Draw a circle of radius 2.7 cm. Draw a tangent to the circle at any point on it.
Rough figure is shown below. We use the property that a line perpendicular to the radius at its outer end is a tangent to the circle.Construction Steps:1.Draw a circle of radius 2.7 cm with centre O. Take any point M on the circle.2. Draw ray PM.3. Draw line l perpendicular to ray PM through point […]
Construct a tangent to a circle with centre P and radius 3.2 cm at any point M on it.
Rough figure is shown below. We use the property that a line perpendicular to the radius at its outer end is a tangent to the circle.Construction Steps:1.Draw a circle of radius 3.2 cm with centre P. Take any point M on the circle.2. Draw ray PM.3. Draw line l perpendicular to ray PM through point […]
PQR ~ LTR. In PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct PQR and LTR, such that PQ/ LT = 3/4 .
Given PQR and LTR are similar.Corresponding angles will be equal.PRQ LRTGiven PQ/LT = 3/4PQ/LT = QR/TR = PR/LR [Corresponding sides of similar triangles]PQ/LT= QR/TR = PR/LR = 3/4Sides of LTR are longer than corresponding sides of PQR.Steps of construction:1. Draw ∆PQR such that PQ = 4.2 , QR = 5.4 , and PR = 4.8. […]
ABC ~ LMN. In ABC, AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. Construct ABC and LMN such that BC /MN = 5/ 4 .
Rough figure is shown below. Given ABC and LMN are similar.their corresponding sides are proportional.AB/LM = BC/MN = AC/LNGiven BC/MN = 5/4AB/LM = BC/MN = AC/LN = 5/4As the sides AB, BC and AC are known we can find lengths of sides LM,MN and LN.Given AB = 5.5 , BC = 6 , CA = […]
In figure 3.91, line PR touches the circle at point Q. Answer the following questions with the help of the figure.
(1)ATSQ is a cyclic quadrilateral.TAQ + TSQ = 180˚ [Opposite angles of cyclic quadrilateral are supplementary](2)Given PR is the tangent.Seg AQ is the secant.AQP = ½ m(arc AQ) [Angle between tangent and secant]ASQ = ½ m(arc AQ) [Inscribed angle]AQP ASQ(3)QTS = QAS [ Angles inscribed in same arc are equal]∠QTS = ½ m(arc QS) [Inscribed angle […]
Prove that any three points on a circle cannot be collinear.
Let O be centre of the circle. Let P, Q, R be any points on the circle.To prove: P,Q ,R cannot be collinear.Proof:OP = OQ [Radii of same circle] O is equidistant from end points P and Q of seg PQ.O lies on perpendicular bisector of PQ. [Perpendicular bisector theorem]In the same way we can […]
Draw circles with centres A, B and C each of radius 3 cm, such that each circle touches the other two circles.
Given radius of each circle = 3cm.If two circles touch each other externally, then the distance between their centres is equal to the sum of their radii.AB = 3+3 = 6AC = 3+3 = 6BC = 3+3 = 6Draw line segment AB = 6cm.With A as centre and radius = 6 cm, mark an arc.With […]
In figure 3.90, seg AB is a diameter of a circle with centre C. Line PQ is a tangent, which touches the circle at point T. seg AP line PQ and seg BQ line PQ. Prove that, seg CP seg CQ.
Given AB is the diameter of the circle with centre C.PQ is the tangent.AP PQBQPQTo prove seg CP seg CQConstruction: Join CP, CT and CQ. Proof:Since PQ is the tangent , CT PQAlso AP PQBQ PQAPCTBQ [Lines which are perpendicular to same lines are parallel]AC/CB = PT/TQ …(i) [Property of three parallel lines and their […]
In figure 3.89, line l touches the circle with centre O at point P. Q is the mid point of radius OP. RS is a chord through Q such that chords RS || line l. If RS = 12 find the radius of the circle.
Given line l is a tangent.Let the radius of circle be r.OP is the radius.OP line l. [Tangent theorem]Given chord RSline l.OP chord RSSince the perpendicular from centre of the circle to the chord bisects the chord,QS = ½ RSQS = ½ ×12 = 6OQ = r/2 [Given Q is the midpoint of OP]In OQSOS2 = OQ2+QS2 [Pythagoras theorem]r2 = (r/2)2+62r2-r2/4 = 36(3/4)r2 = […]
In figure 3.88, circles with centres X and Y touch internally at point Z . Seg BZ is a chord of bigger
Given X and Y are the centres of the circle.Proof:Join YZ In AXZ ,AX = XZ [Radii of same circle]XAZ = XZA….(i) [Isoceles triangle theorem]In BYZ,YB = YZ [Radii of same circle]YBZ = YZB [Isoceles triangle theorem]XZA = YBZ ..(ii) [Y-X-Z, B-A-Z]From (i) and (ii)XAZ = YBZIf a pair of corresponding angles formed by a […]