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Page No S.A.-2.10:

Question 25:

A and B are friends. A is elder to B by 5 years. B's sister C is half the age of B while A's father D is 8 years older than twice the age of B. If the present age of D is 48 years, they find the present ages of A, B and C.

Answer:

Let the present age of B be x yrs. Then,The present age of A=x+5 yrs,The present age of C=x2 yrs andThe present age of D=2x+8 yrsBut the present age of D=48 yrs2x+8=482x=48-82x=40x=402x=20So,The present age of A=20+5=25 years,The present age of B=20 years andThe present of C=202=10 years

Page No S.A.-2.10:

Question 31:

Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Answer:


Given: A rhombus ABCD and a circle with diameter AB.To prove: Point O lies on the circleProof:As, angle formed in the semi-circle is a right angle.AOB=90°Also, diagonals intersect each other at a right angle.AOB=BOC=COD=DOA=90°So, it can be conlcuded that point O lies on the circle.

Hence, the circle drawn with any side of rhombus as diameter passes through the point of intersection of its diagonals.



Page No S.A.-2.2:

Question 1:

The two consecutive  class marks of a distribution are 52 and 57. Find the class limits of the two intervals.

Answer:

As, the two consecutive class marks of the distribution are 52 and 57.

So, the class size or class width of the distribution = 57 - 52 = 5

Now, The lower class limit of the first interval=52-52=52-2.5=49.5;the upper class limit of the first interval=52+52=52+2.5=54.5;the lower class limit of the second interval=57-52=57-2.5=54.5 andthe upper class limit of the second interval=57+52=57+2.5=59.5

So, the class limits of the two intervals are: 49.5-54.5 and 54.5-59.5

Page No S.A.-2.2:

Question 2:

Three cubes each of volume 125 cm3 are joined end-to-end to form a cuboid. Find the total surface area of cuboid.

Answer:

Let the edge of each cube be a.

As,Volume of each cube=125 cm3a3=125a=1253a=5 cmSo, the length of the cuboid, l=3a=3×5=15 cm,the breadth of the cuboid, b=a=5 cm andthe height of the cuboid, h=a=5 cmNow,Total surface area of the cuboid=2lb+bh+hl=215×5+5×5+5×15=275+25+75=2×175=350 cm2

So, the total surface area of the cuboid so formed is 350 cm2.

Page No S.A.-2.2:

Question 3:

For the question given below,four alternative choices have been provided of which only one is correct. You have to select the correct choice:

A triangle and a rhombus are on the same base and between the same parallels. Then, the ratio of area of triangle to that of rhombus is:

(a) 1 : 1

(b) 1 : 2

(c) 1 : 3

(d) 1 : 4

Answer:

We have to find the ratio of area of triangle to that of rhombus, when they lie on the same base.

Let the triangle be ABC and the rhombus be ABDC.

Therefore, by using the properties of a rhombus that the diagonal divides the rhombus in two equal areas, we get

arABCarABDC=arABCarABC+arBDC=arABC2arABC=12

So, the ratio of the area of triangle to that of rhombus is 1 : 2.

Hence, the correct choice is option (b).

Page No S.A.-2.2:

Question 4:

For the question given below,four alternative choices have been provided of which only one is correct. You have to select the correct choice:

In Fig. 1, O is the centre of the circle and ∠OBA = 60°. Then ∠ACB equals:

(a) 60°

(b) 45°

(c) 30°

(d) 90°

 

Answer:

We have, ∠OBA = 60°

In AOB,

As, OA = OB                     (Radii)

So, ∠OAB = ∠OBA           (Angles opposite to equal sides are equal)

Therefore, ∠OAB = 60°

Also, using angle sum property of triangle

AOB=180°-OBA+OAB=180°-60°+60°=180°-120°=60°

Now,

AOB=2ACB    Angle subtended at the center by a chord is twice the angle subtended by the same chord at the remaining part of the circleACB=AOB2ACB=60°2 ACB=30°

Hence, the correct choice is option (c).



