Grade IXMathematics

Exam Preparation Booklet(Chapter Wise

Important Questions-Solutions)

Number Systems_Questions

Q.No.1:

Rationalise the denominator .

Q.No.2:

Find three irrational numbers between and .

Q.No.3:

How can the decimal be expressed in the form ?

Q.No.4: Write one rational and one irrational number between and .

Q.No.5: If a and b are rational numbers and , find the value of

a and b.

√3 √5

= a + b√54+3√5

4−3√5

Polynomials_Questions

Q.No.1: How can the value of (19.9)3 be found using a suitable identity?

Q.No.2: The polynomial p(x) = 2x4 + x3 − 14x2 − ax − 6 is exactly divisible by q(x) = x2

+ 3x + 2. By using the long division method, find the value of a?

Q.No.3: Find the degree of the polynomial .

Q.No.4: If is a zero of the polynomial p(x) = 8x3 – k, then find the value of k.

Q.No.5: Factorize (x3 – 3x2 – 9x – 5)

+ + 4x2 − + 10

x3

4

5x−1

x−3

x

3

−1

2

Coordinate Geometry_Questions

Q.No.1: On a map, Vipul’s house is located at the origin. From his house, he reaches hisschool by moving 5 units on the positive x-axis and then a units on the negativey-axis. However, one day, he reaches his school by moving 4 units on the negativey-axis from his house and then b units on the positive x-axis. Find the respectivevalues of a and b according to the map.

Q.No.2: Which quadrant is enclosed by negative x-axis and positive y-axis? In whichquadrant does the point (−5, −21) lie?

Q.No.3: Among the four points (−2, 5), (1, 2), (3, 0) and (5, 4), which three points lie ona straight line?What are the coordinates of the foot of the perpendicular drawn from theremaining point to the straight line?

Q.No.4: Write the co-ordinates for the points P and N in the following figure:

Q.No.5: The total number of points which are at a distance of 4 cm and 5 cmfrom x-axis and y-axis respectively are _____.

Linear Equations in Two Variables_Questions

Q.No.1: Solve the equation, 2x − 1 = x + 4, and represent the solution on the(i) number line(ii) Cartesian plane

Q.No.2: Draw the graph of the equation y = 2x + 4. From the graph, find the value of xwhen y = −4.

Q.No.3: If point (k, −k) lies on the line x + y + c = 0, then what can be said about c?

Q.No.4: The sum of a two-digit number and the number obtained by interchanging thedigits is 88. Represent this information graphically. Also, graphically find the two-digit number if the digit at the ones place is 2?

Q.No.5: Find the point where 3x + 2y = 12 intersects the y-axis.

Lines and Angles_Questions

Q.No.1: What is the value of x in the given figure?

Q.No.2:

In the given figure, AC is the bisector of ∠BAD.Find the measures of ∠1, ∠2, ∠3, and ∠4.

Q.No.3: In the given figure, AB || DE and EF || BC.

What is the measure of ∠DEF?

Q.No.4:

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

Q.No.5: The maximum number of triangles that can be drawn having angles 55°,63° and 61° is _____.

Triangles_Questions

Q.No.1:

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

Q.No.2: Prove that the angles opposite to equal sides of an isosceles triangle are equal.

Q.No.3: Prove that the angles opposite to equal sides of an isosceles triangle are equal.Use the result to prove the following.In the given figure, BD is equal to DC. What is the measure of ∠ADB?

Q.No.4: Show that in a trapezium, if the diagonals are equal, then the non parallel sidesare also equal.

Q.No.5:

In the given figure, QM = RL, OM = 4 cm and MQ = 6 cm.If the area of ΔROQ is 25 cm2, then what is the area of ΔQRL?

Quadrilaterals_Questions

Q.No.1: The diagonals of a quadrilateral are perpendicular to each other. Prove that thequadrilateral formed by joining the mid-points of the quadrilateral is a rectangle.

Q.No.2:

In the given figure, ALCM is a parallelogram, and L and M are the mid points of ABand CD respectively.Prove that AD || BC

Q.No.3:

In the given figure, ABCD is a rhombus; X and Y are the mid points of sides ABand BC respectively.What is the ratio of area of ΔXBY to the area of pentagon AXYCD?

Q.No.4:

In the given figure, ABCD is a rhombus. L and M are the midpoints of sides ADand CD respectively. The lengths of diagonals AC and BD are 16 cm and 12 cmrespectively. What is the area of ΔDLM?

Q.No.5: Show that the bisectors of angles of a parallelogram form a rectangle.

Circles_Questions

Q.No.1: From the given figure, find ∠ABC.

Q.No.2:

The given figure shows a circle with centre O where the chords AB and CD areparallel. What is the measure of ∠AOC?

Q.No.3:

In the give figure, OA and OC are the angle bisectors of ∠BAD and ∠BCDrespectively.What is the measure of ∠ACD?

Q.No.4:

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

Q.No.5: Two chords AB and CD of a circle are parallel to each other. If the length of thechord CD, whose perpendicular distance from the centre O is 4 cm is given to be 6cm. The length of the chord AB is _____.

