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Grade VIIMathematics

Exam Preparation Booklet

Chapter WiseImportant Questions-Solutions

Data Handling Important Questions

Topic: Mean and Range of Data Sets

1. What is the mean of the data, 10, 6, 19, 25, 32, 24, 31? (2 marks)

2. What is the mean of all the factors of number 28? (2 marks)

3. The given data shows the prices of different designer suits in a shop. Rs 8,250, Rs 7,000, Rs 6,750, Rs 6,900, Rs 7,250, Rs 9,000, Rs 10,000, Rs 11,500, Rs 13,500, Rs 12,000 What is the range of the given data?

(2 marks)

Topic: Mode and Median of Data Sets

4. Find the mode of the following set of numbers. 20, 29, 30, 20, 30, 41, 45, 30, 43, 20, 27, 16

(2 marks)

5. The given data shows the number of babies born in a hospital in different months of a year. 5, 7, 5, 10, 6, 5, 6, 7, 6, 5, 4, 8 What is the mode of the number of babies born in the hospital?

(2 marks)

6. Find the mean and the median of the following data. 35, 25, 13, 21, 20, 32, 29, 4, 10 Are the mean and the median of the data same?

(3 marks)

7. The set of numbers 15, 15, x , 22, y is in the ascending order and has only one mode. The median of this data is 5 more than its mode. If the mean of the data is the same as its median, then what is the value of y ?

(3 marks)

Topic: Double Bar Graphs

8. The following table shows the data of a particular mobile company, which represents the number of mobiles whose parts got defected over the last 2 years.

Part of mobile Keypad Battery Camera Speaker

(3 marks)

Data Handling

2008 28 22 26 10

2007 36 20 30 18

Draw a double bar graph representing the given data.

9. The given data shows the preference of children of a particular class across different schools for home tutor or online tutor.

School A

School B School C

School D

School E

Home tutor 40 15 71 28 36

Online tutor

45 15 39 52 44

Draw a double bar graph for the given data.

(3 marks)

10. The given bar graph shows the number of male employees working in morning and night shifts in five companies A, B, C, D and E.

Read the bar graph and answer the given questions. In which company is the number of male employees working in morning

(3 marks)

Data Handling and night shifts the same?

11.

Study the bar graph and answer the following questions: (a) What information is represented in the bar graph? (b) In which 2 years is the number of foreigners who visited Delhi the same? (c) Name the year in which maximum number of foreigners visited Mumbai. (d) What was the number of foreigners who visited Delhi in the year 2008? (e) What was the greatest increase in the number of foreigners who visited Delhi as compared to the previous year?

(5 marks)

Topic: Probability

12. Check whether the given events are certain to happen or impossible or can happen but are not certain. When a card is drawn out of cards numbered 10 to 20, an even prime number is obtained.

(2 marks)

13. A bag contains three balls, out of which one is red and the other two are blue. A ball is drawn at random. Find the probability of getting a red ball.

(1 mark)

14. A die is rolled. What is the probability of getting a composite number? (1 mark)

Data Handling 15. Cards numbered 10 to 20 are kept in a box. If a card is taken out at

random, then what is the probability of drawing a card with an even prime number written on it?

(2 marks)

Data Handling Solutions

Topic: Mean and Range of Data Sets

1. It is known that,

Thus, the mean of the given data is 21.

2. The factors of number 28 are 1, 2, 4, 7, 14 and 28

Thus, the mean of all the factors of number 28 is 9.33.

3. The range of a data is the difference between its highest and lowest observations. ∴ Range = Highest observation − Lowest observation = Rs 13,500 − Rs 6,750 = Rs 6,750 Thus, the range of the given data is Rs 6,750.

Topic: Mode and Median of Data Sets

4. The mode of a set of observations is the observation that occurs most often. Arranging the given numbers in ascending order, we obtain 16, 20, 20, 20, 27, 29, 30, 30, 30, 41, 43, 45 Here, both the numbers, 20 and 30, occur most often (3 times).

Data Handling Thus, the modes of the given set of numbers are 20 and 30 both.

5. It is known that the mode of a set of observations is the observation that occurs most often. In the given data, it can be observed that numbers 4, 8, and 10 occur once; number 5 occurs four times; number 6 occurs thrice and number 7 occurs twice. It is clear that number 5 occurs most often. Hence, the mode of the number of babies born in the hospital is 5.

6. We first arrange the given data in ascending order as 4, 10, 13, 20, 21, 25, 29, 32, 35 We know that,

Thus, the mean of the given data is 21. Median is the middle value of the data when arranged in ascending order. ∴ Median = 21 It can be seen that the mean and median of the given data are same.

7. The given data is represented in the ascending order as 15, 15, x , 22, y It can be seen that observation 15 is occurring twice. Since the data is in the ascending order, x can be greater than or equal to 15 and y can be greater than or equal to 22. Since the given data has only one mode, Mode = 15 It is given that the median of the data is 5 more than its mode. ∴ Median = 15 + 5 = 20 It is known that median is the middle value. ∴ Median = x = 20 It is given that the mean of the data is the same as its median. ∴ Mean = 20 It is known that

Data Handling

⇒ 67 + y = 20 × 5 (Multiplying both sides with 5) ⇒ 67 + y = 100 ⇒ y = 100 − 67 (Transposing 67 to R.H.S.) ⇒ y = 33 Thus, the value of y is 33.

Topic: Double Bar Graphs

8. We will represent the number of mobile sets on the vertical axis and the parts, which got defected, on the horizontal axis. Here, we take the scale on the vertical axis as: 1 unit = 4 mobiles The double bar graph representing the given information can be drawn as follows.

9. The schools are represented on the horizontal axis, while the number of students is represented on the vertical axis. The scale taken on the vertical axis is: 1 unit = 10 students

Data Handling The double bar graph for the given data can be drawn as

10. Number of male employees of company C working in morning shift = 7500 Number of male employees of company C working in night shift= 7500 Thus, in company C, the number of male employees working in morning shift is equal to the number of male employees working in night shift.

11. (a) The given bar graph shows the number of foreigners who visited Delhi and Mumbai in the last five years. (b) The number of foreigners who visited Delhi was the same in the years 2005 and 2007. (c) The maximum number of foreigners (500) visited Mumbai in the year 2006. (d) 300 foreigners visited Delhi in the year 2008. (e) The increase in the number of foreigners who visited Delhi was the greatest in year 2006 as compared to the previous year 2005 and this increase can be calculated by subtracting the number of foreigners who visited Delhi in the year 2006 and in the year 2005 i.e., 600 − 200 = 400

Topic: Probability

12. (a) When a die is tossed, an odd number may or may not turn up. Hence, this event can happen but it is not certain. (b) When a card is drawn out of cards numbered 10 to 20, an even prime number cannot be obtained because 2 is the only prime number. Hence, this is an impossible event.

Data Handling 13. It is given that the bag contains 3 balls and one of them is red.

∴ Probability of getting a red ball = 31

14. When a die is rolled, there are 6 equally likely possible outcomes and those are 1, 2, 3, 4, 5, and 6. Composite numbers out of 1, 2, 3, 4, 5, 6 are 4 and 6.

∴ Probability of getting a composite number = 62 = 3

1

15. Total number of cards in the box = 11 It is known that number 2 is the only even prime number. However, the cards are numbered from 10 to 20. Hence, the probability of drawing a card with an even prime number written on it is 0.

Lines and Angles Important Questions

Topic: Angles ­ I

1. What is the measure of the supplement of the angle 115°? (2 marks)

2. If an angle is represented by the expression (2 x + 18°) and its supplement is represented by the expression (6 x + 2°), then what is the complement of the smaller of the two angles?

(2 marks)

3. In which of the given figures is ∠1 adjacent to ∠2? Give reasons to justify your answers. (a)

(b)

(2 marks)

4. Identify five pairs of adjacent angles in the given figure. (2 marks)

Lines and Angles

Topic: Angles ­ II

5. Consider the given figure.

Find the measure of angle θ.

(2 marks)

6. In the given figure, what are the values of x and y ?

(2 marks)

Lines and Angles 7. In the given figure, what is the sum of ∠2, ∠3, ∠5 and ∠7?

(2 marks)

8. In the given figure, ∠BOC, ∠COD and ∠DOE are in the ratio 2 : 3 : 4. What are the measures of ∠COE and ∠EOF?

(3 marks)

Lines and Angles Topic: Transversal of Two Lines and Angles Axiom ­ I

9. Use the following information to answer the next question.

In the given figure, a || b . Check whether c is parallel to d . Give reasons to support your answer.

(2 marks)

10. In the given figure, AP || BQ and DC is the angle bisector of both ∠ACB and ∠ADB. If ∠PAC = 70° and ∠ADC = ∠ACQ, then what are the values of x and y ?

(3 marks)

Lines and Angles 11.

In the given figure, arms of ∠AOB are parallel to the respective arms of ∠CO′D as shown in the given figure. If the measure of ∠CO′D is 120°, then what is the measure of ∠AOB?

