CSS Primary Standard “Mathematics” 201 ☻ To explain the addition of measure of two or more line segments. ☻ To explain the subtraction of a line segment from a longer one. ☻ To explain the method of drawing right bisector of gives line segment using compass. ☻ To explain the drawing of perpendicular to given line, form a point on it by using compass. ☻ To explain the method of drawing perpendicular, from point outside the line, using compass. ☻ To explain use of compass to construct a an angle equal and twice of a given angle, bisect the given angle, divide given angle in four equal parts. ☻ To explain the construction of triangle when three sides or two sides and one angle or two angle and one sides is given. ☻ To explain the construction of a right angle triangle when hypothesis and one sides are given. Learning Outcomes: At the end of this unit, students will be able to: ☻ Add measure of two or more line segments. ☻ Subtract measure of a line segment from longer one. ☻ Draw the right bisector of a given line segment by using compasses. ☻ Draw the perpendicular to a given line from point on it using compasses. ☻ Draw the perpendicular to line from a point outside the line, using compass. ☻ Use of compasses to: Construct an angle equal in measure of a given angle. Construct an angle twice in measure of a given angle. Bisect a given angle. Divide a given angle into four equal angles. Construct the following angles: 60°, 30°, 15°, 90°, 45°, 1 22 2 , 75°, 1 67 2 , 120°, 156°, 165°, 135°, 105° ☻ Construct a triangle when three sides (SSS) are given. Caution: Sum of two sides should be greater then third side. ☻ Construct a triangle when two sides and their included angle (SAS) are given. ☻ Construct a triangle when two angles and their included side (ASA) are given. ☻ Construct a triangle when hypotenuse and one sides (RHS) for a light angle triangle are given. Teaching Material: CSS primary standard Mathematics Book 7.
CSS Primary Standard “Mathematics” 202 Writing board. Marker and Eraser. Exercise 10a Q.1 Draw a line segment whose length is equal to (a) the sum of the length given below (b) The difference of length given below. (i) 4.8cm, 2.3cm (a) Construction: (i) Draw a line OA. (ii) With center P on OA , draw an arc of radius 4.8cm, which interest OA at point Q. (iii) With center Q on OA , draw an arc of radius 2.3cm, which intersect OA at point R. (iv) PR is required line segment Hence mPR = mPQ + mQR (v) With help of ruler, measure the required line segment, which is 7.1cm. (b) (i) Draw a line OA (ii) With center P on OA , draw an arc of radius 4.8cm. Which intersect OA at point Q with center Q on OA draw an arc of radius 3.1 cm across the left which intersect OA at point R. PR is required line segment hence PR PQ QR and its length is 2.5cm. (ii) 5cm, 2.3cm (a) Construction: (i) Draw a line OA. (ii) With center P on OA , draw a line segment of radius 5 cm, which intersect OA at point Q. (iii) With center Q on OA , draw an arc of radius 3cm, right which intersect OA at point R. (iv) PR is required line segment. Hence PR PQ QR , the length of is 8cm. (b) (i) Draw a line OA. (ii) With center P on OA , draw an arc of radius 5cm which intersect OA at point Q.
CSS Primary Standard “Mathematics” 203 (iii) With center Q, draw an arc of radius 3cm arcross left, which intersect OA at point R. (iv) PR is required line segment. Hence PR PQ QR . The length of PR is 2cm. (iii) 8.2cm, 1.3cm (a) Construction: (i) Draw a line OA. (ii) With center P on OA , draw an line arc of radius 8.2cm, which intersect OA at point Q. (iii) With center Q on OA , draw an arc of radius 1.3 cm across light, which intersect OA at point R. (iv) PR is required line segment hence PR PQ QR the length PR of is 9.5cm. (b) (i) Draw a line OA. (ii) With center P on OA , draw an arc of radius 8.2cm, which intersect OA at point Q. (iii) With center Q on OA , draw an arc of radius 1.3cm across left, which intersect OA at R. (iv) PR is required line segment. Hence PR PQ QR the length of PR 6.9cm Q.2 Draw the following line segment and bisect them using a compass. (i) 7.5cm Construction: (i) Draw a line segment PQ = 7.5cm. (ii) Adjust the compass to slightly longer than half of line segment and draw arcs, with center P, below and above the line segments. (iii) Keeping the same compass width, draw arcs from Q above and below. These arcs intersect each other at points R and S. Join R with S. RS bisect PQ at point T. (ii) 5.4cm Construction: (i) Draw a line segment PQ = 5.4cm.
CSS Primary Standard “Mathematics” 204 (ii) Adjust the compass to slightly longer than half of line segment and draw arcs from P above and below the line segment. (iii) Keeping the same compass width draw arcs from Q below above. The arcs intersect the previous arcs at point R and S. Join R to S. PQ at point T. (iii) 8.9cm Construction: (i) Draw a line segment PQ = 3.8cm (ii) Adjust the compass slightly longer than half of the line segment length. And draw the arcs from p below and above the line segment. (iii) Keep the same compass width, draw arcs from Q below and above line segment. These arcs intersect the previous arcs at point R and S. (iv) Join R with S. (v) RS bisect PQ at point T. Q.3 Draw a line segment PQ = 6cm. Take any point R outside it. Draw the perpendicular from R on PQ. Construction: (i) Draw PQ = 6cm. (ii) Take a point R outside PQ with centers Q and P, draw arcs of same radius which intersect each other at in point S. Join S with R. RS is required perpendicular on PQ. Q.4 Draw a line segment AB = 5cm. Take any point D on the line segment Draw a perpendicular on AB at point D. Construction: (i) Draw a line segment AB = 5cm. (ii) Selected a point D somewhere in the middle. (iii) With center D, draw arcs of radius 1cm on both sides of d, which intersect AB at P and Q.
CSS Primary Standard “Mathematics” 205 (iv) Adjust the compass and draw an arc with center P on either above or below side of AB (v) With center Q draw an arc of same radius. (vi) Both arcs intersect each other at point P. (vii) Joins R to D. (viii) DR is perpendicular to AB at D. Q.5 If mKL = 4.5cm and mPQ = 3.5cm .Draw AB such that mAB = mKL+ mPQ and mEF = mKL mPQ Solution: Same as question 1. Q.6 m AB = 7cm draw a perpendicular to a point C on AB such that AC = 3cm . Construction: (i) Draw mAB 7cm (ii) Take a point C on AB such that AC 3cm (iii) With center C draw two arcs of radius 1cm on both sides of C which intersect AB at P and Q. With center P, draw arc of some suitable radius on either side above or below of AB. (iv) With same radius draw an arc from Q. (v) Both arcs intersects each other ait R. (vi) Join E to R. (vii) AB is required perpendicular of AB at C. Q.7 A line segment AB is 8cm long. I am intersected to draw a bisector PQ of which is perpendicular to AB . Guide me what steps I will follow. Construction: (i) Draw AB 8cm. (ii) Adjust the compass slightly longer than half of AB and with center a draw the arcs on both sides of AB. (iii) With center B, draw arcs of same radius on both sides of AB which intersect previous arcs at P and Q.