Constructions_Questions

Q.No.1: Construct a ΔABC in which BC = 7 cm, ∠B = 60°, and AB + AC = 12 cm.

Q.No.2: Construct a triangle PQR in which QR = 6.7 cm, ∠Q = 45°, and PQ − PR = 2.8 cm.

Q.No.3: Construct a right-angled triangle whose base is 9 cm and sum of its hypotenuseand other side is 14 cm.

Q.No.4: Construct an angle of 112.5° using a compass.

Q.No.5: Construct a right triangle whose base is 7.5 cm and the difference between itshypotenuse and the remaining side is 4.5 cm.

Heron's Formula_Questions

Q.No.1: In a triangle, the length of the first side is 5 cm less than the length of the secondside. The length of the third side is 7 cm more than the length of the first side. Ifthe first side is of length 18 cm, then what is the semi-perimeter of the triangle?

Q.No.2: A triangular park has three sides, which are in the ratio 12:5:8 such that its areais . The park is to be fenced with wire, leaving 4 m side gates on eachside, at a rate of Rs 30 per metre. Find the cost to fence the wire around the park.

Q.No.3: If each side of a triangle is decreased by 20%, then by what percentage does itsarea decrease?

Q.No.4: The sides of a triangle are in the ratio 2:3:4. If the semi-perimeter of the triangleis 36 cm, then what is its area?

Q.No.5: Anmol decided to find the area of triangle ABC. He measured the sides ofthe triangle and calculated a few things. The observation made by Anmol aregiven as:

, where are the sides of the triangle and s is thesemi-perimeter.The perimeter of this particular triangle is _____ units.

s − a = s − b = s − c = 1 a, b, c

Surface Areas and Volumes_Questions

Q.No.1: Two cubes are joined to form a cuboid. If the total surface area of the cuboid thusformed is 640 cm2, then what is the length of the edge of the cube?

Q.No.2: The curved surface area of a wooden pen stand, which is in the shape of acylinder, is to be varnished. The height and the base radius of the pen stand are14 cm and 3.5 cm respectively. If the cost of varnishing the pen stand is Rs 46.20,then what is the cost of varnishing the top of a wooden table whose area is 3.75m2?

Q.No.3: A sphere has same curved surface area as that of a cone of height 4 cm and baseradius 3 cm. What is the volume of the sphere?

Q.No.4: The curved surface area of a cylinder is 528 cm2 and its volume is 1848 cm3. Theheight of the cylinder is same as the height of a cone of volume 100π cm3. Whatis the cost of painting the cone from outside (including the base) at the rate of 70paise per sq cm?

Q.No.5: What will the new curved surface area of a cylinder be if the radius isdoubled and height is halved?

Statistics_Questions

Q.No.1: The given bar graph shows the numbers of cars and bikes that use a toll road overa particular week.

Study the bar graph and answer the given questions.(a) On which day was the number of cars and bikes using the toll road the same?(b) On which day was the number of cars using the toll road less than thenumber of bikes using the toll road?

Q.No.2: Use the following information to answer the next question.

The given double bar graph shows the number ofstudents of class VI of a particular school and theirpreferences in different types of movies.

Study the given bar graph and answer the following questions.(i) Which type of movie is preferred by most of the girls?(ii) Which type of movie is the least preferred among boys?(iii) Find the total number of students in the class?

(iv) How many girls and how many boys prefer to watch action movies?(v) What scale is used in the given graph?

Q.No.3: The given table lists the daily maximum temperature (°C) of a certain city for themonth of April.32, 33, 32, 31, 32, 29.5, 30.5, 27.5, 28, 29.7, 29.9, 31.2, 31.7, 32, 32.4, 32.5,33, 34.7, 34.9, 35, 34.8, 35, 36, 37, 36.5, 34, 35.5, 36.5 32, 29How can the given information be represented in the form of frequencydistribution table using appropriate class size?

Q.No.4: The given bar graph shows the number of male employees working in morningand 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 and nightshifts the same?

Q.No.5: Study the bar graph given below and answer the following questions:

(a) What information is represented in the bar graph? (b) In which 2 years, the number of foreigners who visited Delhi are 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 Delhias compared to the previous year? (f) In which 2 years, the number of foreigners who visited Mumbai are the same?

Probability_Questions

Q.No.1: A carton contains 240 bulbs, out of which 18 are defective. Find the probabilitythat a randomly chosen bulb is(a) non-defective (1 mark)(b) defective (1 mark)

Q.No.2: Three bags, bag I, bag II, and bag III, contain 40 red balls, 80 red balls, and 30red balls respectively. Find the probability of less than 50 balls in a bag. If a redball is randomly chosen, then find the probability that it is chosen from bag II.

Q.No.3: A survey was held to know the heights of students of a class. The following tablerepresents the heights of the 90 students in the class. Using this table, answer thefollowing questions:Height of students (in cm)Number of students

120−130 2130−140 10140−150 20150−160 36160−170 18170−180 4

(a) Find the probability that the height of a randomly chosen student is more thanor equal to 130 cm, but less than 160 cm.(b) Find the probability that the height of a randomly chosen student is more than180 cm.(c) Find the probability that the height of a randomly chosen student is more thanor equal to 120 cm.