Page No S.A.-2.3:

Question 5:

For the question given below, four alternative choices have been provided of which only one is correct. You have to select the correct choice:

The diameter and height of a right circular cone are 7 cm and 12 cm, respectively. The volume of the cone (in cm3) is:

(a) 88

(b) 112

(c) 154

(d) 616

Answer:

We have,Radius of the cone, r=72=3.5 cm andHeight of the cone, h=12 cmNow,Volume of the cone=13πr2h=13×227×3.5×3.5×12=2221×147=22×7=154 cm3

Hence, the correct choice is option (c).

Page No S.A.-2.3:

Question 6:

For the question given below, four alternative choices have been provided of which only one is correct. You have to select the correct choice:

A fair coin tossed 100 times and the Head occurs 58 times and tail 42 times. The experimental probability of getting a Head is:

(a) 12

(b) 2150

(c) 2950

(d) 4258

Answer:

We have,

The number of times a coin is tossed = 100

The number of times Head occurs = 58 and

The number of times Tail occurs = 42

Let the event of getting a Head be E.

So,PE=Number of favourable outcomesTotal number of possible outcomes=58100=2950

Hence, the correct choice is option (c).

Page No S.A.-2.3:

Question 7:

If the radius and height of a cone both are increased by 10%, then the volume of the cone is increased by:

(a) 10%

(b) 21%

(c) 33.1%

(d) 100%

Answer:

Let the radius and height of the cone be r and h, respectively and the radius and height of the cone after the increase be R and H, respectively.Volume of the cone, v=13πr2hAlso,R=r+10% of r=1.1r andH=h+10% of h=1.1hSo, the volume of the new cone, V=13πR2H=13π1.1r21.1h=1.33113πr2h=1.331vNow, the increase in volume=V-v=1.331v-v=0.331v The percentage increase in the volume of the cone=0.331vv×100=33.1%

Hence, the correct answer is option (c).

Page No S.A.-2.3:

Question 8:

Eleven bags of wheat flour, each marked 10 kg actually contained the following weights (in kg) of flour:

10.05, 10.20, 10.00, 9.75, 10.00, 10.03, 9.95, 10.35, 9.90, 10.00, 10.08

Find the probability that any of these bags chosen at random contains:

(i) more than 10 kg of wheat flour.

(ii) less than 9.5 kg of wheat flour.

Answer:

We have,

The total number of bags = 11


(i) As, the number of bags containing more than 10 kg of wheat flour = 5

So, the probability of choosing a bag containing more than 10 kg of wheat flour = 511

(ii) As, the number of bags containing less than 9.5 kg of wheat flour = 0

So, the probability of choosing a bag containing less than 9.5 kg of wheat flour = 011 = 0

Page No S.A.-2.3:

Question 9:

In the following figure, a circle with centre O is drawn and ∠BAC = 50°. Find x.

Answer:



As, BOC=2BAC         Angle subtended by an arc at the centre is double the angle subtended by the the same arc at any point on the circleBOC=2×50°BOC=100°Now, in BOC, OB=OC    Radii OBC=OCB     Angles opposite to equal sides are equalOBC=OCB=xAlso,BOC+OBC+OCB=180°      Angle sum property of a triangle100°+x+x=180°2x=180°-100°2x=80°x=80°2 x=40°

Page No S.A.-2.3:

Question 10:

The following observations have been arranged in ascending order. If the median of the data is 63, then find the value of x.

29, 32, 48, x − 2, x, x + 2, 72, 78, 84, 95.

Answer:

We have,The observations in ascending order:29, 32, 48, x-2, x, x+2, 72, 78, 84, 95As, the number of observations=10 evenSo, the median=102th observation+102+1th observation2=5th observation+6th observation2=x+x+22=2x+22=x+1but median=63x+1=63x=63-1 x=62

Page No S.A.-2.3:

Question 11:

Express y in terms of x in the equation 2x − 3y = 12. Find the points where the line represented by the equations 2x − 3y = 12 cuts the x-axis and y-axis.

Answer:

We have,2x-3y=123y=2x-12y=2x-123Now,If x=0, then20-3y=12-3y=12y=12-3y=-4So, the line 2x-3y=12 cuts the y-axis at 0,-4.Also,If y=0, then2x-30=122x=12x=122x=6So, the line 2x-3y=12 cuts the x-axis at 6,0.

