Solution: According to the given question, ABC is a triangle with \[AB\text{ }=\text{ }10\text{ }cm,\text{ }BC=\text{ }8\text{ }cm,\text{ }AC\text{ }=\text{ }6\text{ }cm\] Three circles are drawn...

### In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If angle ACO = 30degree, find: angle APB

Solution: In the given fig, \[O\]is the centre of the circle and, \[CA\text{ }and\text{ }CB\]are the tangents to the circle from \[C\]Also, \[\angle ACO\text{ }=\text{ }{{30}^{o}}\] \[P\] is any...

### In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If angle ACO = 30degree, find: (i) angle BCO (ii) angle AOB

Solution: In the given fig, \[O\]is the centre of the circle and, \[CA\text{ }and\text{ }CB\]are the tangents to the circle from \[C\]Also, \[\angle ACO\text{ }=\text{ }{{30}^{o}}\] \[P\] is any...

### ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120degree Calculate: i) ∠BEC ii) ∠BED

i) Join \[OC\text{ }and\text{ }OB\] \[AB\text{ }=\text{ }BC\text{ }=\text{ }CD\] And \[\angle ABC\text{ }=\text{ }{{120}^{o}}~\][Given] So, \[\angle BCD\text{ }=\angle ABC\text{ }=\text{...

### In the given figure, AC = AE. Show that: i) CP = EP ii) BP = DP

Solution: In \[\vartriangle ADC\text{ }and\text{ }\vartriangle ABE\] \[\angle ACD\text{ }=\angle AEB~\] [Angles in the same segment] \[AC\text{ }=\text{ }AE\][Given] \[\angle A\text{ }=\angle...

### ABC is a right triangle with angle B = 90º. A circle with BC as diameter meets by hypotenuse AC at point D. Prove that: i) AC x AD = AB^2 ii) BD^2 = AD x DC.

i) In \[\vartriangle ABC\] We know that \[\angle B\text{ }=\text{ }{{90}^{o}}~and\text{ }BC\]is the diameter of the circle. Hence, \[AB\]is the tangent to the circle at \[B\] Now, as \[AB\] is...

### The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that: ∠ACD + ∠BAC = 90degree

Solution: Join \[OC\] \[BCD\]is the tangent and \[OC\]is the radius. As, \[OC\bot BD\] \[\angle OCD\text{ }=\text{ }{{90}^{o}}\] \[\angle OCD\text{ }+\angle ACD\text{ }=\text{ }{{90}^{o}}~\ldots...

### In the given figure, ABCD is a cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠DCQ = 40degree and ∠ABD = 60degree, find: ∠ADB

Solution: In \[\vartriangle ABD\] \[\angle BAD\text{ }=\text{ }{{90}^{o}}\] [Angle in a semi-circle], \[\angle ABD\text{ }=\text{ }{{60}^{o}}~\] [Given] Or, \[\angle ADB\text{ }=\text{...

### In the given figure, ABCD is a cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠DCQ = 40degree and ∠ABD = 60degree, find: i) ∠DBC ii) ∠BCP

Solution: 15. (i) \[PQ\]is a tangent and \[CD\]is a chord. \[\angle DCQ\text{ }=\angle DBC\] [Angles in the alternate segment] \[\angle DBC\text{ }=\text{ }{{40}^{o}}\] \[[As\angle DCQ\text{...

### In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent. Prove that the line NM produced bisects AB at P.

Solution: From \[P,\text{ }AP\] is the tangent And \[PMN\]is the secant for first circle. \[A{{P}^{2}}~=\text{ }PM\text{ }x\text{ }PN\text{ }\ldots .\text{ }\left( 1 \right)\] Again from \[P,\text{...

### P is the midpoint of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.

Join \[AP\text{ }and\text{ }BP\] As \[TPS\] is a tangent And \[PA\]is the chord of the circle. \[\angle BPT\text{ }=\angle PAB~\] [Angles in alternate segments] But, \[\angle PBA\text{ }=\angle...

### Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.

