Equal Chords and their Distances from the Centre Questions in English

(N/A) Let there be a circle with center $O$. Let $AB$ and $CD$ be two equal chords that intersect at a point $E$ inside the circle.
To prove: $AE = DE$ and $CE = BE$.
Construction: Draw $OM \perp AB$ and $ON \perp CD$. Join $OE$.
Proof:
$1$. Since $AB = CD$ (given),the chords are equidistant from the center. Therefore,$OM = ON$.
$2$. In $\Delta OME$ and $\Delta ONE$:
- $OM = ON$ (Proved above)
- $OE = OE$ (Common side)
- $\angle OME = \angle ONE = 90^\circ$ (By construction)
By $RHS$ congruence criterion,$\Delta OME \cong \Delta ONE$.
$3$. By $CPCT$,$ME = NE$ (Equation $1$).
$4$. Since $OM \perp AB$,$M$ is the midpoint of $AB$,so $AM = MB = \frac{1}{2} AB$.
$5$. Since $ON \perp CD$,$N$ is the midpoint of $CD$,so $CN = ND = \frac{1}{2} CD$.
$6$. Since $AB = CD$,then $\frac{1}{2} AB = \frac{1}{2} CD$,which implies $AM = ND$ (Equation $2$).
$7$. Adding equations $1$ and $2$: $AM + ME = ND + NE$,which gives $AE = DE$.
$8$. Subtracting $AE = DE$ from $AB = CD$,we get $AB - AE = CD - DE$,which implies $BE = CE$.
Thus,the segments of one chord are equal to the corresponding segments of the other chord.

(N/A) Let $AB$ and $CD$ be two equal chords of a circle with centre $O$,intersecting at point $E$ inside the circle.
We need to prove that $\angle OEA = \angle OED$.
Draw $OM \perp AB$ and $ON \perp CD$.
In $\Delta OME$ and $\Delta ONE$:
$1$. $OM = ON$ (Equal chords are equidistant from the centre).
$2$. $OE = OE$ (Common side).
$3$. $\angle OME = \angle ONE = 90^\circ$ (By construction).
Therefore,by $RHS$ congruence criterion,$\Delta OME \cong \Delta ONE$.
By $CPCT$,$\angle OEM = \angle OEN$.
Since $M$ lies on $AB$ and $N$ lies on $CD$,this implies $\angle OEA = \angle OED$.

(B) Let the positions of Reshma,Salma,and Mandip be $R, S,$ and $M$ respectively on the circle with center $O$ and radius $5 \, m$.
Given $RS = SM = 6 \, m$.
Since equal chords subtend equal angles at the center,$\angle ROS = \angle SOM$.
Let $OP$ be the perpendicular from $O$ to $RM$,intersecting $RM$ at $P$ and $OS$ at $K$.
Since $RS = SM$,$OS$ is the perpendicular bisector of $RM$. Thus,$RM \perp OS$ and $RP = PM$.
Let $OK = x$. Then $KS = 5 - x$.
In $\Delta ORK$,$RK^2 = OR^2 - OK^2 = 5^2 - x^2 = 25 - x^2$.
In $\Delta RSK$,$RK^2 = RS^2 - KS^2 = 6^2 - (5 - x)^2 = 36 - (25 - 10x + x^2) = 11 + 10x - x^2$.
Equating the two expressions for $RK^2$: $25 - x^2 = 11 + 10x - x^2$.
$10x = 14 \Rightarrow x = 1.4 \, m$.
Now,$RK^2 = 25 - (1.4)^2 = 25 - 1.96 = 23.04$.
$RK = \sqrt{23.04} = 4.8 \, m$.
Since $RM = 2 \times RK$,$RM = 2 \times 4.8 = 9.6 \, m$.

