Question

$\bold{ \underbrace{ \underline{Urgent \: \: \: Help \: \: \: Request!! }}}$
In the given attachment, ∆PQR is a right angled at Q and points S and T trisects side QR. Prove:
>> $\bold{ {8PT}^{2} = {3PR}^{2} +{5PS}^{2} }$
_
• Quality Answers please!
• No plagiarism​

Answer 1

Let QS = x, then,

QT = 2x

QR = 3x

In ΔPQR,

$PR {}^{2} =PQ {}^{2} +RQ {}^{2}$

$PR {}^{2} =PQ {}^{2} +(3x) {}^{2}$

$PR {}^{2} =PQ {}^{2} +9x {}^{2}      (1)$

In Δ PQT,

$PT {}^{2} =PQ {}^{2} +TQ {}^{2}$

$PT {}^{2} =PQ {}^{2} +2x {}^{2}$

$PT {}^{2} =PQ {}^{2} + 4x {}^{2}$     (2)

In Δ PQS,

$PS {}^{2} =PQ {}^{2} +SQ {}^{2}$

$PS {}^{2} =PQ {}^{2} +x {}^{2}$     (3)

Now Using Equation (1) and (3),

$3PR {}^{2} =5PS {}^{2} +(3PQ {}^{2} + 9x {}^{2} )$ +

$5(PQ {}^{2} + x {}^{2} )$

= $3PQ {}^{2} + 27 {}^{2} + =5PQ {}^{2} + +5x {}^{2}$

= $8PQ {}^{2} + 32{}^{2}$

= $8(PQ {}^{2} + 4x {}^{2} )$

From the Equation (2)

$3PR {}^{2} + 5PS {}^{2} = 8PT {}^{2}$

Hence Proved.

Answer 2

$\\ \mathbb{ \underline { ★ \: GIVEN}} : - \begin{cases} \sf \underline{ \underline{In \: the \: given \: attachment, \: \bf ∆PQR \sf\: is \: a \: right \: angled \: at \: \bf \: Q \sf \: and \: points \: \bf{ S} \: \sf \: and \: \bf \: T \: \sf trisects \: side.}} \end{cases}$

$\\ \mathbb{ \underline {★ \: \: TO \: \: FIND}} : - \begin{cases} \sf \underline{ \underbrace{ \overline{ \overbrace{ \bf{Prove: - } \: \: \tt{ {8PT}^{2} = {3PR}^{2} +{5PS}^{2} }}}}} \end{cases}$

$\huge \bf {\underline{ \underline { \bigstar \: {Solution} : - }}}$

☯ Let,

• QS = x
• QT = 2x
• QR = 3x

$\\\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed {\huge{☯\:\: 1}}$

$\underbrace{ \large \bf{ \purple{\underline{\dag \: In \: \: \triangle\: PQR : - }}}}$

➙ PR² = PQ² + RQ²

➙ PR² = PQ² + (3x)²

➙ PR² = PQ² + 9x²

$\\\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed {\huge{☯\:\:2}}$

$\\\\ \underbrace{ \large \bf{ \blue{\underline{\dag \: In \: \: \triangle\:PQT : - }}}}$

➙ PT² = PQ² + TQ²

➙ PT² = PQ² + 2x²

➙ PT² = PQ² + 4x²

$\\\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed {\huge{☯\:\:3}}$

$\\\\ \underbrace{ \large \bf{ \pink{\underline{\dag \: In \: \: \triangle\: PQS : - }}}}$

➙ PS² = PQ² + SQ²

➙ PS² = PQ² + x²

$\\\\ \underbrace{ \large \bf{\underline{\dag Then, \red{\: use \: equation \: 1 \: and \: 3 ,}}}}$

➙ 3PR² = 5PS² + (3 PQ² + 9x²) + 5(PQ² + x²)

➙ 3PQ² + 27² + = 5PQ² + 5x²

➙ 8PQ² + 32²

➙ 8(PQ² + 4x²)

$\\\\ \underbrace{ \large \bf{\underline{\dag Then, \red{\: from \: equation \:2,}}}}$

➙ 3PR² + 5PS² = 8PT²

$\large{ \boxed{\mathbf{\overbrace{\underbrace{\fcolorbox{lime}{black}{ \pink{Hence \: proved \:!! }}}}}}}$

★ Hope it helps u ★