# MT EDUCARE LTD. MATHEMATICS SUBJECT : Q L M ICSE X. Geometry STEP UP ANSWERSHEET

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1 IS X MT UR LT. SUJT : MTHMTIS Geometry ST U NSWRSHT 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 º [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

2 MT UR LT. 004 X - IS (Maths) roof : 1. Given : In a figure, 30º 30º To prove : 90º [ngle in a semicircle] In, º + 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 º... (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

3 X - IS (Maths) , given MT UR LT cm ~ r(δ) r(δ) Let and : 9 r() k r() 9k r(δ) r(trp. ) r(δ) r(δ) r() k 9k k k 8 k : 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]

4 MT UR LT. X - IS (Maths) cm 9 Now, 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) 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

5 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

6 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 Given, 65º, 70º and 45º (i) Since, is a cyclic quadrilateral. So, the sum of opposite angles of a cyclic quadrilateral is 180º º 65º + 180º 180º 65º 115º (ii) In, º 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, º 90º º 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

7 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) () cm Hence, the area of Q is 70 cm 36

8 MT UR LT. X - IS (Maths) 5. X 5 cm cm 60º cm 4. cm 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

9 X - IS (Maths) MT UR LT. So is in semi-circle. 90º In right, º 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, º + 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, In and, we get F [ommon angle] 38

10 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] (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)] 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

11 X - IS (Maths) MT UR LT (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 (i) 75º (ii) 15º (iii) 105º. (i) 4 m (ii) 1000 cm 3 40

12 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 (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 F 5 F 5 F 8 5 F 3 F 5 F 5 F 5 F 3 41

13 X - IS (Maths) MT UR LT cm (iii) Given, F F [orresponding angles] and F [ommon angle] F ~ Then, ar( F) ar( ) (F) () 5 64 ar(f) : ar() 5 : F º. (i) 1 cm (ii) rea of 36 cm cm 60º 6 cm (iv) 5. cm 4

14 MT UR LT. X - IS (Maths) 4. 4 cm (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( ) Thus, the required ratio is 9:4. 43

15 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, º ( ) 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, º 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)

16 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 (4) Subtracting (1) from (4), Subtracting () from (4), Subtracting (3) from (4), r cm r cm r cm Thus, the radii of the three circles with centres, and are 4 cm, 6 cm and cm respectively. R Q 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 cm 15 Now using ythangoras theorem in triangle, cm Hence, 6 cm and 8 cm M 45

17 X - IS (Maths) MT UR LT cm (i) In, Now, is the angle of semi-circle so is a diameter of the circle. (ii) (angle subtended by the same segment) (i) 80º (ii) T 60º 100º 40º T 46

18 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, cm (iii) ~ ar() : ar() : 6 : 4 36 : 16 9 : (i) 3º (ii) 148º (iii) 3º. onsider the given triangle 47

19 X - IS (Maths) MT UR LT. Given that (i) onsider the triangles and [given] [common] y Similarity, (ii) Hence the corresponding sides are proportional cm.5 cm 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 :

20 MT UR LT. X - IS (Maths) cm 5 cm 6.5 cm N Radius 1.5 cm (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

21 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 [ postulate] (ii) onsider ~... (i)[orresponding sides of similar triangles] onsider y applying ythagoras Theorem, we have cm From equation (1), we have cm 4 lso, 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

22 MT UR LT. X - IS (Maths) cm 3 rea of quad. rea of rea of cm 5. Thus ratio of areas of to quadrilateral cm M T N 105º 5.5 cm R Q (i) 3º (ii) 64º (iii) 58º 51

23 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

24 MT UR LT. X - IS (Maths) S S 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 cm.5 cm M 3 cm 5 cm Q 6. (i) 600 m (ii) m 7. (iii) m 3 (iv) XY 5 cm 53

25 X - IS (Maths) MT UR LT 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 º 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

26 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, º + 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 QR 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

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