CSS Primary Standard “Mathematics” 206 (iv) Join P with Q. (v) is required perpendicular bisector of AB. Q.8 Take a point on each segment and draw perpendicular to the given line segment (use ruler and compass only) (i) 5.3cm Same as question 4 (ii) 7.2cm Same as question 4 (iii) 2.9cm Same as question 4 (iv) 8.9cm Same as question 4 Q.9 Draw CE whose length is equal to difference of length of CD and EF given is figure. Construction: (i) Draw CD = 8cm (ii) With center D, draw an arc of radius 3cm = EF across lift, which intersect CD at point E. hence CE is required line segment whose length is 5cm. Exercise 10b Q.1 Construct the following angles with help of a protector. (i) 56° Construction: (i) Draw a ray AB. (ii) With center A, draw an angle of 56° with help of protector. (iii) CAB = 56° is required angle. (ii) 110° Construction: (i) Draw a ray AB. (ii) with center A, draw an angle of 110° with help of protector.
CSS Primary Standard “Mathematics” 207 (iii) CAB = 110° is required angle. (iii) 85° Construction: (i) Draw a ray AB. (ii) With center A, draw a angle of 85°. (iii) CAB is required angle. (iv) 120° Construction: (i) Draw a ray AB. (ii) With center A, draw an angle of 120° with help of protector. (iii) CAB = 120° is required angle. Q.2 Draw the following angles using protector and bisect them using compass. (i) 135° Construction: (i) Draw a ray AB. (ii) With center A, draw an angle of CAB 135° with help of protector. (iii) With A as enter draw an arc of suitable radius which cut CAB at E and F. With center E draw an arc of radius slightly larger than half of EF. (iv) With center E draw an arc of same radius. (v) Both arcs intersect each other at D. (vi) Join A to D. (vii) AD is requires bisector of CAB. (ii) 90° Construction: (i) Draw an angle CAB = 90° with help of protector. (ii) With center A, draw an arc which intersect CAB at E and F. (iii) With center E, draw an arc of radius slightly larger than half of EF.
CSS Primary Standard “Mathematics” 208 (iv) With center F, draw an arc of same radius. (v) Both arcs intersects each other at D. (vi) Join A to D. (vii) AD is required bisector of CAB. (iii) 30° Same as part (i) and (ii) (iv) 105° Same as part (i) and (ii) (v) 165° Same as part (i) and (ii) (vi) 1 67 ° 2 Same as part (i) and (ii) Q.3 Divide the following angles into four equal. Darks by compass but draw them with protector. (i) 110° Construction: (i) Draw CAB = 110° with help of protector. (ii) Bisect the given angle <CAB into two equal angles <CAD, <DAB by using compass. (iii) AD is bisector of CAB. (iv) With center O and F, draw arcs of suitable radius, which intersect each other at T. Join A to T. (v) With center O and G. Draw arcs of suitable radius which cut each other at S join A to S. Now CAB = 110° has been divided into four equal angles. i.e mCAT = mCAD = mDAS = mSAB (ii) 150° Construction: (i) Draw an angle CAB = 150° with help of protector. (ii) Bisect CAB into two equal angles CAD and DAB using compass. (iii) AD is bisector of CAB. (iv) With center O and F draw arcs of suitable radii which intersect each other at T. Join A to T.
CSS Primary Standard “Mathematics” 209 (v) With center O and G draw arcs of suitable radii which intersect each other at S. Join A to S. (vi) Now given angle CAB = 150° has been divided into four equal angles m CAT = m TAD DAS SAB (iii) 140° Same as (i) and (ii) (iv) 85° Same as (i) and (ii) Q.4 Draw PQ = 5cm. construct angles of 60° at P then construct below angles on Point Q using protector and bisect them using compass. (i) 150° Construction: (i) Draw PQ = 5cm (ii) With center P, draw an angle of 60° with help of protector. (iii) With center Q, draw an angle of 150° with help of protector. (iv) Both angle intersect each other at R. (v) Bisect the angles P and Q with help of compass. (vi) QS is required bisector of Q = 150° and DT is required bisector of P = 60° (ii) 1 67 ° 2 Construction: Same as (i) (iii) 120° Construction: Same as (i) (iv) 105° Construction: Same a (i) Q.5 Draw the following angles. (i) 135° Construction: (i) Draw a point A and make a ray AB. (ii) With center B, draw an arc, of suitable radius, which intersect AB at point E. With
CSS Primary Standard “Mathematics” 210 center E and without changing the radius, draw an arc to intersect the first arc at F. With center F and without changing the radius, draw and arc to intersect the first arc at G. With center G and without changing the radius, draw an arc to intersect the first arc at H. Bisect GH through I. With center I and G draw arcs of suitable radius which intersect each other at C. Join A to C. Hence m CAB = 130° is required angle. (ii) 165° Construction: Draw a point A and make a ray AB. With center A draw an arc of some suitable radius which interest AB at E. With center draw an arc of same radius which intersect the first arc at F. With Center F, draw an arc of same radius which interest the first arc at G. with center G, draw an arc of same radius which intersect the first arc of same radius which intersect the first arc at H. Bisect G H and get point I. With center I and G, draw arcs of some suitable radii which intersect each other at C. Join A to C. mCAB = 165° is required angle. (iii) 75° Construction: Draw a point A and make a ray AB. With center A, draw an arc of suitable radius which intersect AB at E with canter F draw and arc of same radius, which intersect the first arc at F. With center F, draw an arc of same radius which intersect the first arc at G. Bisect FG and get point I. With center I and F, draw arcs of some suitable radii, which intersect each other at C. Joint A to C hence CAB = 75° is required angle. (iv) 105° Construction: (i) Draw a point A and make a ray AB. With center A, draw an arc of some suitable radius, which intersect AB at E with compass. With center E, draw an arc of some radius which intersect first arc at F. With center at F, draw an arc of same radius which intersect the first arc at G. Bisect EG and get point I. With
CSS Primary Standard “Mathematics” 211 centers I and G draw arcs of some suitable radii which intersect each other at C. Join C to A CAB = 105° is required angle. (v) 15° Construction: (i) Draw a point A and make a ray AB. (ii) With center A, draw an arc of suitable radius which intersect AB at E. With center at E, draw an arc of same radius, which intersect the first arc at F. Bisect EF and get point G. Taking centers G and E draw arcs of suitable radius which intersect each other at C. Join C with A. CAB = 15° is required angle. (vi) 1 22 ° 2 Construction: (i) Make DAB = 135° with help of compass. (ii) Bisect DAB with help of compass. (iii) AC is the bisector of DAB . (iv) Hence 1 CAB = 67 2 is required angle. (vii) 120° Construction: (i) Draw a point A and make ray AB. (ii) With center A, draw an arc of suitable radius, which intersect AB at E. (iii) With center E, draw an arc of same radius which cut the first arc at F. (iv) With center F, draw an arc of same radius which cut first arc at G. (v) Through G draw CA . (vi) CAB = 120° is required angle. (viii) 150° Construction: (i) Draw a point A, and make AB. (ii) With center A draw an arc of suitable radius, which cut AB at E.
