IS X MT UR LT. SUJT : MTHMTIS Geometry ST U NSWRSHT 003 1. In QL and RM, LQ MR [Given] LQ RM [Given] QL ~ RM [y axiom of similarity] (i) Since, QL ~ RM QL M L RM QL RM L M (ii) In QL and RQ, we have Q L M R Q Q [ommon angle] QL RM [Given] QL ~ RQ [y axiom of similarity] Then, Q QR Q L Q Q QRQL Hence proved. From the given figure, x 30º 30º [ngle in same segment] lso, 90º [ngle in a semi-circle is 90º] In, we have + + 180º [Sum of all angles in a triangle is 180º] x + 90º + 30º 180º x + 10º 180º x 180º 10º 60º Hence, the value of x is 60º 30
MT UR LT. 004 X - IS (Maths) roof : 1. Given : In a figure, 30º 30º To prove : 90º [ngle in a semicircle] In, + + 180º + 90º + 30º 180º [ is a diameter of a circle and is in semi-circle, so 90º] 180º 10º 60º [ngles in alternate segment] 30º In, is an exterior angle + 60º 30º + 60 30 30º... (ii) In, 30º [From equation (i) and (ii)] [ sides opposite to equal angles are equal] Hence proved. Given : (i) To prove : ~ roof : In and, we have [ommon angles] [orresponding angles as ] ~ [y axiom of similarity] Hence proved. (ii) Since, ~ + [If two triangles are similar, then their corresponding sides are proportional] [ + ] 31
X - IS (Maths) 1 + 1 + 1 1 + 1 1 1, given MT UR LT. 1 + 4.5 4.5 3 1.5 cm ~ r(δ) r(δ) Let and 1 3 1 : 9 r() k r() 9k r(δ) r(trp. ) r(δ) r(δ) r() k 9k k k 8 k 1 8 1 : 8 005 1. Given, 9 cm 3 cm and 4 cm In and, 90º [ and, given] and [ommon angle] ~ [y axiom of similarity] 3 [ If two triangles are similar, then their corresponding sides are proportional]
MT UR LT. X - IS (Maths) 9 3 4 4 3 8 cm 9 Now, 4 8 16 cm. Let 0 be the centre and r be the radius of a circle. Join M, N and S. S We know that the tangent at any point of a r N circle and the radius through the point of contact are perpendicular to each other. Q R M MQ NQ 90º 4 cm lso, NQ QM [Tangents drawn from external point to a circle are equal in length] NQ QM M N r [Say] Thus, MQN is a square. We know that, tangents drawn from an external point are equal in lenngths. N S Q QN S 3 r S... (i) and MR SR QR QM SR 4 r SR n adding equation (i) and (ii), we get 3 r + 4 r S + SR 7 r R 5 [In right QR, R Q + QR (ythagoras theorem) 3 4 9 16 5 5 cm] 7 5 r r r 1 cm [Radius of incircle] 3. Given : In the given figure, M To prove : M (i) M is isosceles. (ii). M roof : (i) Since M [Given] M M... (i) [ngles opposite to equal sides are equal] Since, angle between the tangent and chord is equal to the angle subtended 3 cm 33
X - IS (Maths) by same chord in alternate segment. M M M M M or M M M M Hence, M is an isosceles triangle. (ii) Since, M is tangent to the circle and is a secant. M M [M M] Hence proved MT UR LT. 4. To prove : onstruction : Join and. roof : In and, 90º [ ngle in semi-circle is right angle] and [ommon angle] ~ [y axiom of similarity] Then, Hence proved [In similar triangles sides are proportional] [ ] 5. X 3.5 cm M 10º 6 cm N n measuring, we get 30º 34
