RS Aggarwal 2017 Solutions for Class 9 Math Chapter 17 Summative Assessment II are provided here with simple step-by-step explanations. These solutions for Summative Assessment II are extremely popular among class 9 students for Math Summative Assessment II Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2017 Book of class 9 Math Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2017 Solutions. All RS Aggarwal 2017 Solutions for class 9 Math are prepared by experts and are 100% accurate.

Page No 635:

Question 1:

Mark (✓) against the correct answer in each of the following:
Which of the following equations represents a line parallel to the x-axis?
(a) x + y = 5
(b) 2x + 5 = 8
(c) 2y − 4 = 1
(d) x + 4 = 0

Answer:

(c) 2y − 4 = 1

2y-4=12y=4+12y=5y=2.5
A linear equation that contains only the y variable is parallel to the x-axis.

Page No 635:

Question 2:

Mark (✓) against the correct answer in each of the following:
If the mode of the data 9, 7, x, 10, 12, 11, 14, 15, 13, 12, 10, 15 is 15, then the value of x is
(a) 10
(b) 10.5
(c) 12
(d) 15

Answer:

(d) 15
The mode is 15 and we know that mode is the value which occurs most frequently.

Page No 635:

Question 3:

Mark (✓) against the correct answer in each of the following:
The surface area of a solid hemisphere is
(a) 2πr2
(b) 3πr2
(c) 4πr2
(d) 23πr3

Answer:

(b) 3πr2

Page No 635:

Question 4:

Mark (✓) against the correct answer in each of the following:
The height of a cone of diameter 10 cm and slant height 13 cm is
(a) 12 cm
(b) 13 cm
(c) 69 cm
(d) 194 cm

Answer:

(a) 12 cm

Radius  = Diameter2=102=5 cmWe know thatl2=h2+r2h=l2-r2=132-52=169-25=144=12 cm

Page No 635:

Question 5:

Mark (✓) against the correct answer in each of the following:
Each side of a rhombus is 10 cm long and one of its diagonals measures 16 cm. The length of the other diagonal is
(a) 6 cm
(b) 12 cm
(c) 13 cm
(d) 156 cm

Answer:

(b) 12 cm




We know that the diagonals of a rhombus bisect each other at right angles.
So, BO = OD = 8 cm
In right-angled triangle AOB,AO2+BO2=AB2AO=AB2-BO2=102-82=100-64=36=6 cmSo,AC=2AO=2×6=12 cm

Page No 635:

Question 6:

Mark (✓) against the correct answer in each of the following:
In the given figure, ​∠B < ​∠A and ​∠D > ​C. Then
(a) AD > BC
(b) AD = BC
(c) AD < BC
(d) AD = 2BC

Answer:

(c) AD < BC

 In AOB,
B < A
So, AO < BO .........(i)
                       (Side opposite the larger angle is larger)
 In DOC,
C < D
So, DO  < CO .........(i)
                       (Side opposite the larger angle is larger)
Adding equations (i) and (ii), we get:
AO + DO < BO + CO
Therefore, AD < BC.



Page No 636:

Question 7:

Mark (✓) against the correct answer in each of the following:
In the given figure, ABDC and ABFE are cyclic quadrilaterals. If ACD = 110°, then ∠AEF = ?
(a) 55°
(b) 70°
(c) 90°
(d) 110°

Answer:

(d) 110°

Since ACDE is a cylic quadrilateral, so

B+C=180°B=180°-C=180°-110°=70°
Again, AEFB is a cylic quadrilateral.
So,
B+E=180°E=180°-B=180°-70°=110°
i.e. AEF = 110°

Page No 636:

Question 8:

Mark (✓) against the correct answer in each of the following:
In the given figure, the measure of BCD is
(a) 100°
(b) 70°
(c) 80°
(d) 30°

Answer:

(c) 80°

We know that the angles in the same segment are equal.So,BAC=BDC=30°In DBC, DBC+DCB+BDC=180°  (Angle sum property)BCD=180° -DBC-BDCBCD=180°-70°-30°BCD=80°Hence,BCD=80°

Page No 636:

Question 9:

A dice was rolled 100 times and the number of times that 6 appeared was noted. If the probability of getting a 6 is 25, how many times did 6 come up?

Answer:

P(6 appeared) = Number of favourable outcomesTotal number of outcomes25=Number of favourable outcomes100Number of favourable outcomes=100×25=40
So, 6 appeared 40 times.

Page No 636:

Question 10:

In a circle of radius 10 cm, find the length of a chord which is at a distance of 5 cm from its centre.

Answer:



We know that the perpendicular drawn from the centre of a circle bisects the chord of the circle.
So, AC = BC
Now, in AOC,AO2=AC2+OC2AC=AO2-OC2=102-52=100-25=75=53=5×1.732=8.66 cmAB=2×8.66=17.32 cm

Page No 636:

Question 11:

The radius and slant height of a cone are in the ratio 4 : 7. If its curved surface area is 792 cm2, find its radius.