CSS Primary Standard “Mathematics” 212 (iii) With center E, draw an arc of same radius, which intersect the first arc at point F. (iv) With center F, draw an arc of same radius which intersect the first arc at G. (v) With center G, draw an arc of same radius which intersect the first arc at H. (vi) With center G and H, draw the arcs of same radii, which intersect each other at C. DAB = 150° is required angle. (ix) 45° Construction: (i) Draw a point A and make ray AB. (ii) With canter A, draw an arc of suitable radius, which intersect at E. (iii) With center E, draw an arc of same radius which intersects the first arc at F. (iv) With center F, draw an arc of same radius which intersect the first arc at G. (v) With center F and G, draw the arcs of same radii, which intersect each other at I. each other at I. Join I to A. (vi) AI passes through J. (vii) With center E and J, draw arcs of same radii which intersect each other out C. Join A to C. (viii) CAB = 45° is required angle. (x) 1 22 ° 2 (i) Draw angle of DAB of 45° with help of compass. (ii) Bisect DAB with help of compass. (iii) AC is bisector of DAB. (iv) 1 CAB = 22 2 is required angle. Q.6 Construct angles of the following with help of compass and divide these angles into fur equal angles. (i) 120° Construction: Construction is same as Q# (vii), and divide it into four equal parts by using the procedure explained in question #3. (ii) 60° Construction:
CSS Primary Standard “Mathematics” 213 (i) Draw a point A and make AB. (ii) With center A, draw an arc of suitable radius which cut AB at E. (iii) With center F, draw an arc of same radius, which intersect first arc at F. (iv) CAB = 60° is required angle divide it into four equal parts by using procedure explained in Q # 3. (iii) 150° Construction: Construction is same as Q#5 (viii) and divide it into four equal parts by using procedure explained in Q#3. (iv) 135° Construction is same as Q#5 (i) and divid it into four equal parts by using procedure explained in Q#3. Exercise 10c Q.1 Draw AB = 4cm construct angles of 60° and 45° at point A and B and draw a triangle. Construction: Draw AB 4cm with center A draw an angle of 60° and with center B draw an angle of 45° both angle intersect each other at C. Hence ABC is required triangle. Q.2 Draw AB = 6.4cm. Construct angles of 30 and A and B respectively and draw triangle. Construction: (i) Draw AB = 6.4cm. (ii) With center A, draw angle of 30° and an with center B, draw an angle of 90°. Both angles intersect each other at c. ABC is required triangle. Q.3 Construct a triangle, when: (i) mPQ = 7cm,mQR = 3cmandmRP = 6cm Construction: (i) Draw PQ = 7cm.
CSS Primary Standard “Mathematics” 214 (ii) With center P, draw an arc of 6cm and with center Q draw an arc of 3cm. Both arc intersect each other R. Hence PQR is required triangle. (ii) mAB = 4.6cm, mBC = 3.4cm and mCa = 2.5cm Construction: (i) Draw AB = 4.6cm (ii) With center A, draw an arc of radius 2.5cm. (iii) With center B, draw an arc of radius 3.4cm. (iv) Both arcs intersect each other at C. (v) Hence ABC is required triangle. Q.4 Draw XY =7.2cm. construct angles of 1 22 ° 2 and 7° at X and Y complete the triangle. Construction: (i) Draw XY = 7.2cm. (ii) With center X, draw angle 1 22 ° 2 with help of compass. (iii) With center &, draw an angle of 75° with help of compass. (iv) Both angles intersect each other at Z. (v) Hence XYZ is required angle. Q.5 Construct ABC when: (i) mAB = 5cm,m B = 45°andmBC = 4cm Construction: (i) Draw AB 5cm . (ii) Draw an angle of 45° with help of compass at point B. (iii) With center B, draw an arc of 4cm, which intersect B at C. (iv) Join C with A. (v) ABC is required triangle.
CSS Primary Standard “Mathematics” 215 (ii) mAB = 5.3cm,m A = 60°andmBC = 4.2cm Construction: (i) Draw AB =5cm. (ii) Draw an angle of compass at A. (iii) With center A, draw an arc of radius 4.2cm which intersect B at C. (iv) Join C to B. (v) ABC is required triangle. Review Exercise Q.1 Choose the correct answer and full the circle. i. A closed three sided figure is called and a ____________: angle triangle quadrilateral any of them ii. When two rays meet together they formed a/an ______: angle bisecting angle triangle all iii. A ______is part of line which is made by joining two points together. It has start and end point: line line segment triangle none iv. ____________ is longest side of a night angle triangle. The side opposite of right angle: hypotenuse multi tense bisecting angle all v. When the we bisect 90° . It gives an angle of ____________? 75° 45° 60° 115° vi. When we double the angle we bisects the angle? No yes why none Q.2 Draw PQ and RS such PQ =3.6cm and mPQ = 1.9cm. Draw KL where KL = mPQ + mRS Construction: Take a line OA . With center K on OA , draw an arc of 3.6cm = PQ whic h inters ect OA at point C. With center C, draw an arc of radius 1.9cm = RS which intersect. OA at point L. Join K to L. Hence KL PQ RS is required line segment. Its length is 5.5cm. . mAB = 2cm, mCD = 2.8cmand mEF = 1cm.Draw XY where mXY D = raw AB,CDand F mAB + mCD mEFM Such easu that re XY
CSS Primary Standard “Mathematics” 216 Q.3 Construction: Take a line OL. Draw a point X on OL . With center X, draw an arc of 2cm = mAB which intersect OL at point K across right. With center K, draw an arc of 2.8cm = mCD a cross right which intersect OL at point M with center M draw an arc of radius 1cm = EF across left, which intersect OL at y. Join X to Y. XY is required line segment such that mXY=mAB+mAB mEF . The length of is 3.8cm. Q.4 Draw LM = 5cm long. Cut off LN where mLN = 3cm. Measure length of NM . Construction: Draw LM = 5cm. With center L, draw an arc of radius 3cm which intersect LM at point N. NM LM LN is required line segment and the length NM = 2cm. Q.5 Using ruler and compass only draw a triangle PQR, where: (i) mPQ = 7.5cm, mQR = 4cmandPR = 6cm Construction: (i) Draw PQ 7.5cm . (ii) with center draw an arc of 4cm radius. (iii) With center P, draw an arc of 3.5cm radius. (iv) Both arcs intersect each other at point R. (v) Join R to P and Q. (vi) PQR is required triangle . (ii) mPQ = 4.5cm, mQR = 3.6cmandPR = 2.8cm Construction: Same as Q#5(i) (iii) mPQ = 3.8cm, m P = 30°andPR = 2.3cm Construction: (i) Draw PQ=3.8cm. (ii) With center P, draw and angle of 30° with help of compass.
CSS Primary Standard “Mathematics” 217 (iii) With center of P draw an arc of 2.3cm radius, which intersect P at R. (iv) Join R to Q. (v) Hence PQR is required triangle. (iv) mPQ = 5.2cm,mÐP = 45°andmPR = 2.5cm Construction: Same as Q #5 (iii) (v) mPQ = 7.2cm, m P = 60°andPR = 75° Construction: (i) Draw PQ=7.2cm. (ii) With center P, draw an angle of 60° with help of compass. (iii) With center Q, draw an angle of 75° with help of compass. (iv) Both angle intersect each other at R. (v) Hence PQR is required triangle. Q.6 Draw ABC such that m ABC = 60° using ruler and compass only. (i) Draw the bisector of ABC . Construction: (i) Draw ABC with help of compass. (ii) With center E and F draw arcs of same radii, which intersect each other at D. (iii) Join B to D. (iv) BD is required bisector of ABC (ii) Construct angle in measure of ABC . Construction: Same as Q# 10, 2, i (iii) Construct an angle twice in measure of ABC . Construction: Same as 10, 2, ii (iv) Divid ABC into four equal angles.
