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  • If two chords AB and CD of a circle AYDZBWCX intersect at right angles (see Fig.10.18), prove that arc CXA + arc DZB = arc AYD + arc BWC = semicircle.

If two chords AB and CD of a circle AYDZBWCX intersect at right angles (see Fig.10.18), prove that arc CXA + arc DZB = arc AYD + arc BWC = semicircle.

Answers (1)

Given: In circle AYDZBWCX two chords AB and CD intersect at right angles

To Prove: Arc CXA + Arc DZB = arc AYD + arc BWC = Semicircle

Proof:

Draw Diameter EF || CD having centre M

Since CD || EF

Arc EC = DF … (i)

Arc ECXA = arc EWB  (Symmetrical about diameter of circle)

Arc AF = arc FB … (ii)   (Symmetrical about diameter of circle)

We know that arc ECXAYDF = Semicircle

Arc EA + arc AF = Semicircle

arc EC + arc CXA + arc FB = semicircle  (from ii, Arc AF = arc FB)

arc DF + arc CXA + arc FB = semicircle   (from i, Arc EC = arc DF)

=> arc DF + arc FB + arc CXA = semicircle

Arc DZB + arc CXA = semicircle.  (First statement is proved)

Now, we know that a circle is divided into two semi-circles.

Therefore the remaining portion of the circle (other than, Arc DZB + arc CXA) is also equal to a semi-circle

So, Arc AYD + arc BWC = semicircle

So we get: Arc CXA + Arc DZB = arc AYD + arc BWC = Semicircle

Hence proved

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