(B) Let the circle have centre $O$. Chords $AB$ and $CD$ are parallel,and the distance between them is $6\, cm$. Given $AB = 5\, cm$ and $CD = 11\, cm$.
Let $r$ be the radius of the circle.
Draw $OP \perp AB$ and $OQ \perp CD$. Since the chords are on opposite sides of the centre,$P, O, Q$ are collinear and $PQ = 6\, cm$.
Let $OQ = x\, cm$. Then $OP = (6 - x)\, cm$.
The perpendicular from the centre to a chord bisects the chord.
So,$AP = \frac{1}{2} AB = \frac{5}{2}\, cm$ and $CQ = \frac{1}{2} CD = \frac{11}{2}\, cm$.
In right $\Delta CQO$,$r^2 = CQ^2 + OQ^2 = (\frac{11}{2})^2 + x^2 = \frac{121}{4} + x^2$ --- $(1)$
In right $\Delta APO$,$r^2 = AP^2 + OP^2 = (\frac{5}{2})^2 + (6 - x)^2 = \frac{25}{4} + 36 - 12x + x^2$ --- $(2)$
Equating $(1)$ and $(2)$:
$\frac{121}{4} + x^2 = \frac{25}{4} + 36 - 12x + x^2$
$\frac{121}{4} - \frac{25}{4} - 36 = -12x$
$\frac{96}{4} - 36 = -12x \Rightarrow 24 - 36 = -12x \Rightarrow -12 = -12x \Rightarrow x = 1\, cm$.
Substituting $x = 1$ in $(1)$:
$r^2 = \frac{121}{4} + 1^2 = \frac{121 + 4}{4} = \frac{125}{4}$.
$r = \sqrt{\frac{125}{4}} = \frac{5\sqrt{5}}{2}\, cm$.

(C) Let the circle have centre $O$. Let the parallel chords be $AB = 6\, cm$ and $CD = 8\, cm$.
Draw $OP \perp AB$ and $OQ \perp CD$.
Since the perpendicular from the centre to a chord bisects the chord:
$AP = \frac{1}{2} AB = \frac{1}{2}(6\, cm) = 3\, cm$
$CQ = \frac{1}{2} CD = \frac{1}{2}(8\, cm) = 4\, cm$
Given that the smaller chord is at a distance of $4\, cm$ from the centre,$OP = 4\, cm$.
In right-angled $\Delta OPA$,by Pythagoras theorem:
$OA^2 = OP^2 + AP^2$
$r^2 = 4^2 + 3^2 = 16 + 9 = 25$
$r = 5\, cm$ (where $r$ is the radius of the circle).
Now,in right-angled $\Delta OQC$:
$OC^2 = OQ^2 + CQ^2$
$r^2 = OQ^2 + 4^2$
$5^2 = OQ^2 + 16$
$25 = OQ^2 + 16$
$OQ^2 = 25 - 16 = 9$
$OQ = 3\, cm$.
Thus,the distance of the other chord from the centre is $3\, cm$.

(B) The statement is False.
In a circle,the chord that is closer to the centre is always longer than the chord that is farther from the centre.
Let the lengths of the chords be $L_1 = 10 \, cm$ and $L_2 = 8 \, cm$.
Let their distances from the centre be $d_1 = 8.0 \, cm$ and $d_2 = 3.5 \, cm$.
Since $L_1 > L_2$,the distance $d_1$ must be less than $d_2$ $(d_1 < d_2)$.
However,here $d_1 = 8.0 \, cm$ and $d_2 = 3.5 \, cm$,which implies $d_1 > d_2$.
Therefore,the given information is geometrically impossible for a single circle.

(A) We know that chords equidistant from the centre of a circle are equal in length.
Given that two chords $AB$ and $CD$ of a circle are each at a distance of $4 \ cm$ from the centre,they are equidistant from the centre.
Therefore,according to the theorem,the chords must be equal.
Hence,$AB = CD$.
The given statement is True.

