Class 9 Maths: Perpendiculars from the Centre and Distances of Chords MCQs

25 practice questions on Perpendiculars from the Centre and Distances of Chords from the chapter I'm Up and Down, and Round and Round (NCERT Class 9 Mathematics). Every question includes a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

6 Easy 13 Medium 6 Hard
CBSE Class 9 Maths mock tests and papers

Sample questions

Q1
medium
The point $M$ is the midpoint of the chord $AB$ of a circle with centre $O$, with $OM = 8\ \text{cm}$ and $MA = 6\ \text{cm}$. What is the radius of this circle?
  1. A $12\ \text{cm}$
  2. B $14\ \text{cm}$
  3. C $2\ \text{cm}$
  4. D $10\ \text{cm}$
Show answer and explanation

Correct answer: D - $10\ \text{cm}$

Since $M$ is the midpoint of the chord, $OM$ is perpendicular to $AB$, so triangle $OMA$ has a right angle at $M$ and the radius $OA$ is its hypotenuse. Then $OA = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\ \text{cm}$. Adding the two legs gives $14\ \text{cm}$ and subtracting them gives $2\ \text{cm}$, neither of which is how a hypotenuse is found, and $12\ \text{cm}$ is just double $MA$, which is the whole chord.
Q2
hard
A circle with centre $O$ has radius $13\ \text{cm}$. The chords $AB$ and $CD$ are equal in length, and $AB = 10\ \text{cm}$. How far is $CD$ from the centre?
  1. A $12\ \text{cm}$
  2. B $13\ \text{cm}$
  3. C $5\ \text{cm}$
  4. D $6.5\ \text{cm}$
Show answer and explanation

Correct answer: A - $12\ \text{cm}$

Equal chords are equidistant from the centre, so the distance for $CD$ is the same as for $AB$. Half of $AB$ is $5\ \text{cm}$, and the perpendicular from the centre makes a right triangle with the radius, giving the distance $\sqrt{13^2 - 5^2} = \sqrt{144} = 12\ \text{cm}$. The value $5\ \text{cm}$ is half the chord, $6.5\ \text{cm}$ halves the radius, and $13\ \text{cm}$ is the radius itself.
Q3
medium
In a circle with centre $O$ and radius $8\ \text{cm}$, a chord lies $6\ \text{cm}$ from the centre. What is the exact length of this chord?
  1. A $2\sqrt{7}\ \text{cm}$
  2. B $4\sqrt{7}\ \text{cm}$
  3. C $10\ \text{cm}$
  4. D $4\ \text{cm}$
Show answer and explanation

Correct answer: B - $4\sqrt{7}\ \text{cm}$

Half the chord, the distance $6\ \text{cm}$ and the radius $8\ \text{cm}$ form a right triangle with the radius as hypotenuse, so half the chord is $\sqrt{8^2 - 6^2} = \sqrt{28} = 2\sqrt{7}\ \text{cm}$ and the whole chord is $4\sqrt{7}\ \text{cm}$. Stopping at $2\sqrt{7}\ \text{cm}$ reports only half of it. The value $10\ \text{cm}$ comes from treating the radius as a leg and working out $\sqrt{8^2 + 6^2}$, and $4\ \text{cm}$ doubles the difference $8 - 6$.
Q4
hard
In a circle of radius $13\ \text{cm}$ with centre $O$, the chord $PQ$ is $5\ \text{cm}$ from $O$ and the chord $RS$ is $12\ \text{cm}$ from $O$. How much longer is $PQ$ than $RS$?
  1. A $34\ \text{cm}$
  2. B $7\ \text{cm}$
  3. C $14\ \text{cm}$
  4. D $17\ \text{cm}$
Show answer and explanation

