Question

In a quadrilateral PQRS if the sum is f exterior angle of angle Q and angle R is 270 then the value of Pr^2 -QR^2 is always equal to

Answer 1

In a quadrilateral PQRS if the sum of exterior angle of angle Q and angle R is 270° then the value of PR² -QR² is always equal to ​PS² - QS²

Step-by-step explanation:

Lets extend QP & RS such that it meets at point O

As sum of exterior angle of angle Q and angle R is 270°

so angle at O would be 90°   ( as Sum of Exterior angles = 270 ° = in ΔOQR ∠O + ∠Q + ∠O + ∠R  =  180° + ∠O   => ∠O = 90°)

Now

in Δ OPR

PR² = OP² + OR²

in ΔOQR

QR² = OQ² + OR²

PR² - QR² =  OP² + OR² - (OQ² + OR²)

=> PR² - QR² =   OP²  - OQ²

Similarly

in Δ OPS & Δ OQS

PS²= OP² + OS²

QS² = OQ² + OS²

=> PS² - QS² = OP² - OQ²

Equating with PR² - QR² =   OP²  - OQ²

=> PR² - QR² = PS² - QS²

In a quadrilateral PQRS if the sum of exterior angle of angle Q and angle R is 270° then the value of PR² -QR² is always equal to ​PS² - QS²

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