Class 10 Mathematics Chapter 9: Chords and Arcs of a Circle Online Practice Test (2026) - Apex Rankers Sample Questions & Solved Answers

Question 1

Construct an equilateral triangle with side $5	ext{ cm}$ long. Show that one and only one circle can be drawn through the vertices of the triangle. What is the radius of the circle drawn?

Accepted Answer

Correct Answer: Radius $R = \frac{5\sqrt{3}}{3} \approx 2.89 ext{ cm}$. Solution:

Step 1: Understanding Construction & Theorem 9.1:

According to Theorem 9.1, one and only one circle can pass through three non-collinear points. The three vertices of $ riangle ABC$ are non-collinear.

Construct equilateral $ riangle ABC$ with $AB = BC = CA = 5 ext{ cm}$.
Draw the perpendicular (right) bisectors of sides $AB$ and $BC$.
Let these bisectors intersect at a unique point $O$ (the circumcenter).
Since $O$ lies on the right bisector of $AB$, $OA = OB$. Since $O$ lies on the right bisector of $BC$, $OB = OC$. Hence, $OA = OB = OC = R$.
With center $O$ and radius $R = OA$, draw a circle. It passes through all three vertices $A, B, C$.
Since two distinct straight lines can intersect at only one point, $O$ is unique, proving that one and only one circle can be drawn.

Step 2: Calculating the Circumradius $R$:

For an equilateral triangle with side length $a = 5 ext{ cm}$:

$$ ext{Altitude } h = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} imes 5 = \frac{5\sqrt{3}}{2} ext{ cm}$$

In an equilateral triangle, the circumcenter divides each median/altitude in the ratio $2:1$:

$$R = \frac{2}{3} h = \frac{2}{3} \left(\frac{5\sqrt{3}}{2} ight) = \frac{5\sqrt{3}}{3} = \frac{5}{\sqrt{3}} \approx 2.89 ext{ cm}$$

Final Answer:

One and only one circle passes through the vertices, with radius $R = \frac{5\sqrt{3}}{3} \approx 2.89 ext{ cm}$.

Question 2

Given that $P$ is the centre of each of the circles. Find the values of unknown if line segment drawn from the centre of each circle is perpendicular to the chord:
(i) Chord $= 10 ext{ cm}$, distance $PE = 12 ext{ cm}$, find radius $r$.
(ii) Radius $= 5 ext{ cm}$, distance $= 3 ext{ cm}$, find chord length.

Accepted Answer

Correct Answer: (i) $r = 13 ext{ cm}$, (ii) $ ext{Chord} = 8 ext{ cm}$. Solution:

Part (i): Finding Radius $r$:

Given: Chord length $AB = 10 ext{ cm}$, perpendicular distance $PE = 12 ext{ cm}$.
By Theorem 9.3, the perpendicular from the center to a chord bisects the chord: $$AE = EB = \frac{AB}{2} = \frac{10}{2} = 5 ext{ cm}$$
In right-angled triangle $ riangle PEA$: $$PA^2 = PE^2 + AE^2$$ $$r^2 = 12^2 + 5^2 = 144 + 25 = 169$$ $$r = \sqrt{169} = 13 ext{ cm}$$

Part (ii): Finding Chord Length:

Given: Radius $r = 5 ext{ cm}$, perpendicular distance $d = 3 ext{ cm}$.
In right triangle formed by radius, distance, and half-chord: $$\left(\frac{ ext{Chord}}{2} ight)^2 = r^2 - d^2 = 5^2 - 3^2 = 25 - 9 = 16$$ $$\frac{ ext{Chord}}{2} = \sqrt{16} = 4 ext{ cm}$$
Total Chord Length $= 2 imes 4 = 8 ext{ cm}$.

Final Answer:

(i) Radius $r = 13 ext{ cm}$, (ii) Chord length $= 8 ext{ cm}$.

Question 3

Find the length of diameter $CD$ of the circle when $AB = 10	ext{ cm}$ is a chord perpendicular to diameter $CD$ at $E$, and $PE = 12	ext{ cm}$ (where $P$ is the centre). Also find $CE$.

Accepted Answer

Correct Answer: Diameter $CD = 26 ext{ cm}$, $CE = 1 ext{ cm}$. Solution:

Step 1: Finding Radius $r = PC$:

Chord $AB = 10 ext{ cm}$. Since diameter $CD \perp AB$, $E$ bisects $AB$: $$AE = \frac{AB}{2} = \frac{10}{2} = 5 ext{ cm}$$
Perpendicular distance $PE = 12 ext{ cm}$.
In right $ riangle PEA$: $$r^2 = PA^2 = PE^2 + AE^2 = 12^2 + 5^2 = 144 + 25 = 169 \implies r = 13 ext{ cm}$$

Step 2: Calculating Diameter $CD$ and Sagitta $CE$:

Diameter $CD = 2r = 2 imes 13 = 26 ext{ cm}$.
Since $P$ is center and $C$ is on circumference along diameter $P-E-C$: $$CE = PC - PE = r - PE = 13 - 12 = 1 ext{ cm}$$

Final Answer:

Diameter $CD = 26 ext{ cm}$ and $CE = 1 ext{ cm}$.