Answer:

Suppose, the radius and slant height of the cone are 4x cm and 7x cm.
Curved surface are of the cone  = 792 cm2
πrl = 792227×4x×7x=79288x2 =792x2=9 x=3
 Hence, radius of the cone  = 4x = 4×3 = 12 cm

Page No 636:

Question 12:

Prove that a cyclic parallelogram is a rectangle.

Answer:




Given : ABCD is a cyclic parallelogram.
To prove : ABCD is a rectangle.
Proof :
ABCD is a paralleogram.
 So,ABC=CDA.......(i)                    (Opposite angles of a parallelogram are equal)

ABC+CDA=180°  (ABCD is a cylic quadrilateral)ABC+ABC=180°  2ABC=180°ABC=90°
So, ABC = CDA = 90°
Similarly, DAB =BCD = 90°
Since, all the angles of the parallelogram are 90°, ABCD is a rectangle.

Page No 636:

Question 13:

In the given figure, O is the centre of circle, ABO = 20° and ∠ACO = 30°. If ∠BOC = x°, find the value of x.

Answer:

In AOB,
AOBO (radius of the circle)
So, OBA = BAO = 20°  (Angles opposite equal sides are equal)
In AOC, 
AOCO (Radii of the circle)
So, OCA = CAO = 30°  (Angles opposite equal sides are equal)
BAC = CAO + BAO
= 30°+ 20° 
= 50° 

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point in the remaining part of the circle.So,BOC=2BACBOC=2×50°=100°


BOC = 100° 
x° = 100° 
x = 100

Page No 636:

Question 14:

In the given figure, ABCD is a cyclic quadrilateral with O as the centre of the circle. If BOD = 136°, find ∠BCD.

Answer:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point in the remaining part of the circle.So,BAD=12BODBAD=12×136°=68°
BCD+BAD=180°  (ABCD is a cyclic quadrilateral)BCD=180°-BAD                  =180°-68°                  =112°



Page No 637:

Question 15:

The diameter of a sphere is decreased by 20%. By what per cent does its curved surface area decrease?

Answer:

Let the diameter of the sphere be d.
Curved surface area of the sphere = 4πr2 = 4πd24=πd2
The diameter of the sphere is decreased by 20%.
So, d1=d-20% of d=d-20d100=d-d5=4d5Hence, d1=4d5So r1=4d10=2d5
New curved surface area = 4πr12
=4π×4d225=16πd225
Decrease in curved surface area =πd2 - 16πd225 = 9πd225
Percentage decrease = 9πd225πd2×100=9×10025=36%

Page No 637:

Question 16:

A conical tent is 10 m high and its base radius is 24 m.
Find (i) the slant height of the tent and (ii) the cost of canvas required to make the tent, if the cost of 1 m2 of canvas is Rs 70.

Answer:



(i)
l2=r2+h2=242+102=576+100=676=26 m
So, slant height of the cone = 26 m

(ii) Curved surface area of the cone = πrl

                                                   =  227×24×26 m2
Cost of 1 m2 area of the canvas = Rs 70
Cost of 227×24×26 m2 = 227×24×26 ×70=Rs 137280

Page No 637:

Question 17:

A cuboid has total surface area of 40 m2 and lateral surface area of 26 m2. Find the area of it base.

Answer:

Surface area of top and bottom face of cuboid = Total surface area of cuboid - lateral surface area of cuboid
                                                                          = 40-26 = 14 m2
We know that the area of top and bottom face of a cuboid are equal.
So, area of base = 142=7 m2

Page No 637:

Question 18:

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in square metres.

Answer:

Radius of the cylindrical roller = 842 = 42 cm
Length of the roller = 120 cm
So, curved surface area of the roller = 2πrh
=2×227×42×120=31680 cm2

Area covered by the roller in 1 revolution = 31680 cm2
Area covered by the roller in 500 revolution = 500×31680 = 15840000 cm2

                                                                 = 1584 m2

Page No 637:

Question 19:

Prove that an angle in a semicircle is a right angle.

Answer:

Given: AB is a diameter of a circle C(O,r) and ∠ACB is an angle in the semicircle.
To Prove:ACB = 90°



Proof: We know that the angle subtended by an arc at the centre of a circle is
twice the angle formed by it at any point on the remaining part of the circle.
∴ ∠AOB = 2∠ACB          [Since angle subtended by arc AB = 2 × angle formed by it at C]
⇒ 2∠ACB = 180°             [Since ∠AOB is a straight angle]
⇒ ∠ACB = 90°

Page No 637:

Question 20:

Prove that two angles in the same segment of a circle are equal.

Answer:



Given: C(O,r) is a circle and ∠ACB and ∠ADB are angles in the same segment.
To Prove: ∠ACB = ∠ADB
Proof: We know that the angle subtended by an arc at the centre of a circle is
twice the angle formed by it at any point on the remaining part of the circle.
∴ ∠AOB = 2∠ACB.................(i)                           
                                                  [Since angle subtended by arc AB = 2 × angle formed by it at C]
AOB = 2∠ADB.........(ii) 
                                                  [Since angle subtended by arc AB = 2 × angle formed by it at D]
From equations (i) and (ii):
2∠ACB = 2∠ADB          
⇒ ∠ACB = ∠ADB  

Page No 637:

Question 21:

In ABC, if E is the mid-point of median AD, prove that ar(BED)=14ar(ABC).

