1. (d) Other integer is = 6 2 = (b) 7 ( 15) = = Let the other integer be a. Then 9 + a = 11 a = ( 11) (+ 9) = 11 + ( 9)

Transcription

1 INTEGERS WORKSHEET-. (d) Other integer is (b) 7 ( 5) Let the other integer be a. Then 9 + a a ( ) (+ 9) + ( 9) 0 4. Let other integer be x, then according to the question 9 + x 5 x We can have ( ) + ( 5) 7 and 5 may be the required pair. It may have other ns. 6. (a) Since 5 is odd, so the sign of product of 5 negative integers is negative. (b) Since product of integers is an integer and product of integers and zero is zero. product 0. (c) Sign of product of negative integers is positive, because is even. 7. ( ) 5 5 ( ) 4 4 [( 5) ( )] ( ) [( 4) ( )] ( ) [( ) ( )] ( ) [( ) ( )] ( ) 0 0 [( ) ( )] ( ) ( ) [0 ( )] 8. (i) West 0 km C B 0 km Distance towards East denoted (+ve) sign, Distance towards West denoted ( ve) sign. Since first Rohan goes toward East 0 km So, distance covered by him + 0 km gain, he goes towards West. So, distance covered by him on same road 0 km. So, position of Rohan from km (ii) ddition of Integers. (iii) rolling stone can gather no mass. East P-

2 INTEGERS WORKSHEET-. (b) x ( 8) 6 x 8 6 x ( 8) ( 6) 08. (a) ( 7) ( 7) + ( 7) ( ) We have a ( 50) a + 5 ( ) a 5 ( ) + ( 5) 8 hence a 8 4. (a) ( 0) 0 (b) ( 6) ( 9) (a) ( 8) ( ) 4 (integer) (b) ( 8), which is not an integer (a) 6 ( 48) + ( 6) ( 48) [a ( b)] (a b) ( a) ( b) (a b)] [ (6 48)] + (6 48) ( 48) Or 6 ( 48) + ( 6) ( 48) [6 + ( 6)] ( 48) ( 0) ( 48), by distributive property 480 (b) 5 ( 5) + ( 4) ( 5) [5 + ( 4)] ( 5), by distributive property () ( 5) 75 (c) 7 (50 ) (7 50) (7 ) (distributive property of multiplication over subtraction) 7. L.H.S. ( 0) [ + ( )] ( 0) ( ) ( 0) 0 00 P- M T H E M T I C S -- VII

3 R.H.S [( 0) ] + [( 0) ( )] ( 90) L.H.S. R.H.S., which is verified. 8. (a) Score of Mohan 6 (5) + 4 ( ) 0 + ( 8) 0 8 (b) Rekha s score 5 (5) + ( ) ( 6) (c) Mayank s score 5 (5) + 5 ( ) 5 + ( 0) (d) Jeet s score (5) + 5 ( ) 5 + ( 0) P-

4 INTEGERS WORKSHEET-. ( 5) (8) + ( 5) The additive inverse of Consider the integers ( 5) and ( ). Then ( 5) ( ) ( 5) + Clearly > 5 but is not greater than. 4. (a) [( ) + ] [ + ] ( ) (b) 0 ( ) (a) Since sign of product of odd numbers and negative integers will be negative. Sign of product of 95 negative integers will be negative. gain when even number of ( ve) integers are multiplied, sign of product will be (+ ve). So, when ( ve) sign integer (+ ve) sign integer ( ve) sign integer. (b) 99 ( ) + ( 0) ( 0) (a) We have to subtract 6 from ( ) Since ( ) (b) We have, ( 666) ( ) ( 666) Since temperature on Monday 5ºC Temperature on Tuesday 5ºC ºC 7ºC ccording to the question, temperature on Wednesday 7ºC + 4ºC ºC. 8. (a) Profit on bags of white cement Rs Loss on grey cement bags 5000 ( 5) Rs Income on total transaction Rs. 000 Hence, loss Rs (b) Company sold 6400 bags of grey cement. So, loss Rs. 000 Now this loss is to be fullfilled by the profit on bags of white cement. So, number of bags of white cement Loss Profit per bag So, company has to sell 4000 bags of white cement. P-4 M T H E M T I C S -- VII

5 INTEGERS WORKSHEET-4 Formative ssessment [] nswer the following : , 4., 5.. [B] Match the following : 4 4. (ii). (iii). (iv) 4. (i). [C] Choose the correct nswers : 5 5. b. b. a 4. c 5. c [D] Fill in the Blanks : [E] True / False : 6. T. F. F 4. T 5. T 6. F P-5

6 FRCTIONS ND DECIMLS WORKSHEET-5. (b). (c). cm of 0 and and (a) (b) of (a) 4 of (b) 4 of (a) Or (b) Or P-6 M T H E M T I C S -- VII

7 (c) (a) (b) Since the number of boys of total number of students Number of girls 5 6 Hence, the number of girls th of the total number of students 6 6 th of the total number of students 40 Total number of students 40 / Total number of students 440 Number of boys P-7

8 FRCTIONS ND DECIMLS WORKSHEET-6. (b). (c). Required number to be added (a) (b) (a) (b) (a) (b) (c) For rectangle, rea Length Breadth Since, Length 5 m P-8 M T H E M T I C S -- VII

9 Breadth 0 5 rea m 5 5 m m 0 75 m 8. Distance covered by Dinesh km 7 5 km B 7 km 9 km D 8 km Distance covered by yub km Clearly yub covered more distance which is 0 km. C P-9

10 FRCTIONS ND DECIMLS WORKSHEET-7. (b). (b). It is given that 8 x 4 8x 4 x (a) (b) Cost of pen 8.50 Cost of 48 pens Hence, Cost of 48 pens ` (a) (b) (c) (a) Place value of 5 in 0 5 Place value of in (b) 6 of 4 P-0 M T H E M T I C S -- VII

11 6 of Total money ` students receive money 9. 6 of total money ` 500 Rest money ` ( ) ` 500 Rest student Rest money equally distributed ` 50 P-

12 FRCTIONS ND DECIMLS WORKSHEET-8 Formative ssessment [] nswer the following : ` 5. ` km [B] Match the following : 5 5. (iv). (i). (ii) 4. (iii) 5. (v) [D] Fill in the Blanks : [E] True / False : 0 5. F. F. T 4. F 5. F 6. F 7. T 8. T 9. F 0. T P- M T H E M T I C S -- VII

13 DT HNDLING WORKSHEET-9. (d) years, as range 54 years.. (d), 5 both. (a) Mean (b) Range Mean Total of data Number of data (a) Since data is as follows :, 6, 5,, 0,,, 4, 5,, 4 5 and both occur maximum two times. So, the mode of given data is and 5 both. (b) Given data :, 4, 6,, 4, 4, 6, 4, 0, 4, 8, 4 In the given data frequency of 4 is 6, which is maximum, so mode is Frequency distribution table : Number of shirts sold Tally marks Frequency Total 9 P-

14 7. Marks No. of students f x (x) (f) Total N 8 fx 8480 fx Mean N (i) Mean (ii) Mean of given data. (iii) Girl child has a equal right to get proper education. P-4 M T H E M T I C S -- VII