Question 4

In a circle with centre $O$, $OB = 15	ext{ cm}$ (radius) and $OF = 9	ext{ cm}$. Find the length of chord $CD$ given that $OF \perp CD$.

Accepted Answer

Correct Answer: Chord $CD = 24 ext{ cm}$. Solution:

Step 1: Identify Given Geometric Parameters:

Radius of circle $r = OB = OC = OD = 15 ext{ cm}$.
Perpendicular distance from centre $O$ to chord $CD$ is $OF = 9 ext{ cm}$.

Step 2: Apply Pythagorean Theorem in $ riangle OFC$:

$$ riangle OFC ext{ is a right-angled triangle at } F:$$

$$CF^2 = OC^2 - OF^2$$

$$CF^2 = 15^2 - 9^2 = 225 - 81 = 144$$

$$CF = \sqrt{144} = 12 ext{ cm}$$

Step 3: Chord Bisector Property (Theorem 9.3):

Since $OF \perp CD$, $F$ is the midpoint of $CD$:

$$CD = 2 imes CF = 2 imes 12 = 24 ext{ cm}$$

Final Answer:

The length of chord $CD = 24 ext{ cm}$.

Question 5

Calculate the length of chord $AB$ in a circle of radius $13	ext{ cm}$ if $OC \perp AB$ and $OC = 5	ext{ cm}$ (where $O$ is the centre).

Accepted Answer

Correct Answer: Chord $AB = 24 ext{ cm}$. Solution:

Step 1: Set up Right Triangle $ riangle OCA$:

Radius $OA = 13 ext{ cm}$.
Perpendicular distance $OC = 5 ext{ cm}$.
By Pythagorean theorem in right-angled $ riangle OCA$: $$AC^2 = OA^2 - OC^2 = 13^2 - 5^2 = 169 - 25 = 144$$ $$AC = \sqrt{144} = 12 ext{ cm}$$

Step 2: Calculate Total Chord Length $AB$:

By Theorem 9.3, $OC \perp AB \implies AC = CB$:

$$AB = 2 imes AC = 2 imes 12 = 24 ext{ cm}$$

Final Answer:

Length of chord $AB = 24 ext{ cm}$.

Question 6

Construct a rectangle $EFGH$ such that $EF = 6	ext{ cm}$ and $FG = 4	ext{ cm}$. Draw a circle passing through its vertices and prove that it is the only circle that can be drawn through the vertices.

Accepted Answer

Correct Answer: Radius $r = \sqrt{13} \approx 3.61 ext{ cm}$. Solution:

Step 1: Geometric Analysis of Rectangle Diagonals:

In rectangle $EFGH$, all four interior angles are $90^\circ$.
The diagonals $EG$ and $FH$ are equal in length and bisect each other at a common point $O$: $$EG = \sqrt{EF^2 + FG^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} ext{ cm}$$
Since diagonals bisect each other: $$OE = OF = OG = OH = \frac{EG}{2} = \sqrt{13} \approx 3.61 ext{ cm}$$

Step 2: Proof of Uniqueness (Theorem 9.1):

Any circle passing through vertices $E, F, G$ must have its center at the intersection of the right bisectors of $EF$ and $FG$.
These perpendicular bisectors intersect at the unique center $O$ of the rectangle.
Since two lines intersect at exactly one point, the center $O$ is unique and radius $r = \sqrt{13} ext{ cm}$ is fixed.
Thus, one and only one circle can pass through all four vertices $E, F, G, H$.

Final Answer:

A unique circle passes through all vertices of rectangle $EFGH$ with center at diagonal intersection and radius $r = \sqrt{13} \approx 3.61 ext{ cm}$.

Question 7

Take four non-collinear points $A, B, C$ and $D$ such that $AB = BC = DB = 4	ext{ cm}$. Draw all possible circles that can pass through $A, C$ and $D$.

Accepted Answer

Correct Answer: One unique circle with center $B$ and radius $4 ext{ cm}$. Solution:

Step 1: Conceptual Identification:

Given: Point $B$ is at a distance of $4 ext{ cm}$ from $A$, $C$, and $D$: $$BA = BC = BD = 4 ext{ cm}$$
By definition of a circle, the set of all points equidistant from a fixed center forms a circle.
Here, $B$ is equidistant from the three non-collinear points $A, C, D$.