(N/A) $AB$ and $CD$ are two equal chords of a circle with centre $O,$ which intersect each other at $M.$ We have to prove that:
$(i)$ $MB = MC$ and
$(ii)$ $AM = MD$
Draw $OE \perp AB$ and $OF \perp CD$ from the centre $O.$
Since the perpendicular from the centre to a chord bisects the chord,we have:
$AE = \frac{1}{2} AB$ and $FD = \frac{1}{2} CD$
Since $AB = CD,$ it follows that $\frac{1}{2} AB = \frac{1}{2} CD,$ so $AE = FD$ $...(1)$
Since equal chords are equidistant from the centre,$OE = OF.$
Now,in $\Delta MOE$ and $\Delta MOF$:
$OE = OF$ [Proved above]
$OM = OM$ [Common side]
$\angle OEM = \angle OFM = 90^\circ$ [By construction]
Therefore,$\Delta MOE \cong \Delta MOF$ [By $RHS$ congruence rule]
Thus,$ME = MF$ $...(2)$
Subtracting $(2)$ from $(1),$ we get:
$AE - ME = FD - MF$
$\Rightarrow AM = MD$ [Proved part $(ii)$]
Again,$AB = CD$ and $AM = MD.$
Subtracting these,we get $AB - AM = CD - MD.$
Therefore,$MB = MC$ [Proved part $(i)$]

(A) In a circle,equal chords subtend equal angles at the centre.
Given that $AB$ and $CD$ are equal chords $(AB = CD)$ in a circle with centre $P$.
According to the theorem,equal chords of a circle subtend equal angles at the centre.
Therefore,$\angle APB = \angle CPD$.
Since $\angle APB = 80^{\circ}$,it follows that $\angle CPD = 80^{\circ}$.

(B) $1$. In a circle,equal chords subtend equal angles at the centre. Since chord $AB = CD$,the angles subtended by them at the centre are equal,so $\angle CPD = \angle APB = 100^{\circ}$.
$2$. In $\triangle PCD$,$PC$ and $PD$ are radii of the circle,so $PC = PD$. This makes $\triangle PCD$ an isosceles triangle.
$3$. In an isosceles triangle,the angles opposite to equal sides are equal,so $\angle PCD = \angle PDC$.
$4$. The sum of angles in a triangle is $180^{\circ}$. Therefore,$\angle PCD + \angle PDC + \angle CPD = 180^{\circ}$.
$5$. Substituting the known values: $2 \times \angle PCD + 100^{\circ} = 180^{\circ}$.
$6$. $2 \times \angle PCD = 80^{\circ}$,which gives $\angle PCD = 40^{\circ}$.

(A) In the circle with center $P$,$PA = PB = 8 \ cm$ (radii of the same circle).
Since $PA = PB$,$\triangle PAB$ is an isosceles triangle.
Therefore,$\angle PBA = \angle PAB = 40^{\circ}$.
In $\triangle PAB$,the sum of angles is $180^{\circ}$,so $\angle APB = 180^{\circ} - (40^{\circ} + 40^{\circ}) = 100^{\circ}$.
Since the chords $AB$ and $CD$ are equal and the circles have the same radius $(8 \ cm)$,the chords subtend equal angles at their respective centers.
Thus,$\angle CQD = \angle APB = 100^{\circ}$.

(D) Let $r$ be the radius of the circle.
For chord $AB$ of length $12 \, cm$,the distance from the centre $O$ is $d_1 = 8 \, cm$.
$A$ perpendicular from the centre to a chord bisects the chord. Thus,the distance from the centre,the radius,and half the chord length form a right-angled triangle.
$r^2 = (12/2)^2 + 8^2 = 6^2 + 8^2 = 36 + 64 = 100$.
So,$r = 10 \, cm$.
For chord $CD$ of length $16 \, cm$,let the distance from the centre be $d_2$.
$r^2 = (16/2)^2 + d_2^2$.
$10^2 = 8^2 + d_2^2$.
$100 = 64 + d_2^2$.
$d_2^2 = 36$.
$d_2 = 6 \, cm$.

(A) Let $r$ be the radius of the circle.
Draw a perpendicular from $P$ to chord $AB$ at point $M$ and to chord $CD$ at point $N$.
$PM = 15 \, cm$ and $PN = 8 \, cm$.
Since the perpendicular from the centre bisects the chord,$AM = MB = AB / 2 = 16 / 2 = 8 \, cm$.
In right-angled triangle $\triangle PAM$,by Pythagoras theorem: $r^2 = PM^2 + AM^2 = 15^2 + 8^2 = 225 + 64 = 289$.
So,$r = \sqrt{289} = 17 \, cm$.
Now,in right-angled triangle $\triangle PNC$,by Pythagoras theorem: $r^2 = PN^2 + CN^2$.
$17^2 = 8^2 + CN^2 \implies 289 = 64 + CN^2$.
$CN^2 = 289 - 64 = 225$.
$CN = \sqrt{225} = 15 \, cm$.
Since $N$ is the midpoint of $CD$,$CD = 2 \times CN = 2 \times 15 = 30 \, cm$.