Q.No.4: Use the following information to answer the next question.A survey is conducted across 1000 families to find out the number of boy child ina family. The given data shows the findings of the survey:

Number of boys in the family 0 1 2 3Number of families 205 310 435 50

What is the probability that a family chosen at random has at most 2 boys?

Q.No.5: A recent survey found that the ages of 200 workers in a factory isdistributed as follows:

Age (in years) 20 - 29 30 - 39 40 - 49 50 - 59 60 andabove

Number ofworkers 38 27 86 46 3

If a person is selected at random, find the probability that the person is:(i) 40 years or more(ii) under 40 years(iii) having age from 30 to 39 years(iv) under 60 but over 39 years

Number Systems_SolutionsSolution 1

Multiplying numerator and denominator by , we obtain

Solution 2

A number that has a non-terminating and non-recurring decimal expansion is an

irrational number. Therefore, to find three irrational number between and ,

it is required to search for three numbers that are non-terminating andnon−recurring and also lying between the numbers and 4.898.

Thus, the three irrational numbers between and are

4.46101101110..., 4.67484884888...and 4.7214345414...

Solution 3

Let x = = 1.28888... ... (1)∴ 10x = 12.8888... ... (2)Subtracting equation (1) from equation (2),9x = 11.6

Solution 4

We know that and So, a rational number between and is 2 And, an irrational number between and is 1.801001000100001...

Solution 5

Rationalise the denominator.

√3 = 1. 7320. . . √5 = 2. 2360. . .√3 √5

√3 √5

= a + b√54+3√5

4−3√5

× = a + b√5

⇒ = a + b√5

⇒ = a + b√5

⇒ = a + b√5

⇒ − − = (a + b√5)

4+3√5

4−3√5

4+3√5

4+3√5

(4+3√5)2

(4)2−(3√5)2

16+45+24√5

−29

61+24√5−29

6129

24√5

29

Compare both the sides, we get; a = −   and  b = −61

292429

Polynomials_SolutionsSolution 1

(19.9)3 can also be written as (20 − 0.1)3

Using identity (x − y)3 = x3 − y3 − 3xy (x − y),

Thus, the value of (19.9)3 is 7,880.599.

Solution 2

Using the long division method, p(x) can be divided by q (x).This can be done as:

Since p(x) is exactly divisible by q(x), the remainder must be zero.∴ 10 − a + 9 = 0⇒ a = 19Thus, the value of a is 19.

Solution 3

Given polynomial is This can further be simplified as

+ + 4x2 − + 10x3

45x−1

x−3

x

3

The greatest index of a variable in the given polynomial is 3. Thus, the degree of the given polynomial is 3.

Solution 4

Put in

Since, is a zero of polynomial

Solution 5

Let p(x) = x3 − 3x2 − 9x − 5By trial method, we havep(−1) = (−1)3 − 3(−1)2 − 9(−1) − 5= − 1 − 3 + 9 − 5 = 0Therefore, x + 1 is a factor of this polynomial.Let us find the quotient on dividing x3 − 3x2 − 9x − 5 by x + 1.By long division,

It is known that,Dividend = Divisor × Quotient + Remainder

+ + 4x2 − + 10

= + 5x−1+3 + 4x2 − + 10

= + 5x2 + 4x2 − + 10

= + 9x2 − + 10

x3

45x−1

x−3

x

3

x3

4x

3

x3

4x

3

x3

4x

3

x = − 12

p (x)

− 12

p (x)

⇒ p (− ) = 0

⇒ 8(− )3

− k = 0

⇒ 8 × (− ) − k = 0

⇒ −1 − k = 0

⇒ k = −1

12

12

18

∴ x3 − 3x2 − 9x − 5 = (x + 1)(x2 − 4x − 5) + 0= (x + 1)(x2 − 5x + x − 5)= (x + 1)[(x(x − 5) +1(x − 5)]= (x + 1)(x − 5)(x + 1)= (x − 5)(x + 1)(x + 1)

Coordinate Geometry_SolutionsSolution 1

Vipul’s house is located at the origin (0, 0).It is given that from his house, he reaches his school by moving 5 units on thepositive x-axis and then a units on the negative y-axis.Therefore, his school is located at (5, −a).It is given that he can also reach his school by moving 4 units on the negative y-axis from his house and then b units on the positive x-axis.Therefore, his school's located can also be expressed as (b, −4).It is known that the coordinates of every point on a coordinate plane is unique.∴ (5 , −a) = (b , −4)⇒ a = 4 and b = 5Thus, the respective values of a and b are 4 and 5.

Solution 2

Second quadrant is enclosed by the negative x-axis and the positive y-axis.The point (−5, −21) lies in the third quadrant because the coordinates of a pointin the third quadrant are of the form (−, −).

Solution 3

The given four points can be plotted on the Cartesian plane as:

It can be seen that, (−2, 5), (1, 2) and (3, 0) lie on a straight line.The coordinates of the foot of the perpendicular drawn from (5, 4) to the line are(2, 1).

Solution 4

The coordinates for the points are given by P( 6,2) and N(0,2).