Answer:

Given:  D is the mid-point of BC and E is the mid-point of AD.
To prove: ar(BED) =14 ar(ABC)

Proof: 
Since, D is the mid-point of BC, AD is the median of  ABC and BE is the median of  ABD.
We know that a median of a triangle divides it into two triangles of equal area.
 So, ar ( ABD ) = 12 ar ( ABC).................(i)

So, ar ( BED ) =12ar ( ABD).................(ii)

From (i) and (ii), we have:
          ar ( BED  = 12​​ ⨯ 12⨯​ ar (∆ ABC
                        ∴​     ar (∆ BED )​ =14  ⨯ ar(∆ABC)  

Page No 637:

Question 22:

Diagonals AC and BD of a trapezium ABCD, with AB || DC, intersect at O. Prove that ar(AOD) = ar(BOC).

Answer:

CDA and ​∆ CBD lie on the same base and between the same parallel lines.
So, ar( ​∆ CDA) = ar( CDB)...(i)
Subtracting ar (​​∆ OCD) from both sides of equation(i), we get:
ar(​​∆ CDA) - ar (​​∆ OCD) = ​ ar(​​∆ CDB) -ar (​​∆ OCD)
⇒ ​ ar(​​∆ AOD)  =  ar (​​∆ BOC)

Page No 637:

Question 23:

Draw a cumulative frequency curve for the following frequency distribution.

Class Interval 0−9 10−19 20−29 30−39 40−49 50−59 60−69
Frequency 5 15 20 23 17 11 9

Answer:

 

Class Marks Frequency Cumulative Frequency
4.5 5 5
14.5 15 20
24.5 20 40
34.5 23 63
44.5 17 80
54.5 11 91
64.5 9 100

 We plot the points A(4.5,5),B(14.5,20),C(24.5,40),D(34.5,63),E(44.5,80),F(54.5,91) and G(64.5,100).

Page No 637:

Question 24:

The arithmetic mean of 35 values is 45. What will be the new mean if each of these values is
(a) increased by 7?
(b) multiplied by 5?
(c) divided by 5?

Answer:

Arithmetic mean = 45
(a) If each value is increased by 7, the mean will also be increased by 7.
New mean = 45+7 =52
(b) If each value is multiplied by 5, the mean will also be multiplied by 5.
New mean = 45×5 =225
(c) If each value is divided by 5, the mean will also be divided by 5.
New mean = 45÷5 = 9



Page No 638:

Question 25:

The mean of 16 items was found to be 30. On rechecking, it was found that two items were wrongly taken as 22 and 18 instead of 32 and 28, respectively. Find the correct mean.

Answer:

Given mean of 16 items = 30
Sum of 16 items = 16×30=480
Correct sum of 16 items  = 480-22-18+32+28 = 500
So, correct mean of 16 items = 50016=31.25

Page No 638:

Question 26:

Find the median for the following frequency distribution.

Variate 17 20 22 15 30 25
Frequency 5 9 4 3 10 6

Answer:

First, we will arrange the following data in ascending order.
 

     Variate      Frequency
         15            3
         17            5
         20            9
         22           4
         25           6
        30         10

Now, based on this data, we will make a cumulative frequency table

 

        Variate       Frequency   Cumulative Frequency
          15             3                3
          17             5                8
          20             9              17
          22             4              21
          25             6              27
         30           10              37

Here, n =37, which is odd.
 

Median= value of n+12th observation
Therefore, median=value of 37+12th observation                                 =value of 19th observationHere, we see that the value of 18th to 21th observation is 22.Therefore, value of the 19th observation is 22.Hence, median for the given data is 22.

Page No 638:

Question 27:

The volume of a right circular cone of height 24 cm is 1232 cm3. Find its curved surface area.

Answer:

Height of the cone = 24 cm
Volume of the cone = 1232 cm313πr2h=123213×227×r2×24=1232r2=1232×3×722×24r2=49r=7 cm

Curved surface area of the cone =πrl=227×7×72+242       l=r2+h2=22×625=22×25=550 cm2

Page No 638:

Question 28:

The diameter of a copper sphere is 6 cm. It is melted and drawn into a long wire of uniform circular cross-section. If the length of the wire is 36 cm, find its radius.

Answer:

Diameter of the sphere = 6 cm
So, radius of the sphere = 3 cm
Volume of the sphere = 43πr3
=43×227×3×3×3=7927 cm3
Length of the wire = 36 cm
Since, the sphere is melted and drawn into a wire, the volume of the sphere is equal to the wire.
Volume of the wire = 7927 cm3
πr2h=7927r2=7927×πhr2=792×77×22×36r2=1 r=1 cm

Page No 638:

Question 29:

The curved surface area and volume of a pillar are 264 m2 and 396 m3, respectively. Find the diameter and height of the pillar.

