In the given figure, seg PS is the median

Practice Set 2.2 | Q 3 | Page 43

In the given figure, seg PS is the median of ∆PQR and PT ⊥ QR. Prove that,

(i) PR2 = PS2 + QR × ST + `(“QR”/2)^2`

(ii) PQ2 = PS2 – QR × ST + `(“QR”/2)^2`

Answer:
(i) $$PR^2 = PS^2 + QR times ST + left(frac{QR}{2}right)^2$$
(ii) $$PQ^2 = PS^2 – QR times ST + left(frac{QR}{2}right)^2$$
$$QS = SR = frac{QR}{2}$$ …(1) (S is the midpoint of side QR)
In $$triangle PTS$$, $$angle T = 90°$$ According to Pythagoras theorem, $$PS^2 = PT^2 + ST^2$$ …(2)
(i) In $$triangle PTR$$, $$angle T = 90°$$
According to Pythagoras theorem,
$$PR^2 = PT^2 + RT^2$$
$$therefore PR^2 = PT^2 + (ST + SR)^2$$
$$therefore PR^2 = PT^2 + (ST + frac{QR}{2})^2$$
$$therefore PR^2 = PT^2 + ST^2 + 2ST times left(frac{QR}{2}right) + left(frac{QR}{2}right)^2$$ $$quad quad quad quad quad quad quad quad quad quad quad quad quad$$ [(a + b)^2 = a^2 + 2ab + b^2]
$$therefore PR^2 = (PT^2 + ST^2) + ST times QR + left(frac{QR}{2}right)^2$$
$$therefore PR^2 = PS^2 + ST times QR + left(frac{QR}{2}right)^2$$ $$quad quad quad quad quad quad quad quad quad quad quad quad quad$$ [From 2]
(ii) In $$triangle PTQ$$, $$angle T = 90°$$
According to Pythagoras theorem,
$$PQ^2 = PT^2 + TQ^2$$
$$therefore PQ^2 = PT^2 + (QS – ST)^2$$
$$therefore PQ^2 = PT^2 + left(frac{QR}{2} – STright)^2$$
$$therefore PQ^2 = PT^2 + left(frac{QR}{2}right)^2 – 2 times frac{QR}{2} times ST + ST^2$$ $$quad quad quad quad quad quad quad quad quad quad quad quad quad$$ [(a – b)^2 = a^2 – 2ab + b^2]
$$therefore PQ^2 = (PT^2 + ST^2) – QR times ST + left(frac{QR}{2}right)^2$$
$$therefore PQ^2 = PS^2 – QR times ST + left(frac{QR}{2}right)^2$$ $$quad quad quad quad quad quad quad quad quad quad quad quad quad$$ [From 2]

Answer:- (i) PR² = PS² + QR × ST + (QR/2)^2

(ii) PQ² = PS² – QR × ST + (QR/2)^2

QS = SR = QR/2                 …(1)(S is the midpoint of side QR)

In ΔPTS, 
∠T = 90°                           
According to Pythagoras theorem,  
PS² = PT² + ST²                 …(2)

(i) In ΔPTR, ∠T = 90°  

According to Pythagoras theorem,  

 PR² =  PT² + RT²

∴ PR² = PT² + (ST + SR)²

∴ PR² = PT² + (ST + SR)²

∴ PR² = PT² + (ST + QR/2)^2 

∴ PR² = PT² + ST² + 2ST × (QR/2) + (QR/2)^2    …[(a + b)^2 = a^2 + 2ab + b^2]

∴ PR² = (PT² + ST²) + ST × QR + (QR/2)^2 

∴ PR² = PS² + ST × QR + (QR/2)^2          …[From 2]

(ii) In ΔPTQ, ∠T = 90°  

According to Pythagoras theorem, 

PQ² = PT² + TQ² 

∴ PQ² = PT² + (QS – ST)²

∴ PQ² = PT² + (QR/2 – ST)^2     

∴ PQ² = PT² + (QR/2)^2 – 2 × QR/2 × ST + ST^2 ..[(a – b)^2 = a^2 – 2ab + b^2]

∴ PQ²  = (PT² + ST²) – QR × ST + (QR/2)^2 

∴ PQ² = PS^2 – QR × ST + (QR/2)^2     …[From 2]

Chapter 2 – Pythagoras Theorem- Text Book Solution

Practice Set 2.2 | Q  3 | Page 43

Q 2


Q 3


Q 4