Solution 5

From the figure, it is clear that the total number of points which are at a distanceof 4 cm and 5 cm from x-axis and y-axis respectively are four.

−

Linear Equations in Two Variables_SolutionsSolution 1

The given equation is2x − 1 = x + 4⇒ 2x − x = 4 + 1⇒ x = 5Thus, x = 5 is the solution of the given equation.(i) It can be represented on number line as follows.

(ii) On Cartesian plane, x = 5 represents a line parallel to y-axis at a distance of5 units from the origin.

Thus, the solution of the given equation is represented by line AB on the Cartesianplane.

Solution 2

The given equation isy = 2x + 4The two solutions of this equation are as follows.x1−6y6−8Plotting the points (1, 6) and (−6, −8) on graph paper, we obtain the followinggraph.

It can be seen from the graph that the value of x, corresponding to y = −4, is −4.

Solution 3

It is given that the point (k, −k) lies on the line x + y + c = 0.∴ k + (−k) + c = 0⇒ 0 + c = 0⇒ c = 0Thus, the value of c must be zero.

Solution 4

Let the ones place and tens place digits of the number be x and y respectively.∴ Two-digit number = 10 × y + x = 10y + xWhen the digits of the number are interchanged, then the ones place digit andtens place digit are y and x respectively.∴ New two-digit number = 10 × x + y = 10x + yAccording to the given information,

Four different solutions of equation (1) are given in the table.x 0 1 7 8

y = 8 − x8 7 1 0On plotting and joining points (0, 8), (1, 7), (7, 1) and (8, 0), a straight linerepresenting equation (1) is obtained. This can be done as:

Now, we are required to find the two-digit number whose ones place digit, i.e., xis 2.The value of y for x = 2 can be obtained by drawing a line parallel to y-axis at adistance of 2 units to the right of y-axis; then, from the point of intersection ofthis line and the previously drawn line, a perpendicular is drawn to y-axis. Thisperpendicular intersects y-axis at (6, 0).This means that the value of y corresponding to x = 2 is 6.Thus, the two-digit number is 10 × 6 + 2 = 62.

Solution 5

Since the point of intersection is on the y-axis, so the x coordinate will be 0. Put x = 0 in equation, we get;

So, point of intersection will be .

3 (0) + 2y = 12

⇒ 2y = 12

⇒ y = 6

(0, 6)

Lines and Angles_SolutionsSolution 1

Exterior angle of a triangle is equal to the sum of its two opposite interior angles.∠DBC = ∠BAC + ∠ACB .....(1)∠FCB = ∠CAB + ∠ABC .....(2) From equations (1) and (2) ∠DBC + ∠FCB = ∠BAC + ∠ACB + ∠CAB + ∠ABC3x +2x = ∠BAC + 180°5x = ∠BAC + 180° .....(3)∠BAC + ∠CAE = 180 (Linear pair)∠BAC + 110° = 180°∠BAC = 70° .....(4)From equations (3) and (4),5x = 70° + 180°5x = 250°x = 50°

Solution 2

By applying angle sum property in ΔABC, we obtain∠ABC + ∠BCA + ∠BAC = 180°⇒ 25°+ 134°+ ∠BAC = 180°⇒ ∠BAC = 180°− 159°= 21°Now, AC is the bisector of ∠BAD.∴ ∠BAC = ∠CAD⇒ ∠2 = 21° ( ∠BAC = 21°)We know that the measure of one complete angle is 360°.∴ ∠BCD + ∠BCA + ∠ACD = 360°

⇒ 126°+ 134°+ ∠3 = 360°⇒ ∠3 = 360°− 260°= 100°By exterior angle property of triangles, we have∠1 = ∠2 + ∠3⇒ ∠1 = 21°+ 100°= 121°By applying angle sum property in ΔACD, we obtain∠ACD + ∠CDA + ∠CAD = 180°⇒ 100°+ ∠4 + 21°= 180°⇒ ∠4 = 180°− 121°= 59°

Solution 3

It is given that AB || DE.Here, BC can be regarded as the transversal.∴ ∠ABD = ∠EDC (Corresponding angles)⇒ ∠EDC = 67°It is also given that EF || BC.∴ ∠EDC + ∠DEF = 180° (Interior angles on the same side of thetransversal are supplementary)⇒ 67° + ∠DEF = 180°⇒ ∠DEF = 180° − 67°⇒ ∠DEF = 113°Thus, the measure of ∠DEF is 113°.

Solution 4

Applying angle sum property of triangle in ΔOAB:∠OAB + ∠OBA + ∠AOB = 180°⇒ 90° + 40° + ∠AOB = 180°⇒ 130° + ∠AOB = 180°⇒ ∠AOB = 50°∠COD = ∠AOB (Vertically opposite angles)⇒ ∠COD = 50°Consider ΔOCD:When a side of a triangle is produced, then the measure of the exterior angle soformed is equal to the sum of the measures of its opposite interior angles.∴∠OCE = ∠COD + ∠CDO = 50° + 90° = 140°Thus, the measure of ∠OCE is 140°.

Solution 5

Since, 55° + 63° + 61° = 179° This triangle doesn't satisfy the angle sum property of triangles. Therefore, no such triangle exists. Hence, the required number of triangles is zero.

