1. In the given figure, 0 is the centre of the circle and ∠DCE = 45°. If CD = $$10\sqrt2$$ cm, then what is the length (in cm) of AC . CB = BD






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  • By: anil on 05 May 2019 02.55 pm
    Given : CD = $$10sqrt2$$ cm and BC = BD = $5sqrt2$$ => ∠OBC = 90° Also, ∠DCE = 45°, => ∠OCB = 45°, => OB = BC ---------(i) Now, in $$ riangle$$ OBC, => $$(OC)^2=(OB)^2+(BC)^2$$ => $$(OC)^2=(5sqrt2)^2+(5sqrt2)^2$$
    => $$(OC)^2=50+50=100$$ => $$OC=sqrt{100}=10$$ cm Similarly, in $$ riangle$$ ABC, => $$(AC)^2=(AB)^2+(BC)^2$$ => $$(AC)^2=(10+5sqrt2)^2+(5sqrt2)^2$$ => $$(AC)^2=100+50+100sqrt2+50$$ => $$(AC)^2=100(2+sqrt2)=341.42$$ => $$AC=sqrt{341.42}=18.47approx18.5$$ cm => Ans - (C)
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