In figure 3.58, seg RS is a diameter of the circle with centre O. Point T lies in the exterior of the circle. Prove that RTS is an acute angle.
To prove : RTS is an acute angle.Construction:Join RT and ST. Let RT intersect circle at point A. Join AS. Proof:Given RS is a diameter. O is the centre of the circle.Since RS is the diameter , RAS = 90˚ [Angle in semi circle is right angle]In ATS , RAS is an exterior angle and […]
MRPN is cyclic, R=(5x – 13)°, N=(4x+4)°. Find measures of R and N.
Given R=(5x – 13)°, N=(4x+4)°Opposite angles of a cyclic quadrilateral are supplementary.R+N = 180˚5x -13+4x+4 = 1809x-9 = 1809x = 189x = 189/9x = 21R = 5x-13= 5×21-13= 92˚N=(4x+4)°N = 4×21+4= 84+4= 88N = 88˚Hence the measure of R = 92˚ and N = 88˚.
In figure 3.57, PQRS is cyclic. side PQ side RQ. PSR = 110°, Find
(1)Given PSR = 110˚PQR = 180-110 = 70˚ [Opposite angles of a cyclic quadrilateral are supplementary.](2)PSR = ½ m(arc PQR) [The measure of an inscribed angle is half the measure of the arc intercepted by it]m(arc PQR ) = 2×PSRm(arc PQR ) = 2×110m(arc PQR ) = 220˚(3)Given side PQ side RQarc PQ arc RQ […]
In figure 3.56, in a circle with centre O, length of chord AB is equal to the radius of the circle. Find measure of each of the following.
(1)OA and OB are the radius of circle.Given AB = radius of circleAB = OA = OBOAB is an equilateral triangle.AOB = 60˚ [Angle of equilateral triangle = 60˚](2)ACB = ½ AOB [The measure of an angle subtended by an arc at a point on the circle is half of the measure of the angle […]
In fig 3.38 QRS is an equilateral triangle. Prove that,
(1)Given QRS is an equilateral triangle, sides are equal in measure.QR = RS = QSarc QR = arc RS = arc QS [Corresponding arcs of congruent chords of a circle are congruent]arc RS arc QS arc QR….(i)Hence proved.(2)m(arc RS)+ m(arc QS)+ m(arc QR) = 360˚ [Measure of a complete circle is 360°]Also from (i) arc […]
In figure 3.37, points G, D, E, F are concyclic points of a circle with centre C. ECF = 70°, m(arc DGF) = 200° find m(arc DE) and m(arc DEF).
Given ECF = 70˚m(arc DGF) = 200˚m(arc EF) = 70˚ [The measure of a minor arc is the measure of its central angle.]m(arc DGF)+m(arc EF)+m(arc DE) = 360˚ [Measure of a complete circle is 360°.]200+70+ m(arc DE) = 360m(arc DE) = 360-(200+70)m(arc DE) = 90˚m(arc DEF) = m(arc DE)+m(arc EF) [Property of sum of measures […]
If radii of two circles are 4 cm and 2.8 cm. Draw figure of these circles touching each other – (i) externally (ii) internally.
(i)Circles touching externally If the circles touch each other externally, distance between their centres is equal to the sum of their radii.Distance between the centres = 4+2.8 = 6.8cm(ii)Circles touching internally The distance between the centres of the circles touching internally is equal to the difference of their radii.Distance between the centres = 4-2.8 = […]
Two circles of radii 5.5 cm and 4.2 cm touch each other externally. Find the distance between their centres.
If the circles touch each other externally, distance between their centres is equal to the sum of their radii.We have r1 = 5.5 and r2 = 4.2Distance between their centres = 5.5 +4.2 = 9.7 cm
Two circles having radii 3.5 cm and 4.8 cm touch each other internally. Find the distance between their centres.
The distance between the centres of the circles touching internally is equal to the difference of their radii.We have r1 = 3.5 and r2 = 4.8Distance between their centres = 4.8-3.5 = 1.3 cm.
In the adjoining figure, O is the centre of the circle. From point R, seg RM and seg RN are tangent segments touching the circle at M and N. If (OR) = 10 cm and radius of the circle = 5 cm, then
(1)Given RM and RN are the tangents to the circle. OMR = ONR = 90˚ [Tangent theorem]In OMR , OR = 10 cm , [Given]OM = 5cm [Radius]OR2 = OM2+MR2 [Pythagoras theorem]102 = 52+MR2MR2 = 102-55MR2 = 75MR = 5√3RN = 5√3 [Tangent segments drawn from an external point to a circle are congruent](2)In OMR , OMR = 90˚ [Tangent […]