chapter-header

Selina Solutions Class 9 Mathematics Solutions for Exercise 17(A) in Chapter 17 - Chapter 17- Circles

Question 3 Exercise 17(A)

A chord of length 24 cm is at a distance of 5 cm from the center of the circle. Find the length of the chord of the same circle which is at a distance of 12 cm from the center.

Answer:

Let AB be the chord of length 24 cm and O be the center of the circle.

Let OC be the perpendicular drawn from O to AB.

We know, that the perpendicular to a chord, from the center of a circle, bisects the chord.

AC = CB = 12 cm

Selina Solutions CONCISE Maths - Class 9 ICSE chapter Chapter 17- Circles Question 3 Solution image

Video transcript
"hello students welcome to leader q a video session i am seth your math tutor and the question for today is a chord of length 24 centimeter is at the distance of 5 centimeter from the center of the circle find the length of the pawn of the same circle which is the distance of 12 centimeter from the center now here from the question we know that you can see the figure a b is the chord which is at the distance of 5 centimeter from the center row so let a b be the cord of 24 centimeter which is at distance of 5 centimeter from the o so you have a b 24 centimeter and you also have om that is 5 centimeter now om here is the perpendicular dropped on the chord a b you well know you are well known about that that if you drop the perpendicular from center to the chord then this will bisect the chord so here perpendicular om will bisect the cord and hence you have a m equal to mb and hence your am will be equal to a b upon 2 it is 24 upon 2 and that is 12 centimeter that is your am now consider right triangle amo in the right triangle amo you can apply the pythagoras property in this triangle that is am square plus om square will be equal to ao square now here first of all we need to find the radius of the circle in order to find the length of the second chord that is cd so here the radius of the circle will be ao and that is according to the pythagoras property ao square is equal to am square plus om square hence your am is 12 centimeter that is 12 square plus 5 square this is 169 that is ao square and hence your ao is 13 centimeter hence once you got the radius now as a 13 centimeter you can proceed further with the calculation to find the length of the remaining chord so here consider the chord cd now here there is a same scenario that perpendicular o n which is given that is the distance from the center to the another chord that is 12 centimeter so on is equal to 12 centimeter now this divides the chord cd into two equal halves that is cn and nd therefore c n is equal to n d and hence we can find c n by the help of pythagoras property applying it to triangle ocn in right triangle ocn so first of all if you see the measurement of cn then we need to find the cn because here on is given that is 12 centimeter and you have the radius now that is 13 centimeter so apply the pythagoras property in right triangle ocn you have oc square plus o n o c square is equal to o n square plus c n square and hence your o n square here is 12 square the c n square is what you you need to find so 12 square minus oc square will be equal to minus c n square so there is a here oc that is you have found that 13 square minus 12 square that will be equal to cn square and that is exactly equal to 25 hence your cn comes out to be phi now once you have found the cn you already know that cn will be equal to nd so cd will be equal to cn plus nd and which is 2 times cn so here hence you can say that your called cd will be of measure 2 times cn and you already have the cn which is phi that is 5 centimeter so cd will be equal to 2 into 5 and that is 10 centimeter hence our length of the required chord that is the length of the new cord will be equal to 10 centimeter and this is our final answer so if you have any query regarding this you can drop it down in our comment section and subscribe to lido for more such interesting q a thank you for watching"
Connect with us on social media!
2022 © Quality Tutorials Pvt Ltd All rights reserved