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 = semi-circle.

Solution:

Given, two chords AB and CD intersect at right angles in a circle AYDZBWCX

We have to prove arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle.

Draw a diameter EF parallel to the CD having centre M.

Since, CD||EF

arc EC = arc DF --------------------- (1)

arc ECXA = arc EWB

arc AF = arc BF ---------------------- (2)

We know that arc ECXAYDF = semicircle

Now, arc EA + arc AF = semicircle

From (2),

arc EC + arc CXA + arc BF = semicircle

From (1),

arc DF + arc CXA + arc BF = semicircle

arc DF + arc BF + arc CXA = semicircle

From the figure,

arc DF + arc BF = arc DZB

So, arc DZB + arc CXA = semicircle

We know that the circle divides itself into two semicircles, the remaining portion of the circle is also equal to the semicircle.

Therefore, arc AYD + arc BWC = semicircle.

✦ Try This: In a circle of radius 6 cm, a chord of length 10 cm makes an angle of 110° at the centre of the circle. Find the length of the arc AB.

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10

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NCERT Exemplar Class 9 Maths Exercise 10.4 Problem 6

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 = semi-circle.

Summary:

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

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