Triangles_SolutionsSolution 1

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

Solution 2

Let PQR be an isosceles triangle with PQ = PR

Let PS be the bisector of ∠RPQ.To prove: ∠PQR = ∠QRPIn ΔPQS and ΔPRS,PQ = PR (Given)PS = PS (Common)∠QPS = ∠RPS (By construction)∴ΔPQS ≅ ΔPRS (By SAS congruency criterion)We know that corresponding parts of congruent triangles are equal.Therefore, we have ∠PQR = ∠PRQHence, the angles opposite to equal sides of an isosceles triangle are equal.

Solution 3

Let ABC be an isosceles triangle such that AB = AC.

We have to prove that∠B = ∠CThe median AD is drawn.In ΔADB and ΔADC,AB = AC (Given)BD = DC (Median)AD = AD (Common side)∴ ΔADB≅ΔADC (By SSS congruence rule)∴ ∠B = ∠C (C.P.C.T.)Thus, in an isosceles triangle, angles opposite to equal sides are also equal.

Using the above result,∴ ∠DBC = ∠DCB … (1)In ΔABC, ∠ABC + ∠BAC + ∠ACB = 180°⇒ 84° + 48° + ∠ACB = 180°⇒ ∠ACB = 180° − 132° = 48°From (1), we obtain, ∠DCB = ∠DBC = 48°∠ADB is the exterior angle of ΔBDC.∴ ∠ADB = ∠DCB + ∠DBC⇒ ∠ADB = 2 × ∠DCB = 2 × 48° = 96°Thus, the measure of ∠ADB is 96°.

Solution 4

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]∴ 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 arealso equal.

Solution 5

Comparing ΔRMQ and ΔQLRMQ = 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 cm2

⇒ area (ΔQLR) = × 4 cm × 6 cm + 25 cm2

⇒ area (ΔQLR) = 12 cm2 + 25 cm2

⇒ area (ΔQLR) = 37 cm2

Thus, the area of ΔQLR is 37 cm2.

Quadrilaterals_SolutionsSolution 1

ABCD is a quadrilateral.P, Q, R, and S are mid-points of sides AB, BC, CD, and DA respectively.

Also, AC ⊥BDTo prove:PQRS is a rectangle.Now, by applying mid-point theorem in ΔABC and ΔADC, we obtainPQ||AC and SR||ACThis gives PQ||SR

Also,

⇒ PQ = SRThus, PQRS is a parallelogram.Now, in order to show that PQRS is a rectangle, we just have to prove that themeasure of all the four angles is 90°.Consider the quadrilateral PXOY.We have PY||XO ( PQ||AC)And, PX||OY ( PS||DB)∴ PXOY is a parallelogram.∴ ∠XPY = ∠XOY ( opposite angles of a parallelogram are equal)∴ ∠XPY = 90°i.e., ∠SPQ = 90°Now, PQRS is a parallelogram with one angle, ∠SPQ = 90°Thus, PQRS is a rectangle.

Solution 2

It is given that L and M are mid points of AB and CD.∴ DM = MC … (1)AL = LB … (2)It is known that the opposite sides and the opposite angles of a parallelogram areequal.It is given that ALCM is a parallelogram. Therefore, AL || MC and AM || LC∴ AL = MC … (3)AM = CL … (4)

From (1), (2) and (3),DM = LB … (5)Now, ∠DMA = ∠MAL {Alternate interior angles}∠MAL = ∠BLC {Corresponding angles}∴ ∠DMA = ∠BLC … (6)Using (4), (5) and (6), it can be said that ΔADM ≅ ∠CBL, by SAS CongruencyCriterion.It is known that the corresponding parts of congruent triangles areequal.ΔADM ≅ ΔCBL∴ ∠DAM = ∠BCL … (7)∠ADM = ∠CBLAlso, ALCM is a parallelogram.∴ ∠MAL = ∠MCL … (8)Adding equations (7) and (8),∠DAM + ∠MAL = ∠BCL + ∠MCL⇒ ∠DAL = ∠MCBIt is known that if both pairs of opposite angles of a quadrilateral are equal, thenthe quadrilateral is a parallelogram.Therefore, ABCD is a parallelogram.Thus, AD is parallel to BC.

Solution 3

Join the diagonals AC and BD of the rhombus ABCD such that BD intersects XY atZ, and AC at O.

It is known that the line joining the midpoints of two sides of a triangle is parallelto its third side and its length is half of the length of the third side.Therefore, XY is parallel to AC.

∴ XY =

It is known that the diagonals of a rhombus bisect each other at right angles.

∴ OB =

∠BOC = 90°∴ ∠BZY = 90° {Corresponding angles}According to the converse of midpoint theorem, the line through the midpoint of aside and parallel to the other side, bisects the third side.Therefore, Z is the mid-point of OB.

∴ Area of ΔBXY = × BZ × XY

Area of rhombus ABCD = × AC × BD

∴ Area of pentagon AXYCD = Area of rhombus ABCD − Area of ΔBXY

Thus, the ratio of area of ΔBXY to area of pentagon AXYCD is 1: 7

Solution 4

It is known that the line joining the mid points of two sides of a triangle is parallelto its third side and its length is half of the length of the third side.Therefore, LM is parallel to AC.