15 DT HNDLING WORKSHEET-0. (a) 0 and.. (c) 65.. (a) 4, (b) 4, as 4 occurs maximum times. 4. Since given data is , 48, 56, 75, 76, 8, 85, 85, 90, 95 In the given data Maximum 95 Minimum 9 Range Maximum Minimum Mean Rahul score runs in per inning are as follows : 4, 7, 47, 49, 54, 6 Mean Sum of the data Number of data Thus, the mean runs scored in an innings is (a) First four counting numbers are :,,, 4 Mean (b) rranging the number with same values together, we get,,,,,,,,, 4, 5, 6, 9 In this data maximum frequency of is 6. mode is. P-5

16 (c) Mean In the given question, Number of maximum outcomes 5 (s there are five flash cards) Favourable outcome, a flash card bearing Hence, probability Favourable outcome Total number of outcomes Marks obtained English st term Hindi Maths nd term Subjects Double bar graph (i) Student improve his performance most in Math. (ii) Student improve his performance least in S. Science. (iii) Yes, in Hindi his performance goes down. S. Science Science P-6 M T H E M T I C S -- VII

17 DT HNDLING WORKSHEET-. (c) 5. (b) p p 5 p (i) Here, the data is as follows :, 5, 6, 7, 8, 8, 8, 40, 4, 4, 4, 4, 45, 47, 50 Since the frequency of 4 and 8 is same, which is maximum, i.e.,. Hence, 8 and 4 are two modes. (ii) Yes, there are modes x Mean x x 480 x In the given graph : (a) Most popular pet is cat. (Because the bar representing cat is the tallest) (b) Number of students which has dog as a pet is rranging the observations (marks) in ascending order :,, 5, 5, 5, 8, 8, 8, 8,,,,, 6, 6, 6, 7, 7, 8, 8, 8, 8, 8, 9, 0 Sum of observations Total 4 Number of students 5 P-7

18 Mean marks Sum of the observations Total number of students So, the number of students who scored marks more than mean marks are Frequency Distribution Table : Numbers Tally marks Frequency Total (i) Range (ii) Mode 8, because 8 is repeated maximum 5 times. P-8 M T H E M T I C S -- VII

19 DT HNDLING WORKSHEET- Formative ssessment [] nswer the following : Maximum observation Minimum observation. How many times an observation occurs in data. 4. Median 5. Mode. [B] Match the column :. (ii). (iii). (i). [C] Fill in the blanks : 8 8. Numerical. Original. rray 4. Frequency 5. Tabulation [D] True / False : 4 4. True. False. True 4. False P-9

20 4 SIMPLE EQUTIONS. (d) 6 WORKSHEET- 4. (d). Let the required number be x Then, x x x 6 Hence, the required number is. x 4. x + 4 x x 4 + [On transposing x to L.H.S. and to R.H.S.] x x 6 5 x x 5 6 x Let consecutive numbers be x and x +. Hence, according to question, x + x + 5 x + 5 x 5 On transposing to R.H.S.) x 5 x 5 x 6 Smaller number is Since, x x + x (On transposing to L.H.S.) x x P-0 M T H E M T I C S -- VII

21 7. Let the number be x. ccording to question, (x ) ( x ) x x x x + x + x x. x x 0 x x 0 x x 0 x 0 x (i) Let ages of and B are 5x and x. fter six years their ages will be (5x + 6) and (x + 6) So, according to question, 5x 6 x (5x + 6) 7 (x + 6) 5x + 0 x + 4 5x x 4 0 4x ; x So the present age of 5x 5 5 years and the present age of B x 9 years. (ii) Simple equation. (iii) Time waits for none. P-

22 4 SIMPLE EQUTIONS. (a) n + 5 (n 0) n + 5 9n 0 7n 5 WORKSHEET-4 n (a) x x x x. x + 5 x x x + x x + x 5x 5 x 5x + 0 5x 5x + 5x 0 0x 9 x y + 4 y + y y 5 y y 5. Since, x (a) 5 0 x 0 5 (On transposing 5 to R.H.S.) P- M T H E M T I C S -- VII

23 x 5 x 5 x 0 (b) x 5 7 x Since, x x x 4. (On transposing 5 to R.H.S.) x 6x 5 5(x ) (6x ) 5 5 0x 5 (8 x 6) 5 5 0x 5 8x 6 5 8x 5 8x x 5 4 x Let the lowest marks be x, then the highest marks x + 7 x x 87 7 x 80 x 80 x 40 Hence, the lowest score is Let the number be x. ccording to question, 7 times of x x x x 0 7x x 0 Hence, the number is 0. P-

24 4 SIMPLE EQUTIONS WORKSHEET-5. (c) x x 7 9 9x 9 7x + 7 x x 6 8. (c) 8(x 5) 6 (x 7) 6x 40 8x + 4 x x. Let the number be x Then 4 x x x 5x 0 4 x Hence required number be Let the number be x. ccording to question, 6x x 40 0 (On transposing 0 to R.H.S.) 6x 0 6x x x x x 0.8x x 0.76 x (7x ) x x 4 x x x x x 4 4 P-4 M T H E M T I C S -- VII

25 x x x x 4 4 x x x x 4 4 x x x 4 4 x 4 x x 4 4 7x x 7 x 7. x x 4 x (x ) 5( x ) 4 x x 5x 5 4 x x 5x 5 4 x 0 7 x 4 x (x + ) 0 (4x + ) 9x + 80x + 0 9x 80x 0 x x 8. 5 Quantity of pure alcohol in 400 ml of solution 5% ml. Now, we add x ml of pure alcohol to the sample. So, total pure alcohol (60 + x) ml. But volume of new sample (400 + x) ml. Percentage of pure alcohol in new sample (60 x ) (400 x) 00. which is equal to % 60 x 400 x x 400 x (60 + x) (400 + x) 00x x x x x 6800 x ml. P-5

26 4 SIMPLE EQUTIONS WORKSHEET-6 Formative ssessment [] nswer the following : (i) 5y (ii) y z (iii) ab. 6xy 6x 6y x 6y + xy 5. 0a 4 b 6 c 4 6. a + 5a 0 7. x x 9., years, years [B] Fill in the blanks : 5 5. Monomial. Binomial. Trinomial 4. 5b 5. constant [C] True / False : 5 5. F. F. F 4. T 5. T P-6 M T H E M T I C S -- VII

27 5 LINES ND NGLES WORKSHEET-7. (b) 90º.. (a) 40º.. Given angle measures 5º Let its supplement be xº. Then x + 5º 80º x 80º 5º 55º Hence, the supplement of the given angle measures 55º D B G C E F (a) Given, BC 70º Since B DG and BG is a transversal BC DGC (corresponding angles) DGC 70º (b) gain BC EF and GE is a transversal DGC DEF (corresponding angles) DEF 70º 5. Let the angle be x, Its complement 90º x ccording to question, x 90º x x + x 90º x 90º x 90 x 45º. 6. ( x + 8) ( x ) In the given figure, OB and COB are the angles of linear pair. O B C P-7