(2 marks)

12. In the following figure, the line AB || CD || EF. Find the measures of ∠1, ∠2, ∠3, and ∠4.

(2 marks)

Topic: Transversal of Two Lines and Angles Axiom ­ II

13. (2 marks)

Lines and Angles

In the given figure, ABCD is a trapezium where AB || CD, CX is the bisector of ∠ACB, and ∠BCE = ∠ABD. What is the measure of ∠AEX?

14.

In the given figure, ABCD is a rectangle and DG || EF. What is the value of x ?

( 2 marks)

15.

In the given figure, PQ || RS Find the measures of ∠1, ∠2, ∠3, and ∠4.

(3 marks)

Lines and Angles Solutions

Topic: Angles ­ I

1. Two angles are said to be supplementary if their sum is equal to 180°. ∴ Supplement of angle 115° = 180° − 115° = 65°

2. It is given that: Measure of the angle = 2 x + 18° Measure of its supplement = 6 x + 2° It is known that the sum of measures of two supplementary angles is 180°. ∴ 2 x + 18° + 6 x + 2° = 180° ⇒ 8 x + 20° = 180° ⇒ 8 x = 180° − 20° = 160°

2 x + 18° = 2(20°) + 18° = 58° 6 x + 2° = 6(20°) + 2° = 122° The smaller angle is 58°. It is known that the sum of the measures of two complementary angles is 90°. ∴ Complement of 58° = 90° − 58° = 32° Thus, the complement of the smaller of the two given angles is 32°.

3. Two angles are said to be adjacent if (i) They have a common vertex (ii) They have a common arm (iii) The non­common arms are on either side of the common arm (a) In this figure, ∠1 and ∠2 share a common vertex. However, they do not share a common arm. Thus, ∠1 and ∠2 are not adjacent. (b) In this figure, ∠1 and ∠2 share a common arm. However, they do not share a common vertex. Thus, ∠1 and ∠2 are not adjacent.

4. Two angles are said to be adjacent if

Lines and Angles (i) They have a common vertex (ii) They have a common arm (iii) The non­common arms are on either side of the common arm In the given figure, it can be observed that ∠AOB and ∠BOC; ∠BOC and ∠COD; ∠COD and ∠DOE; ∠DOE and ∠EOF; ∠EOF and ∠FOA are adjacent angles

5. AOB is a straight line. We know that the measure of a straight angle is 180°. Therefore, Φ + 95°= 180° ⇒ Φ = 180°− 95°= 85° Now, ∠AOC, ∠COD, and ∠DOB lie on a straight line. ∴ ∠AOC + ∠COD + ∠DOB = 180° ⇒ θ + 60°+ Φ= 180° ⇒ θ + 60°+ 85° = 180°( Φ= 85°) ⇒ θ = 180°− 145° ⇒ θ = 35°

6. In the given figure POQ is a straight line. ∴ x + 129° = 180° ⇒ x = 180° − 129° = 51° It is known that the sum of the measures of angles lying on a straight line is 180°. ∴ ∠POR + ∠ROS + ∠SOT + ∠TOQ = 180° ⇒ y + 93° + x + y = 180° ⇒ 2 y + 93° + 51° = 180° ( x = 51°) ⇒ 2 y = 180° − 144° = 36°

Thus, the values of x and y are 51° and 18° respectively.

7. The sum of all the angles lying on a straight line is 180°. ∴ ∠5 + ∠6 + ∠7 = 180° ... (1) It can be observed that ∠GOF and ∠BOD are vertically opposite angles.

Lines and Angles ∴ ∠GOF = ∠BOD ⇒ ∠GOF = ∠BOC + ∠COD ⇒ ∠6 = ∠2 + ∠3 Using it in equation (1), we obtain ∠5 + ∠2 + ∠3 + ∠7 = 180° Thus, the sum of ∠2, ∠3, ∠5 and ∠7 is 180°.

8. It can be observed that AE and BF are two intersecting lines. ∠AOB = ∠EOF (Vertically opposite angles) ⇒ ∠EOF = 45° Similarly, ∠AOF = ∠BOE = 135° It is given that ∠BOC, ∠COD and ∠DOE are in the ratio 2:3:4. Let ∠BOC = 2 x , ∠COD = 3 x , ∠DOE = 4 x . ∠BOE = ∠BOC + ∠COD + ∠DOE ⇒ 135° = 2 x + 3 x + 4 x ⇒ 135° = 9 x

∴ ∠COD = 3 x = 3 × 15° = 45° ∠DOE = 4 x = 4 × 15° = 60° ∴ ∠COE = ∠COD + ∠DOE = 45° + 60° = 105° Thus, the measures of ∠COE and ∠EOF are 105° and 45° respectively.

Lines and Angles Topic: Transversal of Two Lines and Angles Axiom ­ I

9.

It is given that a || b . It is known that if two parallel lines are cut by a transversal, then the pair of corresponding angles is equal. Here, ∠WMR and ∠WLP are corresponding angles. ∴∠WMR = ∠WLP ⇒ ∠WMR = 70° (∠WLP = 70°) ∠YNR = ∠SNZ (Vertically opposite angles) ⇒ ∠YNR = 70° (∠SNZ = 70°) Therefore, ∠WMR = ∠YNR = 70° i.e., corresponding angles formed by transversal b on the lines c and d are equal. It is also known that when a transversal cuts two lines, such that the pair of corresponding angles is equal, then the lines are parallel. Hence, line c is parallel to line d .

10. It is given that AP || BQ. AC acts as a transversal to AP and BQ. ∴ ∠PAC = ∠ACB (Alternate interior angles) ⇒ ∠ACB = 70° It is known that the angles on the same side of transversal are supplementary. ∴ ∠PAC + ∠ACQ = 180° ⇒ 70° + ∠ACQ = 180° ⇒ ∠ACQ = 110° It is given that DC bisects ∠ACB. ∴ ∠ACD = ∠DCB

∠ADC = ∠ACQ (Given)

Lines and Angles ⇒ ∠ADC = 110° It is also given that DC bisects ∠ADB. ∴ ∠ADC = ∠BDC = 110° On using angle sum property in ΔADC, we obtain ∠CAD + ∠ADC + ∠ACD = 180° ⇒ y + 110° + 35° = 180° ⇒ y = 180° − 145° = 35° On using angle sum property in ΔBDC, we obtain ∠CBD + ∠BDC + ∠DCB = 180° ⇒ x + 110° + 35° = 180° ⇒ x = 180° − 145° ⇒ x = 35° Thus, the values of x and y are 35° each.

11. It is given that OB || O′D. Here, O′C can be regarded as the transversal. ∴∠CO′D = ∠OMC (Corresponding angles) ⇒ ∠OMC = 120° (∠CO′D = 120°) It is also given that OA || O′C. Here, OB can be regarded as the transversal. It is known that the angles on the same side of a transversal are supplementary ∴∠OMC + ∠AOB = 180° ⇒ 120° + ∠AOB = 180° ⇒ ∠AOB = 180° − 120° ⇒ ∠AOB = 60° Thus, the measure of ∠AOB is 60°.

12. In the given figure, three parallel lines AB, CD, and EF are cut by a transversal PQ. Now, it is given that ∠CYP = 134° ∠CYP + ∠CYQ = 180°(Linear pair) ⇒ 134°+ ∠CYQ = 180° ⇒ ∠CYQ = 180°− 134°= 46° ∴ ∠3 = 46° Now, ∠2 = ∠3 = ∠4 (Corresponding angles) ∴ ∠2 = ∠4 = 46° Now, ∠1 = ∠2 (Vertically opposite angles) ∴ ∠1 = 46° Thus, ∠1 = ∠2 = ∠3 = ∠4 = 46°

13. It is given that AB || CD. ∴∠ABD = ∠BDC (Alternate interior angles) ⇒ ∠ABD = 25°

Lines and Angles ∠BCE = ∠ABD (Given) ⇒ ∠BCE = 25° CX is given as the bisector of ∠ACB. ∴∠ACE = ∠BCE = 25° Consider ΔACE: It is known that when a side of a triangle is produced, then the measure of the exterior angle so formed is equal to the sum of the measures of its interior opposite angles. ∴∠AEX = ∠EAC + ∠ACE ⇒ ∠AEX = 30° + 25° ⇒ ∠AEX = 55° Thus, the measure of ∠AEX is 55°

14. We first extend DG and AB to intersect at point X.

It is known that the opposite sides of a rectangle are parallel. ∴ AB || CD ∴ ∠CDG = ∠GXB (Alternate interior angles) ⇒ ∠GXB = 20° ... (1) It is also given that DG || EF. The interior angles on the same side of the transversal are supplementary. ∴ ∠GXB + ∠BFE = 180° ⇒ 20° + x = 180° [Using (1)] ⇒ x = 180° − 20° ⇒ x = 160° Thus, the value of x is 160°.