∴ LM =

It is known that the diagonals of a rhombus bisect each other at right angles.

∴ OD =

∠DOC = 90°∠DOC = ∠DXM = 90° (Corresponding angles)In ΔODC, M is the mid point of DC and XM is parallel to OC.Therefore, by converse of Mid-point theorem, X is the mid point of OD.

∴ DX =

Now area of a triangle is × base × height.

∴ Area ΔDLM = × LM × DX = × 8 cm × 3 cm = 12 cm2

Thus, the area of ΔDLM is 12 cm2.

Solution 5

In ΔADH, HAD + ADH = Also,

∠ ∠ + = (∠A + ∠D) = × 180° = 90°∠A2

∠D2

12

12

A A A °

(Vertically opposite angles)

Similarly, we can prove that So, EFGH is a quadrilateral in which all angles are 90 Hence, EFGH is a rectangle.

∠HAD + ∠ADH + ∠DHA = 180°

⇒ 90° + ∠DHA = 180°

⇒ ∠DHA = 90°

⇒ ∠GHE = 90°

∠HEF = ∠EFG = ∠FGH = 90°°

Circles_SolutionsSolution 1

We know that angles in the same segment of a circle are equal.∴ ∠ACB = ∠ADB⇒ ∠ACB = 75°By applying angle sum property in ΔABC, we obtain∠ABC + ∠ACB + ∠BAC = 180°⇒ ∠ABC + 75°+ 35°= 180°⇒ ∠ABC = 180°− 110°= 70°Thus, the measure of ∠ABC is 70°.

Solution 2

Sine AB || CD, ∠ABC = ∠BCD [Alternate interior angles]∴∠ABC = 30° [∠BCD = 30°]It is known that the angle subtended by an arc at the centre of the circle is doublethe angle subtended by it at any point on the remaining part of the circle.∴∠AOC = 2 × ∠ABC = 2 × 30° = 60°Thus, the measure of ∠AOC is 60°.

Solution 3

It is known that the angle subtended by an arc at the centre is double the anglesubtended by it at any point on the remaining part of the circle.∴ ∠AOC = 2 × ∠ABC = 2 × 80° = 160° ... (1)OA is given as the angle bisector of ∠BAD.∠BAD = 2 × ∠OAB = 2 × 46° = 92°It is also known that the sum of the either pair of opposite angles of a cyclicquadrilateral is 180°∴ ∠BAD + ∠BCD = 180°⇒ 92° + ∠BCD = 180°⇒ ∠BCD = 88°OC is the angle bisector of ∠BCD.

∴∠OCD = ∠BCD = × 88° = 44° ... (2)

Applying angle sum property of a triangle to ΔOCA:∠OCA + ∠OAC + ∠AOC = 180°⇒ ∠OCA + ∠OCA + ∠AOC = 180° [OA = OC]⇒ 2∠OCA + 160° = 180° [Using equation (1)]⇒ 2∠OCA = 20°⇒ ∠OCA = 10° ... (3)

∠OCD = ∠OCA + ∠ACD⇒ 44° = 10° + ∠ACD [Using equations (1) and (2)]⇒ ∠ACD = 44° − 10°⇒ ∠ACD = 34°Thus, the measure of ∠ACD is 34°.

Solution 4

Join BC.

It is known that the angle subtended by an arc at the centre is double the anglesubtended by it at any point on the remaining part of the circle.∴∠AOB = 2 × ∠ACB

⇒ ∠ACB =

Consider ΔBFC:In a triangle, measure of exterior angle is equal to the sum of the measures ofinterior opposite angles.∠BFA = ∠FBC + ∠FCB⇒ 40° = ∠FBC + 22.5°⇒ FBC = 17.5°⇒ ∠DBC = 17.5°It is also known that angles in the same segment are equal.∠DEC = ∠DBC⇒ ∠DEC = 17.5°Thus, the measure of ∠DEC is 17.5°.

Solution 5

We know that perpendicular from the centre bisects the chord. ∴CQ=CD2=62=3 cm Also, OQ = 4 cm. Therefore, using Pythagoras theorem, we get OC = 5 cm, which is the radius of the circle. Now, AB = The diameter of circle = 10 cm.

Constructions_SolutionsSolution 1

The required triangle can be constructed as follows.1. Draw a line segment BC of length 7 cm.2. Draw a ray BX making an angle of 60° with BC.3. With B as centre and radius equal to 12 cm, draw an arc which cuts BX at D.4. Join DC and draw its perpendicular bisector which intersects BD at point A.5. Join AC. Thus, ΔABC is the required triangle.

Solution 2

The below given steps will be followed to draw the required triangle.Step I:Draw the line segment QR = 6.7 cm and at point Q, make an angle of 45°,say ∠XQR.Step II: Cut the line segment QS = 2.8 cm (equal to PQ − PR) on ray QX.Step III: Join SR and draw the perpendicular bisector AB of SR.Step IV:Let it intersect QX at point P. Join PR. ΔPQR is the required triangle.