Answer:

Curved surface area of the pillar = 264 m2
Volume of  the pillar = 396 m3
Dividing the curved surface area of the pillar by its volume, we get:
2πrhπr2h=2643962r=264396r=396264×2r=3 m
Diameter = 2r = 6 m
So, diameter of the pillar is 6 m.
Curved surface area = 264 m2
2πrh=264h=2642πrh=264×72×22×3h=14 m
Height of the pillar is 14 m.

Page No 638:

Question 30:

In the adjoining figure, ABCD is a parallelogram and X, Y are the points on diagonal BD, such that DX = BY. Prove that CXAY is a parallelogram.

Answer:



Given : ABCD is a parallelogram.
To prove: CXAY is a parallelogram.
Construction : Join AC, meeting BD at O.
Proof: We know that the diagonals of a parallelogram bisect each other.
So, OA = OC and OD = OB.
OD = OB...(i)
DX = BY.........(ii)
Subtracting (ii) from (i),we get:
OD-DX = OB-BY
OX = OY


Since, CXAY is a quadrilateral whose diagonals bisect each other, it is a parallelogram.

Page No 638:

Question 31:

Construct a ABC in which BC = 5.6 cm, ∠B = 30° and the difference between the other sides is 3 cm.

Answer:



Steps of construction:

1. Draw a line segment BC = 5.6 cm.
2. Construct CBX = 30°.
3. Set off BP = 3 cm.
4. Join PC.
5. Draw the right bisector of PC, meeting BP produced to A.
6. Join AC.
ABC is thus constructed.

Page No 638:

Question 32:

In the given figure, calculate the area of quadrilateral ABCD.

Answer:

(i) ​
ar(quadrilateral ABCD) = ar(∆ ABD) + ar(∆ BDC)
ABD and ∆ BDC are right-angled triangles. So,

BD172 - 82  = 225 = 15 cm
and AB152 -92 = 144 = 12 cm
Now, 
ar(∆ ABD) = 12AD ⨯ AB 
12⨯9 ⨯ 12
= 54 cm2
ar(∆ BDC) = 12BD ⨯ BC 
12 ⨯15 ⨯ 8
= 60 cm2

Hence, area of quadrilateral ABCD  =  54 + 60 = 114 cm2.

Page No 638:

Question 33:

In the given figure, AB is a chord of a circle with centre O and AB is produced to C, such that BC = OB. Also, CO is joined and produced to meet the circle at D. If ACD = y° and ∠AOD = x°, prove that x = 3y.

Answer:

OB BC (Given)
So,
BOC∠BCO = y°   (Angles opposite equal sides are equal)
In OBC,
OBA = BOC + BCO  (Exterior angle property)
= (
2y    
Again,
OAOB  (Radii of the circle)
So, OABOBA =(2y (Angles opposite equal sides are equal )
In AOB,
AOD = ∠OAC + ∠ACO  (Exterior angle property)
                           = ∠OAB + ∠BCO
                           = (2y)° + y°
                           = 3y°
Hence, x = 3y.

Page No 638:

Question 34:

In the given figure, AB and CD are straight lines through the centre O of a circle. If AOC = 80° and ∠CDE = 40°, find (i) ∠DCE, (ii) ∠OBC.

Answer:

(i)
CED = 90° (Angle in a semi-circle)
In ΔCED,
CED +EDC + DCE = 180°  (Angle sum property of a triangle)
90° + 40° + DCE = 180°
DCE = (180° – 130°) = 50° ............(i)
DCE = 50°
(ii)
AOC and BOC form a linear pair.
BOC = (180° – 80°) = 100° .............(ii)
In ΔBOC,
OBC + OCB + BOC = 180° (Angle sum property of a triangle)
∠OBC + DCE + BOC = 180°     [Since OCB = DCE]
OBC = 180° – (BOC + DCE)
            = 180° – (100° + 50°)          [from (i) and (ii)]
            = (180° - 150°) = 30°



Page No 640:

Question 1:

Mark (✓) against the correct answer in each of the following:
The equation 2x + 3y = 6 has
(a) a unique solution
(b) only two solutions
(c) no solution
(d) infinitely many solutions

Answer:

(d) infinitely many solutions

If the number of equations are less than the number of variables to be solved, then we get infinitely many solutions.
Here, we have two variables and only one equation. So, we will get infinitely many solutions.