CSS Primary Standard “Mathematics” 218 Construction: Same as (Use of compass to divide a given angle into four equal angles). Q.7 Draw AB = 5.8 cm long. Take a point C on AB such that m AC =2.6cm. At C, draw perpendicular to AB by using ruler and compass only. Construction: (i) Draw AB = 5.8 cm. (ii) With center A, draw an arc of 2.6cm, which intersect AB at C. (iii) With center C draw two arcs of 1cm radii on both sides of C, which intersect AB at D and E. (iv) With center D and E, draw arcs of same radii on both sides of AB. (v) These are intersect each other at points P and Q. (vi) Join P to Q. (vii) AB is required perpendicular of AB passing through C. Q.8 Draw PQ = 4cm long. Take a point R on PQ such that mPR =1.9cm. At R, draw a perpendicular to PQ by using ruler and compass. Construction: Same as Q# 7. Q.9 Draw AB = 5.9cm long. Take a point C outside AB and draw perpendicular to AB form C (use ruler and compass). Construction: Same as Q# 3 (Exercise 10 a) Q.10 Draw CD = 5.2cm long. Draw its right bisector with help of ruler and compass. Construction: Same as Q#7 (Exercise 10a) Q.11 Construct the given angles using ruler and compass only: (i) 135° Construction: Same as Q#5 (i) Exercise 10b. (ii) 105° Construction: Same as Q#5 (iv) Exercise 10b. (iii) 180° Construction: (i) Draw a point A and make ray AB.
CSS Primary Standard “Mathematics” 219 (ii) With center A, draw an arc of suitable radius, which intersect AB at E. (iii) With center E, draw an arc of same radius which intersect the first arc at F. (iv) With center F, draw an arc of same radius, which intersect first arc at G. (v) With center G, draw an arc of same radius, which intersect radius, which intersect first arc at H. (vi) Draw AC passing through H. (vii) CAB = 180° is required angle. (iv) 90° Construction: Same as (construction of an angle of 90°) (v) 45° Construction: Same as Q#5 (x) Exercise 10b. (vi) 15° Construction: Same as Q#5 (v) Exercise 10b. Model Paper 2nd Term Model Paper No. 1 Section – A: Multiple Choice Questions Marks: 40 Time: 50 Minutes Roll No. __________ Choose the correct option. 40 1. The integer was introduced in the year. 1463 1563 1663 1763 2. Additive inverse of 0 is __________. 1 –0 0 1 0 3. |–5| = __________. –5 5 5 5 4. Sum of two negative integers is __________ :
CSS Primary Standard “Mathematics” 220 positive negative both None 6. (+35) + (–25) = 45 –45 10 –10 7. (–13) × (+5) –65 65 45 75 8. 5 × __________ = –30 6 –6 15 10 9. (–169) ÷ (–13): 10 –10 13 –13 10. __________ ÷ (–18) = 25. 450 –450 35 25 11. 12 3 5 7 __________: 89 35 99 35 36 35 17 35 12. 3 7 of 49 is: 63 14 9 21 13. 42.3 – 7.48 35.82 42.82 34.82 36.82 14. 144 100 is __________: 144 36 25 14.4 .144 15. BODMAS stand for: bracket observation, division, multiplication, addition, subtract bracket operation, division, multiplication, addition, subtraction all of them none of them 16. 1.48 + (18.4 – 12.3) = 7.58 5.58 7.52 8.56 17. The simplest form of 75:90 is: 6:5 5:6 5:4 4:6 18. In expression a b a is called __________: denominator antecedent consequent none
CSS Primary Standard “Mathematics” 221 19. 1 1 : 4 12 into the whole number ratio is: 1:3 3:1 4:3 3:4 20. The ratio of Ali’s monthly in came to expenses is 3:5. If the expenses is Rs. 7000, then the income end saving are: 11200, 4200 10000, 3000 12000, 5000 12450, 5450 21. If 30:5::x:4, then x = 1 4 4 37.5 3.7 22. If the cost of 13kg mangoes is Rs.1950,, what is price of 18kg of mangoes. Rs. 2700 Rs. 2500 Rs. 2300 Rs. 2100 23. If 7 girst and 9 boys earn Rs. 630 per day. How much would 9 girls earn per day? 630 540 810 750 24. If 12 men can do a piece of work in 8 days. How many days require to do it by 16 men? 4 days 7 days 6 days 5 days 25. If 13 pencils cost Rs. 312, how many pencils can be bought for Rs. 3367? 12 13 14 15 26. 5 1 4 3 4 3 1 10 12 11 12 27. If 8:x::4:2, then x= 10 4 16 1 4 28. Which of the following is equal to 7 20 ? 3.5 14 to 100 35% None 29. 37.5%is equal to: 375 to 50 375 to 100 .375 4 37.5 100 30. Percentage of 7 to 50 is : 18% 16% 14% 12% 31. 24% =: 7 25 4 8 6 25 9 25 32. Profit % = Profit×100 C.P Profit×100 S.P 100 ×S.P 100+profit% None 33. Hina sold a table for Rs. 2400 at the profit of 700. Find the price at which she bought it: Rs.1600 Rs.1700 Rs.1500 Rs.1800
CSS Primary Standard “Mathematics” 222 34. If a monitor was sold for Rs.5400 at the loss of Rs.600. Find the cost of the monitor. Rs.4800 Rs.6000 Rs.4200 Rs.6200 35. What is added to 4a2 – 3a + 9 to get a2 + 4a –7? –3a2 + 7a –16 3a2 – 7a + 16 3a2 + 7a – 16 –3a2 – 7a + 16 36. 5x is __________ type of allegoric expression: monomial binomial trinomial None 37. 7 times a number decreased by 14 equals to 2: 7x + 14 = 2 7x – 2 = 14 14 + 7x = 2 7x – 14 = 2 38. If 4 8 y + 4 = 10, then y= 12 14 18 8 39. A line perpendicular to a line segment, which divide it into two equal parts is called. perpendicular bisector right bisector None 40. The side opposite to right angle in a right angle triangle is called: perpendiculer base hypotenoes bisecting angle Section – B: Constructed Response Questions Marks: 60 Time: 2 hours 10 minutes Attempt all questions. Each question carries equal marks. Q.1: Find the sum of the following in three steps method. (i) –85316, –42539 (ii) 35467, 39513 (iii) –93568, 53215 Q.2: Simplify the following: (i) (–18) × (–5) × (–7) × 9 × 11 (ii) 25 × (–15) × (–16) × (–3) × (–12) Q.3: simplify the following: 130 – [140 – {13 + (27 + 3 –12)}] Q.4: Sana’s income is Rs. 75,000. She paid 1 10 of her income as house rent, 1 15 of her income on gass bill, 4 7 on other expenditure and remaining she saves calculate what amount she save? Q.5: In a boarding school food for 180 students is sufficient for 24 days. 60 more students join them, for how many days the same food sufficient for the students? Q.6: A certain sum of money is divided among Ali, Amna and Ahmad in the ratio 3:4:5. If Amna’s share is Rs. 24000, find the share of Ali and Ahmad. Q.7: During a summer mega sale an electricity brand reduce the prices of all its goods by 20% calculate the original selling price of the refrigerator which was sold for Rs. 32500 during sale?
