NCERT Exemplar Solutions Class 6 Mathematics Solutions for Exercise in Chapter 4 - Fractions and Decimals

Question 61 Exercise

42.28 - 3.19 = 39.09



Subtracting 3.19 from 42.28,

42.28- 3.19 =39.09

Video transcript
"hello students welcome to lyra q a video session i am sef your math tutor and question for today is if two equal parts of a circle intersects within the circle true that the segments of one of the chord are equal to corresponding segments of other chord so let a b and c d be the two equal chords such that a b is equal to cd so these are the two equal chords which is mentioned in the question now in above question it is given that this a b and c d intersects at a point so let us see in the figure a b and c e those chords intersects at point e over here so and you can see further that it is now bit to be proven that the line segment e equal to d e and c e equal to b so this is given and this is to be proven to proceed with the proof first of all we will construct something in the figure so if i show you over here from the center of the circle we have drawn a perpendicular a b perpendicular to a b and that perpendicular is om so om over here is a perpendicular to our chord a b so this is where it forms 90 degree from the center and in the step two similarly you have to draw draw o n which is perpendicular to cd in the step three you have to join o e so this is the oe so this is the extra construction which we have done now the diagram is there which is shown now from the diagram it is seen that om bisects a b so om is perpendicular to a b as we know that a perpendicular drop from the center to the chord bisects that chord so here we can say that from figure o m perpendicular to a b o m by 6 a b similarly o n also bisects cd now as in this case it is known that a b is equal to cd so from this we can say that am is equal to nd let this be our result number one now another result we can obtain here is mb equal to cn and let this be our result number two now consider triangle ome and o and e so in this triangle o m e and o and e you can see that there can be rhs congruency as per the rhs right hand side congruency angle ome and angle o n e these two angles are equal because they are perpendicular and o e is equal to o e that is the middle common one and it is a common sides that is why it is a equal one now om and on if you see they are also equal because a b and c d are equal and so they are equidistant from the center so c o m and o n this is om and this is o n and because a b and c d are equal chords and so they are equidistant from the center and that is the reason why we are stating that om is equal to o n so let us write it down something over here as per rhs congruency triangle omega and triangle o and e they are similar hence we can say angle ome will be equal to angle o n e because perpendiculars now you can see oe and oe are the common side so oe equal to oe for both the triangle and moreover om will be equal to o n this is because a b and c d are equal chords so they are equidistant from the center now this is this has become our hs congruency and we can say that triangle ome is similar to triangle one i'll write down here they are congruent these two triangles are congruent hence we can say that m e is equal to e n this is because they are congruent parts of the congruent triangle so they are congruent parts and they are corresponding parts of the congruent triangle so cp ct can be the short form for corresponding parts of corresponding congruent triangle now uh let this name be as result number three now from equations one and two we get am plus m e is equal to nd plus en from result 1 and 2 am plus m e is equal to nd plus en and hence we can say a is equal to ed so now from equations 2 and 3 we get mb minus m e is equal to c n minus n so from this we can say that eb is equal to ce hence we have proved if you have any query regarding this you can drop it in our comment section and subscribe to lido for more such q a thank you for watching"
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