Solution 3

The below given steps are followed to construct the required triangle.Step I: Draw line segment AB of 9 cm. Draw a ray AX making 90°with AB.Step II: Cut a line segment AD of 14 cm (as the sum of the other two sides is 14)on ray AX.Step III: Join DB and make an angle DBY equal to ADB.Step IV: Let BY intersect AX at C. Join BC.ΔABC is the required triangle.

Solution 4

The angle 112.5° can be written as

Therefore we first draw angles of 90° and 135°, and then bisect the anglebetween them.The steps of the construction are as follows:Step 1: Draw a line AB and mark a point C on it.

Step 2: Draw an angle of 90° at C using a compass. The angle obtained is ∠QCB= 90º and ∠ACQ = 180° − 90º = 90º.Step 3: With M and Q as centres, and radius more than half of MQ, draw arcsintersecting each other at R. The angle obtained is ∠RCB = 135º.Step 4: With K and Q as centres and radius more than half of KQ, draw arcsintersecting each other at P. Join PC.

Thus, ∠PCB = 112.5° is the required angle.

Solution 5

Let ΔABC be right-angled at B in such that BC = 7.5 cm and AC − AB = 4.5 cm.The steps of construction are as follows:(i) Draw a line segment BC of length 7.5 cm.(ii) At point B, draw ∠XBC = 90°.(iii) Extend XB to Y.(iv) Draw an arc taking B as the centre and radius equal to 4.5 cm such that itintersects BY at point D.(v) Join CD.(vi) Draw the perpendicular bisector of CD intersecting XB at point A.(vii) Join AC.Thus, ΔABC is the required triangle.

Heron's Formula_SolutionsSolution 1

Let a, b and c be the first, second and third sides of the given trianglerespectively.It is given that a = b −5 cm, c = a + 7 cm and a = 18 cm.∴ 18 cm = b−5 cm and c = 18 cm + 7 cm⇒ b = 18 cm + 5 cm = 23 cm and c = 25 cmThus, the semi-perimeter of the triangle =

Solution 2

Let the sides of the triangular park be 12x, 5x, and 8x.

Now, semi-perimeter of the triangular park

Therefore, area of the triangular park

Now, perimeter of the park

Length of wire required to fence the park = 200 m − 3 ×4 m = 188 mIt is given that cost of fencing 1 m of wire = Rs 30Therefore, cost of fencing 188 m of wire = 188 ×Rs 30 = Rs 5640Therefore, it will cost Rs 5640 to fence the wire around the park.

Solution 3

Let the sides of the triangle be a, b and c respectively.

∴ Semi-perimeter, s =

Area, Each side of the triangle is decreased by 20%. Let the new sides be a′, b′, and c′.

Similarly,

∴ New semi-perimeter, s′ =

New area of the triangle,

∴ Decrease in area

∴ Percentage decrease in the area =

Thus, the area of the triangle decreases by 36%.

Solution 4

Let the sides of the triangle be 2x, 3x and 4x.It is given that the semi-perimeter of the triangle is 36 cm.∴ s = 36 cm

Therefore, the sides of the triangle are 8 × 2 = 16 cm, 8 cm × 3 = 24 cm and 8cm × 4 = 32 cm.Let a = 16 cm, b = 24 cm and c = 32 cm

Thus, the area of the triangle is .

Solution 5

The observation made by Anmol are: , where are the sides of the triangle

and s is the semi-perimeter. Adding the three equations, we get

s − a = 1,  s − b = 1,  s − c = 1 a, b, c

And therefore, perimeter = 2s = 6 units.

3s − (a + b + c) = 3

⇒ 3s − 2s = 3                            [∵ s = ]

⇒ s = 3

a+b+c

2

Surface Areas and Volumes_SolutionsSolution 1

Let the length of the edge of the cube be a.Then, the length of the cuboid = 2aBreadth of the cuboid = aHeight of the cuboid = a

It is known that the total surface area of a cuboid is given by 2 (lb + bh + lh),where l, b and h represent the length, breadth and height of the cuboidrespectively.∴ 640 cm2 = 2 (2a.a + a.a + 2a.a)⇒ 640 cm2 = 2 (2a2 + a2 + 2a2)⇒ 10 a2 = 640 cm2

⇒ a2 = 64 cm2

⇒ a = 8 cmThus, the length of the edge of the cube is 8 cm.

Solution 2

It is known that the lateral surface area of a cylinder is given by 2πrh, where r isthe base radius and h is the height of the cylinder.

∴ Area to be varnished = 2πrh

Cost of varnishing 308 cm2 area = Rs 46.20

∴ Cost of varnishing 1 cm2 area = Rs = Rs 0.15

Cost of varnishing 1 m2 area = Rs 1500 Cost of varnishing 3.75 m2 area = Rs 1500 × 3.75 = Rs 5625

Thus, the cost of varnishing the top of the wooden table is Rs 5625.

Solution 3

It is given that:Height of the cone, h = 4 cmBase radius of the cone, r = 3 cm∴ Slant height

Curved surface area of the cone = πrl = π (3 cm) (5 cm) = 15π cm2

Let R be the radius of the sphere.