15. We have PQ || RS and UV is a transversal intersecting the lines PQ and RS. Now, ∠PCV = ∠UDS (Alternate interior angles) ∴∠1 = 95° Now, ∠1 + ∠2 = 180°(Linear pair) ⇒ ∠2 = 180°− 95°= 85° ∠RBT = ∠UBS (Vertically opposite angles) ∴ ∠3 = 85° ∠QAT and ∠SBU form a pair of interior angles made on the same side of the transversal TU. ∴∠QAT + ∠SBU = 180°

Lines and Angles ⇒ ∠QAT + ∠3 = 180° ⇒∠QAT = 180°− 85°= 95° Now, ∠PAU and ∠QAT are vertically opposite angles. ∴ ∠PAU = 95° i.e., ∠4 = 95° Thus, the measures of the required angles are: ∠1 = 95°, ∠2 = 85°, ∠3 = 85°, and ∠4 = 95°

The Triangle and Its Properties Important Questions

Topic: Medians and Altitudes of Triangles

1. State whether the following statements are true or false. (a) In an isosceles triangle, the altitude drawn to the unequal side is the same as one of the medians. (b) In a right triangle, one of the altitudes is the same as one of the medians.

(2 marks)

2. State whether the following statements are true or false? Justify your answer. (a) The altitudes of a triangle always lie inside the triangle. (b) In an isosceles triangle, the altitude drawn to the unequal side is the same as one of the medians. (c) In a right triangle, one of the altitudes is the same as one of the medians.

(3 marks)

Topic: Angle Sum and Exterior Angle Property of Triangles

3.

In the given figure, AB || CD and BE || AC (a) Find ∠ABC. (b) Find ∠EBA and hence show that EB⊥BC.

(2 marks)

The Triangle and Its Properties 4. Use the following information to answer the next question.

In the given figure, ΔCOD is isosceles, where OC = OD. Find the value of unknown x , y , and z .

(2 marks)

5.

In the given figure, AB || CD. What is the value of x ?

(3 marks)

The Triangle and Its Properties 6.

Show that lines BC and EF shown in the given figure are parallel.

(2 marks)

7.

What is the measure of ∠BAC in the given figure?

(2 marks)

Topic: Properties of Different Types of Triangles

8. If the measures of two angles of a triangle are 104° and 38°, then what is the measure of the third angle? Also, name the type of triangle.

(2 marks)

The Triangle and Its Properties 9.

In the given figure, ΔACE is an equilateral triangle whereas ΔACB and ΔECD are isosceles triangles. Find the measures of all the angles in the given figure.

(3 marks)

10. ΔPQR is an equilateral triangle. PX intersects side QR in X and QY intersects side PR in Y.

Prove that PX = QY, QX = RY, and ∠PXQ = ∠QYR

(2 marks)

The Triangle and Its Properties 11. In the given figure, ΔACD is an isosceles triangle with AC = AD. Find

the measures of angles x , y , and z .

(3 marks)

12. The given figure shows a pentagon ABCDE. ΔACD is an isosceles triangle with AC = AD. If BE = CD, then show that 3AE + 2ED > AB.

(3 marks)

Topic: Pythagoras Theorem and Its Converse

13. Use the following information to answer the next question.

What is the length of side BC in the given triangle?

(2 marks)

14. Sonu drives his bike 14 km towards East from place A. He then drives 3 ( 3 marks)

The Triangle and Its Properties km towards North, then 5 km to West and then 9 km towards North to reach place B. What is the shortest distance between places A and B?

15. Determine whether a triangle with sides 7 cm, 24 cm and 25 cm is a right angled triangle or not.

(2 marks)

The Triangle and Its Properties Solutions

Topic: Medians and Altitudes of Triangles

1. (a) True Consider an isosceles ΔABC with AB = AC.

In can be seen that AD is the altitude of ΔABC, also AD is the median of ΔABC. Thus, the given statement is true. (b) False Consider a right ΔABC.

It can be seen that AB, BC and BD are the altitudes of ΔABC and neither of them acts as median. Thus, the given statement is false.

2. (a) True A median connects a vertex of a triangle to the mid­point of the opposite side. The medians of a triangle always lie inside the triangle.

The Triangle and Its Properties

AD, BE and CF are the three medians of ΔABC and they lie inside the triangle. Thus, the given statement is true. (b) True Consider an isosceles ΔABC with AB = AC.

In can be seen that AD is the altitude of ΔABC, also AD is the median of ΔABC. Thus, the given statement is true. (c) False Consider a right ΔABC.

It can be seen that AB, BC and BD are the altitudes of ΔABC and neither of them acts as

The Triangle and Its Properties median. Thus, the given statement is false.

Topic: Angle Sum and Exterior Angle Property of Triangles

3. (a) In the given figure, AB||CD ∴ ∠BAC = ∠ACD (Alternate interior angles) ∴ ∠BAC = 41° (b) In ΔABC, ∠ABC + ∠BCA + ∠BAC = 180° (By angle sum property) ⇒ ∠ABC + 90°+ 41°= 180° ( ∠BAC = 41°) ⇒ ∠ABC = 180°− 131° ⇒ ∠ABC = 49° Now, BE||AC ∴ ∠EBA = ∠BAC (Alternate interior angles) ⇒ ∠EBA = 41° Now, ∠EBA + ∠ABC = 41°+ 49°= 90° Thus, EB is perpendicular to BC.

4. In the given figure, it is seen that ∠ODC and ∠ODE form a linear pair. ∴ ∠ODC + ∠ODE = 180° ⇒ ∠ODC + 115° = 180° ⇒ ∠ODC = 180° − 115° = 65° It is given that ΔCOD is isosceles, where OC = OD ∴∠ODC = ∠OCD ⇒ x = 65° By applying angle sum property of triangles in ΔCOD, we obtain ∠OCD + ∠CDO + ∠COD = 180° ⇒ 65° + 65° + ∠COD = 180° ⇒ 130° + ∠COD = 180° ⇒ ∠COD = 180° − 130° ⇒ ∠COD = 50° ∠AOB = ∠COD (Vertically opposite angles) ⇒ y = 50° Again, by applying angle sum property of triangles in ΔAOB, we obtain

The Triangle and Its Properties ∠AOB + ∠ABO + ∠BAO = 180° ⇒ 50° + 40° + z = 180° ⇒ 90° + z = 180° ⇒ z = 180° − 90° = 90° Thus, x = 65°, y = 50°, and z = 90°

5. It is given that AB || CD. ∴ ∠EGF = ∠GFH (Alternate interior angles) ⇒ ∠EGF = 60° Applying angle sum property for a triangle in ΔEGF: ∠EFG + ∠EGF + ∠GEF = 180° ⇒ ∠EFG + 60° + 70° = 180° ⇒ ∠EFG + 130° = 180° ⇒ ∠EFG = 50° Now, ∠EFG = ∠FGH = 50° Here, ∠EFG and ∠FGH are alternate interior angles with respect to the lines FI and GJ. It is known that if the alternate interior angles formed by a transversal with two lines are equal, then the lines are parallel. ∴ FI || GJ It is known that if a transversal intersects two parallel lines, then the angles on the same side of transversal are supplementary. ∴ ∠JHK + ∠HKX = 180° ⇒ 80° + x = 180° ⇒ x = 180° − 80 ⇒ x = 100° Thus, the value of x is 100°.

6. Consider ΔEGH: When a side of a triangle is produced, then the measure of the exterior angle so formed is equal to the sum of the measures of its opposite interior angles. ∴∠AGE = ∠GEH + ∠EHG ⇒ 121° = 19° + ∠EHG ⇒ ∠EHG = 121° − 19° ⇒ ∠EHG = 102°

The Triangle and Its Properties ∠BHF = ∠EHG (Vertically opposite angles) ∴ ∠BHF = 102° It can be observed that ∠BHF + ∠HBC = 102° + 78° = 180°. Here, ∠BHF and ∠HBC lie on the same side of transversal BH. If the interior angles on the same side of the transversal are supplementary, then the lines are parallel. ∴ BC || EF Thus, lines BC and EF are parallel.

7. It is known that when a side of a triangle is produced, then the measure of the exterior angle so formed is equal to the sum of the measures of its interior opposite angles. ∴∠ADE = ∠DCE + ∠DEC ⇒ 150° = ∠DCE + 25° ⇒ ∠DCE = 150° − 25° = 125° Now, ∠ABC + ∠BAC = ∠DCE ⇒ 65° + ∠BAC = 125° ⇒ ∠BAC = 125° − 65° ⇒ ∠BAC = 60° Thus, the measure of ∠BAC is 60°.

Topic: Properties of Different Types of Triangles

8. The measures of two angles of the triangle are given as 104° and 38°. On using angle sum property, we obtain 104° + 38° + Third angle = 180° ⇒ Third angle = 180° − 142° = 38° Thus, the three angles of the triangle measure 104°, 38° and 38°. It can be observed that the two angles are of the same measure. Thus, the triangle is an isosceles triangle.