Page No 640:

Question 2:

Mark (✓) against the correct answer in each of the following:
In a ||gm ABCD, DL ​⊥ AB and BM ​⊥ AD. If AB = 10 cm, DL = 6 cm and BM = 8 cm, then AD = ?
(a) 8 cm
(b) 9 cm
(c) 7 cm
(d) 7.5 cm

Answer:

(d) 7.5 cm
Area of a parallelogram  = Base ×Height
Area of parallelogram ABCD = AB ×DL
AD×MB = AB ×DL
AD×8 = 10×6
AD×8 = 60
AD = 7.5 cm

Page No 640:

Question 3:

Mark (✓) against the correct answer in each of the following:
How many bricks will be required to construct a wall 13.5 m long, 6 m high and 22.5 cm thick if each brick measures (27 cm × 12.5 cm × 9 cm)?
(a) 6000
(b) 7500
(c) 5000
(d) 3750

Answer:

(a) 6000

Length of the wall = 13.5 m = 1350 cm                Height of wall = 6 m = 600 cm                Breadth of wall =22.5 cmVolume of wall = Length×Breadth×Height=1350×22.5×600=18225000 cm3Volume of a brick = Length×Breadth×Height=27×12.5×9=3037.5 cm3Number of bricks=Volume of wall Volume of a brick=182250003037.5=6000

Page No 640:

Question 4:

Mark (✓) against the correct answer in each of the following:
The radius of a sphere is 21 cm. What is the surface area of the sphere?
(a) 9702 cm2
(b) 5544 cm2
(c) 12932 cm2
(d) 4312 cm2

Answer:

(b) 5544 cm2

Surface area of sphere=4πr2=4×227×21×21=5544 cm2

Page No 640:

Question 5:

Mark (✓) against the correct answer in each of the following:
In the given figure, ABCD is rhombus in which A = 70°. Then, ∠CDB = ?
(a) 35°
(b) 20°
(c) 55°
(d) 50°

Answer:

(c) 55°

ABCD is a rhombus.So, CDAB.DAB+ADC =180°  (Angles on the same side of a transversal)ADC =180°-DAB= 180°-70°=110°We know that the diagonals of a rhombus bisect its angles.So,CDB= 12 ADC =12×110°=55°



Page No 641:

Question 6:

Mark (✓) against the correct answer in each of the following:
In the given figure, O is the centre of a circle and BAC = 40°. Then, ∠OBC = ?
(a) 40°
(b) 50°
(c) 80°
(d) 20°

Answer:

(b) 50°

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the circumference of the circle.So,BOC=2BACBOC=2×40°=80°Now, BO=OC. So,BOC is an isoceles triangle.In BOC, BOC+OBC+BCO=180°  (Angle-sum property)80°+OBC+OBC=180°  (OBC=BCO)2OBC=180°-80°2OBC=100°OBC=50°

Page No 641:

Question 7:

Mark (✓) against the correct answer in each of the following:
In the given figure, ABC = 70° and ∠ACB = 30°. Then, ∠BDC = ?
(a) 70°
(b) 60°
(c) 80°
(d) 90°

Answer:

(c) 80°

We know that angles in the same segment are equal.So,BAC=BDCIn ABC, ABC+ACB+BAC=180°  (Angle sum property)BAC=180° -ABC-ACBBAC=180°-70°-30°BAC=80°Hence, BAC=BDC=80°.

Page No 641:

Question 8:

Mark (✓) against the correct answer in each of the following:
In a medical examination of students of a class, the following blood groups were recorded.

Blood group A AB B O
No. of students 10 12 13 5
A student is selected at random. The probability that the chosen student's blood group is AB is
(a) 14

(b) 18

(c) 310

(d) 1340

Answer:

(c) 310

Number of students with blood group AB = 12
Total number of students = 10 + 12 + 13 + 5 = 40
P(AB)=Number of favourable outcomesTotal number of outcomes=1240=310

Page No 641:

Question 9:

Find the area formed by the line x + 3y = 12 with the coordinate axes.

Answer:

We can draw the graph of the given line by using the table below .

x 0 3
y 4 3


Area of triangle AOB = 12×Base×Height=12×12×4=24 square units

Page No 641:

Question 10:

How many metres of cloth, 5 m wide, will be required to make a conical tent of base radius 7 m and height 24 m?

Answer:

Radius of conical tent, r  = 7 m
Height of conical tent, h = 24 m
Therefore, slant height, l = r2 + h2                                            =72 + 242                                            =49 +576                                            =625 = 25 m
Curved surface area of conical tent =πrl                                                  =227×7×25                                                  =22×25                                                  =550 m2

Length of cloth = Curved surface area of tentWidth of cloth=550 m25 m=110 m

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Question 11:

Find the curved surface area of a cylinder whose base radius is 14 m and height is 30 cm.

Answer:

Radius of cylinder, r = 14 m =1400 cm
Height of cylinder, h  = 30 cm
Curved surface area of cylinder = 2πrh=2×227×1400×30=264000 cm2

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Question 12:

The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find these angles.

Answer:

Suppose the angles of the quadrilateral are 2x, 4x, 5x and 7x.
So, 2x+4x+5x+7x = 360°18x = 360°x = 20°Hence, the angles of the quadrilateral are2x=2×20°=40°4x=4×20°=80°5x=5×20°=100°7x=7×20°=140°

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Question 13:

In the given figure, O is the centre of a circle and ADC = 130°. Find ∠BAC.

