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
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Step-by-step explanation:
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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°
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
=> 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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