9. It is given that ΔACE is an equilateral triangle. ∴ ∠ACE = ∠CEA = ∠EAC = 60° Now, consider ΔABC. AC = CB ∴ ∠ABC = ∠BAC Now, ∠BCD is an exterior angle of ΔABC. Therefore, by exterior angle theorem, we have

The Triangle and Its Properties ∠BCD = ∠ABC + ∠BAC ⇒ 2∠ABC = 100°( ∠ABC = ∠BAC)

Thus, ∠ABC = ∠BAC = 50° By applying angle sum property in ΔABC, we obtain ∠ABC + ∠BCA + ∠BAC = 180° ⇒ 50°+ ∠BCA + 50°= 180° ⇒∠BCA = 180°− 100°= 80° Now, ∠BCD + ∠DCE + ∠ECA + ∠ACB = 360°( Measure of one complete angle is 360°) ⇒ 100°+ ∠DCE + 60°+ 80°= 360° ⇒ ∠DCE = 360°− 240°= 120° Now, consider ΔECD. We have, EC = CD Therefore, ∠CED = ∠CDE Using angle sum property, we obtain ∠CED + ∠CDE + ∠DCE = 180° ⇒ 2∠CED = 180°− 120°= 60°

∴∠CED = ∠CDE = 30°

10. In ΔPQX and ΔRQY, ∠PQX = ∠QRY = 60°(Since ΔPQR is an equilateral triangle) PQ = QR (Since ΔPQR is an equilateral triangle) Now, ∠QPR = ∠PQR ( ΔPQR is an equilateral triangle)

⇒ ∠QPX = ∠YQR ∴ΔPQX ≅ ΔQRY (By ASA congruency criterion) We know that corresponding parts of congruent triangles are equal. Therefore, we have PX = QY QX = RY ∠PXQ = ∠QYR

The Triangle and Its Properties Hence, proved

11. We have, ∠ADE + ∠ADC = 180°(Linear pair) ⇒ ∠ADC = 180°− 132° ⇒ ∠ADC = 48° Now, it is given that AC = AD ∴ ∠ACD = ∠ADC (Angles opposite to equal sides of a triangle are equal) ∴ y = 48° In ΔACD, ∠CAD + ∠ACD + ∠CDA = 180°(Angle sum property of triangles) ⇒ x + 48°+ 48° = 180° ⇒ x = 180°− 96° ⇒ x = 84° Observe that ΔADE is an exterior angle of ΔABD. ∴ By using exterior angle property, we obtain ∠ADE = ∠BAD + ∠ABD ⇒ 132°= ∠BAD + 30° ⇒ ∠BAD = 132°− 30° = 102° Now, ∠BAD = ∠CAD + ∠CAB ⇒ 102°= x + z ⇒ z = 102°− 84°( x = 84°) ⇒ z = 18° Thus, the measures of angles are given by x = 84°, y = 48°, and z = 18°

12. It is known that the sum of the lengths of any two sides of a triangle is greater than the third side. In ΔABC: AB + AC > BC ... (1) In ΔACD: AC + AD > CD ⇒ 2AC > CD ... (2) ( AC = AD) In ΔAED: AE + ED > AD ⇒ AE + ED > AC ⇒ 2AE + 2ED > 2AC ⇒ 2AE + 2ED > CD ... (3) Using equation (2) It is known that the difference between the lengths of any two sides of a triangle is always smaller than the third side. In ΔABE: AB − AE < BE

The Triangle and Its Properties

⇒ CD > AB − AE ... (4) ( BE = CD) From equations (3) and (4), 2AE + 2ED > AB − AE ⇒ 2AE + 2ED + AE > AB ⇒ 3AE + 2ED > AB Hence proved.

Topic: Pythagoras Theorem and Its Converse

13. We know that the sum of all the interior angles of a triangle is 180°. In the given triangle, ∠A + ∠B + ∠C = 180° ⇒ ∠A + 37°+ 53°= 180° ⇒ ∠A = 180°− (37°+ 53°) = 180°− 90°= 90° Thus, ΔABC is right­angled at A. By applying Pythagoras theorem in ΔABC, we obtain BC 2 = AB 2 + AC 2 = (8 cm) 2 + (6 cm) 2 = 64 cm 2 + 36 cm 2 = 100 cm 2 = (10 cm) 2 ∴BC = 10 cm Hence, the length of the side BC is 10 cm.

14. The path travelled by Sonu from place A to place B can be diagrammatically represented as

Draw a line EF through E such that it is perpendicular to AC. ∴ EF = CD = 3 km

The Triangle and Its Properties FC = ED = 5 km ∴ AF = AC − FC = 14 km − 5 km = 9 km BF = BE + EF = 9 km + 3 km = 12 km On applying Pythagoras theorem to ΔABF, we obtain AB 2 = BF 2 + AF 2 ⇒ AB 2 = (12 km) 2 + (9 km) 2 ⇒ AB 2 = 144 km 2 + 81 km 2 ⇒ AB 2 = 225 km 2 ⇒ AB 2 = (15 km) 2 ∴ AB = 15 km Thus, the shortest distance between places A and B is 15 km.

15. The lengths of the sides of the triangle are given as 7 cm, 24 cm and 25 cm. It can be observed that: (7 cm) 2 + (24 cm) 2 = 49 cm 2 + 576 cm 2 = 625 cm 2 = (25 cm) 2 Here the sum of the squares of two sides of the triangle is equal to the square of its third side. Thus, by the converse of Pythagoras theorem, it can be said that the given triangle is a right angled triangle.

Congruence of Triangles Important Questions

Topic: SSS Congruence Rule

1. In the given figure, ΔABC and ΔDBC are isosceles, where AB = AC and DB = DC. Show that AD is the angle bisector of ∠BAC.

(2 marks)

2.

Is ΔADC ≅ ΔABC? Give reasons in support of your answer.

(2 marks)

3. Two triangles ABC and ADC are given below. Is ΔABC ≅ ΔADC? Give reasons in support of your answer.

(2 marks)

Congruence of Triangles

4.

In the given figure, AB = AC. If AM is the median, then prove that AM is also the altitude and the bisector of ∠BAC.

(3 marks)

Topic: SAS Congruence Rule

5. In the given figure, AB and CD bisect each other. Prove that AC || BD.

(2 marks)

Congruence of Triangles 6.

PQR is an isosceles triangle with PR = QR. A line RX through R is drawn perpendicular to side PQ. Prove that PX = QX. Also, find ∠QRX, if ∠QPR = 55°.

(3 marks)

7.

In the given figure, AB = CF, EF = BD and ∠EFC = ∠ABD. Prove that ΔAFE ≅ ΔCBD.

(2 marks)

Congruence of Triangles 8.

In the given figure, ΔABD and ΔBCE are equilateral triangles. Prove that AE = CD.

(2 marks)

9.

In the given figure, ABCD and AXYZ are squares. Prove that DZ = BX.

(2 marks)

Topic: ASA Congruence Rule

Congruence of Triangles 10.

In ΔPQR, PS is the bisector of ∠RPQ. Prove that PU = PV.

(2 marks)

11.

In the given figure, ∠ABC = ∠ACB, AD = BD and AE = EC. Show that BE = DC.

(3 marks)

12. Use the following information to answer the next question .

In the given figure, D and E are the mid­points of AB and AC respectively. F and G are the mid­points of BE and CD respectively.

If AB = 6 cm, AC = 7 cm, and FG = 2 cm, then find the perimeter of ΔABC.

( 3 marks)

Topic: RHS Congruence Rule

Congruence of Triangles 13.

In the given figure, QM = RL, OM = 4 cm and MQ = 6 cm. If the area of ΔROQ is 25 cm 2 , then what is the area of ΔQRL?

(3 marks)

14.

In the given figure, BF and CE are of equal lengths and are perpendicular to AD. If AB = DE, then prove that ∠ABC = ∠DEF.

(3 marks)

15. Show that in a trapezium, if the diagonals are equal, then the non parallel sides are also equal.

(3 marks)

Congruence of Triangles Solutions

Topic: SSS Congruence Rule

1. Comparing ΔADB and ΔADC, AB = AC (given) BD = CD (given) AD = AD (common) ∴ΔADB ≅ ΔADC [SSS congruence rule] ⇒∠BAD = ∠CAD (corresponding parts of congruent triangles are equal) Thus, AD is the angle bisector of ∠BAC.

2. In ΔADC and ΔABC, AD = AB (Given) DC = BC (Given) AC = AC (Common) Thus, ΔADC ≅ ΔABC (By SSS congruency criterion)

3. In ΔABC and ΔADC: AB = AD = 9 cm AC = AC = 4 cm BC = CD = 5 cm Thus, by SSS congruence rule, ΔABC ≅ ΔADC.

4. Given, AM is the median of ΔABC. ∴ BM = MC ...(1) Comparing ΔAMB and ΔAMC: AB = AC [Given] BM = MC [From (1)] AM = AM [Common] ∴ ΔAMB ≅ ΔAMC [SSS congruence rule] ⇒ ∠BAM = ∠CAM [Corresponding parts of congruent triangles are equal] Thus, AM is the angle bisector of ∠BAC. Also, ∠AMB = ∠AMC [Corresponding parts of congruent triangles are equal] Since BC is a straight line, ∴ ∠AMB + ∠AMC = 180°. ⇒ 2∠AMB = 180° [∠AMB = ∠AMC]

Congruence of Triangles ⇒ ∠AMB = 90° Thus, AM is the altitude of ΔABC.

