please solve Q.no. 40 in the given photo...
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Let chord DC be x
In ΔPAC and ΔPDB,
∠P = ∠P(Common) (i)
∠BAC = ∠BDP(Ext. angle = Int. opp. angle) (ii)
From (i) and (ii),
∴ ΔPAC ~ ΔPDB by AA
∴ PA/PD = PC/PB
PA×PB = PC×PD
(3+5)×3 = (4+x)×4
8×3 = 16+4x
24 = 16+4x
8 = 4x
2 = x
∴ Chord DC = 2 cm
In ΔPAC and ΔPDB,
∠P = ∠P(Common) (i)
∠BAC = ∠BDP(Ext. angle = Int. opp. angle) (ii)
From (i) and (ii),
∴ ΔPAC ~ ΔPDB by AA
∴ PA/PD = PC/PB
PA×PB = PC×PD
(3+5)×3 = (4+x)×4
8×3 = 16+4x
24 = 16+4x
8 = 4x
2 = x
∴ Chord DC = 2 cm
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