WBBSE Class 10 Maths Chapter 3 Exercise 3.1 Solutions

Section 1: Core Concepts & Figures

Question 1: Look at the adjoining figure of the circle with center O and write the names of the radii that are situated in the segment PAQ.

Answer: In the circle with center O, the line segments joining the center to the boundary points are the radii. In the specific segment region PAQ, the corresponding radii are OP, OA, OC, and OQ.

Question 2: Fill in the blanks with appropriate geometric terms:

1. In a circle, there are an infinite number of points.
2. The greatest chord of a circle is its diameter.
3. A chord divides a circular region into two segments.
4. All diameters of a circle pass through the centre.
5. If two chords are equal in length, then their corresponding arcs are equal in length.
6. The sector of a circular region is the region enclosed by an arc and its two bounding radii.
7. The length of a line segment joining a point outside a circle to its center is greater than the length of the radius.

Question 3: Draw a circle with the help of a compass and indicate its centre, chord, diameter, radius, major arc, and minor arc on it.

Geometric Labels:

• Centre: $O$
• Chord: $CD$ (not passing through the center)
• Diameter: $AB$ (passing through center $O$)
• Radius: $OA$, $OB$, or $OE$
• Minor Arc: $\widehat{XY}$ (shorter path along the boundary)
• Major Arc: $\widehat{XPY}$ (longer path along the boundary)

Figure 1: Standard parts of a Circle

Section 2: True or False Statements

1. A circle is a plane figure.
   Answer: True
2. A segment of a circle is a plane region.
   Answer: True
3. A sector of a circle is a plane region.
   Answer: True
4. A chord is a line segment.
   Answer: True
5. An arc is a line segment.
   Answer: False
6. There are a finite number of chords of the same length in a circle.
   Answer: False
7. One and only one circle can be drawn by taking a fixed point as its center.
   Answer: False
8. The lengths of the radii of two congruent circles are equal.
   Answer: True

Section 3: Structural Geometric Proofs

Question 1 (Proof): I draw a chord PQ of a circle with centre A, which is not a diameter. I draw a perpendicular from center A onto the chord PQ meeting at M. Prove with reason that PM = MQ.

Given: $PQ$ is a chord of a circle with centre $A$. $AM \perp PQ$.

To prove: $PM = MQ$.

Figure 2: Perpendicular AM on Chord PQ

Construction: Join lines $AP$ and $AQ$.

Proof: In right-angled triangles $\Delta APM$ and $\Delta AQM$: $$\angle AMP = \angle AMQ = 90^\circ \quad [\because AM \perp PQ]$$ $$\text{Hypotenuse } AP = \text{Hypotenuse } AQ \quad [\text{Radii of the same circle}]$$ $$AM = AM \quad [\text{Common side}]$$ $$\text{Therefore, } \Delta APM \cong \Delta AQM \quad [\text{By RHS rule}]$$ $$\implies PM = MQ \quad [\text{By C.P.C.T.}]$$

Proved: PM = MQ.

Question 2 (Proof): Prove Theorem 33 (The straight line joining the center of a circle to the midpoint of a chord, which is not a diameter, is perpendicular to the chord) with the help of the S-S-S axiom of congruency.

Given: In a circle with centre $O$, $D$ is the midpoint of chord $AB$. Therefore, $AD = DB$.

To prove: $OD \perp AB$.

Construction: Join $OA$ and $OB$.

Proof: In triangles $\Delta OAD$ and $\Delta OBD$: $$OA = OB \quad [\text{Radii of the same circle}]$$ $$AD = DB \quad [\text{Given}]$$ $$OD = OD \quad [\text{Common side}]$$ $$\text{Therefore, } \Delta OAD \cong \Delta OBD \quad [\text{By S-S-S rule}]$$ $$\implies \angle ODA = \angle ODB \quad [\text{By C.P.C.T.}]$$ Since line $OD$ stands on straight line $AB$: $$\angle ODA + \angle ODB = 180^\circ \implies 2 \cdot \angle ODA = 180^\circ \implies \angle ODA = 90^\circ$$ $$\implies OD \perp AB$$

Proved: OD is perpendicular to chord AB.