CSS Primary Standard “Mathematics” 223 Q.8: The length of a rectangular pool is 8cm more than its width and perimeter of rectangle is 38 cm. What is its length and width? Q.9: Simplify the following expression: (3x2 yz + 7yz) + [4yz + {13x2 y + y + 9y2 – (3x2 yz + 7)}] Q.10: Draw AB = 4cm. Construct angles of 60° and 45° at point A and B and draw a triangle. Model Paper No. 2 Section – A: Multiple Choice Questions Marks: 40 Time: 50 Minutes Roll No. __________ Choose the correct option. 40 1. Negative numbers were accepted into number system in __________. 17th 18th 15th 19th 2. (–3) + (–5) = –8 8 –2 2 3. (+20) – (+15) = 35 –35 5 –5 4. (–4) × (–17) : 64 –64 72 84 5. __________ × (–2) = 24 –12 12 6 8 6. 4 × (–3) × 8 = –84 76 88 –96 7. (–300) ÷ 2 = 150 200 –150 100 8. 0 ÷ 400 = __________ 440 0 1 None 9. __________ ÷ 12 = –15. 160 –180 –150 3 2 10. 2:4 ÷ 0.3 0.8 0.08 8 4 11. 1 8 of 48,000: 5,000 6,000 4,000 7,000 12. 34.4 + 26.8
CSS Primary Standard “Mathematics” 224 61.2 62.8 7.6 63.2 13. [ ] is called __________: vinculum square bracket braces parranthese 14. 3.54 – (12.5 + 3.4 – 1.2) = 11.16 11.19 –11.16 11.30 15. 1800 100 180 18 1.8 .18 16. 7.8 × 1.3 = 12.14 11.14 10.14 10.24 17. 3 1 5 5 0.675 0.565 0.575 0.535 18. The simplest form of 40:80 is: 4:8 8:4 1:2 2:1 19. In expression a b , b is called: numerator antecedent consequent none 20. 1 1 : 16 20 into the whole number ratio is __________ : 4:5 5:4 3:4 5:3 21. If in a book shop the ratio of science books to Urdu books, is 4:5. If there to are 2000 books of Urdu what will be the number of science books. 1800 1600 1400 1200 22. If x:20::7:10, then x = 7 14 1 7 45 23. If Irfan earns Rs.1015 per week. In how many days he earns Rs. 2610. 16 14 18 22 24. If the cost of 14 milk packs, each weighing 1kg, is Rs. 1540. Find the cost of 16 milk packs of the same kind and weight. 1460 1560 1660 1760 25. Percentage 12 to 25 is: 38% 48% 52% 40% 26. 48% = 9 25 6 25 12 25 14 25
CSS Primary Standard “Mathematics” 225 27. Loss % = Loss 100 C.P Loss×100 S.P 100 C.P 100 Loss% None 28. Ahmad sold a camera worth Rs. 30,000 at a loss of Rs. 12,000. Find the selling price of camera. Rs. 42,000 Rs. 18,000 Rs. 15,000 Rs. 48,000 29. At Eid, Anoral gave gifts to 36 out of 100 friends and Alishaba gave 9 out of 25 friends what percent of each friend received gift. 42% 30% 36% 9% 30. If 20% discount on a shalwar kameez is Rs. 1500, find the marked price of it. 6000 7500 7200 8000 31. 123xy + 21xy __________ 144xy: > < = None 32. 7×30 __________ 200: > < = None 33. 3a + 7y is an allegoric: expression sentence equation terms 34. 3 items a number increased by 5: 3x + 5 5x + 3 3x – 5 3 – 5x 35. Equation 3x + 5 = 29 for x =? 7 6 8 9 36. A mathematical sentence with an equality sign is called: allegoric expression equation symbol 37. We bisects < 270°. It gives an angle of __________? 175 105 135 95 38. When two rays meet together they form a/an __________: angle triangle bisecting all 39. The distance between two point on a line is called: triangle ray line line segment 40. A closed three sided figure is called __________: rectangle square triangle angle Section – B: Constructed Response Questions Marks: 60 Time: 2 hours 10 minutes Attempt all questions. Each question carries equal marks. Q.1: For the following numbers write ascending and descending order and also arrange the absolute. Value of integer in descending order. (i) 845,739,910 – 1100 – 500 (ii) 4578,4138,4258 – 7853,5678 Q.2: Find the sum of the following in three steps method:
CSS Primary Standard “Mathematics” 226 (i) 25317, –59321 (ii) 95473, 843712 (iii) –43876, 63791 Q.3: simplify the following: 875.32 + 12.8 × [4.32 + {1.5 (7.8 × 4 –3)}] Q.4: Humaira had ot travel from Islamabad 10 Lahore which is 392km. She travel 7 9 of the distance by bus, 1 13 by train and remaining on local van. Find the distance. She travelled by bus train and van. Q.5: 35 men got a project to construct a house in 24 days. 13 of them went on leave, in how many days the remaining men will construct house? Q.6: Anoral has pencils of Rs. 10, Rs. 25, Rs. 35. The ratio of these pencils is 3:5:2 respectively and total amount of all kind of pencils is Rs. 200. Find the number of pencils of each kind. Q.7: A company finds that 3 5 % 6 of their bike made in the year 2016 to 2018 has some problems in FGS. The company made 15000 bikes in this time duration. How many bikes were defective. Q.8: Find three consecutive numbers in which x is middle number and their sun is 99. Find the numbers. Q.9: Subtract the following: (i) 3x2 y + 4y2 z – 9zx from –9x2 y –7y2 z + 6zx (ii) a 4 –9b2 + 6a2 b from 7a4 – 3b2 + 3a2 b (iii) –5a from a + 5 Q.10: Draw CD = 5.2cm long. Draw its right bisector with help of ruler and compass. Model Paper No. 3 Section – A: Multiple Choice Questions Marks: 40 Time: 50 Minutes Roll No. __________ Choose the correct option. 40 1. Europe started using the negative number in __________. 1445 1554 1545 1645 2. (+17) + (–19) = 2 –2 –36 36
CSS Primary Standard “Mathematics” 227 3. (–135) – (–455)= –590 –320 320 None 4. 24 × 12 = 268 288 320 358 5. –7 ×__________ = –82 –12 14 16 12 6. 3 × (–5) × (18) × (–4) = 1000 –1080 1080 1320 7. (–369) ÷ 3 = 125 –123 129 139 8. –1600 ÷ __________ = 40 40 –40 30 50 9. __________ ÷ 19 = –25. –675 575 –475 375 10. Integers were denoted by N W I Z 11. Absolute value is the distance of a number form __________: total sum 1 zero number line 12. 2 9 of 63 is: 10 21 14 27 13. 32.1 + 4.81: 37.91 36.81 36.91 39.81 14. 7 1 8 9 70 72 52 72 60 70 71 72 15. 3.8 × 9.3 = 34.34 42.36 35.34 24.92 16. 154 100 is 1.54 0.154 15.4 154 17. 2.16 – (3.2 + 1.9) = 0 – 2.94 8.6 0.86 18.6 18. 1 3 of 9000: 3000 300 4500 4200 19. BODMAS stand for: bracket observation, division, multiplication, addition, subtraction