Page No S.A.-2.3:

Question 12:

D is the mid-point of side BC of ΔABC and E is the mid-point BD. If O is the mid point of AE, then prove that ar(ΔBOE) = 18 ar(ΔABC).

Answer:


Given: In ABC, D is the mid-point of BC, E is the mid-point of BD and O is the mid-point of AE.To prove: arBOE=18arABCProof:In ABC, AD is median       As, D is the mid-point of BC arABD=12arABC    .....i          Median divides the triangle into two parts equal in areaSimilarly, in ABD, AE is median       As, E is the mid-point of BD arABE=12arABD    .....ii          Median divides the triangle into two parts equal in areaAlso, in ABE, BO is median       As, O is the mid-point of AE arBOE=12arABE                     Median divides the triangle into two parts equal in areaarBOE=1212arABD          Using iiarBOE=14arABDarBOE=1412arABC          Using i arBOE=18arABC

Page No S.A.-2.3:

Question 13:

Construct a ΔABC with perimeter 11 cm and base angles 90° and 60°.

Answer:

Given: In ΔABC, AB + BC + CA = 11 cm, B = 60° and C = 90°.

To construct: ΔABC

Steps of construction:

1. Draw a line segment XY = 11 cm.
2. At point X, make PXY = 30°.
3. At point Y, make QYX = 45° and let PX and QY meet at point A.
4. Draw perpendicular bisector ST of AX and let ST meets XY at point B.
5. Draw perpendicular bisector UV of AY and let UV meets XY at point C.
6. Join AB and AC.
Thus, ABC is the required triangle.



Page No S.A.-2.4:

Question 11:

A three-wheeler scooter charges Rs. 10 for the first kilometer and Rs. 4.50 each for every subsequent kilometer. For a distance of x km, an amount of Rs. y is paid. Write the linear equation representing the above information.

Answer:

As,

For every first kilometer the scooter charges Rs 10.

For every subsequent kilometer the scooter charges Rs 4.50 each.

To find a distance of x km, an amount of Rs y is paid.

Therefore,

x km can be written as

x = 1 + (x - 1)

The price paid for x km is y can be written as

y=10+4.50x-1y=10+92x-12y=20+9x-12y=20+9x-99x-2y+11=0

So, the linear equation representing the given information is 9x - 2y + 11 = 0.

Page No S.A.-2.4:

Question 12:

ABCD is parallelogram. The angle bisectors of ∠A and ∠D intersect at O. Find the measures of ∠AOD.

Answer:


In parallelogram ABCD,A+D=180°                Adjacent angles of a parallelogram are supplementary12A+12D=180°2OAD+ODA=90°   .....i       As, OA and OD are angle bisectors of A and D, respectivelyNow, in AOD,AOD+OAD+ODA=180°          Angle sum property of triangleAOD+90°=180°       Using iAOD=180°-90° AOD=90°

Page No S.A.-2.4:

Question 13:

In Fig. 2, ABCD is a quadrilateral in  which P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that PQRS is a parallelogram.

Answer:

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

To prove: PQRS is a parallelogram.

Construction: Join AC.


Proof:

In ACD,

As, R and S are mid-points of DC and DA, respectively.

So, by using the Mid-point theorem − The segment connecting the mid-points of two sides of a triangle is parallel to the third side and is half the length of the third side, we get

RSAC and RS = 12AC           .....(1)

Similarly, in ABC, by using the mid-point theorem, we get

PQAC and PQ = 12AC           .....(2)

Therefore, from (1) and (2), we get

PQRS and PQ = RS

But this is a pair of opposite sides of the quadrilateral PQRS.

Therefore, PQRS is a parallelogram.

Page No S.A.-2.4:

Question 14:

What length of canvas 3 m wide will be required to make a conical tent of height 8 m and radius of base 6 m? (use π = 3.14)

Answer:

We have,Height of the conical tent, h=8 m,Radius of the conical tent, r=6 m andWidht of the canvas, B=3 mLet the length of the canvas required be L.Also, the slant height of the conical tent, l=h2+r2=82+62=64+36=100=10 cmNow,The curved surface area of the conical tent=πrl=3.14×6×10=188.4 cm2Area of the canvas required=188.4 cm2LB=188.4L×3=188.4L=188.43 L=62.8 cm

So, the length of the canvas required to make the conical tent is 62.8 cm.

