\text{AT}}=\frac{\text{AP}}{\text{AS}}$
$\Rightarrow\frac{6}{6}=\frac{x}{3}$
$\Rightarrow\text{x}=\frac{6\times3}{6}=3$
And $\frac{\text{AQ}}{\text{AT}}=\frac{\text{PQ}}{\text{ST}} $
$\Rightarrow\frac{6}{6}=\frac{\text{x}}{4}$
$\Rightarrow\text{y}=\frac{4\times6}{6}=4$
$\Rightarrow$ Therefore $,\text{x}=3, \text{ y}=4$
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