ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.
[question] Question. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD. [/question] [solution] Solution: It can be observed that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is 180°. $\angle A E C+\angle C B A=180^{\circ}$ $\angle \mathrm{AEC}+\angle \mathrm{AED}=180^{\circ}($ Linear pair $)$ $\angle A E D=\angle C B A \ldots(1)$ For a parallelogram, opposite angles are equal. $\ang...
Read More →Prove that the circle drawn with any side of a rhombus as diameter passes
[question] Question. Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals. [/question] [solution] Solution: Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc. $\therefore \angle C O D=90^{\circ}$ Also, in rhombus, the diagonals intersect each other at $90^{\circ}$. $\angle A O B=\angle B O C=\angle C...
Read More →The lengths of two parallel chords of a circle are 6 cm and 8 cm.
[question] Question. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre? [/question] [solution] Solution: Let AB and CD be two parallel chords in a circle centered at O. Join OB and OD. Distance of smaller chord AB from the centre of the circle = 4 cm $O M=4 \mathrm{~cm}$ $\mathrm{MB}=\frac{\mathrm{AB}}{2}=\frac{6}{2}=3 \mathrm{~cm}$ In $\triangle O M B$ $\mathrm{OM}^{...
Read More →Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre.
[question] Question. Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle. [/question] [solution] Solution: Draw $O M \perp A B$ and $O N \perp C D$. Join $O B$ and $O D$. $\mathrm{BM}=\frac{\mathrm{AB}}{2}=\frac{5}{2}$ (Perpendicular from the centre bisects the chord) $\mathrm{ND}=\frac{\mathrm{CD}}{2}=\frac{11}{2}$ Let $O N$ be $x$. Therefor...
Read More →Prove that line of centres of two intersecting circles subtends
[question] Question. Prove that line of centres of two intersecting circles subtends equal angles at the two points of intersection. [/question] [solution] Solution: Let two circles having their centres as $O$ and $O^{\prime}$ intersect each other at point $A$ and $B$ respectively. Let us join $O O^{\prime}$. In $\triangle \mathrm{AO} \mathrm{O}^{\prime}$ and $\mathrm{BO} \mathrm{O}^{\prime}$, OA $=$ OB (Radius of circle 1) $\mathrm{O}^{\prime} \mathrm{A}=\mathrm{O}^{\prime} \mathrm{B}$ (Radius ...
Read More →Prove that a cyclic parallelogram is a rectangle.
[question] Question. Prove that a cyclic parallelogram is a rectangle. [/question] [solution] Solution: Let ABCD be a cyclic parallelogram. $\angle \mathrm{A}+\angle \mathrm{C}=180^{\circ}$ (Opposite angles of a cyclic quadrilateral) ... (1) We know that opposite angles of a parallelogram are equal. $\therefore \angle A=\angle C$ and $\angle B=\angle D$ From equation (1), $\angle \mathrm{A}+\angle \mathrm{C}=180^{\circ}$ $\Rightarrow \angle \mathrm{A}+\angle \mathrm{A}=180^{\circ}$ $\Rightarrow ...
Read More →ABC and ADC are two right triangles with common hypotenuse AC.
[question] Question. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD. [/question] [solution] Solution: In $\triangle \mathrm{ABC}$, $\angle A B C+\angle B C A+\angle C A B=180^{\circ}$ (Angle sum property of a triangle) $\Rightarrow 90^{\circ}+\angle B C A+\angle C A B=180^{\circ}$ $\Rightarrow \angle B C A+\angle C A B=90^{\circ} \ldots(1)$ In $\triangle \mathrm{ADC}$ $\angle C D A+\angle A C D+\angle D A C=180^{\circ}$ (Angle sum property of a triangle) $\...
Read More →If circles are drawn taking two sides of a triangle as diameters,
[question] Question. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side. [/question] [solution] Solution: Consider a $\triangle \mathrm{ABC}$. Two circles are drawn while taking $A B$ and $A C$ as the diameter. Let they intersect each other at $D$ and let $D$ not lie on $B C$. Join AD. $\angle A D B=90^{\circ}$ (Angle subtended by semi-circle) $\angle \mathrm{ADC}=90^{\circ}$ (Angle subtended by semi-circl...
Read More →Two circles intersect at two points B and C.