MT UR LT. 6 ngle at centre is twice the angle at circumference. x ut 160º [Given] 160º x x 160º 80º lso, xº + yº 180º 80º + y 180º y 180º 80º yº 100º Now, 3y x 3 100º 80º 300º 160º 140º Hence proved X - IS (Maths) y 160º x 006 1. Given, 65º, 70º and 45º (i) Since, is a cyclic quadrilateral. So, the sum of opposite angles of a cyclic quadrilateral is 180º. + 180º 65º + 180º 180º 65º 115º (ii) In, + + 180º 65º + 70º + 180º 180º 135º 45º + 45º + 45º 90º i.e. is in semi-circle. So is a diameter. 45º 65º 70º. (i) Given, is a diameter of circle. 90º [ ngle in a semi-circle is 90º] In, + + 180º 90º + + 34º 180º 180º 14º 34º 56º (ii) Since, Q is a tangent to the circle. Q 34º [lternate segment property] Since, Q is a straight line. + Q 180º Q 35
X - IS (Maths) MT UR LT. 56º + Q 180º Q 180º 56º 14º Now in Q, Q + Q + Q 180º 14º + Q + 34º 180º Q Q º 3. Since, RT is a tangent at R and QR is a chord. S QR RSQ yº [lternate segment property] Since, SQ is a diameter of a circle. QRS 90º [ngle in semi-circle] In RS, SR + RS + RS 180º xº + (yº + 90º) + yº 180º xº + yº 180º 90º xº + yº 90º Hence proved Q xº yº R T [Sum of all angles of a triangle is 180º] 4. Given, 6 cm, Q 9 cm and ar() 10 cm In Q and, we have Q 90º [ and Q ] and Q [Vertically opposite angles] Q ~ [y axiom of similarity] Then, ar( Q) ar( ) ar(q) (Q) () 9 6 10 8110 36 70 cm Hence, the area of Q is 70 cm 36
MT UR LT. X - IS (Maths) 5. X 5 cm cm 60º cm 4. cm 007 1. X 5.8 cm 60º 6.4 cm Radius 1.5 cm. Given, T 16 cm and 1 cm We know that, chord and tangent at meet at point T, then T T T 16 (T ) T 16 (16 1) T T 16 4 T 64 T 8 cm 3. Given 45º alculate the value of Q Since, is a diameter of circle, T 37
X - IS (Maths) MT UR LT. So is in semi-circle. 90º In right, + + 180º 90º + 45º + 180º 45º 135º + 180º 180º 135º 45º Q Q 45º [ngles in the same segment are equal] Hence, the value of Q is 45º 4. In, + + 180º + 6º + 43º 180º 180º 105º 75º Since, points,, and are the points on a circle, so quadrilateral is a cyclic quadrilateral. We know that, sum of opposite angles is 180º [ Sum of all the angles of a triangle is 180º] 6º + 180º a + 75º 180º [ 75º] a 105º F [xterior angle property] c 6º In F, F + F + F 180º 105º + b + 6º 180º b 180º (105º + 6º) 180º (167º) 13º Hence, a 105º, b 13º and c 6º a 43º c [y angle sum property of a triangle] b F 5. Given : and 3 (i) Now, 3 3 3 3 5 In and, we get F [ommon angle] 38
MT UR LT. X - IS (Maths) and [orresponding angles as ] ~ [y axiom of similarity] Then, [ If two triangles are similar, then their corresponding sides are proportional] 3 5... (i) (ii) To prove : F ~ F roof : In F and F, we have F F [lternate interior angles as ] and F F [Vertically opposite angles] F ~ F [y axiom of similarity] Then, (iii) Since, F F F F 3 5 F ~ F ar( F) ar( F) [ If two triangles are similar, then their corresponding sides are proportional] [From equation (i)] 3 5 9 5 Hence, ar(f) : ar(f) 9 : 5 [From equation (i)] 6. Q R X (ii) 9.4 cm 4 cm 45º 5 cm M 39
X - IS (Maths) MT UR LT. 008 1. (i) area : 4 : 5 (ii) area : Q 4 : 9. (i) Q 11º (ii) Q 68º 3. 1 cm 4. X 8 cm Q Y F XY is a square 009 1. (i) 75º (ii) 15º (iii) 105º. (i) 4 m (ii) 1000 cm 3 40
