Question

in the following figure find the value of X and Y​

Answer 1

Let Inner ∆ be PRQ,

And Outer ∆ be PST

Given that

$\\\\\begin{gathered}\frak{Given}\begin{cases}\sf PQ=PR \\\sf PT=2x+4 \\ \sf PS=3y+8 \\ \sf ∠PQT=∠PRS \\ \sf ∠QTP=∠SPR\end{cases}\end{gathered}$

To Find

• Value of x and y

Solution

• PR=PQ

Now ,

$\\\\ \sf{→∠PQR=∠PRQ \: \: \: \: \: - - - -given }$

$→ \text{Adding both sides 180° }$

$\sf{→∠PQR + 180=∠PRQ + 180}$

$\sf{→∠PQT=∠PRS}$

Now In ∆PQT and ∆PRS We have

$\\\\ \sf{→∠QPT=∠RPS}$

$\sf{→PQ=PR}$

$\sf{→∠PQT=∠PRS}$

$\\\text{So by AAS Criterion of congruence we have,}$

$\\\sf→∆PQT ≅ ∆PRS$

$\\\\\sf{\therefore PT=PS \: \: \: \: \: \: \: \: \: \: (by \: CPCT )_{(\: Congruence \: part \: of \: Congruence \: Triangle)}}$

So x=2y _(1)

$\\ \\ \sf{\therefore PT=PS \: \: \: \: \: \: \: \: \: \: (by \:CPCT )_{(\:Congruence \: part \: of \: Congruence \: Triangle)}}$

$→2x + 4 = 3y + 8$

⁣$\\\text{Putting the Value of x=2y from Equation 1.}$

$\\\sf{→2×(2y) + 4 = 3y + 8 }$

$→\sf4y + 4 = 3y + 8$

$→\sf4y - 3y = 8 - 4$

$→\sf y = 4$

$→\sf{x= 2y=8}$

• Therefore x = 8 and y= 4