[question] Question. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD. [/question] [solution] Solution: Join chords AP and DQ. For chord AP, $\angle P B A=\angle A C P$ (Angles in the same segment)...(1) For chord DQ, $\angle D B Q=\angle Q C D$ (Angles in the same segment) ... (2) $A B D$ and $P B Q$ are line segments intersecting at $B$. $\therefo...
Read More →If the non-parallel sides of a trapezium are equal,
[question] Question. If the non-parallel sides of a trapezium are equal, prove that it is cyclic. [solution] Solution: Consider a trapezium $A B C D$ with $A B|| C D$ and $B C=A D$. Draw AM $\perp C D$ and BN $\perp C D$. In $\triangle \mathrm{AMD}$ and $\triangle \mathrm{BNC}$, $A D=B C$ (Given) $\angle \mathrm{AMD}=\angle \mathrm{BNC}\left(\right.$ By construction, each is $\left.90^{\circ}\right)$ $\mathrm{AM}=\mathrm{BN}$ (Perpendicular distance between two parallel lines is same) $\therefor...
Read More →If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral,
[question] Question. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle. [/question] [solution] Solution: Let $A B C D$ be a cyclic quadrilateral having diagonals $B D$ and $A C$, intersecting each other at point $O$. $\angle B A D=\frac{1}{2} \angle B O D=\frac{180^{\circ}}{2}=90^{\circ}$ (Consider BD as a chord) $\angle B C D+\angle B A D=180^{\circ}($ Cyclic quadrilateral $)$ $\angle B C D=180^{\circ}-90^{...
Read More →ABCD is a cyclic quadrilateral whose diagonals intersect at a point E
[question] Question. $A B C D$ is a cyclic quadrilateral whose diagonals intersect at a point $E$. If $\angle D B C=70^{\circ}, \angle B A C$ is $30^{\circ}$, find $\angle B C D$. Further, if $A B=B C$, find $\angle E C D$. [/question] [solution] Solution: For chord CD, $\angle C B D=\angle C A D$ (Angles in the same segment) $\angle C A D=70^{\circ}$ $\angle B A D=\angle B A C+\angle C A D=30^{\circ}+70^{\circ}=100^{\circ}$ $\angle B C D+\angle B A D=180^{\circ}$ (Opposite angles of a cyclic qu...
Read More →In the given figure, A, B, C and D are four points on a circle.
[question] Question. In the given figure, $A, B, C$ and $D$ are four points on a circle. $A C$ and $B D$ intersect at a point $E$ such that $\angle B E C=130^{\circ}$ and $\angle E C D=20^{\circ}$. Find $\angle B A C$. [/question] [solution] Solution: In $\triangle C D E$, $\angle C D E+\angle D C E=\angle C E B$ (Exterior angle) $\Rightarrow \angle C D E+20^{\circ}=130^{\circ}$ $\Rightarrow \angle C D E=110^{\circ}$ However, $\angle B A C=\angle C D E$ (Angles in the same segment of a circle) $...
Read More →In the given figure, ∠ABC = 69°,
[question] Question. In the given figure, $\angle A B C=69^{\circ}, \angle A C B=31^{\circ}$, find $\angle B D C$. [/question] [solution] Solution: In $\triangle A B C$ $\angle B A C+\angle A B C+\angle A C B=180^{\circ}$ (Angle sum property of a triangle) $\Rightarrow \angle B A C+69^{\circ}+31^{\circ}=180^{\circ}$ $\Rightarrow \angle B A C=180^{\circ}-100^{\circ}$ $\Rightarrow \angle B A C=80^{\circ}$ $\angle B D C=\angle B A C=80^{\circ}$ (Angles in the same segment of a circle are equal) [/s...
Read More →In the given figure, ∠PQR = 100°, where P,
[question] Question. In the given figure, $\angle P Q R=100^{\circ}$, where $P, Q$ and $R$ are points on a circle with centre $O$. Find $\angle O P R$. [/question] [solution] Solution: Consider PR as a chord of the circle. Take any point S on the major arc of the circle. PQRS is a cyclic quadrilateral. $\angle \mathrm{PQR}+\angle \mathrm{PSR}=180^{\circ}$ (Opposite angles of a cyclic quadriateral) $\Rightarrow \angle \mathrm{PSR}=180^{\circ}-100^{\circ}=80^{\circ}$ We know that the angle subtend...