Page No S.A.-2.4:

Question 15:

Heights in cm of 12 students are given as:

154, 158, 155, 160, 165, 145, 148, 150, 147, 151, 166, 152

Make a frequency table taking 145-150 as first class interval.

Answer:

The frequency distribution table of the given data is as follows:
 

Class Interval Tally Marks Frequency
145-150 ||| 3
150-155 |||| 4
155-160 || 2
160-165 | 1
165-170 || 2
Total   12
 

Page No S.A.-2.4:

Question 16:

Write 7y = 2x as a linear equation of the form ax + by + c = 0. Also, write the values corresponding to a, b, c. Does the graph of this linear equation pass through origin? Give your answer in 'yes' or 'no'.

Answer:

We have,7y=2x2x-7y+0=0So, a=2, b=-7 and c=0Also,If x=0, then7y=207y=0y=07y=0

Yes, the graph of the given linear equation passes through origin.

Page No S.A.-2.4:

Question 17:

A sphere and a cube have the same surface area. Show that the ratio of the volume of the sphere to that of the cube is 6: π.

Answer:

Let the radius of the the sphere be r and the edge of the cube be a.As,Surface area of the sphere=Surface area of the cube4πr2=6a2r2a2=64πra2=32πra=32π           .....iNow,The ratio of the volume of the sphere to that of the cube=Volume of the sphereVolume of the cube=43πr3a3=43πra3=43π32π3             Using i=43π332π2π=232π=23×22π=6π=6:π

So, the ratio of the ratio of the volume of the sphere to that of the cube is 6: π.

Page No S.A.-2.4:

Question 18:

In the following figure, ABCD is a parallelogram. If E is the mid-point of BC and AE is the bisector of ∠A, then prove that AB = 12 AD.

Answer:



Given: In a parallelogram ABCD, E is the mid-point of BC and AE is the angle bisector of A.To prove: AB=!2ADProof:As, AE is the angle bisector of A DAE=BAEbut DAE=AEB            Alternate interior anglesSo, BAE=AEBAB=BE        .....i      Sides opposite to equal angles are equalAlso, E is the mid-point of BCBE=12BCAB=12BC          Using ibut AD=BC     Opposite sides of a parallelogram ABCD AB=12AD

Page No S.A.-2.4:

Question 19:

E and F are mid-points of sides AB and CD, respectively of a parallelogram ABCD. AF and CE intersect diagonal BD in P and Q, respectively. Prove that diagonal BD is trisected at P and Q.

Answer:



Given: In a parallelogram ABCD, E and F are mid-points of sides AB and CD, respectively.To prove: Diagonal BD is trisected at P and Q.Proof:As, E and F are mid-points of sides AB and CD, respectively.AE=12AB and CF=12CDBut AB=CD and ABCD     Opposite sides of parallelogram ABCDAE=CF and AECF AECF is a parallelogram.AFEC              .....iNow, in ABP,As, E is the mid-point of AB and QEAP From iSo, BQ=PQ        .....ii     Using converse of  mid-point theoremSimilarly, in DQC,As, F is the mid-point of DC and PFQC From iSo, DP=PQ        .....iii     Using converse of  mid-point theoremFrom ii and iii, we getDP=PQ=BQ Diagonal BD is trisected at P and Q.

Page No S.A.-2.4:

Question 20:

Consider the following frequency distribution which gives the weight of 38 students of a class:
 

Weights (in kg) 31-35 36-40 41-45 46-50 51-55 56-60 61-65 66-70 Total
No. students 9 5 14 3 1 2 2 2 38

(i) Find the probability that the weight of a student in the class lies between 36-45 kg.

(ii) Give 2 events in the context, one having probability 0 and the other having probability 1.