Section 4: Numerical Computations

Question 3: The perpendicular distance of a chord from the center of a circle is $8\text{ cm}$. If the radius of the circle is $17\text{ cm}$ in length, let us calculate the total length of the chord.

Solution: Let $AB$ be the chord, and $OD$ be the perpendicular distance dropped from center $O$ to $AB$.
Given, Radius ($OB$) = $17\text{ cm}$ and Perpendicular distance ($OD$) = $8\text{ cm}$.

Figure 3: Calculating Chord Length using Right Triangle ODB

In right-angled triangle $\Delta ODB$, applying Pythagoras Theorem: $$OB^2 = OD^2 + BD^2 \implies 17^2 = 8^2 + BD^2$$ $$289 = 64 + BD^2 \implies BD^2 = 225 \implies BD = 15\text{ cm}$$ Since perpendicular from center bisects the chord: $$\text{Chord } AB = 2 \times BD = 2 \times 15 = 30\text{ cm}$$

Answer: Total length of the chord = 30 cm

Question 4 (Proof): Prove with reasons that two equal chords of any circle are equidistant from its centre.

Given: $AB$ and $CD$ are two equal chords of a circle with center $O$ ($AB = CD$). Let $OE \perp AB$ and $OF \perp CD$.

To prove: $OE = OF$.

Figure 4: Equal Chords AB & CD with Perpendiculars OE & OF

Proof: Perpendicular from center bisects the chord: $$AE = \frac{1}{2}AB \quad \text{and} \quad CF = \frac{1}{2}CD$$ $$\text{Since } AB = CD \text{ (Given)}, \text{ we have: } AE = CF$$ In right-angled triangles $\Delta AOE$ and $\Delta COF$: $$\text{Hypotenuse } OA = \text{Hypotenuse } OC \quad [\text{Radii of the same circle}]$$ $$AE = CF \quad [\text{Proved above}]$$ $$\text{Therefore, } \Delta AOE \cong \Delta COF \quad [\text{By RHS criterion}]$$ $$\implies OE = OF \quad [\text{By C.P.C.T.}]$/p>

Proved: Two equal chords are equidistant from the center.

Question 5 (Proof): Let us prove that the perpendicular bisector of a chord of a circle passes through its centre.

To prove: The perpendicular bisector of chord $AB$ passes through center $O$.

Figure 5: Perpendicular Bisector passing through O

Proof: Let $AB$ be a chord and $D$ be its midpoint. Suppose the perpendicular bisector does not pass through center $O$, but passes through another point $O'$.
By Theorem 33, the line segment joining the center $O$ to the midpoint of a chord $AB$ must be perpendicular to it ($\angle ODB = 90^\circ$).
But according to our assumption, $O'D \perp AB$, meaning $\angle O'DB = 90^\circ$.
This implies $\angle ODB = \angle O'DB$, which is only structurally possible if line segment $OD$ and $O'D$ coincide completely.
Thus, the perpendicular bisector must pass directly through center $O$.

Proved: The perpendicular bisector passes through the center.

Question 6 (Proof): Let us prove that a straight line cannot intersect a circle at more than two points.

Proof: Let us assume a straight line intersects a circle with center $O$ at three distinct points: $A$, $B$, and $C$.

Figure 6: Contradiction of Line Intersecting at points A, B, and C

Since $A$, $B$, and $C$ lie on the boundary of the same circle, the distances from the center $O$ to these points must be equal to the radius ($R$): $$OA = OB = OC = R$$ This implies that triangles $\Delta OAB$ and $\Delta OBC$ are isosceles triangles.
If we drop a single perpendicular line from $O$ to the intersecting line, it must uniquely bisect the base of an isosceles triangle.
This implies the midpoint of segment $AB$ and segment $BC$ would share matching geographic bounds along a single line projection, creating a geometric contradiction.
Hence, a straight line cannot cut a circle at more than two points.

Proved: Intersection cannot exceed two distinct points.