CSS Primary Standard “Mathematics” 228 bracket operation, division, multiplication, addition, subtraction all of them none of them 20. 25% of 600 : 120 150 180 300 21. Find the ratio of marks in maths of Anoral and Alishaba. Anoral got 65 marks and Alishabs got 80 marks. 12:16 16:13 14:16 13:16 22. The lowest form of ratio 630 700 is: 10 9 5 7 9 10 8 7 23. Which of the following is equal to 6 25 ? 4.3 24% 12 to 100 none 24. Selling price = 100 ×S.P 100+profit% 100 ×S.P 100 +Loss% 100 Loss%×C.P 100 none 25. Hina sold a table for Rs.2400 at the profit of 700. Find the price at which she bought it. Rs. 1600 Rs. 1700 Rs. 1500 Rs. 1800 26. If a monitor was sold for Rs. 5400 at the loss of Rs. 600. Find the cost of monitor. Rs. 4800 Rs. 6000 Rs. 4200 Rs. 6200 27. 32% = 7 25 9 25 8 25 6 25 28. Percentage of 8 out of 40 is. 15% 30% 20% 25% 29. 28.5% = 2.85 285 to 100 57 to 200 none 30. Ali gets Rs. 25,000 salary per month. If his salary is increased by 10%, how much will be get? 25,750 26,500 27,500 27,000 31. 2P + 5P __________ 7P: > < = None 32. 72 ÷ 8 __________ 7: < > = None 33. What is added to 3a + 5 to get 7a – 9: 4a – 6 3a – 14 4a – 14 10a – 4 34. 3x + 5 = 7 is an algebraic __________. expression equation sentence term
CSS Primary Standard “Mathematics” 229 35. A number multiplied by 3 increased by 8: 3x + 8 3 + 8x x 8 3 3(x + 8) 36. Two fifth of a number subtracted from 7: 2x 7 5 2 7x 5 2x 7 5 2 7 x 2 37. If 5x – 5 = 20, then a = ? 4 15 3 5 38. The word triangle cones from __________: Greek French English Latin 39. An angle 15° is obtain by dividing 60° into __________ equal parts: 2 3 4 5 40. The longest side in a right angle triangle opposite two right angle is called: base perpendicular hypotenuse none Section – B: Constructed Response Questions Marks: 60 Time: 2 hours 10 minutes Attempt all questions. Each question carries equal marks. Q.1: Simplify the following: (i) 25 × (–13) × (–9) × (–7) × 8 × 2 (ii) (–12) × (+20) × (–15) × 3 × (–8) × 17 Q.2: For the following numbers write ascending and descending order and also arrange the absolute value of integers in descending order. (i) 235, 453, 310, –900, –750 (ii) 2315, 2158, 2437, –3415, –1100 Q.3: simplify the following: 1 1 8 12 100 12 13 9 8 3 7 5 Q.4: Alishba has Rs. 15000 she gave 3 7 to Anoral. Anoral gave 1 8 of her won share to her friend Sana. How much anoint Sana got? Q.5: The ratio of numbers of male and female employees in a company is 4:5. If there are 2400 male employees, find the number of female employees. Q.6: If 2 women by a micro for Rs. 6000 each. First woman shell the micro after one year for Rs. 500 gain and second woman sell the micro in Rs. 1000 Loss. Find the selling. Q.7: Age of father is 3 times the age of his daughter. If father is 18 years old then his daughter, then find the age of father and daughter? Q.8: Ibrahim purchased 32 dozens of pencils at rate of Rs. 288 per dozen. He sold each one of them at rate of Rs. 27, what was his percentage profit?
CSS Primary Standard “Mathematics” 230 Q.9: Simplify the following expression: 3x2 y – [7xy + {3y2 + x + 9z – 2 (3x2 y + 4xy – 2x)}] Q.10: Construct ABC , when: mAB=4cm, mBC=3.5, m =60° Week Unit: 11 Perimeter and Area Exercise 11a Q1. Complete this table. Length Width Area Perimeter A 7cm 10.1cm 34.2cm B 3.1cm 5.3cm2 C 4.1cm 44.8cm2 D 2.8cm 8.4cm2 For A Area = length × width = 7 × 10.1 = 70.1cm2 For B Area = length × width 5.3 = 3.1 × width 5.3 3.1 = width 1.71 width width = 1.71cm Perimeter = 2(length + width) = 2 (3.1 + 1.71) = 9.62 For C Area = length × width
CSS Primary Standard “Mathematics” 231 44.8 = length × 4.1 44.8 4.1 = length 10.93 = length length = 10.93 Perimeter = 2(length + width) = 2(10.93 + 4.1) = 2(15.03) = 30.05 For D Area = length × width 8.4 = length × 2.8 8.4 2.8 = length 3 = length length = 3 Perimeter = 2(length = width) = 2(3 + 2.8) = 2(5.8) = 11.6 Q2. Find the area and perimeter of each square. (i) x = 3.3mm Area of square = x2 = 3.32 = 10.89 Perimeter of square = 4x = 413.37 = 13.2 (ii) x = 2.8mm Area = x2 = 2.82 Perimeter = 4x = 4(2.8) (iii) x = 4.8 Area = x2 = 4.82 Perimeter = 4x
CSS Primary Standard “Mathematics” 232 = 4(4.8) (iv) x = 8.8mm Area = x2 = 8.82 = mm 2 Perimeter = 4x = 4(8.8) = 30.2mm (v) x = 7.3mm Area = x2 = (7.3)2 = mm 2 Perimeter = 4x = 4(7.3) = mm (vi) x = 4.3m Area = x2 = 4.32 = m 2 Perimeter = 4x = 4(4.3) = m Q3. Complete the following rectangle. (i) A = P = l = 5cm w = 3cm As area = l × w = 5 × 3 = 15cm2 P = 2 (l × w) = 2(5 + 3) = 2(8) = 16cm (ii) A = P = 20
CSS Primary Standard “Mathematics” 233 w = 4cm l = As P = 2(l + w) 20 = 2(l +4) 10 20 2 4 l 10 = l + 4 10 – 4 = l 6 = l l = 6 A = l × w = 6 × 4 = 24cm2 (iii) A = 20cm2 P = w = 4cm l = As A = l × w 20 = l × w 20 4 = l 5 = l l = 5 P = 2(l + w) = 2(5 + 4) = 2(9) = 18 (iv) P = 18cm A = l = 7cm w = As P = 2(l + w) 18 = 2(7 + w)
CSS Primary Standard “Mathematics” 234 9 18 2 7 w 9 – 7 = w w = 2 Aria = l × w = 7 × 2 = 14 Exercise 11b Q1. How many square tile of size 81cm will be needed to fit in a square floor if a bathroom of side of 720cm. Find the cost of tiling of Rs. 75 per tile Solution Area of tiles = a2 = 81cm2 Area of floor = 7202 = 51.8400cm2 No. of tiles = 518400 8 = 6400 Cost of tilling = Rs. 75 × 6400 = Rs. 480000 Q2. If it cost Rs. 1600 to fence a rectangular park of length 20m at the rate of Rs. 25 per m2. Find the width of park and its perimeter. Also find the area of field. Solution Length of park = 20m Cost of fencing the park = Rs. 25 × perimeter Rs. 1600 = Rs. 25 × perimeter Primeter = 1600 25 Perimeter = 64 perimeter = 64 As perimeter = 2(length + width) 64 = 2(length + width) 32 = 20 + width width = 32 – 20 width = 12 Now Area = length × width = 20 × 12 = 240m2 Q3. Danial has a rectangle rose garden that measure 8m by 10m. One bag of fertilizer can cover 16m2 . How many bags will be needed to cover the entire garden?