MT UR LT. X - IS (Maths) 3. X 5.5 cm 3.4 cm Y 4.9 cm 4. Given : 7 cm and 9 cm 7 cm 9 cm (i) To prove : ~ roof : In and, [ommon angles] [nlges in alternate segment] ~ [y axiom of similarity] (ii) Since, chord and tangent at intersect each other at point. 16 9 144 (1) 1 cm 5. Given : and F : 5 : 8 F (i) To prove : F ~ F roof : In F and F, F F F F [Vertically opposite angles] [lternate angles, since ] F ~ F [y axiom of similarity] (ii) Given : 6 cm Since, F ~ F F F F 5 8 8 1 1 8 F 5 F 5 F 8 5 F 3 F 5 F 5 F 5 F 3 41
X - IS (Maths) MT UR LT. 6 5 3 5 6 3 5 10 cm (iii) Given, F F [orresponding angles] and F [ommon angle] F ~ Then, ar( F) ar( ) (F) () 5 64 ar(f) : ar() 5 : 64 5 8 F 5 8 010 1. 40º. (i) 1 cm (ii) rea of 36 cm 3. 3.5 cm 60º 6 cm (iv) 5. cm 4
MT UR LT. X - IS (Maths) 4. 4 cm 011 1. (i) In and, 8 cm [both 90º] [common angle] 18 cm [ similarity criterion] In and, [both 90º] [common angle] [ similarity criterion] If two triangles are similar to one triangle, then the two triangles are similar to each other. or (ii) Since the corresponding sides of similar triangles are proportional. x 18 x 8 1 cm (iii) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. So, r( ) r( ) 144 64 9 4 Thus, the required ratio is 9:4. 43
X - IS (Maths) MT UR LT.. Q 6 cm 3.5 cm M R Q 4.8 cm 3. In, 30º, [Given] 90º [radius is prependicular to the tangent at the point of contact] y angle sum property, + + 180 0 30º 180 0 (90 0 + 30 0 ) 60 0 onsider and [radii] [tangents to a circle from an external point are equal in length] [ommon]. (i) 30º (ii) 60º + 10º (iii) So, 1 44 1 10º 60º 4. Let the radii of the circles with,, and as centres be r 1, r and r 3 respectively. ccording to the given information, 10 cm r 1 + r...(1)
MT UR LT. X - IS (Maths) 8 cm r + r 3...() 6 cm r 1 + r 3...(3) dding equations (1), () and (3), (r 1 + r + r 3 ) 4 r 1 + r + r 3 1...(4) Subtracting (1) from (4), Subtracting () from (4), Subtracting (3) from (4), r 3 1 10 cm r 1 1 8 4 cm r 1 6 6 cm Thus, the radii of the three circles with centres, and are 4 cm, 6 cm and cm respectively. R Q 01 1. 11.5 cm.. (i) 50º (ii) 40º 3. In and M (i) M [each 90 0 ] M [common] M [y similarity] Since the triangles are similar, we have (ii) Taking, M M M 1 10 15 1 10 15 1 10 8 cm 15 Now using ythangoras theorem in triangle, + 10 8 36 6 cm Hence, 6 cm and 8 cm M 45
X - IS (Maths) MT UR LT. 4. 5. 3 cm 013 1. (i) In, + + 180 0 65 0 + 70 0 + 180 0 180 0 70 0 65 0 45 0 Now, + 45 0 + 45 0 90 0 is the angle of semi-circle so is a diameter of the circle. (ii) (angle subtended by the same segment) 45 0 45 0 65 0 70 0. (i) 80º (ii) T 60º 100º 40º T 46
MT UR LT. X - IS (Maths) 3. X 3.5 cm 10º 6 cm n measuring 30º 4. (i) In and, 90º (perpendiculars to ) (ommon) ( criterion) (ii) Since, 6 15 4 6 60 60 6 10 cm (iii) ~ ar() : ar() : 6 : 4 36 : 16 9 : 4 014 1. (i) 3º (ii) 148º (iii) 3º. onsider the given triangle 47