Answer:

We have,

Total number of students = 38

(i) As, the number of students in the class whose weight lies between 36-45 kg = 5 + 14 = 19

So, the probability that the weight of a student in the class lies between 36-45 kg = 1938=12

(ii) As, the number of students in the class whose weight is less than 31 kg = 0

So, the probabilty that the weight of a student in the class is less than 31 kg = 038=0


Also, the number of students in the class whose weight lies between 31-70 kg = 38

So, the probabilty that the weight of a student in the class lies between 31-70 kg = 3838=1

Hence, the two events, one having probabilty 0 is " A student whose weight is less than 31 kg" and the other having probability 1 is " A student whose weight lies between 31-70 kg".



Page No S.A.-2.5:

Question 16:

The curved surface area of a cylinder is 176 cm2 and its base area is 38.5 cm2. Find the volume of the cylinder and justify your answer.
[Use π = 227]

Answer:

We have,Curved surface area of a cylinder=176 cm2,Base area of the cylinder=38.5 cm2Let the base radius be r and the height of the cylinder be h.As, the area of the base=38.5 cm2πr2=38.5227×r2=38.510227×r2=772r2=772×722r2=494r=494r=72 cmAlso,The curved surface area of the cylinder=176 cm22πrh=1762×227×72×h=17622h=176h=17622h=8 cmNow,Volume of the cylinder=πr2h=227×72×72×8=308 cm3

So, the volume of the cylinder is 308 cm3.

Page No S.A.-2.5:

Question 17:

A hemispherical bowl is made of steel 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. [Use π = 227]

Answer:

We have,Inner radius of the hemispherical bowl, r=5 cm andThickness of the bowl=0.25 cmAlso, the outer radius of the bowl, R=5+0.25=5.25 cmNow,The outer curved surface area of the bowl=2πR2=2×227×5.25×5.25=173.25 cm2

So, the outer curved surface area of the hemispherical bowl is 173.25 cm2.

Page No S.A.-2.5:

Question 18:

Find the mean of the first ten prime numbers.

Answer:

As, the first ten prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.So, the mean=Sum of observationsTotal number of observations=2+3+5+7+11+13+17+19+23+2910=12910=12.9

Hence, the mean of the first ten prime numbers is 12.9.

Page No S.A.-2.5:

Question 19:

Draw the graph of two lines, whose equations are 3x − 2y + 6 = 0 and x + 2y − 6 = 0 on the same graph paper. Find the area of triangle formed by the two lines and x-axis.

Answer:

The points satisfying the equation 3x − 2y + 6 = 0 are: 

And, the points satisfying the equation x + 2y − 6 = 0 are:

The graph of the given two lines is as follows:

The vertices of the triangle formed by the given lines and x-axis are B(6, 0), A(−2, 0) and C(0, 3).

Now,

arABC=12×AB×OC=12×8×3=12 square units

So, the area of the triangle so formed is 12 square units.

Page No S.A.-2.5:

Question 20:

If the number of hours for which a labourer works is x and y are his wages (in rupees) and y = 2x − 1, then draw the graph of work-wages equation. From the graph, find the wages of the labourer if he works for 6 hours.

Answer:

The values of x and y satisfying the given equation y = 2x − 1 are:

x

0

1

2

3

4

y

–1

1

3

5

7

The graph of the given equation y = 2x − 1 is as follows:

In the graph, a point A(6, 11) lies on the line.

So, at x = 6, y = 11

Hence, the wages of the labourer if he works for 6 hours is Rs 11.

Page No S.A.-2.5:

Question 22:

Construct a ΔABC in which BC = 4.5 cm, ∠C = 45° and sum of sides AB and AC is 8 cm. Justify your construction.

Answer:

Given: In ΔABC, BC = 4.5 cm, ∠C = 45° and AB + AC = 8 cm

To construct: ΔABC

Steps of construction:

1. Draw a line segment BC = 4.5 cm.
2. At point C, make ∠XCB = 45°.
3. With C as centre, draw an arc of radius 8 cm to cut the ray CX at point D.
4. Join BD.
5. At point B, make ∠YBD = ∠BDC and let the ray BY intersect CD at point A.
Thus, ABC is the required triangle.


