Where will I get the NCERT Solutions for Class 10 Maths PDF?

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2. https://www.entrancei.com/ CIRCLES AND CONSTRUCTIONS CIRCLES 1. IMPORTANT TERMS Circle: A circle is a collection of all points in a plane which are at a constant distance [radius] from a fixed point [centre]. 2. 3. Tangent: A tangent to a circle is a line that touches the circle at exactly one point. Point of contact: The point at which the line meets the circle is called its point of contact and the tangent is said to touch the circle at this point. 4. 5. Secant: A line which intersects the circle at two points is called a secant of the circle. Length of tangent: The length of the segment of the tangent from the external point and the point of contact with the circle is called the length of the tangent from the external point to the circle. TANGENTS AND THEIR PROPERTIES 1. 2. 3. There is no tangent to a circle passing through a point lying inside the circle. There is one and only one tangent to a circle passing through a point lying on the circle. There are exactly two tangents to a circle through a point lying outside the circle. THEOREM 1: The tangent at any point of a circle is perpendicular to the radius through the point of contact. Given: A circle with centre O and a tangent AB at a point P of the circle. O To prove: OP⊥AB R Construction: Take a point Q on AB. Join OQ. A P Q B Proof:Q is a point on the tangent AB, other than the point of contact P.  Q lies outside the circle. Let OQ intersect the circle at R. Then, OR < OQ But, OP = OR From (i) and (ii)  OP < OQ OP is the shortest distance between the point O and the line AB. But, the shortest distance between a point and a line is the perpendicular distance.  OP⊥AB. ..... (i) ..... (ii) [radii of the same circle]. For example: https://www.entrancei.com/ncert-solutions-class-10-maths

3. https://www.entrancei.com/ In the figure O is the centre of a circle; PA and PB is the pair of tangents drawn to the circle from point P outside the circle; If AOB = 117º, find . In quadrilateral AOBP, A = B = 90º [Radius is ⊥ to tangent at the point of contact]  AOB + APB = 180º  117º +  = 180º   = 180º – 117º = 63º   = 63º A  P O 117º B [ sum of angles of quad. is 360º] THEOREM 2: The lengths of tangents drawn from an external point to a circle are equal. Given: Two tangents AP and AQ are drawn from a point A to a circle with centre O. To prove: AP = AQ Construction: Join OP, OQ and OA. Proof: AP is a tangent at P and OP is the radius through P.  OP ⊥ AP Similarly, OQ⊥AQ In the right OPA and OQA, we have P A O Q OP = OQ [radii of the same circle] OA = OA [common] OPA = OQA [both 90º] OPAOQA [by RHS-congruence] Hence, AP = AQ. [cpct] For example: In figure PA and PB are tangents to the circle drawn from an external point P. CD is another tangent touching the circle at Q. If PA = PB = 12 cm and QD = 3 cm, find the length of PD. Point D outside the circle. DQ and DB is the pair of tangents drawn to the circle from the point D.  DB = DQ  DB = 3 cm cm 3 ( = = DQ QD  Now, PD + DB = PB  PD + 3 cm = 12 cm  PD = 9 cm For example: A C P Q D B is given) = ( PB 12 cm is given)  A quadrilateral ABCD is drawn so that D = 90º, BC = 38 cm and CD = 25 cm. A circle is inscribed in the quadrilateral and it touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If BP = 27 cm, find the radius of the inscribed circle. https://www.entrancei.com/ncert-solutions-class-10-maths

4. https://www.entrancei.com/ Let O be the centre and r be the radius of the inscribed circle. Join OR and OS. Here, D = 90º 25 cm R ORD = OSD = 90º D C OR = OS r Q Therefore, ORDS is a square. S O r  [1] BP and BQ are tangents to the circle from point P. RD = SD = r ..... (i) 38 cm A P 27 cm  BQ = BP B  BQ = 27 cm Then CQ = BC–BQ = 38 cm – 27 cm.  CQ = 11 cm  CR = 11 cm ..... (ii) Now, RD + CR = CD  r + 11 cm = 25 cm By (i) and (ii)  r = 14 cm. GEOMETRICAL CONSTRUCTIONS 1. 2. The perpendicular bisector of a chord passes through the centre. The tangent of a circle is perpendicular to the line joining the centre and the point of contact. 3. The similar triangles or quadrilaterals have their corresponding angles equal and corresponding sides in same ratio. The point of concurrence of bisectors of angles of a triangle is called the incentre of the circle inscribed in the triangle. 4. 5. The point of concurrence of perpendicular bisectors of a triangle is called the circumcentre of the circle circumscribed through the vertices of the triangle. 6. The lengths of two tangents drawn from an external point to a circle are equal. For Example: Construct one isosceles triangle whose base is 8 cm and altitude 4 cm and then 1 1 construct another triangle whose sides are times the corresponding sides of the 2 isosceles triangle. https://www.entrancei.com/ncert-solutions-class-10-maths

5. https://www.entrancei.com/ Steps of Construction: 1. Take BC = 8 cm. A A 2. Draw DA, the right-bisector of BC and equal to 4 cm. 4 cm 3. Join BA and CA. ABC is the isosceles triangle of given data. 8 cm B C D C X 4. Let-any acute angle be CBX (< 90). 5. Mark the three points B1, B2 and B3 such that BB1 = B1B2 = B2B3· B1 B2 B3 X 6. Joint B2C. Draw a line through B3 parallel to B2C which meets BC extended at C'. 7. Through C' draw C'A' parallel to CA which meets BA produced at A'. 3 times the corresponding sides Then A'BC is the required triangle whose sides are 2 of the given triangle For example: Draw a line segment AB of length 6 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle. Steps of construction: 1. Draw a line segment AB = 6 cm. 2. With centre A and radius = 3 cm, draw a circle. 3. With centre B and radius = 2 cm, draw another circle. R P 4. With M, the midpoint of AB, as centre and radius AM = MB, draw the circle intersecting the circle of radius 3 cm at R and S and intersecting the circle of radius 2 cm at P and Q. B A 2 cm 3 cm M Q 5. Join AP, AQ, BR and BS. S Then AP and AQ are tangents to the circle with centre B and radius 2 cm from point A BR and BS are tangents to the circle with centre A and radius 3 cm from point B. https://www.entrancei.com/ncert-solutions-class-10-maths

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