Answer:

We know that ABCD is a cyclic quadrilateral.
Hence, B+D = 180°B= 180°- DB= 180°-130°B=50°

In ABC, ABC+ACB+BAC=180°  (Angle-sum property)BAC=180° -ABC-ACBBAC=180°-50°-90°  (ACB =90°, as angle in a semi-circle is a right angle)BAC=40°



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Question 14:

In the given figure, O is the centre of a circle, such that OA = 5 cm, AB = 8 cm and OD AB, meeting AB at C. Find the length of CD.

Answer:

ODAB.We know that the perpendicular drawn from thecentre of a circle to a chord bisects the chord.Hence, OD will bisect AB.So, AC=12AB=12×8=4 cmOCA is right-angled triangle.So, by Pythagoras' theorem:OC=OA2-AC2=52-42=25-16=9=3 cmand CD = OD-OC =5-3 = 2 cm

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Question 15:

Fill in the blanks.
(i) For the line 4x + 3y = 12, x-intercept = ...... and y-intercept = ......
(ii) If x = 4, y = 3 is a solution of 2x + ky = 14, then k = ......
(iii) If the point P(p, 4) lies on the line 3x + y = 10, then p = ......

Answer:

(i) For the line 4x + 3y = 12, x-intercept = 3 and y-intercept= 4.

We will draw the graph of the number line by using the values of x and y axes given below.

x 0 3
y 4 0


So, the line intersects the x-axis at 3 and y-axis at 4.
(ii) If x = 4, y = 3 is a solution of 2x + ky = 14, then k = 2.

By putting the values of x and y we can find the value of .
2x+ky=142(4)+k(3)=148+3k=143k=14-83k=6k=2
(iii) If the point P(p, 4) lies on the line 3x + y = 10, then p = 2.

Since the point P(p,4) lies on the given line, it will satisfy the given equation of the line.
3x+y=103(p)+4=103p=10-43p=6p=2

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Question 16:

Which of the following statements are true?
I. If two non-parallel sides of a trapezium are equal, then it is cyclic.
II. A cyclic parallelogram is a rectangle.
III. The opposite angles of a cyclic quadrilateral area equal.
(a) I only
(b) II only
(c) I and II
(d) II and III

Answer:

(c) I and II
(i)

Given : ABCD is an isosceles trapezium.
To prove : ABCD is cyclic.
Construction : Draw CE parallel to DA.
Proof :
AEDC and ADEC.
Hence, ABCD is a parallelogram.
So, ADC=AEC.......(i)                    (Opposite angles of a parallelogram are equal)
AD = EC  (Opposite sides of a parallelogram are equal)
and AD = BC (ABCD is an isosceles trapezium)
So, BC = CE
and
CEB=EBC...........(ii)                           (Angles opposite equal sides are equal)
AEC+CEB=180°  (Linear pair)From equations (i) and (ii), we get: ADC+ABC=180°  Similarly, A+C=180°.
Hence, ABCD is cyclic.

(ii)


Given : ABCD is a cyclic parallelogram.
To prove : ABCD is a rectangle.
Proof :
ABCD is a parallelogram.
 So,ABC=CDA.......(i)                    (Opposite angles of a parallelogram are equal)

ABC+CDA=180°  (ABCD is a cylic quadrilateral)ABC+ABC=180°  2ABC=180°ABC=90°
So, ABC = CDA = 90°
Similarly, DAB =BCD = 90°
Since, all the angles of the parallelogram ABCD are 90°, it is a rectangle.
(iii) We know that the opposite angles of a cyclic quadrilateral are supplementary.
They are not necessarily equal.

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Question 17:

Prove that an isosceles trapezium is always cyclic.

Answer:


Given : ABCD is an isosceles trapezium.
To prove : ABCD is cyclic.
Construction : Draw CE parallel to DA.
Proof :
AEDC and ADEC.
Hence, ABCD is a paralleogram.
So, ADC=AEC.......(i)                    (Opposite angles of a parallelogram are equal)
AD = EC  (Opposite sides of a parallelogram are equal)
and AD = BC (ABCD is an isosceles trapezium)
So, BC = CE.
And
CEB=EBC...........(ii)                           (Angles opposite equal sides are equal)
Also,
AEC+CEB=180°  (Linear pair)From equations (i) and (ii) we get: ADC+ABC=180°  Similarly, DAE+BCD=180°
Hence, ABCD is cyclic.

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Question 18:

If the diagonals of a quadrilateral bisect each other, prove that the quadrilateral is a parallelogram.

Answer:



Given: ABCD is a quadrilateral whose diagonals bisect each other.To prove: ABCD is a paralleogram.Proof : In AOD and BOC, AO=CO  (Given) DO=OB (Given)AOD=COB (Vertically-opposite angles)So,AOD COB (By SAS congruency)AD=BC (c.p.c.t)
In AOB and DOC AO=CO  (Given) BO=OD (Given)AOB=COD (vertically opposite angle)So,AOB COD (By SAS congruency)AB=DC (c.p.c.t)
Since, the opposite sides of the quadrilateral are equal, ABCD is a parallelogram.

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Question 19:

Heights of two cones are in the ratio 1 : 3 and the radii of their bases are in the ratio 3 : 1. Show that their volumes are in the ratio 3 : 1.

