perpendiculars-from-the-centre-and-distances-of-chords MCQs for UPSC Prelims

25 practice questions on perpendiculars-from-the-centre-and-distances-of-chords from the I'm Up and Down, and Round and Round section of the UPSC Prelims syllabus. 25 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

6 Easy 13 Medium 6 Hard

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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