Read More →In the given figure, ∠PQR = 100°,
[question] Question. In the given figure, $\angle P Q R=100^{\circ}$, where $P, Q$ and $R$ are points on a circle with centre $O$. Find $\angle O P R$. [/question] [solution] Solution: Consider PR as a chord of the circle. Take any point S on the major arc of the circle. PQRS is a cyclic quadrilateral. $\angle P Q R+\angle P S R=180^{\circ}$ (Opposite angles of a cyclic quadrilateral) $\Rightarrow \angle \mathrm{PSR}=180^{\circ}-100^{\circ}=80^{\circ}$ We know that the angle subtended by an arc ...
Read More →A chord of a circle is equal to the radius of the circle.
[question] Question. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc. [/question] [solution] Solution: In $\triangle O A B$ $A B=O A=O B=$ radius $\therefore \triangle O A B$ is an equilateral triangle. Therefore, each interior angle of this triangle will be of $60^{\circ}$. $\therefore \angle A O B=60^{\circ}$ $\angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOB}=\frac{1}{2}\left(60^{...
Read More →If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D
[question] Question. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see figure 10.25). [/question] [solution] Solution: Let us draw a perpendicular OM on line AD. It can be observed that BC is the chord of the smaller circle and AD is the chord of the bigger circle. We know that perpendicular drawn from the centre of the circle bisects the chord. $\therefore B M=M C \ldots(1)$ And, AM = MD ... (2) On subtracting equa...
Read More →If two equal chords of a circle intersect within the circle
[question] Question. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords. [/question] [solution] Solution: Let PQ and RS are two equal chords of a given circle and they are intersecting each other at point T. Draw perpendiculars OV and OU on these chords. In $\triangle O V T$ and $\triangle O U T$, OV = OU (Equal chords of a circle are equidistant from the centre) $\angle O V T=\angle ...
Read More →If two equal chords of a circle intersect within the circle
[question] Question. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord [/question] [solution] Solution: Let PQ and RS be two equal chords of a given circle and they are intersecting each other at point T. Draw perpendiculars OV and OU on these chords. In $\triangle O V T$ and $\Delta O U T$, OV = OU (Equal chords of a circle are equidistant from the centre) $\angle O V T=\angle O U T\left(\righ...
Read More →Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm.
[question] Question. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord. [solution] Solution: Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively. $O A=O B=5 \mathrm{~cm}$ $O^{\prime} A=O^{\prime} B=3 \mathrm{~cm}$ OO' will be the perpendicular bisector of chord $A B$. $\therefore A C=C B$ It is given that, $O O^{\prime}=4 \mathrm{~cm}$ Let $O C$ be $x$. Therefore, $O^{\prime} ...
Read More →If two circles intersect at two points,
[question] Question. If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord. [/question] [solution] Solution: Consider two circles centered at point O and O’, intersecting each other at point A and B respectively. Join AB. AB is the chord of the circle centered at O. Therefore, perpendicular bisector of AB will pass through O. Again, AB is also the chord of the circle centered at O’. Therefore, perpendicular bisector of AB will...
Read More →Suppose you are given a circle.
[question] Question. Suppose you are given a circle. Give a construction to find its centre [solution] Solution: The below given steps will be followed to find the centre of the given circle. Step1. Take the given circle. Step2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these chords. Step3. Let these perpendicular bisectors meet at point O. Hence, O is the centre of the given circle....
Read More →Draw different pairs of circles. How many points does each pair have in common?
[question] Question. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points? [/question] [solution] Solution: Consider the following pair of circles. The above circles do not intersect each other at any point. Therefore, they do not have any point in common. The above circles touch each other only at one point Y. Therefore, there is 1 point in common. The above circles touch each other at 1 point X only. Therefore, the circles ...
Read More →Prove that if chords of congruent circles subtend equal angles at their centres,
[question] Question. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. [/question] [solution] Solution: Let us consider two congruent circles (circles of same radius) with centres as O and O'. In $\triangle \mathrm{AOB}$ and $\triangle C O^{\prime} D$, $\angle \mathrm{AOB}=\angle \mathrm{CO}^{\prime} \mathrm{D}($ Given $)$ $O A=O^{\prime} C$ (Radii of congruent circles) $O B=O^{\prime} D$ (Radii of congruent circles) $\therefore \triangle...
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