Topic: SAS Congruence Rule

5.

In ΔAOC and ΔBOD, OA = OB OC = OD (AB and CD bisect each other) ∠AOC = ∠BOD (Vertically opposite angles) ∴ ΔAOC ≅ ΔBOD (By SAS congruence rule) ∴ ∠OAC = ∠OBD (C.P.C.T.) ∠OCA = ∠ODB ∠OCA and ∠ODB are alternate angles. Therefore, AC is parallel to BD.

6. In ΔPXR and ΔQXR, PR = QR (Given) XR = XR (Common) ∠PXR = ∠QXR (Each equal to 90°, since RX⊥PQ) Also, ∠QPR = ∠PQR ( angles opposite to equal sides are equal) Therefore, we must also have ∠XRP = ∠QRX ( Sum of all angles of a triangle equal 180°) ∴ ΔPXR ≅ ΔQXR (By SAS congruency criterion) We know that corresponding parts of congruent triangles are equal. ∴ PX = QX Also, we know that ∠QPR = ∠PQR

Congruence of Triangles

∴ ∠PQR = 55°( ∠QPR = 55°) Now, in ΔQRX, ∠QRX + ∠RXQ + ∠XQR = 180° (By angle sum property of triangles) ⇒ ∠QRX + 90°+ 55°= 180° ( ∠PQR = 55°) ⇒ ∠QRX = 180°− 145°= 35°

7. It is given that AB = CF ∴ AB + BF = CF + BF ⇒ AF = BC … (1) It is also given that ∠EFC = ∠ABD ∴ 180° − ∠AFE = 180° − ∠CBD ⇒ ∠AFE = ∠CBD … (2) Comparing ΔAFE and ΔCBD: AF = BC [Using (1)] ∠AFE = ∠CBD [Using (2)] EF = BD [Given] ∴ ΔAFE ≅ ΔCBD [SAS congruency criterion]

8. Measure of each angle of an equilateral triangle is 60°. ∴ ∠ABD = ∠CBE = 60° ⇒ ∠ABD + ∠ABC = ∠CBE + ∠ABC ⇒ ∠DBC = ∠ABE ...(1) Comparing ΔDBC and ΔABE, BD = AB [Sides of an equilateral triangle ABD] BC = BE [Sides of an equilateral triangle BCE] ∠DBC = ∠ABE [Using (1)] ∴ ΔDBC ≅ ΔABE [SAS congruence rule] ⇒ CD = AE [Corresponding parts of congruent triangles are equal] Hence, AE = CD

9. Since ABCD and AXYZ are squares, ∠DAB = ∠XAZ = 90°. ∴∠DAZ = ∠DAB + ∠BAZ = ∠XAZ + ∠BAZ = ∠BAX ... (1) Comparing ΔDAZ and ΔBAX: AD = AB (Sides of square ABCD) ∠DAZ = ∠BAX Using equation (1) AZ = AX (Sides of square AXYZ) ∴ ΔDAZ ≅ ΔBAX [SAS congruence rule] The corresponding parts of congruent triangles are equal.

Congruence of Triangles ∴ DZ = BX

Topic: ASA Congruence Rule

10. In ΔPUS and ΔPVS, ∠UPS = ∠VPS ( PS is the bisector of ∠RPQ) PS = PS (Common) ∠PUS = ∠PVS (Each equal to 90°) ∴ ΔPUS ≅ ΔPVS (By ASA congruency criterion) ∴ By CPCT, PU = PV Hence, proved

11. ∠ABC = ∠ACB (Given) ...(1) It is known that, the sides opposite to equal angles of a triangle are equal. ∴ AB = AC ...(2) It is also given that AD = BD. ∴ ∠BAD = ∠ABD ...(3) Similarly, ∠CAE = ∠ACE ...(4) From equations (1), (3) and (4), we have ∠BAD = ∠ABC = ∠CAE = ∠ACE ...(5) Comparing ΔABD and ΔACE, ∠ABD = ∠ACE [From (5)] AB = AC [From (2)] ∠BAD = ∠CAE [From (5)] ∴ ΔABD ≅ ΔACE [ASA congruence rule] ⇒ BD = EC (Corresponding parts of congruent triangles are equal) ⇒ BD + DE = EC + DE ⇒ BE = DC Hence, the result is proved.

12. By joining EG and extending it up to BC such that it intersects BC at H, we obtain the following figure.

Let the length of BC be x .

Congruence of Triangles In ΔABC, by mid­point theorem, we obtain

In ΔDGE and ΔCGH, DG = CG ∠DGE = ∠CGE (Vertically opposite angles) ∠GDE = ∠GCH (Alternate angles) ∴ΔDGE ΔCGH (By ASA congruency criterion)

DE = CH (By C.P.C.T.)

Similarly, in ΔEBH, we obtain

∴ x = 8 cm

BC = 8 cm ∴ Perimeter of ΔABC = AB + BC + AC = 6 cm + 8 cm + 7 cm = 21 cm Thus, the perimeter of ΔABC is 21 cm.

Topic: RHS Congruence Rule

13. Comparing ΔRMQ and ΔQLR MQ = LR [Given] ∠RMQ = ∠QLR [Each is a right angle] QR = QR [Common] ∴ ΔRMQ ≅ ΔQLR [RHS congruence rule] It is known that congruent triangles are equal in area. ∴ area (ΔQLR) = area (ΔRMQ) = area (ΔOMQ) + area (ΔOQR)

⇒ area (ΔQLR) = × OM × MQ + 25 cm 221

⇒ area (ΔQLR) = × 4 cm × 6 cm + 25 cm 221

⇒ area (ΔQLR) = 12 cm 2 + 25 cm 2 ⇒ area (ΔQLR) = 37 cm 2 Thus, the area of ΔQLR is 37 cm 2 .

Congruence of Triangles 14. Comparing ΔABF and ΔEDC:

∠BFA = ∠ECD (Each is 90°) AB = DE (Given) BF = CE (Given) ∴ΔAFB ≅ ΔDCE [RHS congruence rule] The corresponding parts of congruent triangles are equal. ∠BAC = ∠EDF ... (1) ∴ AF = CD ⇒ AF + FC = CD + FC ⇒ AC = FD ... (2) Now, comparing ΔABC and ΔDEF: AB = DE (Given) ∠BAC = ∠EDF From equation (1) AC = FD From equation (2) ∴ ΔABC ≅ ΔDEF [SAS congruence rule] ⇒∠ABC = ∠DEF (Corresponding parts of congruent triangles) Hence, the result is proved.

15. Let ABCD be the trapezium where AD || BC. Given, AC = BD. It is required to show that AB = DC. Draw AL⊥BC and DM⊥BC.

Comparing ΔALC and ΔDMB, ∠ALC = ∠DMB [Each is 90° by construction] AC = BD [Given] AL = DM [AD || BC] ∴ ΔALC ≅ ΔDMB [RHS congruence criterion] ⇒ ∠ACL = ∠DBM [Corresponding parts of congruent triangles] ...(1) Now, comparing ΔABC and ΔDCB, AC = BD [Given] ∠ACB = ∠DBC [Using (1)] BC = BC [Common]

Congruence of Triangles ∴ ΔABC ≅ ΔDCB [SAS congruence criterion] ⇒ AB = DC [Corresponding parts of congruent triangles] Thus, in a trapezium, if the diagonals are equal, then the non parallel sides are also equal.

Practical Geometry Important Questions

Topic: Construction of Parallel Lines

1. Draw a line segment PQ of length 4.7 cm. Take a point A on PQ. Draw a perpendicular on point A. On the perpendicular line, take a point B at a distance of 3.5 cm away from A. At point B construct a line l parallel to PQ.

(3 marks)

2. Take three non collinear points P, Q, and R. Join them to form ΔPQR. Through Q draw a line parallel to PR.

(3 marks)

3. Construct ΔPQR with PQ = 6 cm, QR = 5 cm and ∠Q = 70°. Through each of the vertices of ΔPQR draw a line parallel to the opposite side such that each pair of parallel lines meet together to form another ΔSTU.

(4 marks)

Topic: Construction of Triangles

4. Construct ΔJKL such that JK = 3 cm, KL = 4 cm, and JL = 5 cm. (3 marks)

5. Construct an isosceles ΔPQR with perimeter 16 cm, and the length of the unequal side 6 cm.

(3 marks)

6. Construct a ΔLMN with MN = 5.6 cm, NL = 6.5 cm and ∠N = 40°. Measure the side LM. What type of triangle is it?

(3 marks)

7. Construct an isosceles triangle in which the lengths of each of its equal sides is 7 cm and the angle between them is 120°.