Curved surface area of the sphere = 4πR2

According to the given information:∴ 4πR2 = 15π cm2

∴ Volume of the sphere

Thus, the volume of the sphere is 9.75π cm3.

Solution 4

Let r and h respectively be the radius and the height of the cylinder.It is known that:Curved surface area of the cylinder = 2πrhVolume of the cylinder = πr2h

∴ Volume of the cylinder = πr2h

⇒ 154h cm2 = 1848 cm3

It is given that the height of the cylinder is same as the height of a cone.∴ Height of the cone, h = 12 cmLet R be the radius of the cone.

It is known that the volume of a cone is given by .

∴ Slant height of the cone,

∴ Total surface area of the cone

∴ Area of the cone to be painted

Cost of painting 1 cm2 = 70 paise = Rs 0.70

∴ Cost of painting the cone from outside =

Thus, the cost of painting the cone from outside is Rs 198.

Solution 5

Let the radius of the cylinder be r and its height be h. Its curved surface area is given by New radius, r' = 2r and new height, h' =

So, the new curved surface is , which is the same

as compared to that of the original cylinder.

2πrhh

2

2πr'h' = 2π (2r)( ) = 2πrhh

2

Statistics_SolutionsSolution 1

(a) It can be seen in the graph that the heights of the bars corresponding to bikesand cars are equal on Tuesday.Thus, the number of bikes and cars using the toll road was the same on Tuesday.(b) It can be seen in the graph that on Sunday, the bar corresponding to cars isshorter than the bar corresponding to bikes.Thus, on Sunday, the number of cars using the toll road was less than the numberof bikes using the toll road.

Solution 2

(i)The length of the bars shows the number of girls or boys who prefer to watchdifferent types of movies.It can be seen from the bar graph that comedy movies are preferred by most ofthe girls.(ii)It can be seen from the bar graph that among the boys, the least preferred typeof movie is drama.(iii)Total number of girls in the class = 30 + 12 + 12 + 20 = 74Total number of boys in the class = 28 + 14 + 26 + 12 = 80Therefore, total number of students in the class = 74 + 80 = 154(iv)Number of girls who prefer to watch action movies = 12Number of boys who prefer to watch action movies = 26(v)The scale used in the given bar graph is 1 unit = 4 students

Solution 3

It can be observed that the given data is quite large. Here, the maximumtemperature varies from 27.5°C to 36.5°C. Thus, one can condense it into groupssuch as 27 − 30, 30 − 33, 33 − 36, and 36 − 39. Then, tally marks table is drawnby marking for the daily maximum temperature in the respective class interval.Here, five tally marks are represented by , where the fifth tally mark is markedacross the remaining four tally marks to bunch them into five.Then, the required frequency table can be constructed as:Class interval (°C)Tally marksFrequency

27 − 30 630 − 33 1133 − 36 9

36 − 39 4Total 30

Solution 4

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

Solution 5

(a) The given bar graph shows the number of foreigners who visited Delhi andMumbai during the years 2004-2008.(b) The number of foreigners who visited Delhi was the same in the years 2005and 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 inyear 2006 as compared to the previous year 2005. The increase can be calculated by subtracting the number of foreigners whovisited Delhi in the year 2005 from the number of foreigners who visited Delhi inthe year 2006 i.e. 600 − 200 = 400(f) The number of foreigners who visited Mumbai was the same in the years 2004and 2008.

Probability_SolutionsSolution 1

(a) Number of bulbs in the carton = 240Number of defective bulbs in the carton = 18∴ Number of non-defective bulbs in the carton = 240 − 18 = 222Probability that a randomly chosen bulb is non-defective

(b) Probability that a randomly chosen bulb is defective

Solution 2

Number of red balls in bag I, bag II, and bag III are 40, 80, and 30. ∴ Total number of red balls in three bags = 40 + 80 + 30 = 150 Out of the three bags, only two bags (bag I and bag III) contain less than 50balls.

Thus, probability of less than 50 balls in a bag

Bag II contains 80 red balls. Total number of red balls = 150

Thus, probability that a randomly chosen red ball is from bag II

Solution 3

Total number of students = 90(a) Number of students with height more than or equal to 130 cm, but less than160 cm = 10 + 20 + 36 = 66∴ Probability that height of a randomly chosen student is more than or equal to130 cm, but less than 160 cm

(b) There is no student with height more than 180 cm.∴ Probability that height of a randomly chosen student is more than 180 cm

(c) The height of all the students is either more than or equal to 120 cm.∴ Number of students with height more than or equal to 120 cm = 90Probability that height of a randomly chosen student is more than or equal to 120cm

Solution 4

Total number of families = 1000Number of families having at most 2 boys = 205 + 310 + 435 = 950

Thus, required probability

Solution 5

The total number of workers = 200. We know that

(i) P(40 years or more) = (ii) P(under 40 years) = (iii) P(Age from 30 to 39 years) = (iv) P(Under 60 but over 39 years) =

P(an   event) =number   of   favourable   outcomes

number   of   possible   outcomes

= =86+46+3

200

135

200

27

40

= =38+27

200

65

200

13

40

27

200

= =86+46

200

132

200

33

50