Step 2: Conclusion via Theorem 9.1:

Through three non-collinear points $A, C, D$, there exists one and only one circle. Its center is uniquely $B$ and its radius is $r = 4 ext{ cm}$.

Final Answer:

Only 1 circle can be drawn through points $A, C$, and $D$. Its center is $B$ and radius is $4 ext{ cm}$.

Question 8

In a circle with centre $O$, chord $AB = 16 ext{ cm}$ and $RS = 10 ext{ cm}$ (perpendicular sagitta/segment from chord to outer circle boundary where $OU = r - 10$ or $16 - OU = r$).
(i) Express $OU$ in terms of radius $r$.
(ii) Form an equation in $r$ and solve it to find radius $r$.

Accepted Answer

Correct Answer: (i) $OU = 16 - r$ (or $r - 10$), (ii) $r = 10 ext{ cm}$. Solution:

Step 1: Express Distance in terms of Radius $r$:

Chord $AB = 16 ext{ cm} \implies$ half-chord $AU = 8 ext{ cm}$.
Given geometric relation from diagram: Perpendicular distance from center to chord $OU = 16 - r$.

Step 2: Set up and Solve Pythagorean Equation:

In right triangle $ riangle OUA$:

$$OA^2 = OU^2 + AU^2$$

$$r^2 = (16 - r)^2 + 8^2$$

$$r^2 = 256 - 32r + r^2 + 64$$

$$r^2 - r^2 + 32r = 320$$

$$32r = 320 \implies r = \frac{320}{32} = 10 ext{ cm}$$

Final Answer:

(i) $OU = 16 - r$, (ii) Equation: $r^2 = (16 - r)^2 + 64 \implies r = 10 ext{ cm}$.

Question 9

In a circle, chord $AB = 16	ext{ cm}$ and the sagitta (height of arc from chord midpoint $E$ to arc midpoint $D$) $DE = 4	ext{ cm}$. Find the radius of the circle.

Accepted Answer

Correct Answer: Radius $r = 10 ext{ cm}$. Solution:

Step 1: Set up Geometric Relations:

Let $O$ be the centre and $r$ be the radius of the circle.
Chord $AB = 16 ext{ cm} \implies AE = \frac{16}{2} = 8 ext{ cm}$.
Point $D$ lies on the circle, so $OD = r$.
Since $E$ is on radius $OD$ and $DE = 4 ext{ cm}$, the distance from centre to chord is: $$OE = OD - DE = r - 4$$

Step 2: Apply Pythagorean Theorem in $ riangle OEA$:

$$OA^2 = OE^2 + AE^2$$

$$r^2 = (r - 4)^2 + 8^2$$

$$r^2 = r^2 - 8r + 16 + 64$$

$$8r = 80 \implies r = 10 ext{ cm}$$

Final Answer:

The radius of the circle is $r = 10 ext{ cm}$.

Question 10

Given that diameter and chord of a circle are $10	ext{ cm}$ and $8	ext{ cm}$ long respectively. The diameter bisects the chord and the distance between chord and centre of the circle is $3	ext{ cm}$. Show that the diameter bisects the chord perpendicularly.

Accepted Answer

Correct Answer: Verified: By converse of Pythagorean theorem, $\angle = 90^\circ$. Solution:

Step 1: Verify Side Lengths in Triangle Formed by Center, Midpoint, and Endpoint:

Diameter $= 10 ext{ cm} \implies$ Radius $r = OA = 5 ext{ cm}$.
Chord $AB = 8 ext{ cm}$. Since diameter bisects the chord at $M$, $AM = 4 ext{ cm}$.
Distance from centre $O$ to chord midpoint $M$ is $OM = 3 ext{ cm}$.

Step 2: Check Pythagorean Identity in $ riangle OMA$:

$$OM^2 + AM^2 = 3^2 + 4^2 = 9 + 16 = 25$$

$$OA^2 = 5^2 = 25$$

Since $OM^2 + AM^2 = OA^2$, by the Converse of Pythagoras' Theorem, $ riangle OMA$ is a right-angled triangle with $\angle OMA = 90^\circ$.

Therefore, the diameter is perpendicular to the chord at its midpoint.

Final Answer:

Since $3^2 + 4^2 = 5^2$, $\angle OMA = 90^\circ$, confirming the diameter bisects the chord perpendicularly (Theorem 9.2).

Class 10 Mathematics - Ch 9: Chords and Arcs of a Circle

Class 10 Mathematics - Ch 9: Chords and Arcs of a Circle

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Class 10 Mathematics - Ch 9: Chords and Arcs of a Circle

Class 10 Mathematics - Ch 9: Chords and Arcs of a Circle

🎯 Question Types & Curriculum Breakdown

💡 Strategic Preparation & Exam Hall Guidelines

📝 Pre-Rendered Solved Sample Questions & Detailed Solutions

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