Jump to

- Rational and Irrational Numbers
- Compound Interest
- Expansions
- Factorization
- Simultaneous Linear Equations
- Problems on Simultaneous Linear Equations
- Quadratic Equations
- Indices
- Logarithms
- Triangles
- Mid Point Theorem
- Pythagoras Theorem
- Rectilinear Figures
- Theorems on Area
- Circle
- Mensuration
- Trigonometric Ratios
- Trigonometric Ratios and Standard Angles
- Coordinate Geometry
- Statistics

A chord of length 16 cm is at a distance of 6 cm from the centre of the circle. Find

the length of the chord of the same circle which is at a distance of 8 cm from the

centre.

Answer:

AB is a chord a circle with center O and

OA is the radius of the circle and OM ⊥ AB

AB = 16 cm, OM = 6 cm

OM ⊥ AB

AM = ½ AB = ½ × 16 = 8 cm

Now in right ∆OAM

\begin{aligned} &\mathrm{OA}^{2}=\mathrm{AM}^{2}+\mathrm{OM}^{2}\\ &\text { (By Pythagoras Axiom) }\\ &=(8)^{2}+(6)^{2}\\ &64+36=100=(10)^{2} \end{aligned}

Now CD is another chord of the same circle

ON ⊥ CD and OC is the radius.

In right ∆ONC

\begin{aligned} &\mathrm{OC}^{2}=\mathrm{ON}^{2}+\mathrm{NC}^{2}\\ &\text { (By Pythagoras Axioms) }\\ &(10)^{2}=(8)^{2}+(N C)^{2}\\ &100=64+\mathrm{NC}^{2}\\ &\mathrm{NC}^{2}=100-64=36=(6)^{2} \end{aligned}

NC = 6

But ON ⊥ AB

N is the mid-point of CD

CD = 2 NC = 2 × 6 = 12 cm

"hello and welcome idea students
i'm surprised you don't know leader and
i'm here
the new question it is a chord of length
16 centimeter is at a distance of 6
centimeter from the center of this
circle
find the length of the chord of the same
circle which is at a distance of 8
centimeter from the
center so without wasting much time
let's have a look at the solution so we
know that
a b is required of the circle
at center o and o is the
radius of the circle
and also
om is perpendicular to a b
we know that a b is 16 centimeter
and om is six centimeter
since we know that om is perpendicular
to a b
so this implies that
am is half of a b
which is nothing but half of 16.
nothing but equals to 8 centimeter
so therefore in triangle om four a
square is equals to
om square plus
am square
which further implies that
osb is equals to om square which is 6
which is nothing but equal to 36 whereas
am square am is 8 am square is 64
so which means that
2a square is 100 and oa
is nothing but equals to
10 centimeter
and since we know that
o and oc are equal because both these
are the radius of the same circle so oc
is also
10 centimeter
now cd is another chord
of the same circle where o n
is perpendicular to c d
and o c is the radius
so in right triangle
i am writing rt for right o
nc we get
cozy square which is equals to
on square plus
cn square which further implies that
oc is 10 so this is 10 square
o n is 8 so 8 square plus
c n square which further implies that c
n is
equals to root over 100 minus 64
which is nothing but equal to root over
which is nothing but equals to six
centimeter
and since we know that
so you know but o n is perpendicular to
c d which implies
that n is the midpoint of c d
therefore cd is equals to
twice of cn
which further implies that cd is equals
to twice of
six which is equal to 12
centimeter so i hope my students got
this question well
do subscribe our channel for the regular
updates you can also share your doubt in
the comment section
see you next time bye"

Related Questions

Chapters

Lido

Courses

Quick Links

Terms & Policies

Terms & Policies

2022 © Quality Tutorials Pvt Ltd All rights reserved