Join \[OL,\text{ }OM\text{ }and\text{ }ON\] Suppose, \[D\text{ }and\text{ }d\] be the diameter of the circumcircle and incircle. Also, \[R\text{ }and\text{ }r\] be the radius of the circumcircle and...

### In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20degree, find angle AOD.

Solution: Join \[OB\] In \[\vartriangle OBC\] we have \[BC\text{ }=\text{ }OD\text{ }=\text{ }OB\][Radii of the same circle] \[\angle BOC\text{ }=\angle BCO\text{ }=\text{ }{{20}^{o}}\] And ext....

### Bisectors of vertex angles A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angle EDF = 90degree – ½ ∠A

Join \[ED,\text{ }EF\text{ }and\text{ }DF\] And \[BF,\text{ }FA,\text{ }AE\text{ }and\text{ }EC\] \[\angle EBF\text{ }=\angle ECF\text{ }=\angle EDF\text{ }\ldots ..\text{ }\left( i \right)\] [Angle...

### Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Join \[AD\] And \[AB\] is the diameter. We have \[\angle ADB\text{ }=\text{ }90{}^\text{o}~\][Angle in a semi-circle] But, \[\angle ADB\text{ }+~\angle ADC\text{ }=\text{ }180{}^\text{o}~\][Linear...

### In cyclic quadrilateral ABCD, ∠A = 3 ∠C and ∠D = 5 ∠B. Find the measure of each angle of the quadrilateral.

Given, cyclic quadrilateral \[ABCD\] So, \[\angle A\text{ }+\angle C\text{ }=\text{ }{{180}^{o}}~\][Opposite angles in a cyclic quadrilateral is supplementary] \[3\angle C\text{ }+\angle C\text{...

### In the given figure, C and D are points on the semi-circle described on AB as diameter. Given angle BAD = 70degree and angle DBC = 30degree, calculate angle BDC.

Solution: \[ABCD\] is a cyclic quadrilateral, \[\angle BCD\text{ }+~\angle BAD\text{ }=\text{ }{{180}^{\circ }}\] [Opposite angles of a cyclic quadrilateral are supplementary] \[\angle BCD\text{...

### ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110o and angle BAC = 50o. Find angle DAC and angle DCA.

Given, \[ABCD\] is a cyclic quadrilateral in which \[AD\text{ }||\text{ }BC\] And, \[\angle ADC\text{ }=\text{ }{{110}^{o}},\angle BAC\text{ }=\text{ }{{50}^{o}}\] We know that, \[\angle B\text{...

### The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the center of the circle?

We have, \[AB\]as the diameter and \[AC\]as the chord. Now, draw \[OL\bot AC\] Since\[OL\bot AC\] and hence it bisects \[AC\] \[O\] is the centre of the circle. Therefore, \[OA\text{ }=\text{...

### Two chords AB and AC of a circle are equal. Prove that the center of the circle, lies on the bisector of the angle BAC.

Given: \[AB\text{ }and\text{ }AC\] are two equal chords of \[C\text{ }\left( O,\text{ }r \right)\] Required to prove: Centre, \[O\] lies on the bisector of \[\angle BAC\] Construction: Join \[BC\]...

### Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.

We know that, If two circles touch internally, then distance between their centres is equal to the difference of their radii. So, \[AB\text{ }=\text{ }\left( 5\text{ }-\text{ }3 \right)\text{...

### OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O. i) If the radius of the circle is 10 cm, find the area of the rhombus. ii) If the area of the rhombus is 32√3 cm2, find the radius of the circle.

(i) Given, \[radius\text{ }=\text{ }10\text{ }cm\] In rhombus \[OABC\] \[OC\text{ }=\text{ }10\text{ }cm\] So,...

### Prove that, of any two chords of a circle, the greater chord is nearer to the center.

Given: \[A\] circle with center \[O\] and radius \[r\] \[AB\text{ }and\text{ }CD\]are two chords such that \[AB\text{ }>\text{ }CD\] Also, \[OM\bot AB\text{ }and\text{ }ON\bot CD\] Required to...