Justification:

In ΔABD,
As, ∠ABD = ∠BDC    (By construction)
So, AB = AD    (Sides opposite to equal angles are equal)

Now,
CD = 8 cm    (By construction)
AC + AD = 8 cm
So, AC + AB = 8 cm    (Proved above)

Page No S.A.-2.5:

Question 21:

ABCD is a parallelogram. AP the bisector of ∠A and CQ the bisector of ∠C meet the opposite sides in P and Q, respectively. Prove that
AP || CQ.

Answer:



Given: In a parallelogram ABCD, AP is bisector of A and CQ is the bisector of C.To prove: APCQProof:As, PAQ=12BAD          Since, AP is bisector of AAnd, PCQ=12BCD        Since, CQ is bisector of CBut BAD=BCD                 Opposite angles of parallelogramABCDor 12BAD=12BCDSo, PAQ=PCQBut PCQ=BQC                 Alternate interior anglesPAQ=BQCBut these are corresponding angles for the pair of lines AP and CQ.Hence, APCQ

Page No S.A.-2.5:

Question 23:

Write a linear equation in two variables to represent the statement ''Four times the cost of a table is equal to seven times the cost of a chair''. From the graph of above linear equation, find the cost of one chair if the cost of the table is ₹1050.

Answer:

Let the cost of a table and a chair be x and y, respectively.

A.T.Q, we get4x=7y 4x-7y=0, is the required linear equation.

The graph of the linear equation is as follows:


Since, at x = 1050, y =  600

So, the cost of one chair is â‚¹600.

Page No S.A.-2.5:

Question 24:

Rain water which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius
20 cm. What will be the height of water in the cylindrical vessel if the rainfall is 1 cm. Give your answer to the nearest whole number.
(Use π = 3.14)

Answer:

We have,Length of the rectangular surface, l=6 m=600 cm,Breadth of the rectangular surface, b=4 m=400 cm,Height of the rainfall, h=1 cm andRadius of the cylindrical vessel, R=20 cmLet the height of water in the cylindrical vessel be H.Now,Volume of the water in the cylindrical vessel=Volume of the rainfall in the rectangular surfaceπR2H=lbh3.14×202×H=600×400×11256H=240000H=2400001256H=191.08 cm H191 cm

So, the height of water in the cylindrical vessel is 191 cm.

Page No S.A.-2.5:

Question 25:

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

Answer:

Theorem: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Given: Arc AB of a circle with centre O, subtending AOB at the centre and APB at a point P on the remaining part of the cirlce.

To prove: AOB = 2APB

In the given condition, clearly we have three cases:
Case I: Arc AB is a minor arc
Case II: Arc AB is a semicircle
Case III: Arc AB is a major arc

Construction: Join PO and produce it to a point Q.


Proof:

In AOP, AOQ is an exterior angle
As, an exterior angle is equal to the sum of the interior opposite angles.
So, AOQ = APO + PAO         .....(i)

Also, AO = PO           (Radii)
So, APO = PAO   (Angles opposite to equal sides are equal)       .....(ii)

From (i) and (ii), we get
AOQ = APO + APO
AOQ = 2APO     .....(iii)

Similalry,
In BOP, BOQ is an exterior angle
As, an exterior angle is equal to the sum of the interior opposite angles.
So, BOQ = BPO + PBO         .....(iv)

Also, BO = PO           (Radii)
So, BPO = PBO   (Angles opposite to equal sides are equal)       .....(v)

From (iv) and (v), we get
BOQ = BPO + BPO
BOQ = 2BPO     .....(vi)

Adding (iii) and (vi), we get
AOQ + BOQ = 2APO + 2BPO
AOB = 2(APO + BPO)
So, AOB = 2APB

Note: In Case III, where AB is a major arc, AOB is replaced by reflex AOB.
So, reflex AOB = 2APB

Page No S.A.-2.5:

Question 26:

Prove that the opposite angle of an isosceles trapezium are supplementary.

Answer:



Given: In trapezium ABCD, AB = CD and AD || BC.

Construction: Draw AEBC and DFBC.

