PS is perpenducular to QR if PQ =a QS=c PR=b RS=d then prove that. (a+b)(a-b)=(c-d)(c+d).
Attachments:
Answers
Answered by
13
Answer:
Step-by-step explanation:
Attachments:
SamanthSD:
bhai wrong answer
(a+d) is where
In Right Δ P SQ
P Q²= PS² + SQ²→→[By Pythagoras theorem]
a²= PS² + c²
→PS² = a² - c²-------(1)
In Right Δ P SR
PR²= PS² + SR²→→[By Pythagoras theorem]
b²= PS² + d²
PS²= b² - d² ------(2)
From (1) and (2)
b²- d²= a² - c²
a² - b²= c² - d²
(a -b)(a+b)= (c-d)(c+d)→→Using the identity, A²-B²= (A-B)(A+B)
Hence proved.
Read more on Brainly.in - https://brainly.in/question/7535007#readmore
P Q²= PS² + SQ²→→[By Pythagoras theorem]
a²= PS² + c²
→PS² = a² - c²-------(1)
In Right Δ P SR
PR²= PS² + SR²→→[By Pythagoras theorem]
b²= PS² + d²
PS²= b² - d² ------(2)
From (1) and (2)
b²- d²= a² - c²
a² - b²= c² - d²
(a -b)(a+b)= (c-d)(c+d)→→Using the identity, A²-B²= (A-B)(A+B)
Hence proved.
Read more on Brainly.in - https://brainly.in/question/7535007#readmore
Find bhai
Thnx
Answered by
0
Answer:T
The proof is explained step-wise below :
Step-by-step explanation :
PS is perpendicular to QR
So, using Pythagoras theorem in ΔPSR
PR² = PS² + RS²
⇒ PS² = PR² - RS²
⇒ PS² = b² - d² .........(1)
Now, using Pythagoras theorem in ΔPSQ
PQ² = PS² + QS²
⇒ PS² = PQ² - QS²
⇒ PS² = a² - c² .........(2)
From equation (1) and equation (2)
⇒ b² - d² = a² - c²
⇒ a² - b² = c² - d²
⇒ (a + b)(a - b) = (c + d)(c - d)
Hence Proved.
Similar questions