(3 marks)

8. Construct ΔUVW where UV = 7 cm, m ∠UVW = 120°, and m ∠VWU = 40°.

(3 marks)

9. Examine whether you can construct ΔPQR such that QR = 4.3 cm, m ∠Q = 85°, and m ∠R = 115°. Justify your answer.

(3 marks)

10. Construct an isosceles right­angled triangle PQR, where m ∠PRQ = 90° and PR = QR = 4.7 cm.

(3 marks)

Practical Geometry Solutions

Topic: Construction of Parallel Lines

1. The steps of the construction are as follows: ( i ) Draw a line segment, PQ, of length 4.7 cm. Take a point A on PQ.

( ii ) Taking A as centre and with a convenient radius draw an arc cutting PQ at points X and Y.

( iii ) With X and Y as centres and radius more than half of XY draw two arcs which intersect each other at Z. Join AZ. Let AZ intersects XY at point C.

( iv ) With A as centre and radius 3.5 cm draw an arc to cut AZ at point B.

Practical Geometry

( v ) With B as centre and radius the same as in step ( ii ), draw an arc to cut AZ at points U and V.

( vi ) Place the pointed tip of the compass at Y such that the pencil tip is at C. With this opening draw an arc with V as centre to cut the arc UV at W. Join BW to form a line l parallel to PQ.

Practical Geometry

2. The steps of the construction are as follows: ( i ) Take three non­collinear points P, Q, and R. Join them to form ΔPQR.

( ii ) With R as centre, draw an arc to intersect PR and QR at points X and Y respectively.

( iii ) With Q as centre and radius same as in step ( ii ), draw an arc to intersect QR at Z.

Practical Geometry ( iv ) With Z as centre and radius equal to the length of arc XY, draw an arc to intersect the arc drawn in step ( iii ) at A.

( v ) Join QA to form a line l parallel to PR.

3. The steps of the construction are as follows: ( i ) Draw a line segment, QR, of length 5 cm. At point Q, draw a ray QV making an angle of 70° with QR.

Practical Geometry

( ii ) With Q as centre and radius 6 cm, draw an arc to cut the ray QV at P. join PR to form ΔPQR.

( iii ) Taking R as centre and with a convenient radius, draw an arc to intersect QR and PR at points A and B respectively.

( iv ) With P as centre and the same radius as in step ( iii ) draw an arc to cut PR at C.

Practical Geometry

( v ) Measure arc AB and taking C as centre and radius equal to arc AB draw an arc cutting the arc drawn in step ( iv ) at point D. Join PD to get line l parallel to QR.

( vi ) Similarly, through Q and R respectively draw line m parallel to PR and line n parallel to PQ.

Practical Geometry

Thus, the parallel lines meet together to form ΔSTU.

Topic: Construction of Triangles

4. The steps of construction are as follows: ( i ) Draw a line segment KL of length 4 cm.

( ii ) Taking point L as centre, draw an arc of 5 cm radius.

( iii ) Taking point K as centre, draw an arc of radius 3 cm to meet the previous arc at point J.

Practical Geometry

( iv ) Join J to L and J to K.

ΔABC is the required triangle.

5. It is given that ΔPQR is an isosceles triangle. Let PQ = PR It is given that the length of unequal side is 6 cm. ∴ QR = 6 cm Perimeter = 16 cm ⇒ PQ + QR + PR = 16 cm ⇒ PQ + 6 cm + PQ = 16 cm ⇒ 2PQ = 16 cm − 6 cm ⇒ 2PQ = 10 cm

∴ PQ = PR = 5 cm The steps of the construction are as follows: ( i ) Draw a line segment, QR, of length 6 cm.

( ii ) Taking Q as centre and radius as 5 cm draw an arc.

Practical Geometry

( iii ) Taking R as centre, and radius as 5 cm draw another arc to intersect the previous arc at P.

( iv ) Join PQ and PR to obtain the required isosceles ΔPQR.

6. The steps of the construction are as follows: ( i ) Draw a line segment, MN, of length 5.6 cm.

( ii ) Draw a ray, NZ, making an angle 40° with MN.

Practical Geometry

( iii ) With N as centre and radius 6.5 cm, draw an arc to intersect NZ at L. Join ML.

It can be measured that the length of side LM is 3.3 cm. Using protractor ∠M is measured as 90°. Thus, ΔLMN is a right angled triangle.

7. We have to construct an isosceles triangle PQR with PQ = QR = 7 cm and ∠PQR = 120°. The steps of construction are as follows: ( i ) Draw the line segment QR of length 7 cm.

( ii ) At point Q, draw a ray QX making an angle 120° with QR.

Practical Geometry

( iii ) Taking Q as centre, draw an arc of 7 cm radius. It intersects QX at point P.

( iv ) Join P to R to obtain the required triangle PQR.

Practical Geometry

8. A rough sketch of the required ΔUVW is drawn below:

In order do construct ΔUVW, we have to know the measure of ∠WUV. According to the angle sum property of triangles, we have ∠UVW + ∠VWU + ∠WUV = 180º 120º + 40º + ∠WUV = 180º 160º + ∠WUV = 180º ∠WUV = 180° − 160° = 20° The steps of construction are as follows: ( i ) Draw a line segment UV of length 7 cm.

( ii ) At U, draw a ray UX making an angle of 20º with UV.

( iii ) At point V, draw a ray VY making an angle of 120º with UV.

Practical Geometry

( iv ) The point W has to lie on both the rays UX and VY. Therefore, W is the point of intersection of these two rays.

Hence, the construction of required triangle is completed.

9. Given that: m ∠Q = 85° and m ∠R = 115° Hence, m ∠Q + m ∠R = 85° + 115° = 200° However, according to the angle sum property of triangles, we should have m ∠P + m ∠Q + m ∠R = 180° Thus, angle sum property is not followed by the given triangle. And thus, we cannot construct ΔPQR with the given measurements.

Practical Geometry

Also, it can be observed that point P should lie on both rays QX and RY for constructing the required triangle, but both rays do not seem to intersect each other. Hence, the required triangle cannot be formed.

10. In an isosceles triangle, the lengths of any two sides are equal. Let in ΔPQR, PR = QR = 4.7 cm. Rough sketch of this ΔPQR is shown as below:

The steps of construction are as follows: ( i ) Draw a line segment PR of length 4.7 cm.

( ii ) At point R, draw a ray RX making 90º with PR.

Practical Geometry

( iii ) Taking point R as centre, draw an arc of 4.7 cm radius to intersect RX at point Q.

( iv ) Join P to Q to obtain the required isosceles ΔPQR.

Practical Geometry

Perimeter and Area Important Questions

Topic: Perimeter of Different Shapes

1. If the cost of fencing a square­shaped field at the rate of Rs 7 per m is Rs 840, then find the side of the field.

(2 marks)

2. A wire which was bent in the shape of a regular octagon of side 14 cm is stretched into a straight line and then a piece of length 2 cm is cut out. The remaining wire is then bent into the shape of a regular decagon. What is the length of each side of the decagon so formed?

(2 marks)

3. If the sum of the length and the breadth of a rectangular land is 90 m, then what is the perimeter of the field?

(2 marks)

4. A farmer wants to fence a rectangular garden of dimensions 14 m ×12 m with three rounds of barbed wires excluding an entrance gate of width 2 m. If the cost of barbed wire is Rs 15 per metre, then how much amount of money is required for the fencing?

(2 marks)

Topic: Area ­ I

5. The perimeter of a rectangular metal sheet is 280 cm. If its length is 80 cm, then what is the area of the metal sheet?

(2 marks)

6. In the given figure, ABCD is a rectangle with AB = 24 cm and BC = 18 cm. A rhombus PQRS is formed inside ABCD by joining the midpoints of AB, BC, CD and DA. What is the area of the rhombus PQRS?

(3 marks)

Perimeter and Area 7. What is the area of hexagon ABCDEF?

(4 marks)

Topic: Area ­ II

8.

The side AB of parallelogram ABCD is of length 16 cm. The ratio of the area of parallelogram ABCD to the area of ΔADE is 16 : 5. If the area of ΔADE is 60 cm 2 , then find the area of trapezium ABCE.

(4 marks)

9. The given figure shows a parallelogram ABCD. What is the area of trapezium ABCX?

(3 marks)

Perimeter and Area

Topic: Area ­ III

10. In the given figure, PQRS is a rectangle. Find the area of the shaded region.

(3 marks)

11. In the given figure a right ΔBAC is shown. D is the midpoint of AC. What is the length of DE?

(3 marks)

Perimeter and Area Topic: Circumference and Area of Circle

12. If the ratio of the circumferences of two circles is 5:7, then what is the ratio of the radii of the circles?

(2 marks)

13. The length of a chord of a circle passing through the centre is 5.6 cm. What is the circumference of the circle?

(2 marks)

14. What is the area of the shaded regions in the given figure?

(4 marks)

15. In the given figure, ABCD is a rectangle and CDEF is a parallelogram. Three semicircles are drawn with sides AB, BC and AD as their diameters. What is the area of the given figure?

