Class 9 Mathematics · Ganita Manjari Part 1

Class 9 Maths Chapter 5 – End-of-Chapter Exercise Solutions

This page gives original, step-by-step explanations for all 26 end-of-chapter problems. The question prompts below are shortened/paraphrased so the page focuses on the mathematical solution.

Q1. Chord 5 cm from centre, radius 13 cm

Half the chord is $x$. Then $13^2=5^2+x^2$, so $x=12$ cm.

Q2. Central angle 70°

The angle at the centre is twice the angle at a point on the circle.

Q3. Diameter 26 cm and chord 24 cm

Radius $=13$ cm and half-chord $=12$ cm.

Q4. Radius 15 cm and distance of chord 9 cm

Q5. Prove that the perpendicular bisector of a chord passes through the centre

Let $M$ be the midpoint of chord $AB$. In $\triangle OMA$ and $\triangle OMB$, $OA=OB$, $AM=MB$ and $OM$ is common. Hence SSS congruence gives equal adjacent angles at $M$. They form a linear pair, so each is $90^\circ$.

Thus $OM\perp AB$, so the perpendicular bisector of $AB$ passes through $O$.

Hence proved.

Q6. Angle subtended by a diameter

A diameter subtends a straight angle $180^\circ$ at the centre. The corresponding angle on the circle is half of this.

Q7. Opposite angles 75° and 110° in a cyclic quadrilateral

Q8. Opposite angles $(2x+10)^\circ$ and $(3x-20)^\circ$

Therefore

Answer: $\boxed{x=38}$, $\boxed{\angle P=86^\circ}$, $\boxed{\angle R=94^\circ}$.

Q9. Chord 16 cm, distance 6 cm

Half-chord $=8$ cm.

Q10. Cyclic quadrilateral with sides 5, 5, 12, 12

Semiperimeter:

Using Brahmagupta’s formula for a cyclic quadrilateral:

Q11. Locate the circumcentre relative to a cyclic quadrilateral

Inspect the angles. If all four interior angles are acute, the circumcentre is inside. If the quadrilateral has an obtuse angle, the circumcentre lies outside. A right-angle case places the circumcentre on the relevant diagonal/side boundary configuration.

Answer: The angle test is the practical way to decide without first drawing the circle.

Q12. Equal intersecting chords

Let equal chords $AB$ and $CD$ intersect at $P$. Perpendiculars from centre $O$ to the chords meet them at $M$ and $N$.

Equal chords are equidistant from the centre, so $OM=ON$. In right triangles $OPM$ and $OPN$, $OP$ is common and both have equal perpendicular sides. Hence RHS congruence gives $PM=PN$.

Since the perpendicular from the centre bisects each chord, the corresponding half-chords are equal. Combining these equal parts gives

Hence proved.

Q13. Construct a circle with a 6 cm chord 3 cm from the centre

1. Draw $AB=6$ cm.
2. Bisect $AB$ at $M$.
3. Draw a perpendicular through $M$.
4. Mark $O$ on it with $OM=3$ cm.
5. Draw the circle with centre $O$ through $A$ (or $B$).

Here $AM=3$ cm, so

Answer: Required radius $=\boxed{3\sqrt2\text{ cm}}$.

Q14. Show that the only parallelogram inscribed in a circle is a rectangle

In a cyclic quadrilateral, opposite angles sum to $180^\circ$. In a parallelogram, opposite angles are equal.

All angles are therefore right angles.

Answer: The parallelogram must be a rectangle.

Q15. Diagonals of an inscribed rectangle meet at the centre

Let the diagonals meet at $O$. A rectangle has equal diagonals which bisect one another. Hence the four half-diagonals from $O$ to the vertices are equal:

Thus $O$ is equidistant from all four points on the circle.

Answer: $O$ is the centre of the circle.

Q16. Locus of midpoints of all fixed-length chords

Let every chord have fixed length $x$ in a circle of radius $r$. Its midpoint is at distance $d$ from the centre, where

Both $r$ and $x$ are fixed, so $d$ is fixed.

Answer: The midpoints form a circle concentric with the given circle.

Q17. Equal chords $AB$ and $AC$

Join $OA,OB,OC$. In $\triangle AOB$ and $\triangle AOC$, $OB=OC$ (radii), $OA$ is common and $AB=AC$ is given. Hence SSS congruence gives

Therefore $AO$ bisects $\angle BAC$.

Hence proved.

Q18. Parallel chords 10 cm and 24 cm, same side, 7 cm apart

Let their distances from the centre be $d_1$ and $d_2$. The longer chord is nearer the centre, so

Half-chords are 5 cm and 12 cm. Thus

Hence

So $d_1+d_2=17$. Solving gives $d_1=12$, $d_2=5$.

Q19. Regular hexagon in a circle of radius $r$

Each central angle is $60^\circ$. Therefore the triangle formed by two radii and one side is equilateral.

For the distance $d$ of a side from the centre:

Q20. A cyclic quadrilateral with one diameter

The angles $\angle MOP$ and $\angle MNP$ are opposite angles of the cyclic quadrilateral, so

Thus they are supplementary.

Q21. Exterior angle of a cyclic quadrilateral

For cyclic $ABCD$:

The exterior angle $\angle CDE$ forms a linear pair with $\angle ADC$:

Comparing the two equations gives

Hence proved.

Q22. Why no chord is longer than the diameter

For a chord at distance $d>0$ from the centre:

But $2r$ is the diameter.

Answer: Every non-diameter chord is shorter than the diameter, so the diameter is the longest chord.

Q23. Shortest chord through an interior point

For a fixed interior point $A$, the chord is shortest when its distance from the centre $O$ is greatest. Among all lines through $A$, this distance is greatest when the line is perpendicular to $OA$.

Answer: The shortest chord through $A$ is the chord perpendicular to $OA$.

Q24. Justify that an angle in a semicircle is 90°

Let $BC$ be a diameter and $A$ a point on the semicircle. Join $OA$. Since $OA=OB=OC$, triangles $AOB$ and $AOC$ are isosceles. If their equal base angles are represented by $a$ and $b$, then

Using the angle sum of $\triangle ABC$:

Therefore

Q25. Midpoints of two chords perpendicular to a diameter

Let $AB$ be the diameter. The two chords $CC'$ and $DD'$ are both perpendicular to $AB$, so they are parallel to each other.

Let $M$ and $M'$ be the midpoints of $CD$ and $C'D'$, respectively. Then $CDD'C'$ is a trapezium whose parallel sides are $CC'$ and $DD'$.

Since $M$ and $M'$ are the midpoints of the non-parallel sides $CD$ and $C'D'$ of this trapezium, the trapezium mid-segment theorem gives

But $CC'\perp AB$. Therefore a line parallel to $CC'$ is also perpendicular to $AB$.

Hence proved.

Q26. Sum of opposite angles of a cyclic quadrilateral

Join the centre to all four vertices. The radii make isosceles triangles. Using the central-angle result, the two opposite angles together subtend the complete $360^\circ$ around the centre, so their sum is half of $360^\circ$.

Similarly,

Hence proved.