Question

QUESTION:-

PQR is a triangle right angled at P. If PS is perpendicular drawn from P to QR, then which of the following is true:
(a). 1/PQ^2=1/QS^2+1/RS^2
(b). 1/PR^2=1/PS^2+1/QS^2
(c). 1/PQ^2=1/PS^2-1/PR^2
(d). 1/PS^2=1/PQ^2-1/QR^2


NO SPAMMING ALLOWED....NEED URGENTLY...

Answer 1

Answer:

Step-by-step explanation:

Let, x=1.3  

23

.

Then, 10x=13.  

23

(1)

                =13.23  

23

 

and 1000x=1323.  

23

(2)

Now, subtracting  (1) from (2) we get,

990x=1310

⇒x=  

990

1310

​

=  

99

131

​

 

Which is the required  

q

p

​

form

Answer 2

Answer:

hi it's my new I'd dear

Step-by-step explanation:

☮In ΔPQR 

In ΔPQR ∠PQR=90∘ [Given] 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 =QS2=20 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 =QS2=20 ⇒QS=25cm 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 =QS2=20 ⇒QS=25cm QS2=PS×SR  (i) 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 =QS2=20 ⇒QS=25cm QS2=PS×SR  (i) ⇒(25)2=4×SR 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 =QS2=20 ⇒QS=25cm QS2=PS×SR  (i) ⇒(25)2=4×SR ⇒420=SR 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 =QS2=20 ⇒QS=25cm QS2=PS×SR  (i) ⇒(25)2=4×SR ⇒420=SR ⇒SR=5cm 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 =QS2=20 ⇒QS=25cm QS2=PS×SR  (i) ⇒(25)2=4×SR ⇒420=SR ⇒SR=5cm Now , QS⊥PR 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 =QS2=20 ⇒QS=25cm QS2=PS×SR  (i) ⇒(25)2=4×SR ⇒420=SR ⇒SR=5cm Now , QS⊥PR ∴∠QSR=90∘ 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 =QS2=20 ⇒QS=25cm QS2=PS×SR  (i) ⇒(25)2=4×SR ⇒420=SR ⇒SR=5cm Now , QS⊥PR ∴∠QSR=90∘ ⇒QR2=QS2+SR2 [By Pythagoras theorem] 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 =QS2=20 ⇒QS=25cm QS2=PS×SR  (i) ⇒(25)2=4×SR ⇒420=SR ⇒SR=5cm Now , QS⊥PR ∴∠QSR=90∘ ⇒QR2=QS2+SR2 [By Pythagoras theorem] =(25)2+52 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 =QS2=20 ⇒QS=25cm QS2=PS×SR  (i) ⇒(25)2=4×SR ⇒420=SR ⇒SR=5cm Now , QS⊥PR ∴∠QSR=90∘ ⇒QR2=QS2+SR2 [By Pythagoras theorem] =(25)2+52 =20+25 

In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR  (i)  [By theorem] Now , in ΔPSQ we have QS2=PQ2−PS2   [By Pythagoras theorem] =62−42 =36−16 =QS2=20 ⇒QS=25cm QS2=PS×SR  (i) ⇒(25)2=4×SR ⇒420=SR ⇒SR=5cm Now , QS⊥PR ∴∠QSR=90∘ ⇒QR2=QS2+SR2 [By Pythagoras theorem] =(25)2+52 =20+25 ⇒Q

don't forget to fol low me for more answers dear ir☯️