(4 marks)

Perimeter and Area Solutions

Topic: Perimeter of Different Shapes

1. Cost of fencing the square field = Rs 840 Rate of fencing = Rs 7 per m It is known that, Cost of fencing the square field = Perimeter of the field × Rate of fencing

It is also known that, Perimeter of square = 4 × Side

Thus, the side of the field is 30 m.

2. Since the wire was bent in the shape of the regular octagon of side 14 cm, the length of the wire is the perimeter of the octagon. Perimeter of the octagon = 8 × Side = 8 × 14 cm = 112 cm Therefore, original length of the wire is 112 cm. Since 2 cm long piece of wire is cut from the wire, length of remaining wire = 112 cm − 2 cm = 110 cm It is given that the remaining wire is re­bent into the shape of a regular decagon. ∴ Perimeter of regular decagon = 110 cm ⇒ 10 × Side = 110 cm

Thus, length of the side of the regular decagon is 11 cm.

3. Perimeter of rectangle = 2(Length + Breadth) i.e., perimeter of rectangle is twice the sum of its length and breadth The sum of length and breadth of the rectangular field is given as 90 cm. Thus, perimeter of the rectangular field = 2 × 90 m = 180 m

4. Length of rectangular garden = 14 m

Perimeter and Area Width of rectangular garden = 12 m Perimeter = 2 ×(Length + Breadth) = 2 ×(14 m + 12 m) = 2 ×26 m = 52 m Since an entrance gate of width 2 m is excluded in fencing, ∴ In one round of fencing, the farmer requires (52 − 2) m = 50 m of barbed wire. Thus, cost of three round fencing of the garden = Rs (3 × 50 ×15) = Rs 2,250

Topic: Area ­ I

5. Perimeter of the rectangular metal sheet is given as 280 cm. This implies that twice the sum of its length and breadth is 280 cm. ∴ Sum of length and breadth = 280 cm ÷ 2 = 140 cm The length of the rectangular sheet is 80 cm ∴ Breadth = 140 cm − 80 cm = 60 cm ∴ Area of the rectangular metal sheet = length × breadth = 80 cm × 60 cm = 4800 cm 2

6. It is given that P, Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively. AB = 24 cm ∴ AP = 12 cm AD = BC = 18 cm ∴ AS = 9 cm ∴ Area of ΔAPS = × AP × AS = × 12 cm × 9 cm = 54 cm 22

121

Similarly, Area of ΔPBQ = Area of ΔRCQ = Area of ΔRDS = 54 cm 2

Area of rectangle ABCD = AB × BC = 24 cm × 18 cm = 432 cm 2

∴ Area of the rhombus PQRS = Area of rectangle ABCD − Area of the four triangles = 432 cm 2 − (54 cm 2 + 54 cm 2 + 54 cm 2 + 54 cm 2 ) = 432 cm 2 − 216 cm 2

= 216 cm 2

Thus, the area of the rhombus PQRS is 216 cm 2 .

7. Area of the hexagon = Area of ΔABF + Area of rectangle BCDF + Area of ΔDEF On applying Pythagoras theorem to ΔABX, we obtain AB 2 = AX 2 + BX 2

Perimeter and Area ⇒ BX 2 = (5 cm) 2 − (4 cm) 2 ⇒ BX 2 = 25 cm 2 − 16 cm 2 ⇒ BX 2 = 9 cm 2 ⇒ BX 2 = (3 cm) 2 ⇒ BX = 3 cm Since AB = AF = 5 cm, ΔABF is isosceles. ∴ BX = XF = 3 cm BF = BX + XF = 3 cm + 3 cm = 6 cm

Area of ΔABF = × BF × AX = × 6 cm × 4 cm = 12 cm 221

21

Area of rectangle BCDF = BF × BC = 6 cm × 10 cm = 60 cm 2

Since ED = EF, ΔDEF is isosceles. Hence, perpendicular EY bisects side DF. ∴ DY = YF = 5 cm On applying Pythagoras theorem to ΔDEY, we obtain EY 2 = DE 2 − DY 2

⇒ EY 2 = (13 cm) 2 − (5 cm) 2 ⇒ EY 2 = 169 cm 2 − 25 cm 2 ⇒ EY 2 = 144 cm 2 ⇒ EY 2 = (12 cm) 2 ⇒ EY = 12 cm ∴ Area of ΔDEF = × DF × EY = × 10 cm × 12 cm = 60 cm 22

121

∴ Area of hexagon ABCDEF = 12 cm 2 + 60 cm 2 + 60 cm 2 = 132 cm 2 Thus, the area of hexagon ABCDEF is 132 cm 2 .

Perimeter and Area Topic: Area ­ II

8. We know that:

Area of parallelogram ABCD = Base × Height = AB × BF = 16 cm × BF Area of ΔADE = × Base × Height2

1 = × DE × BF2

1 Area of parallelogram: Area of triangle = 16 : 5 (Given) Therefore, we have

Now, area of ΔADE = 60 cm 2 = × DE × BF2

1 ⇒ 60 cm 2 = × 10 cm × BF2

1 ⇒ BF = 12 cm ∴Area of parallelogram ABCD = AB × BF = 16 cm × 12 cm = 192 cm 2

∴Area of trapezium ABCD = Area of parallelogram ABCD − Area of ΔADE = 192 cm 2 − 60 cm 2

= 132 cm 2

Perimeter and Area 9. Area of parallelogram ABCD = Base × Height

= CD × AX = 15 cm × 12 cm = 180 cm 2 It is known that the opposite sides of a parallelogram are equal. ∴ AD = BC = 13 cm On applying Pythagoras theorem to ΔADX, we obtain DX 2 = AD 2 − AX 2 = (13 cm) 2 − (12 cm) 2 = 169 cm 2 − 144 cm 2 = 25 cm 2 = (5 cm) 2 ⇒ DX = 5 cm ∴ Area of ΔADX = × Base × Height = × 5 cm × 12 cm = 30 cm 22

121

∴ Area of trapezium ABCX = Area of parallelogram ABCD − Area of ΔADX = 180 cm 2 − 30 cm 2 = 150 cm 2 Thus, the area of trapezium ABCX is 150 cm 2 .

Topic: Area­III

10. Area of the shaded region = Area of rectangle PQRS − Area of ΔPQY − Area of ΔXYR − Area of ΔPSX Now, area of rectangle PQRS = Length × Breadth = 16 m × 10 m = 160 m 2 Area of ΔPQY = × Base × Height2

1 = × 7 m × 10 m = 35 m 22

1

Area of ΔXYR = × Base × Height21

= × (16 m − 7 m) × (10 m − 4 m)21

= × 9 m × 6 m21

= 27 m 2 Area of ΔPSX = ×Base × Height2

1 = × 16 m × 4 m2

1 = 32 m 2 ∴ Area of shaded region = 160 m 2 − 35 m 2 − 27 m 2 − 32 m 2 = 66 m 2

11. Applying Pythagoras theorem in ΔABC, BC 2 = AB 2 + AC 2 ⇒ AC 2 = (20 cm) 2 − (12 cm) 2

Perimeter and Area = 400 cm 2 − 144 cm 2 = 256 cm 2 = (16 cm) 2 ⇒ AC = 16 cm It is given that D is the mid­point of AC. ∴ AD = DC = 8 cm It is known that the area of a triangle is given by × Base × Height.2

1 Area of ΔABC = × AB × AC = × 12 cm × 16 cm = 96 cm 22

121

Area of ΔABD = × AB × AD = × 12 cm × 8 cm = 48 cm 221

21

∴ Area of ΔBDC = Area of ΔABC − Area of ΔABD = 96 cm 2 − 48 cm 2 = 48 cm 2 Also, area of ΔBDC = × BC × DE2

1 ∴ 48 cm 2 = × 20 cm × DE2

1 ⇒ 10 cm × DE = 48 cm 2

Thus, the length of DE is 4.8 cm.

Topic: Circumference and Area of Circle

12. It is known that the circumference of a circle with radius r is given by2π r . Let r 1 and r 2 be the radii of the two circles. It is given that the ratio of the circumferences of the two circles is 5:7.

Thus, the ratio of the radii of the two circles is 5:7.

13. It is given that the length of a chord of a circle passing through the centre is 5.6 cm. ∴ Diameter of the circle = 5.6 cm Radius of the circle,

∴ Circumference of the circle = 2π r

Thus, the circumference of the circle is 17.6 cm.

Perimeter and Area 14.

Diameter of the semicircle I = 3.5 cm Radius of the semicircle I

∴ Area of the semicircle I

Diameter of the semicircle II = 2.1 cm Radius of the semicircle II

∴ Area of the semicircle II

Diameter of the semicircle III = 3.5 cm + 2.1 cm = 5.6 cm Radius of the semi­circle III = 2.8 cm ∴ Area of the semicircle III =

∴ Area of the shaded region = Area of semicircle I + Area of semicircle III − Area of semicircle II = 4.81 cm 2 + 12.32 cm 2 − 1.73 cm 2 = 15.4 cm 2

15. Area of parallelogram CDEF = Base × Height = EF × CG = 14 cm × 12 cm = 168 cm 2 It is known that the opposite sides of a parallelogram are equal.

