Question

In the figure 2.35,ΔPQR is an equilatral triangle. Point S is on seg QR such that QS=1/3QR. Prove that : 9PS²=7PQ²

Answer 1

Steps:
1) Let the side of an equilateral triangle be '2a' units . 
Then ,
Height of an equilateral triangle is given by  :
$PT= \frac{ \sqrt{3} }{2} *(2a) = \sqrt{3}a \:\: units$ 

2) I think PT is perpendicular to QR . 
=> QT=TR= 'a' units

Since, we have  
$QS= \frac{1}{3}QR \\ \\ =\ \textgreater \ QS = \frac{2a}{3} units$
      
=>ST = a/3 units .

3) In right ΔPST ,
By Pythagoras Theorem , 

[tex] PS^{2} = PT^{2} + ST^{2} \\ \\ =\ \textgreater \ PS^{2} = ( \sqrt{3}a)^{2} + (\frac{a}{3})^{2} \\ \\ =\ \textgreater \ PS^{2} = \frac{28a^{2}}{9} \\ \\ =\ \textgreater \ PS^{2} = \frac{7}{9} * (2a)^{2} \\ \\ =\ \textgreater \ PS^{2} = \frac{7}{9}*PQ^{2} \\ \\ =\ \textgreater \ 9 PS^{2} =7 PQ^{2} [/tex]

Hence Proved :

Answer 2

Answer:

Let the side of an equilateral triangle be '2a' units .  

Then ,

Height of an equilateral triangle is given by  :

PT= Root 3/ 2 x(2a)= root 3a units

I think PT is perpendicular to QR .  

=> QT=TR= 'a' units

Since, we have    

      =>ST = a/3 units .

3) In right ΔPST ,

By Pythagoras Theorem ,

Hence Proved :

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