Answer:

Let the volume of the two cones be V1 and V2 , their radii be 3r and  and their heights be h and 3h.
Volume of cone=13πr2hSo, V1=13π3r2hV2=13πr2×3hRatio of volumes:V1V2=13π3r2h13πr2×3h                                            =9r2×hr2×3h                                            =31

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Question 20:

How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?

Answer:

Radius of bullet = Diameter2=42=2 cm
Side of cube  = 44 cm
Volume of cube= (Side)3=443=85184 cm3Volume of spherical bullet = 43πr3=43×227×23=70421cm3Number of bullets = Volume of cubeVolume of spherical bullet=85184704×21=2541

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Question 21:

The mean of 40 numbers was found to be 38. Later on, it was detected that the number 56 was misread as 36. Find the correct mean of the given numbers.

Answer:

Given mean of 40 numbers = 38
Sum of 40 numbers = 38×40=1520
Correct sum of 40 numbers = 1520-36+56 =1540
So, correct mean of 40 numbers = 154040=38.5

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Question 22:

In the given figure, PQ is the diameter of a circle with centre O. If PQR = 65°, ∠SPR = 40° and ∠PQM = 50°, find (i) ∠QPR (ii) ∠QPM and (iii) ∠PRS.

Answer:

We know that the angle in a sem- circle is 90°.So,QRP=PMQ=90°(i) In QPR, QPR+QRP+PQR=180°  (Angle-sum property)QPR=180° -QRP-PQRQPR=180°-90°-65°QPR=25°Hence,QPR = 25°.

(i) In QPM, QPM+QMP+PQM=180°  (Angle-sum property)QPM=180° -QMP-PQMQPM=180°-90°-50°QPM=40°Hence,QPM = 40°.

(iii) PQRS is a cyclic quadrilateral. So, SPQ+QRS=180°.QPR+SPR+QRP+PRS=180°PRS=180°-QPR-SPR-QRPPRS=180°-25°-40°-90°PRS=25°Hence,PRS = 25°.

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Question 23:

In a quadrilateral ABCD, M is the midpoint of AC. Prove that ar   ABMD = ar(   DMBC).

Answer:

Since, M is the mid-point of AC, AM is the median of ADC.
We know that a median divides a triangle in two triangles of equal area.
Hence,ar(AMD) = ar(CMD) ......(i)
Also, BM is the median of ABC.
Hence,ar(AMB) = ar(CMB).....(ii)
By adding equations (i) and (ii), we get:
ar(AMD)+ar(AMB) = ar(CMD) +ar(CMB)
ar ABMD = ar( DMBC)

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Question 24:

The curved surface area of a cylinder is 4400 cm2 and the circumference of the base is 110 cm. Find the volume of the cylinder.

Answer:

Circumference of base of cylinder = 110 cm2πr=1102×227×r=110r=110×744=17.5 cm

Curved surface area of cylinder = 4400 cm22πrh = 4400h = 44002πr=4400×72×22×17.5=40 cm
Volume of cylinder =π r2h=227×17.5×17.5×40=38500 cm3



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Question 25:

Calculate the mode for the following data:

Marks 10 15 20 25 30 35 40
No. of students 8 12 36 35 28 18 9

Answer:

Marks (xi) No. of students (fi) Cumulative frequency      (xi )(fi)
10 8 8 80
15 12 20 180
20 36 56 720
25 35 91 875
30 28 119 840
35 18 137 630
40 9 146 360
  N = 146   fi xi = 3685

Number of observations is an even number, i.e. N = 146So, median =12size of N2th term+size of N2+1th term =12size of 73rd term+size of 74th term=1225+25=25

Mean  = fixifi= 3685146= 25.23

 Mode = 3(Median) -2(Mean)
or, Mode  = 3(25) - 2(25.23)
= 75-50.46
= 24.54

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Question 26:

Prove that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Answer:

Given : A circle C(O,r) in which arc AB subtends AOB at the centre and ACB at any point C on the remaining part of the circle.
   
To prove: AOB = 2ACB when AB is a minor arc and or a semi-circle.
Reflex AOB = ACB when AB is a major arc.
Construction: Join AB and CO and produce CO to a point D outside the circle.
Proof: There are three cases:
Case-I  AB is a minor arc (Figure -a)
Case-II  AB is a semicircle (Figure -b)
Case-III  AB is a major arc (Figure -c)
Now, in AOC,
OA = OC (Radii of the circle)
1 = 3  (Angles opposite equal sides)
5 = 1 + 3 (Exterior angle property)
5 = 1 + 1
5 = 21...................(i)
Now, in BOC,
OB = OC (Radii of the circle)
2 = 4  (Angles opposite equal sides)
6 = 2 + 4   (Exterior angle property)
6 = 2 + 2
6 = 22...................(ii)
Adding equations (i) and (ii), we get:
In figures (a) and (b),
 5 + 6 = 21 + 22
5 + 6 = 2(1 + 2)
AOB = 2ACB
In figure (c),
5 + 6 = 2(1 + 2)
Reflex AOB = 2ACB

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Question 27:

The mean of 100 items was found to be 64. Later on, it was discovered that two items were misread as 26 and 9 instead of 36 and 90, respectively. Find the correct mean.