CSS Primary Standard “Mathematics” 235 Solution Area of rectangle garden = 8 × w = 80m2 One bag of fertilizer cover = 16m2 Then the bags needed to cover the entire garden = 80 16 = 5 Q4. The area of rectangle is 117 sq meter the width is 9m, what is its length? Solution Given area of rectangle = 117m2 width = 9m As area of rectangle = length × width 117 = length × 9 Length = 117 9 = 13m Q5. The perimeter of a square country yard is 144m. Find the cost of cementing it at the rate of Rs. 50 per m2 . Solution Cost of cementing = 50 × 144 = Rs. 7200 Q6. A square swimming pool has length 20m. Find the area and perimeter of pool. Solution Area of square swimming pool = length × length = 20 × 20 = 400m2 Perimeter of square swimming pool = 4 × 20 = 80m Q7. A man has square space that measures 37m on each side. His house is 9m by 11m. If the rest of space is a plot of grass. What area of grass does have now? Solution Area of square space = 37 × 37 = 1369m2 Area of house = 9 × 11 = 99m2 Area of rest of space = 1369 – 99 = 1270m2 Area of grass = 1270m2 Q8. Ali wants to run a wire around the space to keep his sheep in. if the space measures 110m by 165m, how much wire does he need? Solution
CSS Primary Standard “Mathematics” 236 Area of space = 110 × 165 = 18150 Perimeter of space = 2(110 + 165) = 550 The required wire = 550m Exercise 11c Q1. Find the Area of each triangle (i) Area of = 1 2 w × h 2 4 3 1 2 = 6 (ii) Area of = 1 2 w × h = 1 2 5.5 × 4 = 1 2 20.0 = 10 (iii) Area of = 1 2 w × h = 1 2 3.5 × 3.5 = 6.125 Q2. A parallelogram has sides 12cm and 9cm. If the shortest distance between its shortest sides is 8cm. Find the shortest distance between the longer sides. Area of parallelogram = 9 × 8 = 72 Also area of parallelogram = BC × BE 72 = 12 × BE BE = 72 12 = 6
CSS Primary Standard “Mathematics” 237 Shortest distance between longer sides = 6cm. Q3. Copy and complete the following. CD BF AD BE Area of ABCD (i) 9 6 7 6 (ii) 16 7 11 (iii) 13 9 11 w × h (i) Area of ABCD = Cd × BF = 9 × 6 = 54 Also area of ABCD = AD × BE 54 = 7 × BE BE = 54 7 = 7, 714 (ii) Area of ABCD = AD × BE = 7 × 11 = 77 Also area of ABCD = CD × BF 77 = 16 × BF BF = 77 16 = 4, 8125 (iii) Area of ABCD = AD × BF = 13 × 19 = 117 Also area of ABCD = AD × BF 117 = AD × 11 AD = 117 11 = 10.64 Q4. Find are of each parallelogram. (i) Area of parallelogram = b × h
CSS Primary Standard “Mathematics” 238 = 3.5 × 2.5 = 8.75cm2 (ii) Area of parallelogram = b × h = 2.5 × 4 = 10cm2 Exercise 11d Q1. The length of the parallel side of a trapezium are in the ratio 3:2, the distance between them is 10cm. If the area of trapezium is 325cm2 . Find the length of parallel side. Solution The ratio of length of parallel sides = 3:2 Distance between parallel sides = 10cm Area of trapezium = 325cm2 As area of trapezium = 1 2 height × sum of parallel sides 325 = 1 2 (10) × sum of parallel sides 325 = 5 × sum of parallel sides 325 5 = sum of parallel sides Sum of parallel sides = 65cm As the ratio of parallel sides are 3:2 So the length of first parallel sides = 3 5 × 65 = 39 Length of 2nd parallel sides = 3 5 × 65
CSS Primary Standard “Mathematics” 239 = 26 Q2. Find the area of trapezium whose parallel sides are CD = 12cm AB = 36cm and non-parallel sides are BC = 15cm and AD = 15cm and the distance between parallel sides is 9cm. Solution Sum of parallel sides = 12 + 36 = 48cm h = the distance between parallel sides = 9cm Area of trapezium = 1 2 h × sum of parallel sides = 1 2 9 × 48 = 216cm2 Q3. Copy and complete the area of each trapezium. b1 b2 h Area of ABCD (i) 6 9 7 ? (ii) 4 10 ? 42 (iii) 7 11 10 ? (i) Area of trapezium = 1 2 h × (b1 + b2) = 1 2 7 × (6 + 9) = 1 2 7 × (15) = 52.5 (ii) Area of trapezium = 1 2 h × (b1 + b2) 42 = 1 2 h × (4 + 10) 42 = 1 2 h × 14 42 = h × 7 h = 42 7 h = 6 Area of trapezium = 1 2 h × (b1 + b2)
CSS Primary Standard “Mathematics” 240 = 1 2 10 × (7 + 11) = 1 2 10 × 18 = 90 Review Exercise 11 Q1. Choose the correct answer and fill the circle. i. Area can be measured in: centimeter kilometer square meter fit ii. perimeter of a square is multiplication of: 3 2 4 5 iii. Area of foot path has always _________ area on the outer side as compare to its inner area: grater lesser equal none iv. An altitude of a geometry figure is the _______ distance from its top to its opposite side: longer shorter bigger longest v. A parallelogram is a quadrilateral in which the opposite pairs of sides are parallel and ________: non equal equal bisects none of them vi. A trapezium is a quadrilateral that has _______ of parallel sides: none only one pair both none of them Q2. A square park has sides of 250m. Fid the area of a jogging track 5m wide which is constructed on the inner side of its boundary. Solution Area of park = 250 × 250 = 62500cm2 Area of park without jogging area = 240 × 240 = 57600cm2 Area of jogging track = 62500 = 57600 = 57600 = 4900cm2 Q3. Find the area and perimeter of square for the following. Solution (i) l = 4.3cm Area = l × l
CSS Primary Standard “Mathematics” 241 = 4.3 × 4.3 = 18.49cm2 Perimeter = 4 × l = 4 × 4.3 = 17.2cm (ii) l = 73.3cm Area = l 2 = 17.32 = 299.29cm2 Perimeter = 4l = 4(17.3) = 69.2cm (iii) l = 3.2 Area = l 2 = 3.22 = 10.24cm2 Perimeter = 4l = 4(3.2) = 12.4cm (iv) l = 18.2cm Area = l 2 = 18.22 = 331.24cm2 Perimeter = 4l = 4(18.2) = 72.8cm (v) l = 5.7cm Area = l 2 =5.72 = 32.49cm2 Perimeter = 4l = 4(5.7) = 22.8cm (vi) l = 10.3cm Area = l 2 = 10.32 = 106.09cm2 Perimeter = 4l
CSS Primary Standard “Mathematics” 242 = 4(10.3) = 41.2cm Q4. If we have two parallel sides of a figure of length 4cm and 7cm respectively and their height is 5cm. Find the area of sign figure. Solution Sum of parallel sides = 4 + 7 = 11cm Height between them = 5cm Area of trapezium = 1 2 h × sum of parallel sides = 1 2 × 5 × 11 = 55 2 = 27.5cm2 Q5. If we have area of parallelogram 84cm2 and its height is 7cm. Find base of parallelogram. Solution Area of parallelogram = height × base 84 = 7 × base base = 84 7 base = 12cm Week Unit: 12 Three Dimensional Sides Exercise 12a Q1. Find the volume of each of the following cuboids. Solution (i) length = 14cm, width = 12cm, height = 10cm
