In the figure $P S = 3, S Q = 6. Q R = 5, P T = x$ and $T R = y$ .Give any two pairs of value of $x$ and $y$ such that line $S T | | Q R$

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Solution

If $S T | | S R$ , then $Δ P S T \sim Δ P Q R$ .

$⟹ \frac{P T}{P R} = \frac{P S}{P Q}$
$⟹ \frac{x}{x + y} = \frac{3}{9} = \frac{1}{3}$
$⟹ 3 x = x + y$
$⟹ 2 x = y$
Two pairs satisfying the above are:
$x = 3, y = 6$
$x = 4, y = 8$

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In the figure below, if the line segment ST is parallel to line segment QR such that $\frac{P S}{S Q} = \frac{P T}{T R}$ . This provided data is not sufficient to prove that
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Q. In

△ P Q R, S T ∥ Q R

and

\frac{P S}{S Q} = \frac{3}{5}

P R = 6

cm find

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Q. In the given figure, in a triangle PQR, ST

| |

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and

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In the figure PS=3,SQ=6.QR=5,PT=x and TR=y.Give any two pairs of value of x and y such that line ST||QR

If ST||SR, then ΔPST∼ΔPQR.⟹PTPR=PSPQ⟹xx+y=39=13⟹3x=x+y⟹2x=yTwo pairs satisfying the above are:x=3,y=6x=4,y=8

In the figure $P S = 3, S Q = 6. Q R = 5, P T = x$ and $T R = y$ .Give any two pairs of value of $x$ and $y$ such that line $S T | | Q R$

If $S T | | S R$ , then $Δ P S T \sim Δ P Q R$ .

$⟹ \frac{P T}{P R} = \frac{P S}{P Q}$
$⟹ \frac{x}{x + y} = \frac{3}{9} = \frac{1}{3}$
$⟹ 3 x = x + y$
$⟹ 2 x = y$
Two pairs satisfying the above are:
$x = 3, y = 6$
$x = 4, y = 8$