Answer:

Given mean of 100 items = 64
Sum of 100 items = 64×100=6400
Correct sum of 100 items  = 6400-26-9+36+90 =6491
So, correct mean of 100 items = 6491100=64.91

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Question 28:

Construct a ABC whose perimeter is 14 cm and the sides are in the ratio 2 : 3 : 4.

Answer:

Steps of construction:
1. Draw a line segment XY = 14 cm.
2. Construct an acute angle on XY and drawn in the downward direction
3. Starting from  X , mark (2+3+4) = 9 equal distances along XZ.
4. Mark points L,M and N, such that XL = 2 units, LM = 3 units and MN = 4 units.
5. Join YN.
6. Through L and M, draw LBYN and MCYN, cutting XY at B and C, respectively.
7. With B as centre and radius XB, draw an arc.
8. With C as centre and radius CY, draw an arc, cutting the previously drawn arcs at point A.
9. Join AB  and AC.
 ABC, the required triangle, is thus drawn.

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Question 29:

Prove that there is one and only one circle passing through three given points.

Answer:

Given: Three non-collinear points P, Q and R.
To prove: There is one and only one circle passing through the points P, Q and R.
Construction: Join PQ and QR.
Draw perpendicular bisectors AB of PQ and CD of QR.
Let the perpendicular bisectors intersect at the point O.
Now join OP, OQ and OR. A circle is obtained passing through the points P, Q and R.

Proof:
We know that each and every point on the perpendicular bisector of a line segment is equidistant from its ends points. Thus,
OP = OQ           (Since O lies on the perpendicular bisector of PQ)
OQ = OR           (Since O lies on the perpendicular bisector of QR)
OP = OQ = OR
Let OP = OQ = OR = r      (Radius of a circle)
Now, draw a circle C (O, r) with O as centre and r as radius.
Then, circle C(O, r) passes through the points P, Q and R.

We have to show that this circle is the only circle passing through the points P, Q and R.
If possible, suppose there is another circle C(O′, t) which passes through the points P, Q and R.
Then, O will lie on the perpendicular bisectors AB and CD.
But O was the intersection point of the perpendicular bisectors AB and CD.
  O must coincide with the point O.  (Since, two lines can not intersect at more than one point)
As, O′P = t , OP = r  and O coincides with O,
t = r
C(O, r) and C(O', t) are congruent.
Hence, there is one and only one circle passing through the given non-collinear points.

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Question 30:

In the adjoining figure, O is the centre of a circle. Chord CD is parallel to diameter AB. If ABC = 25°, calculate ∠CED.

Answer:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.So,AOC=2ABCAOC=2×25°=50°AOC=DCO  (Alternate angles)Now, since OD=OC, DOC is an isoceles triangle.In DOC, DOC+ODC+DCO=180°  (Angle sum property)DOC=180° -ODC-ODC DOC=180°-50°-50° (ODC=DCO)DOC=80°
and
CED=12COD=12×80°=40°

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Question 31:

If O is the centre of the given circle, such that AOC = 110° and side AB is produced to D, find ∠CBD.

Answer:

MajorAOC=360°-110°=250°  (complete angle)We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.So,CBA=12MajorAOCCBA=12×250°CBA=125°Now,CBA+CBD=180°   (Linear pair)CBD=180°-CBA CBA=180°-125°CBA=55°

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Question 32:

In the given figure, O is the centre of a circle and BOD = 150°. Find the values of x and y.

Answer:

MajorBOD=360°-150°=210°  (complete angle)We know that the angle subtended by an arc of a circle at the centreis double the angle subtended by it at any point on the remaining part of the circle.So,BAD=12MajorBODx°=12×210°x°=105°x=105 ABCD is a cylic quadrilateral. So,A+C=180° x°+y°=180° y°=180°-105°y°=75°y=75

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Question 33:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes and the ratio of their curved surfaces.

Answer:

Let there be two cylinder whose volumes are V1 and V2 and curved
surfaces are S1and S2. Let their radii be 2r and 3r and their heights be 5h and 3h.
Volume of a cylinder=πr2hSo, V1=π2r2×5hV2=π3r2×3hRatio of their volumes is V1V2=π2r2×5hπ3r2×3h=20r2h27r2h=2027

Curved surface area of cylinder=2πrhSo, S1=2π2r×5hS2=2π3r×3hRatio of their curved surface areas is S1S2=2π2r×5h2π3r×3h=20rh18rh=109

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Question 34:

How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?

Answer:

Radius of a bullet = Diameter2=42=2 cm
Side of cube  = 44 cm
Volume of cube= (Side)3=443=85184 cm3Volume of spherical bullet = 43πr3=43×227×23=70421cm3Number of bullets = Volume of cubeVolume of spherical bullet=85184704×21=2541



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