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...
Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that: ∠CPA = ∠DPB
Solution: Let’s draw a tangent \[TS\text{ }at\text{ }P\]to the circles given. As \[TPS\]is the tangent and \[PD\]is the chord, we have \[\angle PAB\text{ }=\angle BPS\text{ }\ldots .\text{ }\left( i...
Two circles with centres O and O’ are drawn to intersect each other at points A and B. Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O’ at A. Prove that OA bisects angle BAC.
Solution: Join \[OA,\text{ }OB,\text{ }OA,\text{ }OB\text{ }and\text{ }OO\] \[CD\]is the tangent and \[AO\]is the chord. \[\angle OAC\text{ }=\angle OBA\text{ }\ldots \text{ }\left( i \right)\] ...
Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that triangle PQR is isosceles.
Let \[DE\]be the tangent to the circle at \[P\] And, \[DE\text{ }||\text{ }QR\][Given] \[\angle EPR\text{ }=\angle PRQ\][Alternate angles are equal] \[\angle DPQ\text{ }=\angle PQR\][Alternate...
AB is diameter and AC is a chord of a circle with centre O such that angle BAC=30º. The tangent to the circle at C intersects AB produced in D. Show that BC = BD.
Join \[OC\] \[\angle BCD\text{ }=\angle BAC\text{ }=\] \[{{30}^{o}}~\] [Angles in the alternate segment] It’s seen that, arc \[BC\]subtends \[\angle DOC\]at the center of the circle And \[\angle...
If PQ is a tangent to the circle at R; calculate: i) ∠PRS ii) ∠ROT When O is the centre of the circle and ∠TRQ = 30degree
Solution: (i) As \[PQ\] is the tangent and \[OR\]is the radius. So, \[OR\bot PQ\] \[\angle ORT\text{ }=\text{ }{{90}^{o}}\] \[\angle TRQ\text{ }=\text{ }{{90}^{o}}-\text{ }{{30}^{o}}~=\text{...
In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If ∠ADB = 30degree and ∠CBD = 60degree; calculate: ∠CDB
Solution: As \[BD\] is the diameter, we have \[\angle BCD\text{ }=\text{ }{{90}^{o}}\] [Angle in a semi-circle] Now in \[\vartriangle BCD\] \[\angle CDB\text{ }+\angle CBD\text{ }+\angle BCD\text{...
In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If ∠ADB = 30degree and ∠CBD = 60degree; calculate: i) ∠QAB ii) ∠PAD
Solution: (i) Given, \[PAQ\]is a tangent and \[AB\]is the chord \[\angle QAB~=\angle ADB\text{ }=\text{ }{{30}^{o}}~\] [Angles in the alternate segment] (ii) \[OA\text{ }=\text{ }OD\][radii of the...
In the given figure, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find (i) AB. (ii) the length of tangent PT.
Solution: (i) \[PA\text{ }=\text{ }AB\text{ }+\text{ }BP\text{ }=\text{ }\left( AB\text{ }+\text{ }4 \right)\text{ }cm\] \[PC\text{ }=\text{ }PD\text{ }+\text{ }CD\] \[=\text{ }5\text{ }+\text{...
In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.
Solution: As \[PAB\]is the secant and \[PT\]is the tangent, we have \[P{{T}^{2}}~=\text{ }PA\text{ }x\text{ }PB\] \[{{12.5}^{2}}~=\text{ }10\text{ }x\text{ }PB\] \[PB\text{ }=\text{ }\left(...
(i) In the given figure, 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP. (ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4cm. Find CD.
(i) (ii) Solution: (i) As the two chords \[AB\text{ }and\text{ }CD\] intersect each other at \[P\] We have \[AP\text{ }x\text{ }PB\text{ }=\text{ }CP\text{ }x\text{ }PD\] \[4.5\text{ }x\text{...
From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that: i) ∠AOP = ∠BOP ii) OP is the ⊥ bisector of chord AB.
According to the given question, i) In \[\vartriangle AOP\text{ }and\text{ }\vartriangle BOP\], we have \[AP\text{ }=\text{ }BP\] \[\left[ Tangents\text{ }from\text{ }P\text{ }to\text{ }the\text{...
Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if – i) they touch each other externally. ii) they touch each other internally.
According to the given question \[Radius\text{ }of\text{ }bigger\text{ }circle\text{ }=\text{ }6.3\text{ }cm\]and \[smaller\text{ }circle\text{ }radius\text{ }=\text{ }3.6\text{ }cm\] i) The two...
In the figure, if AB = AC then prove that BQ = CQ.
Solution: From point \[A\], we know that \[AP\text{ }and\text{ }AR\]are the tangents to the circle So, we have \[AP\text{ }=\text{ }AR\] Similarly, we also have \[BP\text{ }=\text{ }BQ\text{...
From the given figure prove that: AP + BQ + CR = BP + CQ + AR. Also, show that AP + BQ + CR = ½ x perimeter of triangle ABC.
Solution: From point \[B,\text{ }BQ\text{ }and\text{ }BP\]are the tangents to the circle We have, \[BQ\text{ }=\text{ }BP\text{ }\ldots \ldots \ldots ..\left( 1 \right)\] Similarly, we also get...
If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.
Solution: Suppose a circle touch the sides \[AB,\text{ }BC,\text{ }CD\text{ }and\text{ }DA\]of parallelogram \[ABCD\text{ }at\text{ }P,\text{ }Q,\text{ }R\text{ }and\text{ }S\]respectively. Now,...
If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.
Solution: Suppose, a circle touch the sides \[AB,\text{ }BC,\text{ }CD\text{ }and\text{ }DA\] of quadrilateral \[ABCD\text{ }at\text{ }P,\text{ }Q,\text{ }R\text{ }and\text{ }S\]respectively. As,...
Three circles touch each other externally. A triangle is formed when the centers of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.
Suppose, \[ABC\]be the triangle formed when centres of 3 circles are joined. Given, \[AB\text{ }=\text{ }6\text{ }cm,\text{ }AC\text{ }=\text{ }8\text{ }cm\text{ }and\text{ }BC\text{ }=\text{...
Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.
Solution: Given, \[OS\text{ }=\text{ }5\text{ }cm\text{ }and\text{ }OT\text{ }=\text{ }3\text{ }cm\] In right triangle\[OST\], we have \[S{{T}^{2}}~=\text{ }O{{S}^{2}}~\text{ }O{{T}^{2}}\] \[=\text{...
Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Solution: Suppose, \[Q\]be the point on the common tangent from which, two tangents \[QA\text{ }and\text{ }QP\]are drawn to the circle with \[centre\text{ }O.\] So, \[QA\text{ }=\text{ }QP\text{...
Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.
Solution: Suppose, \[Q\]be the point from which, \[QA\text{ }and\text{ }QP\]are two tangents with centre \[O\] So, \[QA\text{ }=\text{ }QP\text{ }\ldots ..\left( a \right)\] Similarly, from point...
In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.
Solution: Given, \[AB\text{ }=\text{ }15\text{ }cm,\text{ }AC\text{ }=\text{ }7.5\text{ }cm\] Suppose,the radius of the circle to be \[r.\] So, \[AO\text{ }=\text{ }AC\text{ }+\text{ }OC\text{...
The radius of a circle is 8cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10cm from its centre.
Given, a circle with \[centre\text{ }O\]and \[radius\text{ }8\text{ }cm.\] An external point \[P\]is the point, from where a tangent is drawn to meet the circle at \[T.\] \[OP\text{ }=\text{...