Question

5) In a rectangle PQRS, the diagonal PR and OS intersects at right angle.
Prove that PQ2 + RS2 = PS2+ QR2​

Answer 1

Answer:

PQ2+RS2=PS2+QR2

IT IS THEOREM OF DIAGONAL

Answer 2

Step-by-step explanation:

Given: The diagonals of a convex □ABCD intersect at right angles at O.

To Prove: PQ

2

+RS

2

=PS

2

+QR

2

Proof: In ΔPOS,m∠O=90

o

∴PS

2

=OP

2

+OS

2

......... (i)

In ΔQOR,m∠O=90

o

∴QR

2

=OQ

2

+OR

2

......... (ii)

In ΔPOQ,m∠O=90

o

∴PQ

2

=OP

2

+OQ

2

......... (iii)

In ΔROQ,m∠O=90

o

∴QR

2

=OR

2

+OS

2

........ (iv)

Adding results (3) and (4)

PQ

2

+QR

2

=OP

2

+OQ

2

+OR

2

+OS

2

∴PQ

2

+QR

2

=(OP

2

+OS

2

)+(OQ

2

+OR

2

)

∴PQ

2

+RS

2

=PS

2

+QR

2

......... [from (i) and (ii)]

solution