28 7. OB + COB 80 (x + 8 ) + (x ) 80 x x x OB (x + 8) ( ) (6 + 8) 4. t 4 l 5 6 m 8 7 Since l m and t is transversal and 0º 0º (vertically opposite angles) + 80º (linear pair) 0º + 80º 80º 0º 50º Similarly, 4 (vertically opposite angles) 4 50º gain, 5 (corresponding angles) So, 5 0º 5 7 (vertically opposite angles) 7 0º Further, 4 6 (alternate interior angles) 6 50º and º (vertically opposite angles) 8. Let O be the initial position of the man. Let he cover O 4 m due east and then B 0 m due north. Finally, he reaches the point B. Join B. N Now, in right OB, we have OB (O + B B ) {(4) + (0) }m 0m ( ) m 676 m W O 4m E OB 676 m 6 m S Hence, the man is at a distance of 6 m from his initial position. P-8 M T H E M T I C S -- VII

29 5 LINES ND NGLES WORKSHEET-8. (c) 0º.. (b) x 5º.. Let given angle x then x + x 80º 5x 80º x 80º 5 6º So, smaller angle x 6 7º 4. In the given figure, 0º Here, (vertically opposite angles) 0º gain and make a linear pair. + 80º 0º + 80º 80º 0º 50º. 5. (a) No, two acute angles cannot be supplementary because their sum is less than 80. (b) No, two obtuse angles cannot be supplementary because their sum is greater than C D 60 x 70 ccording to the figure, O B 60º + x + 70º 80º x + 0º 80º x 80º 0º 50º Thus, COD 50º. P-9

30 7. ( x + 0 ) O C ( x 0 ) ccording to the figure, OC and BOC are complete linear angles. So, OC + BOC 80º (x + 0º) + (x 0º) 80º x x 0 80º 5x 0 80º 5x x 00 x 00 5 x 40º. 8. Let B be the street and C be the foot of the ladder. Let D and E be the windows at the heights of 8m and 5m respectively from the ground. E B D 8 cm 7 cm 7 cm 5 cm B C Then CD and CE are the two positions of the ladder. From right DC, we have C + D CD C (CD D ) {(7) (8) }m (7 + 8) (7 8) m (5 9) m 5 m C 5 m 5 5 m 5 m From rightcbe, we have CB + BE CE CB (CE BE ) {(7) (5) } m [(7 + 5) (7 5)] m ( ) m 64 m CB 64 m 8 8 m 8 m Hence, width of the stret B (C + CB) (5 + 8) m m P-0 M T H E M T I C S -- VII

31 5 LINES ND NGLES WORKSHEET-9. (d) greater than the third side.. (d) always bisect each other at right angle.. x + y 80 x x x L 55 x y P x M 006 N O MPO LPN (Vertically opposite angles) x 55º gain, y (linear pair) y Let the angle be x and its complement be 90º x. ccording to question, x (90 x) x 90 x 6. (i) E x + x 60º x x 60º 5x 80º x 80º F 05 G B C H P-

32 7. (ii) Since EF GH and C is transversal. CF + CH 80 (interior angles on same side of transversal) CF º CF º (iii) Since EF is a straight line. BC + CF BC BC 80º 5º 45. C D 70º E F B BD BE + ED 40º + 0º 70º and CD 70 BD CD But they form a pair of alternate angles. B CD...(i) lso, BE + EF 40º º But they form a pair of interior opposite angles. B EF...(ii) From (i) and (ii), we get B CD EF CD EF b c d t a p f e 5 q Since e + 5º 80º(linear pair) e 80º 5º e 55º e f P- M T H E M T I C S -- VII

33 (vertically opposite angles) f 55º lso, a e (corresponding angles) a 55º gain, b 5º [alternate exterior angles] Since b and c form a linear pair b + c 80º 5º + c 80º c 80º 5º 55º Now, b and d are vertically opposite angles d b 5º [ b 5º] Thus, the required measures are : a 55º, b 5º, c 55º d 5º, e 55º, f 55º. P-

34 5 LINES ND NGLES Formative ssessment WORKSHEET-0 [] nswer the following : (i) BOD 50 (ii) BOC 0 4. x 4, (i) OC (ii) BOC C x 48, y 7, z cm [B] Fill in the blanks : greater. B, C 4. BC [C] True / False : 4 4. T. T. F 4. T [D] Match the Columns : 4 4. (ii). (i). (iv) 4. (iii). P-4 M T H E M T I C S -- VII

35 6 THE TRINGLE ND ITS PROPERTIES WORKSHEET-. (c) 80º.. (c).. Clearly, we have ( ) > 4.6 ( ) > 0. ( ) > 5.8 Thus, the sum of any tur of those numbers in greater than the third. Hence, it is possible to draw a triangle whose sides are 0. cm, 5.8 a and 4.6 a B 5 Since given triangle is a right-angled triangle. Hence, (C) (B) + (BC) (7) (B) + (5) (B) (7) (5) 89 5 (B) 64 B (a) x 55 B 90 C D Sum of interior opposite angles Exterior angle x x (b) C B 0º 90 C x D P-5

36 6. Similarly, 0º + 90º x x 0º ' 5 m B m In a right-angled triangle BC, C B + BC C C 69 m. Since, C m Total length of tree B m. 7. In a triangle sum of the lengths of the sides is always greater than the third side. Let the third side be x. We can say + 5 > x 7 > x x < x > x > 5 x > gain, x + > 5 x > 5 x > We know > x > and x < 7 So, the length of third side should fall between cm and 7 cm. 8. C E B F C D P-6 M T H E M T I C S -- VII

37 We know that the exterior angle is equal to the sum of its opposite interior angles. + B...(i) B + C...(ii) C +...(iii) So by [(i) + (ii) + (iii)], we get B + B + C + C + + B + C + + ( + B + C) gain we know that, + B + C 80º º º. (angle sum property) P-7

38 6 THE TRINGLE ND ITS PROPERTIES WORKSHEET-. (d) 0º.. (d) 80º.. Let the length of the third side be x cm we know that the sum of any two sides of a triangle is less than the third side (8 6) < x x > Thus, < x < 4 Hence, the length of the third side must be larger than cm and smaller than 4 cm. 4. In this right angled triangle, let the acute angles be x and x. By angle sum property 90º + x + x 80º 90º + x 80º x 80º 90º x 90º x 90 0º. Hence, acute angles are 0º and 60º. 5. Let the equal angles be x x + x + 0º 80º (By angle sum property) x + 0º 80º x 80º 0º x 70º x 70 x 5º Hence, equal angles are of 5º. 6. (i) Bigger side is 6 5 It may be assumed in this way (6 5) (6) + ( 5) , which is true Hence, the sides of lengths 6 5, 6, and 5 make a right-angled triangle. (ii) Bigger side is 5 cm. It may be assumed in this way (5) () + () , which is not true. Hence, these may not be the sides of right-angled triangle. (iii) Bigger side is 5 It may be assumed in this way ( 5) () + ( 5) So, these may be the sides of right-angled triangle. P-8 M T H E M T I C S -- VII