X - IS (Maths) MT UR LT. Given that (i) onsider the triangles and [given] [common] y Similarity, (ii) Hence the corresponding sides are proportional 8 5 4 4 5 5 cm.5 cm 8 8 5 4 8 4 5 3 5 cm 6.4 cm (iii) We need to find the ratios of the area of the triangles and. Since the triangles and are similar triangles, we have Thus, rea(δ) rea(δ) So, 3. (i) 1 cm (ii) 8 cm rea( ) rea(δ) rea(δ) rea(δ) 5 5 : 64 64 5 8 48
MT UR LT. X - IS (Maths) 4. 5.5 cm 5 cm 6.5 cm N Radius 1.5 cm 015 1. (i) x 5º (ii) y 50º (iii) z 40º. 5 cm F x x 3. onstruction : Join and. In and [ngles in the same segment] 49
X - IS (Maths) MT UR LT. [ngles in the same segment] [y ostulate] [orresponding sides of similar triangles] 4. (i) onsider and [ommon] m m 90 0. [ postulate] (ii) onsider ~... (i)[orresponding sides of similar triangles] onsider y applying ythagoras Theorem, we have + + 5 13 1 cm From equation (1), we have 4 1 13 5 1 3 13 13 3 cm 4 lso, 1 5 0 5 cm 1 3 (iii) We need to find the area of and quadritateral. 4 cm 13 cm 1 cm 9 cm 5 cm 50 rea of 1
MT UR LT. X - IS (Maths) 1 4 5 3 10 cm 3 rea of quad. rea of rea of 1 10 3 1 5 1-10 3 10 30 3 90 10 3 80 3 cm 5. Thus ratio of areas of to quadrilateral 10 3 80 3 1 8 6 cm M T N 105º 5.5 cm R Q 016 1. (i) 3º (ii) 64º (iii) 58º 51
X - IS (Maths) MT UR LT.. (i) ( 4, 4) ( 3, 0) (0, 3) (ii) The figure is an irregular hexagon (or arrow head). 3. X L N M Y F 5 cm S 4. (i) Since QRS is a cyclic quadrilateral, R 5 Q RS RQT... (i)[xterior angle property] In TS and TRQ, TS RTQ [ommon angle] RS RQT [From (i)] TS ~ TRQ [ similarity criterion] (ii) Since TS ~ TRQ, S QR T TR T
MT UR LT. X - IS (Maths) S 4 18 6 S 18 4 6 S 1 m (iii) Since TS ~ TRQ, ar( TS) ar( TRQ) 7 ar( TRQ) S RQ ar(trq) 7 9 ar(trq) 3 cm ar(qrs) ar(ts) ar(trq) ar(qrs) 7 3 ar(qrs) 4 cm 1 4 5..5 cm.5 cm M 3 cm 5 cm Q 6. (i) 600 m (ii) m 7. (iii) 175500000 m 3 (iv) XY 5 cm 53
X - IS (Maths) MT UR LT. 017 1. Steps of construction : (i) raw line 7 cm and 60º. ut off 5 cm. Join. is the required triangle. N (ii) raw angle bisectors of and. M (iii) isector of meets at M and bisector of meets at N. 60º (iv) is the point which is 7 cm equidistant from, and. (v) raw a perpendicular from point to and let it intersect at point (vi) With as the radius, draw a circle touching the three sides of the triangle (incircle.) 5 cm. (i) Q 30º Since is the bisector of Q Q 30º is the bisector of and Q + + Q 180º + + 60º 180º 180º 60º 60º So, + 60º + 30º 90º So, is the diameter of the circle. Q (ii) Q (lternate segment theorem) 30º In, 30º ( is the bisector of Q) (Sides opposite to equal angles are equal) is an isosceles triangle. 3. (i) 70º Since is cyclic quadrilateral [xterior angle property] 70º 54
MT UR LT. X - IS (Maths) (ii) [ngle at centre is twice the angle at the circumference] (70º) 140º (iii) In, y ngle Sum property, + + 180º + 140º 180º 40º 0º 4. (i) In QR and SR, we have QR SR (Given) RQ RS (ommon) QR ~ SR (.. xiom) Q S QR R R SR... (i) 8 cm 6 cm (ii) Now, QR R R SR QR 6 6 3 QR 6 6 3 1 cm Q... [From (i)] S 3 cm R lso, Q S 8 S 8 S R SR 6 3 S 8 4 cm...[from (i)] (iii) rea of QR rea of SR Q S 8 4 64 16 4 55