(D) Let the centre of the circle be $P$ and the radius $r = 25 \, cm$.
Draw perpendiculars from $P$ to the chords $AB$ and $CD$,meeting them at points $M$ and $N$ respectively.
Since the perpendicular from the centre to a chord bisects the chord,we have:
$AM = MB = \frac{AB}{2} = \frac{48}{2} = 24 \, cm$
$CN = ND = \frac{CD}{2} = \frac{40}{2} = 20 \, cm$
In right-angled triangle $\triangle PMA$:
$PM^2 + AM^2 = PA^2$
$PM^2 + 24^2 = 25^2$
$PM^2 + 576 = 625$
$PM^2 = 49 \implies PM = 7 \, cm$
In right-angled triangle $\triangle PNC$:
$PN^2 + CN^2 = PC^2$
$PN^2 + 20^2 = 25^2$
$PN^2 + 400 = 625$
$PN^2 = 225 \implies PN = 15 \, cm$
Since the centre $P$ lies between the chords,the distance between $AB$ and $CD$ is $PM + PN = 7 \, cm + 15 \, cm = 22 \, cm$.

(A) Let $P$ be the center of the circle with radius $r = 13 \, cm$. Let $M$ and $N$ be the midpoints of chords $AB$ and $CD$ respectively. Since the perpendicular from the center to a chord bisects the chord,$AM = MB = 12 \, cm$ and $CN = ND$. In $\triangle PMA$,by the Pythagorean theorem,$PM^2 + AM^2 = PA^2$,so $PM^2 + 12^2 = 13^2$,which gives $PM^2 = 169 - 144 = 25$,so $PM = 5 \, cm$. There are two cases: chords are on the same side or opposite sides of the center. Case $1$: Chords are on opposite sides. The distance $MN = PM + PN = 17 \, cm$. Since $PM = 5 \, cm$,$PN = 17 - 5 = 12 \, cm$. In $\triangle PNC$,$PN^2 + CN^2 = PC^2$,so $12^2 + CN^2 = 13^2$,which gives $CN^2 = 169 - 144 = 25$,so $CN = 5 \, cm$. Thus,$CD = 2 \times CN = 10 \, cm$. Case $2$: Chords are on the same side. The distance $MN = |PM - PN| = 17 \, cm$. Since $PM = 5 \, cm$,$PN$ would be $22 \, cm$ or $-12 \, cm$,which is impossible as the radius is $13 \, cm$. Thus,$CD = 10 \, cm$.

(B) Let the radius $r = 50 \, cm$. Let $M$ and $N$ be the midpoints of chords $AB$ and $CD$ respectively.
Since $AB = 80 \, cm$,$AM = MB = 40 \, cm$.
In $\triangle PMA$,by Pythagoras theorem,$PM^2 + AM^2 = PA^2$.
$PM^2 + 40^2 = 50^2 \implies PM^2 = 2500 - 1600 = 900 \implies PM = 30 \, cm$.
Since $P$ does not lie between the chords,the distance between the chords is $PN - PM = 10 \, cm$.
$PN = 10 + PM = 10 + 30 = 40 \, cm$.
In $\triangle PNC$,$PN^2 + NC^2 = PC^2$.
$40^2 + NC^2 = 50^2 \implies 1600 + NC^2 = 2500 \implies NC^2 = 900 \implies NC = 30 \, cm$.
Since $N$ is the midpoint of $CD$,$CD = 2 \times NC = 2 \times 30 = 60 \, cm$.

(A) According to the theorem in geometry,if two chords of a circle subtend equal angles at the centre,then the chords are equal in length.
Given that the chords $PQ$ and $RS$ subtend equal angles at the centre $O$.
Therefore,$PQ = RS$.
Since $PQ = 12 \text{ cm}$,it follows that $RS = 12 \text{ cm}$.