39 7. D C O B In OB, O + OB > B...(i) In OBC, OB + OC > BC...(ii) In ODC, OD + OC > CD...(iii) In OD, O + OD > D...(iv) dding (i) to (iv), we get O + OB + OB + OC + OD + OC + O + OD > B + BC + CD + D O + OB + OC + OD > B + BC + CD + D (O + OB + OC + OD) > B + BC + CD + D (O + OC + OB + OD) > B + BC + CD + D (C + BD) > B + BC + CD + D or B + BC + CD + D < (C + BD). 8. (a) True (b) False (c) False (d) False +++ P-9

40 6 THE TRINGLE ND ITS PROPERTIES WORKSHEET-. (c) 7 cm.. (a) 5º.. Length of hypotenuse (9) () cm. 4. Since, sides are : 0 cm, 5 8 cm, 4 5 cm (i) > 4 5 (ii) > 5 8 (iii) > 0 Since, sum of any two sides is greater than third sides. Hence, there may be a triangle with these sides. 5. D C 4 cm x 6. In right-angled triangle BC, 40 cm B C B + BC (4) (40) + x x (4) (40) x 8 9 cm Now, perimeter of rectangle (l + b) (40 + 9) cm. 5 C x B P-40 M T H E M T I C S -- VII

41 7. In right-angled triangle BC, C B + BC (5) () + x x (5) () x 8 9 Hence, required distance 9 m. B 7 7 In BD, D 90º D x C ( median of isosceles triangle) Hence, BD is right-angled triangle B BD + D (7) () + (x/) (x/) x x 4 5 x 4 5 x (a) y + 0º 80º (linear pair) y 80º 0º 60º x + y + 50º 80º x + 60º + 50º 80º x 80º 0º 70º (b) y 80º (vertically opposite angles) x + y + 60º 80º x + 80º + 60º 80 x 80º 40º 40 (c) y + 50º º y 80º 0º 70º x + y 80º (linear pair) x + 70º 80º x 80º 70º 0º (d) y + 50º 80º (linear pair) y 80º 50º 0º x º 80º x + 65º 80º x 80º 65º 5º. P-4

42 6 THE TRINGLE ND ITS PROPERTIES WORKSHEET-4 Formative ssessment [] nswer the following : 7 7. C 45. D 5. Y , 60, , [B] Fill in the blanks : 5 5. Obtuse. Complementary Hypotenuse [C] True / False : 4 4. T. T. F 4. T [D] Match the column : 4 4. (iii). (iv). (ii) 4. (i). P-4 M T H E M T I C S -- VII

43 7 CONGRUENCE OF TRINGLES WORKSHEET-5. (b) Sides.. (a) X. Since BC is an isosceles triangle so C BC 5 cm Hence B (5) (5) In the given triangles, BC and QPR 5 cm B QP 5 C QR 5 BC PR 7 So, there exists S.S.S. congruency. Hence, BC QPR, 5. Given, B C and D is bisector by CB, so B D C (i) In BD and CD B C (given) D D (ii) In BD and CD, SS congruency exists. BD DC. P-4

44 6. B C C B 7. (i) In the given triangles BC and CB, B C C B (ii) s all the three corresponding sides of BC and CB are equal. Hence, by S.S.S. congruency BC CB (iii) By C.P.C.T. B C. B C D E Since, B C (given) BD CE (given) B + BD C + CE D E Now in DC and EB D E (we have just proved) C B (given) (common) So, by S..S. congruency DC EB by C.P.C.T., CD BE 8. Three pairs of equal parts in BC and CB are as follows : In BC and CB B C (given) BC CB (same side of both triangles) C B (given) Hence, BC CB (by SSS congruence rule) P-44 M T H E M T I C S -- VII

45 7 CONGRUENCE OF TRINGLES WORKSHEET-6. (a) DEF.. (c) Two corresponding sides and angle included are equal.. In BD and BC, C BD hypotenuse (given) B 90 (given) B B (common side) D C 4. B D C B In the above figure, it is given that D BC and D BC Now in DC and BC D BC (given) DC CB (s D BC and C is transversal DC and CB are alternate angles) C C (common) So by S..S. congruency, DC BC by C.P.C.T., B CD 5. In BC and DCB, D B C P-45

46 6. C DB (given) BC CDB 90 (given) BC BC (common) So, by R.H.S. congruency BC DBC by C.P.C.T., B DC. 7. B D C Given : triangle BC in which B C and D is the bisector of. To prove : DB DC 90 and BD DC Proof : In DB and DC, we have B C (given) BD CD (given) D D (common) DB DC (SS congruence property) So BD DC and DB DC But DB + DC 80 (linear pair property) DB DC 90 Hence, D bisects BC at right angles. O B C In OB, O + OB > B...(i) Similarly, in BOC, OB + OC > BC...(ii) and in OC, O + OC > C...(iii) By adding (i), (ii) and (iii), we get O + OB + OB + OC + O + OC > B + BC + C O + OB + OC > B + BC + C (O + OB + OC) > B + BC + C. P-46 M T H E M T I C S -- VII

47 7 CONGRUENCE OF TRINGLES WORKSHEET-7. (c) Q.. Since, O is the pairs of bisection of B and CD. Therefore, O OB OC OD OC BOD Hence, three pairs of equal parts in OC and BOD are O OB, OC OD and OC BOD.. (i) In BCD and BCE E D B C 4. BD DE (given) BEC CDB 90 (given) BC BC (common) (ii) In BCD and BCE, R.H.S. congruency exists. BCD BCE (iii) s BCD BCE by C.P.C.T. DCB EBC. B m D 5 m E 4 m 9 m 9 m C In the above figure, B and CD are two poles whose height s are 9 m and 4 m respectively. ` B EC 9 m and BD m DE m Now in right BDE, BD BE + DE BE + 5 P-47

48 BE () (5) 69 5 BE 44 BE 44 BE m. Hence, distance between their feets m. 5. Three pairs of equal parts in triangles BC and DC are BC DC (Z bisects DB) BC DC (Z bisects DCB) C C (common side) BC DC CD CB 6. Let B is the street and C be the foot of ladder. Let D and E be windows at heights of 8 m and 5 m respectively from the ground. E D C B Then CD and CE are the positions of ladder. From the right DC, CD C + D C CD D C 5 5m gain from right EBC, Width of street, CE BC + BE BC CE BE CE 64 8 m B C + BC m P-48 M T H E M T I C S -- VII

49 7 CONGRUENCE OF TRINGLES WORKSHEET-8 Formative ssessment [] nswer the following : 5 5. SSS, CB DEF. RHS, RPQ LNM. SSS, YXZ TRS 4. S, DEF PNM 5. S, CB CD. [B] Fill in the blanks : 5 5. The same length. The same measure. The same side length 4. The same radius 5. The same length and same breadth [C] True / False : 0 0. F. T. F 4. F 5. F 6. T 7. F 8. T 9. F 0. T P-49

50 8 COMPRING QUNTITIES WORKSHEET-9. (b) 7 0. (c) ` 76.. Let number be x then x x (a) 90% of (b) Let x% of hour 6 seconds x x 6 6 x x% % 5. Since, a b a b and 6b 5c b c 5 6 a b 5 b c 6 a c Let the C.P. be x S.P. Rs. 540, loss 5% S.P. C.P. loss C.P. 5% of C.P. 540 x 5 x x x x x x 0 9x x P-50 M T H E M T I C S -- VII