Perimeter and Area ∴ DE = CF = 13 cm CD = EF = 14 cm On applying Pythagoras theorem to ΔCGF, we obtain FG 2 = CF 2 − CG 2 ⇒ FG 2 = (13 cm) 2 − (12 cm) 2 ⇒ FG 2 = 25 cm 2 ⇒ FG 2 = (5 cm) 2 ⇒ FG = 5 cm ∴ Area of ΔCGF = × Base × Height = × FG × CG = × 5 cm × 12 cm = 30 cm 22

121

21

Area of rectangle ABCD = AD × CD = 7 cm × 14 cm = 98 cm 2 Area of the semicircle with AB as diameter

Area of the semicircle with AD as diameter = Area of the semicircle with BC as diameter

∴ Area of the given figure = Area of CDEF + Area of ΔCGF + Area of ABCD + Area of the three semicircles = 168 cm 2 + 30 cm 2 + 98 cm 2 +77 cm 2 + 19.25 cm 2 + 19.25 cm 2 = 411.5 cm 2 Thus, the area of the given figure is 411.5 cm 2 .

Symmetry Important Questions

Topic: Rotational Symmetry

1. Use the following information to answer the next question.

What is the order of rotational symmetry of the given figure?

(2 marks)

2. What is the order of the rotational symmetry of the given figure?

(2 marks)

3. Draw a figure which has no lines of symmetry, but order of rotational symmetry of two.

(2 marks)

4. Find the order of rotational symmetry of the following figure.

(2 marks)

Symmetry 5. A figure is given as:

(a) Locate the centre of rotation. (b) Find the order of rotational symmetry. (c) What is the angle of rotation?

(3 marks)

6. In the following figure, consider the given dotted line as the line of symmetry and hence mark the other hole (s).

(2 marks)

7. Is the order of rotational symmetry of the given figure the same as the number of its lines of symmetry?

(2 marks)

Symmetry 8. A figure is given as:

(a) Draw all the lines of symmetry of the given figure. (b) Find the order of rotational symmetry and the angle of rotation.

(3 marks)

9. Name the quadrilateral(s) which has both line of symmetry and rotational symmetry of order more than 1. Justify your answer by giving suitable explanations.

(3 marks)

10. Which letter of the word “NIGHT” does not possess rotational symmetry and line symmetry both? Also, find what are the order of rotational symmetry and number of lines of symmetry of each alphabet?

(3 marks)

Symmetry Solutions

Topic: Rotational Symmetry

1. The order of rotational symmetry of a figure is the number of different positions of the figure in which the figure looks exactly the same, when it is rotated about its centre. To find the order of rotational symmetry of the given figure, a dot is placed at the corner of the figure to identify a different position. Then, the figure is rotated on a tracing paper about its centre.

Here, it is seen that there are two different positions in which the figure looks exactly the same. Thus, the order of rotational symmetry of the given figure is 2.

2. In a complete turn of 360°, the number of times an object looks exactly the same is called the order of rotational symmetry. The order of the rotational symmetry of the given figure is 1 as it looks exactly the same only after rotating 360°.

3.

Parallelogram has no lines of symmetry. However, it has a rotational symmetry of order 2. This can be represented as:

4. Order of rotational symmetry of a figure is the number of different positions of the figure in which it looks exactly the same when it is being rotated about its centre. The given figure has exactly 2 positions in which it looks exactly the same when it is being

Symmetry rotated about its centre. It can be represented by the following figure.

Hence, the given figure has a rotational symmetry of order 2.

5. (a) The centre of rotation of the given figure is its centre.

(b) It can be observed that the given figure has rotational symmetry of order 6.

(c) The angle of rotation is 60°.

6. The other holes should be marked in such a way that the obtained figure is symmetric with respect to the given dotted line i.e., the line of symmetry.

Symmetry Thus, the given figure can be completed in the following way:

7. It is known that if, after a rotation, an object looks exactly the same, then it has a rotational symmetry. The given figure has a rotational symmetry of order 2.

The given figure has two lines of symmetry which can be shown by dotted lines as:

Thus, the order of rotational symmetry of the given figure is the same as the number of its lines of symmetry.

8. (a) Lines of symmetry of the given figure can be represented by dotted line as:

Symmetry (b) It can be observed that the given figure has a rotational symmetry of order 8 and angle of rotation of 45°. This can be represented as:

9. It is known that a square has 4 lines of symmetry. They can be shown by dotted lines as:

Also, when a square is rotated about its centre, then at four different positions it looks exactly the same in a complete rotation of 360°. Thus, a square has a rotational symmetry of order 4.

Symmetry

A rectangle has 2 lines of symmetry. They can be shown by dotted lines as:

Also, when a rectangle is rotated about its centre, then at two different positions it looks exactly the same in a complete rotation of 360°. Thus, a rectangle has a rotational symmetry of order 2.

A rhombus has 2 lines of symmetry. They can be shown by dotted lines as:

Also, when a rhombus is rotated about its centre, then at two different positions it looks exactly the same in a complete rotation of 360°. Thus, a rhombus has a rotational symmetry of order 2.

Thus, square, rectangle and rhombus has both lines of symmetry and rotational symmetry of order more than 1.

Symmetry 10. Consider the letter N.

The letter N has no line of symmetry. When it is rotated about its centre, then at two different positions it looks exactly the same in a complete rotation of 360°. Thus, it has a rotational symmetry of order 2.

Consider the letter I. The letter I has 2 lines of symmetry. They can be shown by dotted lines as:

Also, the letter I has a rotational symmetry of order 2.

Now, consider the letter G. It has no line of symmetry. However, it has a rotational symmetry of order 1.

∴ The letter G does not possess rotational symmetry and line symmetry both. Now, consider the letter H. It has 2 lines of symmetry. They can be shown by dotted lines as:

Also, it has a rotational symmetry of order 2.

Symmetry

Now, consider the letter T. It has one line of symmetry. It can be shown by a dotted line as:

Also, it has a rotational symmetry of order 1.

Visualising Solid Shapes Important Questions

Topic: Three Dimensional Figures ­ I

1. Find the dimensions of the cuboid formed by folding the following net.

(1 mark)

2. Draw the net for the following figure.

(2 marks)

3. Draw the oblique sketch of the figure given in isometric dot paper.

(2 marks)

4. The isometric sketch of a solid has been shown in the given figure. Draw the oblique sketch of the solid.

(2 marks)

Visualising Solid Shapes

5. Draw an oblique and isometric sketch of a cuboid of length 6 units, breadth 4 units, and height 3 units.

(3 marks)

6. How many cubes are there in the given figure?

(2 marks)

7. How many cubes are there in the given figure?

(3 marks)

Topic: Three Dimensional Figures ­ II

8. Identify the cross­sections obtained in the given cases. (a) When a cylinder is cut vertically

(4 marks)

Visualising Solid Shapes

(b) When a cuboid is cut horizontally

(c) When a cone is cut horizontally

(d) When a cube is cut vertically

9. When a torch is kept in front of, and above a cone, then what will be the (2 marks)

Visualising Solid Shapes shapes of the resulting shadows?

10. Draw the front view and the side view of the given solid.

(3 marks)

Visualising Solid Shapes Solutions

Topic: Three Dimensional Figures ­ I

1. By folding the given net, the following cuboid can be formed.

∴The dimensions of the obtained cuboid are: 3 units × 2 units × 1 unit

2. The net for the given three­dimensional figure can be drawn as:

3. The required oblique sketch is:

Visualising Solid Shapes

4. The oblique sketch of the solid can be drawn as:

5. Oblique sketch

Visualising Solid Shapes Isometric sketch

6. The lower layer of the given figure comprises 9 cubes.

The middle layer of the given figure comprises 2 cubes.

The upper layer of the given figure comprises 1 cube.

Thus, total number of cubes in the given figure = 9 + 2 + 1 = 12

7. The lower layer of the given figure comprises 18 cubes.

The middle layer of the given figure comprises 17 cubes.

Visualising Solid Shapes

The upper layer of the given figure comprises 3 cubes.

Thus, total number of cubes in the given figure = 18 + 17 + 3 = 38

Topic: Three Dimensional Figures ­ II

8. (a) When a cylinder is cut vertically, a rectangle is obtained.

(b) When a cuboid is cut horizontally, a rectangle is obtained.

(c) When a cone is cut horizontally, a circle is obtained.

Visualising Solid Shapes

(d) When a cube is cut vertically, a square is obtained.

9. When a torch is kept in front of a cone, a triangular shadow will be formed. When the torch is above the cone, the resulting shadow will be circular in shape. This can be shown as:

Visualising Solid Shapes

10. The side view of the given solid can be drawn as:

The front view of the given solid can be drawn as: