Question

In the given figure, O is the center of the circle and DCE = 45o. If CD = 10 cm, then What is the length (in cm) of AC? (CB = BD)

Answer 1

where is the diagram

Answer 2

Solution:-

Now we have < DCE = 45°

< OCB = 45°

hence we can say ∆ OBC is a right angle isosceles triangle means BC = OB = 10√2 /2 cm = 5√2

in right angle isosceles triangle OBC we can have

OC^2 = BO^2 + BC^2 ( By phythagoras theorem)

=> OC^2 =(5 √2 ) ^2 + (5√2)^2

=> OC^2 = 100 => OC= 10cm

OC= AB = 10cm ( OC= AB radius of same circle )

Now and right angle ∆ ABC we have

AC^2 = BC^2 + AB^2

AC^2 = (5√2)^2 + (OB+ AO)^2

AC^2 = 50+50 +2× 5 √2 ×10 +100

AC= √314.42 = 18.5 cm

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@GauravSaxena01