51 Hence, C.P. of item Rs Let C.P. be x S.P. Rs. 80 Profit 5% S.P. C.P. + Profit 80 x + 5% of x 80 5 x x 00 x x 0 80 x 0 x C.P. of article Rs (i) Rupesh secured 50 out of 500 in st term and in nd term he scored 80% of 500 His marks in nd term His improvement in marks Hence, he improved by %. (ii) Percentage. (iii) Hard work is the key to success. P-5

52 8 COMPRING QUNTITIES WORKSHEET-0. (a) x (b) 6 days. (c) 60. Sum of ratio terms s share 50 9 ` 00 B s share 50 9 ` 450 C s share ` Let B C 4 5 k k B 4k, C 5k : B : C k : 4k : 5k : 4 : 5 5. S.I. P R T 00 50, S.I. Rs., (a) 9% of Rs Rs. 6 (b) 6 0 % of 75 m % of 75 m m 5 75 m 5 m (c) 7 5% of 80 kg (d) 8% of 5 litres kg kg litres litre. 00 P-5 M T H E M T I C S -- VII

53 7. Let the cost price be ` x. x Then, Gain ` 6 SP ` x x ` 7 x 6 But, SP ` 0 (given) 7x 0 6 x CP ` 760 and SP ` 0 Gain ` (SP CP) ` (0 760) ` 460 Gain Gain 00 CP % % 50 % 6 6 % 8. (i) Mr. Narayan saves 0% of his salary His expenses % 80 His expenses 0, Rs. 6,000 His expenses Rs. 6,000. (ii) Percentage. (iii) Saving is very important. P-5

54 8 COMPRING QUNTITIES WORKSHEET-. (b) :. (d) 0%. Let x% of.5 litres 700 ml. Then x ml 700 ml 5x 700 x Hence, 0% of.5 litres is 700 ml. 4. C.P. Rs. 40 Profit 0% S.P.? Profit S.P. C.P. + Profit Rs P Rs. 90,000 R 5 5 T years S.I. P R T 00 90, ,850 P + S.I. 90, ,04,850 mount () Rs.,04, Let the angles be x, x and 4x x + x + 4x 80 ngle sum property) 9x 80 x 80 9 x 0 ngles are x 0 40 x x P-54 M T H E M T I C S -- VII

55 7. Total of parts + 5 Son got 5 part So his percentage share 00 60% 5 His daughter got 5 part So, her percentage share %. 8. Let Mr. Saxena s income be 00 Money spent on house rent 0 Rest of income Now 60% of 70 is spent household expenses. Household expenses 60% of Savings 70 4 Rs. 8 If he saves Rs. 8 his monthly income be Rs. 00. If he saves Rs. 600 his monthly income Rs.,500. P-55

56 8 COMPRING QUNTITIES WORKSHEET- Formative ssessment [] nswer the following : 0 5. ` ` ` ` ` ` ` ` % p.a. [B] Fill in the blanks : [C] True / False : 6 6. T. T. F 4. T 5. T 6. T [D] Match the column : 4 4. (c). (d). (a) 4. (b). P-56 M T H E M T I C S -- VII

57 9 RTIONL NUMBERS. (b). (c). 4 5 Hence 4 ( ) WORKSHEET- 4. (a) (b) 5. (a) (b) 6. Given sequence is L.C.M. of, 9 9 Sequence is Since ( 7) ( ) ( ) ,, ,, < P-57

58 So, ascending order is 7. (a), 5 L.C.M. of and 6 (b) and Since 4 In the same way, Clearly 7 and 4 L.C.M. of 4 and 44 4,, > > > 8 44 > (i) Sequence is, 7, 8, L.C.M. of 5, 0, 5, 0 0 Sequence be 6 7 8,,, or 6,,, Its ascending order is 6 < or 5 < (ii) L.C.M. and to find ascending order. (iii) In a class the students should stand in ascending order of their height. P-58 M T H E M T I C S -- VII

59 9 RTIONL NUMBERS WORKSHEET-4. (c) 4. (a) 5 4. The given number is 5. HCF of and 5 is 7 so Hence 5 (in standard form) (a) (b) ( ) (a) Rational number equivalent to 7 are 6, and (b) Rational numbers equivalent to are and P-59

60 6. Sequence is 7 5,, L.C.M. of 5, 0 and ,, ,, ,, ,, Since, < Hence, sequence in ascending order is 5 7. Total length of the rope 0 m 6 < 7 0 < 5 Length of a piece of rope m 4 Numbers of pieces 5 m 4 Total length of the rope Length of a piece Hence 8 pieces can be cut from the rope. 8. Given pattern is 4,,,, Here, ( ) ( ) ( ) and 4 ( ) 4 4 Hence, next four numbers are ( ) ( ) ( ) ( ) P-60 M T H E M T I C S -- VII

61 9 RTIONL NUMBERS. (a). (c) ( 6) 5 9 ( 6) ( 5) 6 5 () (9) 9 WORKSHEET (a) (b) ( ) ( ) (a) s is a negative number While is positive number 9 7 So, 9 ( ) (b) since ( ) and are same 6. (a) ( ) P-6

62 (b) 7. Let the missing number be a b, then 5 ( 8) ( 8) a b b a b 5 7 a b 5 6 a b a 0 9 a 9 Number is. b 0 8. L.C.M. of 5 and and Hence, five numbers are to be inserted between and < < < Hence numbers are 7, 4,, , P-6 M T H E M T I C S -- VII

63 9 RTIONL NUMBERS WORKSHEET-6 Formative ssessment [] nswer the following :.. 6 < 5 < < 4 < km ` [B] Fill in the blanks : [C] True / False : 5 5. F. T. T 4. F 5. T P-6

64 0 PRCTICL GEOMETRY WORKSHEET-7. (d) 4.. (c) 65º.. Steps of Construction : (i) Draw a line B and mark point C outside it. (ii) Mark another point D on B and join CD. (iii) With D as centre and a convenient radius, draw an arc cutting B at E and CD at F. (iv) With C as centre and radius equal to DE draw an arc, cutting CD at G. (v) With G as centre and radius (opening) equal to EF, mark a point H on the above arc. (iv) Join CH to draw a line m Thus, m is required line such that m B H C m F G D E B 4. Steps of Construction : (i) Draw a line segment QR 8 cm. (ii) t Q, construct RQX 90º. (iii) With centre at R and radius 0 cm, draw an arc to cut QX at P. (iv) Join PR. Thus, PQR is the required right-angled triangle. X P 0 cm Q Steps of Construction : (i) Draw a line QR 5 cm. (ii) Draw an arc of radius 4 5 cm from the point Q. (iii) Draw an arc of radius 6 cm from the point R. (iv) Both the arcs cut each other at a point P. So it is the point of intersection. 8 cm R P-64 M T H E M T I C S -- VII