Correct answer: C - $14\ \text{cm}$

Half of $PQ$ is $\sqrt{13^2 - 5^2} = 12\ \text{cm}$, so $PQ = 24\ \text{cm}$, and half of $RS$ is $\sqrt{13^2 - 12^2} = 5\ \text{cm}$, so $RS = 10\ \text{cm}$. The difference is $24 - 10 = 14\ \text{cm}$, and the chord nearer the centre is indeed the longer one. The value $7\ \text{cm}$ compares only the half chords, $34\ \text{cm}$ adds the chords instead of subtracting, and $17\ \text{cm}$ adds the two distances.
Q5
hard
A chord $AB$ of length $16\ \text{cm}$ lies at a distance of $6\ \text{cm}$ from the centre $O$ of a circle. The perpendicular from $O$ meets $AB$ at $N$ and is then extended to meet the circle at the point $C$. How long is $NC$?
  1. A $10\ \text{cm}$
  2. B $16\ \text{cm}$
  3. C $4\ \text{cm}$
  4. D $6\ \text{cm}$
Show answer and explanation

Correct answer: C - $4\ \text{cm}$

The perpendicular from the centre bisects the chord, so $AN = 8\ \text{cm}$, and the right triangle $ONA$ gives the radius $OA = \sqrt{6^2 + 8^2} = 10\ \text{cm}$. The segment $OC$ is also a radius, $10\ \text{cm}$ long, and $ON = 6\ \text{cm}$, so $NC = 10 - 6 = 4\ \text{cm}$. The value $10\ \text{cm}$ is the radius, $6\ \text{cm}$ repeats the distance from the centre, and $16\ \text{cm}$ is the chord.
Q6
medium
The perpendicular drawn from the centre $O$ of a circle meets the chord $AB$ at the point $N$. The two pieces measure $AN = 2x + 1$ and $NB = 3x - 4$, in centimetres. How long is $AB$?
  1. A $11\ \text{cm}$
  2. B $22\ \text{cm}$
  3. C $44\ \text{cm}$
  4. D $5\ \text{cm}$
Show answer and explanation

Correct answer: B - $22\ \text{cm}$

The perpendicular from the centre bisects the chord, so $AN = NB$, giving $2x + 1 = 3x - 4$ and therefore $x = 5$. Each piece is then $11\ \text{cm}$, so $AB = 11 + 11 = 22\ \text{cm}$. Stopping at $11\ \text{cm}$ reports only half the chord, $5\ \text{cm}$ is the value of $x$ rather than a length, and $44\ \text{cm}$ doubles the chord one extra time.
Q7
medium
Two equal chords $AB$ and $CD$ are drawn in a circle with centre $O$. The distance of $AB$ from $O$ is $3x - 4$ and the distance of $CD$ from $O$ is $x + 6$, both measured in centimetres. What is the value of $x$?
  1. A $10$
  2. B $1$
  3. C $2.5$
  4. D $5$
Show answer and explanation

Correct answer: D - $5$

Equal chords of a circle are equidistant from the centre, so $3x - 4 = x + 6$, giving $2x = 10$ and $x = 5$. Halving that to $2.5$ divides once too often. Putting $x = 10$ makes the two distances $26$ and $16$ centimetres, which are not equal, and $x = 1$ makes one distance $-1\ \text{cm}$, and no distance is negative.
Q8
hard
Two parallel chords of a circle of radius $17\ \text{cm}$ lie on opposite sides of the centre. One chord is $16\ \text{cm}$ long and the other is $30\ \text{cm}$ long. How far apart are the two chords?
  1. A $34\ \text{cm}$
  2. B $46\ \text{cm}$
  3. C $7\ \text{cm}$
  4. D $23\ \text{cm}$
Show answer and explanation

Correct answer: D - $23\ \text{cm}$

Half of the $16\ \text{cm}$ chord is $8\ \text{cm}$, so that chord is $\sqrt{17^2 - 8^2} = 15\ \text{cm}$ from the centre, and half of the $30\ \text{cm}$ chord is $15\ \text{cm}$, so that one is $\sqrt{17^2 - 15^2} = 8\ \text{cm}$ from the centre. They lie on opposite sides, so the gap is $15 + 8 = 23\ \text{cm}$. A gap of $7\ \text{cm}$ would apply if both chords were on the same side, $34\ \text{cm}$ is the diameter, and $46\ \text{cm}$ doubles the correct answer.

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