CSS Primary Standard “Mathematics” 243 volume = length × breadth × height = 14 × 12 × 10 = 1680cm3 (ii) length = 10dm, width = 6dm, height = 80dm volume = length × breadth × height = 10 × 6 × 80 = 4800cm3 (iii) length = 6m, width = 4.5m, height = 2.5m volume = length × breadth × height = 6 × 4.5 × 2.5 = 67.5cm3 (iv) length = 4m, width = 1.2m, height = 2m volume = length × breadth × height = 4 × 1.2 × 2 = 9.6cm3 (v) length = 3m 50cm, width = 2.2m, height = 1m 50cm volume = length × breadth × height = 3.5 × 2.5 × 1.5 = 8.75cm3 (vi) length = 10cm, width = 7cm, height = 9cm volume = length × breadth × height = 10 × 7 × 9 = 630cm3 Q2. Find the volume of each of cubes whose edges are given belwo. Solution (i) l = 2.5m volume = l 3 = 2.53 = 15.625m3 (ii) l = 80m volume = l 3 = 803 = 152000m3 (iii) l = 22 3 m
CSS Primary Standard “Mathematics” 244 volume = l 3 = 3 22 3 = 10648 27 m 3 (iii) l = 150dm volume = l 3 = 21503 = 3375000dm 3 (iv) l = 17dm volume = l 3 = 173 = 4913dm 3 Q3. A water storage room contains liquid if the dimension of the room are 6m × 5m × 4m, find how many littres of liquid it holds. Solution volume of room = 6m × 5m × 4m = 120m3 = 120 × 1m3 = 120 × 1000 littre = 120000 littres Q4. Find the cost of painting a refrigerator whose length, width and height are respectively 3m, 2m, 1m at rate of Rs. 10 per meter. Solution Surface area of refrigerator = 2(length + width + height) = 2(3 × 2 + 2 × 1 + 1 × 3) = 2(6 + 2 + 3) = 22 Cost of painting = 10 × 22 = 220 Q5. A cuboid has dimension 7m × 5m × 4m. is its surface area greater or smaller then a cube has a side of length 6m. Solution Surface area of cuboid = 2(7 × 5 + 5 × 4 + 4 × 7) = 2(35 + 20 + 28)
CSS Primary Standard “Mathematics” 245 = 2(83) = 166 Surface area of cube = 6l 2 = 6(6)2 = 216 The surface area of cuboid is smaller than surface area of cube. Q6. Find the surface area of a cuboid power bank with length 12cm, width 8cm and height 6cm. Solution Surface area of cuboid = 2(length + width + height) = 2(12 × 8 + 8 × 6 + 6 × 12) = 2(96 + 48 + 72) = 2(216) = 432 Q7. The cost of the paint is Rs. 36per kg .If 1kg of paint cover 16sq feet, how much will it cost to paint outside of a cube having 8 feet each side. Solution Cost of paint = Rs. 36per kg 1kg paint covers = 16sq feet Surface area of cube = 6l 2 = 6(8)2 = 384 sq feet Paint required for area of 384 sq feet = 386 16 = 24 kg Cost of paint outside the cube = Rs. 36 × 29 = Rs. 864 Review Exercise 12 Q1. Choose the correct answer and fill the circle. i. A cuboid is a______ D figure: 2D 3D all none ii. All solid bodies are: 2D 4D 3D none iii. Volume of cuboid is______: l × l × w 6l 2 l × B × H l 3 iv. Surface area of cube is:
CSS Primary Standard “Mathematics” 246 l × 6l 2 l 2 6l 2 none v. Volume of is a measure of how much________ a particular object occupies: space fluid liquid all vi. Three cubes each of side 5cm are joined and to end. What is the surface area of resulting cuboid? 300cm2 350cm2 375cm2 400cm2 vii. If a cuboid is 3.2cm high, 8.9cm large and 4.7cm wide then the total surface area is: 170.7cm2 180cm2 205.7cm2 325.8cm2 Q2. Identify the missing elements in the following sets of cuboids. No. Length Width Height Volume Surface area 1 14 672 2 11 8 616 3 6 5 270 Fro 1 As volume = L × W × H 672 = 14 × W × 6 672 = 84 W 672 84 = W 8 = W W = 8 Surface area = 2(LW + WH + HL) = 2(14 × 8 + 8 × 6 + 6 × 14) = 2(112 + 48 + 84) = 488 For 2 volume = L × W × H 616 = 11 × 8 × H 616 = 88 H 616 88 = H 7 = H H = 7 Surface area = 2(LW + WH + HL) = 2(11 × 8 + 8 × 7 + 7 × 11) = 2(88 + 56 + 77)
CSS Primary Standard “Mathematics” 247 = 2(221) = 442 For 2 volume = L × W × H 270 = L × 6 × 5 270 = L × 30 270 30 = L L = 9 Surface area = 2(LW + WH + HL) = 2(9 × 6 + 6 × 5 + 5 × 9) = 2(54 + 30 + 45) = 2(129) = 258 Q3. A cuboidal water tank is 6m long, 5m wide and 4.5m deep. How many litres of water can it hold. (1m3 = 1000l) volume = L × W × D = 6 × 5 × 4.5 = 135m3 = 135 × 1000l = 135000l Q4. A cooking oil contains in a tin can with the dimensions of 30cm × 25cm × 15cm. Find how many litres of oil it holds. 1000cm2 = 1 litre volume = 30cm × 25cm × 15cm = 11250cm3 = 11.25 litre Q5. How many cubes of 0cm edge can be put in a cubical box of 1m edge. Surface area of cube = 6(10)2 = 600cm2 Surface area of cubical box = 6(100)2 cm 2 1m = 100cm = 60000cm2 No. of cube put in cubical box = 60000 600 = 100 Q6. How many bricks each measuring 25cm × 112cm × 6cm will be need to build a wall 8m × 6m × 225m.
CSS Primary Standard “Mathematics” 248 volume of bricks = 25cm × 11.25cm × 6cm = 1687.5cm3 volume of wall = 8m × 6m × 22.5m = 1080m3 = 1080 × 1000000cm3 = 1080000000cm3 No. of bricks = volume of wall volume of brick = 1080000000 1687.5 = 640000 Unit: 13 Information Handling Exercise 13 Q1.(i) Define and differentiate between primary and secondary sources with some examples (ii) What is the difference between group and ungrouped data explain it. Solution (i) answer on page 176 (ii) answer on page 176 and 178 Q2 Given data is from ungroup data arrange it in ascending order. (i) 67,56,89,34,65,87,5,8,11,35,26,47,59,89,94,100,12 5,8,11,12,26,34,35,47,56,59,65,67,87,89,89,94,100 (ii) 100,106,804,651,854,987,367,724,679,111,298,176,333,456 Solution (i) answer on page 176 (ii) answer on page 176 and 178 (iii) Q3. Given below the ungrouped data of 35 person’s ages. Ans: Class Interval Tally Marks Frequency 1-20 |||| ||| 8 21-40 |||| |||| 10
CSS Primary Standard “Mathematics” 249 41-60 |||| |||| 10 61-80 |||| 5 81-100 || 2 Exercise 13 Q1. The marks of Alin in different subject are given below. Ans: Q.2 The below table shows the favorite cricket players of 250 students of a school. Name of player Imran khan Wasim akram Shahid Afraid Youna Khan Umer Gull No. of students 70 40 50 30 50 Ans: Horizontal Bar Graph Vertical Bar Garph
CSS Primary Standard “Mathematics” 250 Vertical Bar Graph Q.3 The following table shows the year wise strength of a medical college. Years 2006-07 2007-08 2008-09 2009-10 2010-11 No. of students of Medical college 800 975 1100 1400 1625 Ans: Q.4 Ali ha there types of food items is his store as shown in pie graph read the graph and answer this following questions: (i) Which types of food item is greater in numbers?