65 (v) Join PQ and PR. We obtain PQR. P 4 5 cm 6 cm Q 5 cm 6. Steps of Construction : (i) Draw a line B of length 7 cm. (ii) t, draw a line X making an angle 60 with B. (iii) t B, draw a line BY making an angle 0 with B. (iv) Both lines X and BY cut each other at point C. So it is the point of intersection. (v) We obtain BC. R Y C X cm B 7. Steps of Construction : (i) Draw C 5 5 cm. (ii) t, construct CX 45º. (iii) t C, construct CY 45º, such that CY meets X at B. Y B X cm C Then, BC is the required isosceles triangle, in which B CB and B 90º. P-65

66 0 PRCTICL GEOMETRY WORKSHEET-8. (d) 8, 5, 7.. (b) 50º.. Steps of Construction : (i) Draw a line segment BC 4 5 cm. (ii) Construct CBX 60º at B. (iii) Construct BCY 45º at C. (iv) Let the rays BX and CY intersect at. Thus, BC is the required triangle. Y X B 4 5 cm C 4. Steps of Construction : (i) Draw a line l and mark a point on it. (ii) Construct an angle of 90º at to draw B perpendicular at l. (iii) Mark a point X on B such that X cm. (iv) t X, construct an angle of 90º to draw a perpendicular to B. Thus, m is the required line through X such that m l. B X m cm l 5. Steps of Construction : (i) Draw PQ of length cm. (ii) t Q, draw QX PQ. (iii) With P as centre, draw an arc of radius 4 cm, which cut line OX at R. So the distance from P to R is 4 cm. P-66 M T H E M T I C S -- VII

67 (iv) Now, we obtain PQR, which is right angled at Q. R X 4 cm Q 0 cm P 6. Construction : First we draw a rough sketch of BC, as shown. Then, we draw BC in following steps. Steps of Construction :. Draw a line segment BC 4.8 cm.. Construct CBX 60º and BCY 75º.. Let BX and CY intersect at. Then, BC is the required triangle. Y X 60º 75º B 4.8 cm C P B 60º 4.8 cm 75º C 7. Steps of Construction : (i) Draw a line CB 4 cm. (ii) t B, draw an angle 45. (iii) With C as centre, draw an arc of radius 4 cm. which cut the line BX at. So the distance from to C is 4 cm. (iv) Now, we obtain BC, which is right angled at C. X 4 cm B 4 cm C P-67

68 0 PRCTICL GEOMETRY WORKSHEET-9. (c) 0, 6, 4.. (b) 55º.. Steps of Construction : (i) Draw a line segment PQ 5 cm. (ii) t P, construct QPX 45º. (iii) t Q. construct PQY 75º. (iv) Let the rays PX and QY intersect at R. Thus, PQR is the required triangle. Y R X P 5 cm Q 4. Construction : First we draw a rough sketch of BC, as shown. Then, we draw BC in following steps. Steps of Construction :. Draw line segment BC 4.8 cm.. Construct CBX 90º.. With C as centre and with radius 6. cm, draw an arc, cutting BX at. 4. Join C. Then, BC is the required triangle. X 6. cm 6. cm B 4.8 cm C B 4.8 cm C P-68 M T H E M T I C S -- VII

69 5. Steps of Construction : Part I : Construction of BC : (i) Draw BC 5 4 cm. (ii) Construct CBX 0º. (iii) From BX cut off B 4 5 cm. (iv) Join and C. Now BC is the required triangle. Part II : Construction of D BC : (i) Produce CB through B to Y. (ii) With centre and a sufficient radius, draw an arc intersecting BY and BC at U and M respectively. (iii) With U as centre and a radius more than half of UM, draw an arc opposite to side of. (iv) With the same radius and centre M, draw another arc, cutting the previous arc at T. (v) Joint T such that it meets YB at D. Then D BC. X Y U D B M 5 4 cm C T 6. Steps of Construction : Part I : To construct the BC. (i) Draw BC 8 5 cm. (ii) Construct CBX 90º. (iii) With centre C and radius equal to 9 5 cm, draw an arc to cut BX at. (iv) Join and C. Thus BC is the required triangle. Part II : Perpendicular bisector of B and BC. (i) With centre at and radius more than half of B, draw two arcs on both sides of B. (ii) With centre at B and the same radius as in step (i), draw two arcs intersecting the arcs drawn in step (i) at R and S. (iii) Join R and S and extended it on both sides to P and Q. Now PQ is the perpendicular bisector of B. (iv) Similarly, draw NM perpendicular bisector of BC. P-69

70 We note that PQ and NM meet at a point D, which is at C. X N R S D P Q B C M 7. We know that the sum of angles of triangle be 80º. Let us consider a right BC in which 90º, hypotenuse BC 5 6 cm and B 0º. We know that the sum of angles of a triangle is 80º + B + C 80º 90º + 0º + C 80º C 80º 0 60º. Steps of Construction : (i) Draw a line segment BC 5 6 cm. (ii) Construct CBX 0º and BCY 60º. (iii) Let BX and CY intersect at point. Then, BC is the required triangle. Y X 0 60 B 5 6 cm C P-70 M T H E M T I C S -- VII

71 0 PRCTICL GEOMETRY WORKSHEET-9 Formative ssessment [] nswer the following : C D 5 cm B. m 4 cm l 4. Steps of Construction : (i) Draw a line segment BC 5 5 cm. (ii) With B as a centre, draw an arc of radius 5.5 cm. (iii) With C as a centre, draw an arc of radius 5.5 cm. (iv) Both arcs cut at point to each other. (v) Meet point and B, and C. (vi) We obtain equilateral triangle BC. 5 5 cm 5 5 cm B 5 5 cm C 5. Steps of Construction : (i) Draw a line segment QR 5 cm. (ii) With Q as a centre, draw an arc of radius 4 cm. (iii) With R as a centre, draw an arc of radius 4 cm. (iv) P has to be on both the are drawn. So it is the point of intersection. Join PQ and PR. (v) Since, two sides arr of equal length. Thus PQR is an isosceles triangle. P 4 cm 4 cm Q 5 cm R 6. Steps of Construction : (i) Draw a line segment BC of length 6 cm. P-7

72 5 cm 6 5 cm B 6 cm C (ii) With B as a centre, draw an arc of radius.5 cm. (iii) With C as centre, draw an arc of radius 6.5 cm. (iv) The point of intersection of arcs is point. Join B and C. (v) We obtain BC. B Steps of Construction : (i) Draw a line segment BC of length 6 5 cm. (ii) t B, draw BX making an angle of 0 with BC. (iii) With B as a centre, draw an arc of radius 6.5 cm. It cuts BX at the point. (iv) Join C. The isosceles BC is obtained. 6 5 cm 0 B 6 5 cm C 8. By angle sum property, B m + mb + mc mc 80 mc Now, Steps of Construction : (i) Draw a line segment C of length 7 cm. 7 cm C (ii) Draw an angle of measure 60 at point and an angle of 7 at point C. (iii) Both the rays intersect each other at point B. (iv) Join B and C, we obtain BC. [B] Fill in the blanks : 5 5. alternate angles, corresponding angle equal [C] True / False : 7 7. T. F. T 4. T 5. F 6. F 7. T. P-7 M T H E M T I C S -- VII

