Question

plz solve it .........

Answer 1

from the figure PQ ║AB ║CD


ABCD is a rectangle , it means ABPQ and

PQDC are also rectangle

For , ABPQ ,

AP = BQ [opposite sides are equal ]

For, PQDC

PD = QC [ opposite sides are equal ]

now, for ∆OPD,

OD² = OP² + PD² ------(1)

For, ∆OQB,

OB² = OQ² + BQ² --------(2)

add equations (1) and (2),

OB² + OD² = (OP² + PD²)+ (OQ² + BQ²)
= (OP² + CQ²) + (OQ² + AP²)

as you can see figure,

∆OPA and ∆OQC are also right angled triangles

For, ∆OCQ ⇒OQ² + CQ² = OC²

For,∆OPA ⇒OP² + AP² = OA² , put it above

Now ,
OB² + OD² = OC² + OA

hence proved that

I hope it is helped you

Answer 2

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