73 PERIMETER ND RE WORKSHEET-4. (b) (l + b).. (b) base height. here r 0 cm Circumference r cm. 4. Perimeter of square 4 side 4 side 00 side m rea of park (side) (50) m. 5. Perimeter of rectangle (l + b) (5 + b) b 00 b b Breadth 5 m. D E 5 cm 5 cm cm F C 5 cm 8 cm H G B rea of margin [rea of BCD rea of EFGH] [(8 5) (5 )] cm Hence, area of margin 0 cm. P-7

74 7. 90 m B P Q E m F m m H G m D C 60 m S R (i) rea of roads (rea of BCD + rea of PQRS) (rea of EFGH) [( ) ( )] [( ) (9)] cm (ii) Cost of construction 44 0 Rs. 48,50 Hence, cost is Rs. 48, (i) D 0 m C H G E B Side of outer square BCD 0 m Side of inside square EFGH 0 ( + ) 0 8 m. Hence area of path rea of square BCD rea of square EFGH (0) (8) m rea of remaining portion rea of square EFGH m Cost of planting grass Rs,60 (ii) rea of square. (iii) Logic is required for anything. F P-74 M T H E M T I C S -- VII

75 PERIMETER ND RE WORKSHEET-4. (b) r.. (a) r. Given r r 8.5 r r.6.5 cm. 4. rea of parallelogram base height In the question, B base 8 cm. and height 5 cm. rea cm. 5. Given, area of circular garden is 4 m and the radius of the area covered by the sprinkler is m. Hence, area of watered area r m Since, 45 6 > 4 watered area > area of park Hence, sprinkler can water the entire field. 6. C D B 8 rea of the triangle BC base height B BC 8 4 P-75

76 gain, also area of BC 48 m. C BD 48 BD BD 96 BD cm. 7. ccording to the question, Circumference Diameter 0 r r 0 r( ) 0 r 7 0 r r 4 7 cm. 8. m (4)() 44 () 4 () 8 5 m 0 m (i) rea of whole land l b m (ii) rea of flower bed r 4 () m (iii) rea of lawn excluding flower bed rea of lawn rea of flowed bed m (iv) Circumference of flower bed r 4 56 m P-76 M T H E M T I C S -- VII

77 PERIMETER ND RE WORKSHEET-4. (c) 4 side.. (c) 9 cm.. Let the required height h Then 5 h 0 h Hence height of the triangle 6 cm 4. Given figure is a semi-circle. 0 cm d Here, d 0 cm, r 0 5 cm. Its perimeter r + d m. 5. Let the radii of circles are r and r. ccording to the question, r r 4 r r 4 Ratio of their areas r r r r : rea of circle r r r 8 5 P-77

78 r r 5 r 5 cm. Circumference r 7 5 cm. 7. Radius of circular park 84 4 m. Width of the road 5 m radius of outer circle m 45 5 m 4 m 5 m rea of the road (rea of outer circle) (rea of inner circle) (45 5) (4) {(45 5) (4) } {( ) (45 5 4)} m Cost of the road Rs.,9, rea enclosed the copper wire in square shape (side) (side) side cm. Hence length of wire 4 44 cm Now this length circumference of the circle r 44 7 r 44 r 44 7 r 7 m Hence, area enclosed by the wire when it is bent in circular shape r 7 (7) m. P-78 M T H E M T I C S -- VII

79 PERIMETER ND RE WORKSHEET-44 Formative ssessment [] nswer the following : cm. base m, altitude 6 m. 84 cm cm [B] Fill in the blanks : 5 5. d d. (l + b) h a [C] True / False : 6 6. F. T. F 4. F 5. F 6. T [D] Match the column : 4 4. (d). (c). (a) 4. (b). P-79

80 LGEBRIC EXPRESSIONS WORKSHEET-45. (a) 6. (b). Required sum (5x 7x + ) + ( 8x + x 5) + (7x x ) 5x 7x + 8x + x 5 + 7x x ( ) x + ( 7 + )x + ( 5 ) 4x 6x 4 4. (a) 5x y (5) () () ( ) 0 + (b) x + y () () + ( ) 6 + ( ) x + x (x 5) x + x x 0 x + x + 4x x + 5x. If x, given expression () + 5() (a) We have (4m mn + 8) ( m + 5mn) 4m mn m 5mn 4m + m mn 5mn + 8 5m 8mn + 8 (b) We have (5x 0) ( x + 0x 5) 5x 0 + x 0x + 5 x + 5x 0x x 5x Sum of x y + and y (x y + ) + ( y ) x y + y x y y + x y Subtraction (x y) (x y ) x y x + y + x x y + y + y +. P-80 M T H E M T I C S -- VII

81 8. (i) Total amount deposited in account Balance amount (ii) ddition of algebraic expression. (iii) Economy is everywhere. (5x + x ) + (x x + 5) 5x + x + x x + 5 5x + x + x x + 5 6x + x + (6x + x + ) (x x ) 6x + x + x + x + 6x x + x + x + + 4x + x + 6. P-8

82 LGEBRIC EXPRESSION WORKSHEET-46. (b) 5. (a) 0. 9 x y (x + y) 9 9 x y x x y y 9 9 x x y x y y 9 9 x y x y 9 9 x y x y 4. We have, b (5 a) a (b 5) 5b ab ab + 5a 5a + 5b + ( ) ab 5a + 5b ab 5. (a) mn mn + mn + 8 mn + 6 (b) m + n + m n m+ 0 m 6. If n, (a) 5n (b) 5n + 5n 5() + 5() If z 0, z (z 0) (0) (0 0) and 5z z (0) P-8 M T H E M T I C S -- VII

83 a 4 b c a 5 b c 4 + a 7 b 5 c 4 6 a a 5 a 4 + b 5 b 7 b c c 5 c 5 6 8a 9a 0a 6b 50b 5b 8c 0c 5c a b c P-8

84 LGEBRIC EXPRESSION WORKSHEET-47. (b). (c) x + y (x + y ) 6x 4 + x y + 9x y + y 4 6x 4 + x y + y 4 4. If m, we have (a) m 0 (b) 5m Expression mn + m n + 8mn + 9 Term I : mn degree + Term II : m n degree + Term III : 8mn degree Term IV : 9 degree x x 7 + x x + x x x 4 x x 5 x + x x x + 5 x x x x x y 4 y 5 y y 7 7 y 4 y 5 y y 7 7 y y 5 6 y y y 6 y 7 7 P-84 M T H E M T I C S -- VII

85 6 () () 7 [ y ] Sum of (8m 7n + 6p ) and ( m 4n p ) (8m 7n + 6p ) + ( m 4n p ) 8m 7n + 6p m 4n p 8m m 7n 4n + 6p p 5m n + 5p Sum of (m + 4n p ) and ( m n p ) (m + 4n p ) + ( m n p ) m + 4n p m n p m m + 4n n p p m + n 4p Subtraction of (5m n + 5p ) from (m + n 4p ) (m + n 4p ) (5m n + 5p ) m + n 4p 5m + n 5p m 5m + n + n 4p 5p 4m + 4n 9p. P-85

86 LGEBERIC EXPRESSION WORKSHEET-48 Formative ssessment [] nswer the following : 0 0. a + b + 4c. 4m + 4n 9p. 4 + a a 4. x 0y x + y 6. y y a5 b 8. 6 x y [B] Fill in the blanks : x y 4. x x 5 x b 4. (xy + y) 5. 6x 4 + x y + y 4 [C] True / False : 5 5. T. F. T 4. F 5. F P-86 M T H E M T I C S -- VII

87 EXPONENTS ND POWERS. (b) 8 WORKSHEET-49. (a) 6. Let the required number be x. Then 9 x x Hence the required number is , , , , , , (a) ( 5) ( ) (b) x 0 x ( 5) x x 4 P-87

88 8 5 x 5 x x x4 4 5 x 4 5x (a) 79404,00, , ,00, , (b) 006,00, , ,00, , (5) n (n ) (5 ) n (n ) 5 n n 5 n 5 n 00 or 5 n 5 n 00 5 n (5 ) 00 5 n n n 5 s base is same on both the sides n n + n 4 n 4. P-88 M T H E M T I C S -- VII

89 EXPONENTS ND POWERS WORKSHEET-50. (c). (c) (5) ( 8) a5 a a8 5 8 a a a a (5 + 8) a a a a a Since, 5 and > 5 > ( ) ( ) ( ) () P-89

90 7. (a) 5 n n s base 5 is same on both sides. n + 9 n 9 n 6 n 6 8. n + (b) 8 n + n+ 5 n n s base is same on both sides n n n n (7) n 5 ( ) 7 n n n ( ) ( ) 5 n n n ( ) ( ) 5 n n n 5 nn n 5 n n 5 n ( ) 5 n5 (9 ) n n 5 s base are same on both sides, so n 5 n + 5 n 4. P-90 M T H E M T I C S -- VII

91 EXPONENTS ND POWERS. (b). (b) 4 8 WORKSHEET-5. ( ) 4 ( 4)( ) 4 4 ( ) 4 () ( 4) Prime factors ( + 6) ( + ) Let the number ( 5) should be divided by x to get the quotient ( 5). ( 5) x ( 5) x 5 x 5 P-9

92 5x 5 5x 5 5x 5 5 x 5 x x 7. (a) (6 8 ) + ( ) (b) 6 8. (a) ( 4) ( 4) ( 4) ( 4) ( 4) (b) ( 4) () ( 8) ( 8) ( 8) ( 8) ( 8) 5 5 ( 8) () 8 5 P-9 M T H E M T I C S -- VII

93 EXPONENTS ND POWERS WORKSHEET-5 Formative ssessment [] nswer the following : 7 7. (a) 4 5 (b) 6 (c) 5 (d) (i) 0 6 (ii) 6,40, n a [B] Fill in the blanks : b. 6 a 4. [C] True / False : 9 9. T. F. F 4. F 5. F 6. F 7. F 8. F 9. T P-9

94 4 SYMMETRY WORKSHEET-5. (a) no line of symmetry.. (c) a line joining the mid points of its opposite sides.. 4. (a) C B (b) D C 5. Yes, there is one line of symmetry. B D B C 6. Isosceles triangle has only one line of symmetry. 7. There are two lines of symmetry. 8. fter rotating by 60 about a centre, a figure looks exactly the same, as its original position. This will happen for the figure at angles 0, 80, 40, 00, 60 respectively. 4 P-94 M T H E M T I C S -- VII

95 4 SYMMETRY WORKSHEET-54. (d) Four line of symetry.. (b) Each of its diagonals.. Perpendicular bisector of the diameter. No 4. ngle by which the body rotates is called, the angle of rotation. 5. (a) Median of an isosceles triangle. (b) Diameter of a circle. 6. No, generally trapeziums have no line of symmetry leaving isosceles trapezium. In Isosceles Tropezium, D BC. So, there is one line of symmetry. D C B 7 square has a rotational symmetry of order 4 about its centre, In this case : (i) The centre of rotation is the centre of the square. (ii) The angle of rotation is 90º. (iii) The direction of rotation is clockwise. (iv) The order is 4. 90º 90º (i) 90 (ii) 80 90º 90º (iii) 70 (iv) 60 (v) P-95

96 8. (a) (b) P-96 M T H E M T I C S -- VII

97 4 SYMMETRY WORKSHEET-55. (d) n unlimited number of lines of symmetry... (a) regular hexagon has six lines of symmetry. (b) quadrilateral in general has no line of symmetry. 4. (a) (b) 5. (a) (b) an equilateral triangle Circle Both the figures have line of symmetry as well as rotational symmetry º, 60º. 7. n scalene triangle. 4 B C P-97

98 4 SYMMETRY WORKSHEET-56 Formative ssessment [] nswer the following : 6 6. Point of rotation. Line of symmetry. (three) 4. order 5. n isosceles triangle, parallelogram. 6. The letter Z of the english alphabet. [B] Fill in the blanks :., H, I, M, O, T, U, V, W, X, Y. B, C, D, E, H, I, K, O, X [C] True / False : 8 8. F. T. F 4. F 5. T 6. F 7. T 8. T [D] Match the column : 4 4. (iii). (iv). (ii) 4. (i). P-98 M T H E M T I C S -- VII

99 5 VISULISING SOLID SHPES WORKSHEET-57. (c) Cuboid.. (b) Sphere.. (a) Triangular pyramid (b) Six 4. No, because one pair of opposite faces will have and 4 on them whose total is not equal to 7. nother faces are having and 6 on them, whose total is also not equal to (a) Cuboid (b) Cylinder (c) Cuboid (d) Cylinder Three cubes each with cm edge are placed side by side to form a cuboid, then Length of cuboid cm Breadth of cuboid cm Height of cuboid cm cm cm cm cm cm cm 4 P-99

100 5 VISULISING SOLID SHPES WORKSHEET-58. (c) 4 cm.. (d) Triangle.. For a pyramid, F E + V 4 E + 4 (F 4 and V 4) 8 E E 8 6 Hence, there are 6 edges. 4. (a) For a brick : Vertical cut is Rectangle and Horizontal cut is rectangle. (b) For an ice-cream cone. Vertical cut is triangle and horizontal cut is circle (a) (ii) (b) (iii) (c) (iv) (d) (i) P-00 M T H E M T I C S -- VII

101 5 VISULISING SOLID SHPES WORKSHEET-59. (c) five faces (a) (b) P-0

102 Fold the given faces in such a way that lies opposite 4; lies opposite 5; lies opposite 6; Thus, a cube is formed. 4 P-0 M T H E M T I C S -- VII

103 5 VISULISING SOLID SHPES WORKSHEET-60 Formative ssessment [] Fill in the blanks : 6 6. Four.. Two 4. 6 faces. 5. No Vertex 6. Vertices. [B] True / False :. T. F. T [C] Match the column : 5 5. (c). (d). (e) 4. (b) 5. (a). [D] Complete the following table : 66 Name of solid No. of Faces No. of Vertices No. of Edges Square Prism 6 8 Triangular Prism Tetrahedron Hexagonal Pyramid 7 7 Cuboid 6 8 Pentagonal Pyramid P-0