To prove: A+C=180° and B+D=180°

Proof:

In ABE and DCF,AB=CD    GivenAE=DF     Perpendicular distance between the parallel linesAEB=DFC=90° By RHS congruence criterionABEDCFABE=DCF    By CPCTABC=DCB    .....iNow, ADBC and AB is a transversalABC+BAD=180°      Consecutive interior anglesDCB+BAD=180°       using i C+A=180°Also, ADBC and CD is a transversalADC+DCB=180°      Consecutive interior anglesADC+ABC=180°       using i D+B=180°

Hence, the opposite angle of an isosceles trapezium are supplementary.

Page No S.A.-2.5:

Question 27:

Sketch the graph of the equation 3x + 5y = 15. Find the area of the figure formed by this line and the two axes.

Answer:

The graph of the equation 3x + 5y = 15 is as follows:


The figure formed by the line and the two axes is a right angled AOB.

In AOB, AOB=90°, AO=3 units and BO=5 unitsNow,arAOB=12×AO×BO=12×3×5=152=7.5 square units

Page No S.A.-2.5:

Question 28:

In the following figures, ABCD is a parallelogram. A circle through A, B, C intersects CD or CD produced at E. Prove that AE = AD.

Answer:




Given: A parallelogram ABCD and a circle through A, B and C.To prove: AE=ADConstruction: Join AEProof:From figure i;As, ABC=ADC                  Opposite angles of parallelogram ABCDBut ABC+AEC=180°      Opposite angles of cyclic quadrilateral AECBSo, ADC+AEC=180°ADC+AED=180°       .....1Also, ADC+ADE=180°     Linear pair    .....2From 1 and 2, we getADC+AED=ADC+ADEAED=ADE AD=AE      Sides opposite to equal angles are equalSimilalry, from figure ii;As, ABC=ADC                  Opposite angles of parallelogram ABCDBut ABC+AEC=180°      Opposite angles of cyclic quadrilateral AECBSo, ADC+AEC=180°ADE+AEC=180°       .....1Also, AEC+AED=180°     Linear pair    .....2From 1 and 2, we getADE+AEC=AEC+AEDADE=AED AD=AE      Sides opposite to equal angles are equal



Page No S.A.-2.6:

Question 23:

Construct a Δ ABC, in which base BC = 3 cm, ∠B = 30° and AB + AC = 5.2 cm.

Answer:

Given: In ΔABC, BC = 3 cm, ∠B = 30° and AB + AC = 5.2 cm

To construct: ΔABC

Steps of construction:

(1) Draw a line segment BC = 3 cm

(2) At point B, make an angle ∠PBC = 30°.

(3) With B as centre, draw an arc of radius 5.2 cm which cuts the ray PB at point D.

(4) Join DC.

(5) Draw perpendicular bisector of DC which cuts DB at point A.

(6) Join AC.

Thus, ΔABC is the required triangle.

Page No S.A.-2.6:

Question 24:

A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, then how much soup the hospital has to prepare daily to serve 250 patients?

Answer:

We have,Base radius of the cylindrical bowl, r=72 cm andHeight of the cylindrical bowl, h=4 cmNow,Volume of the cylindrical bowl=πr2h=227×72×72×4=154 cm3Also,Volume of the soup in 250 bowls=250×154=38500 cm3=38.5 L     As, 1000 cm3=1 L

So, the volume of the soup the hospital has to prepare daily to serve 250 patients is 38.5 litres.

 

Page No S.A.-2.6:

Question 25:

A heap of wheat is in the form of a cone, whose diameter is 10.5 m and height 7 m. Find the volume of wheat in the heap. The heap is to be covered by canvas to protect it from rain, find the area of the canvas required.

Answer:

We have,Base radius of the conical heap, r=10.52=214 m andHeight of the conical heap, h=7 mNow,Volume of wheat in the heap=13πr2h=13×227×214×214×7=16178 m3=20218 m3Also,the slant height of the heap, l=h2+r2=2142+72=44116+49=122516=354 mSo, the area of the canvas required to protect the heap of wheat from rain=CSA of the heap=πrl=227×214×354=11558 m2=14438 m2

Page No S.A.-2.6:

Question 26:

Find the mean of the following data by shortcut method.
 

Marks 20 22 25 30 35 39 45 50 Total
Frequency 4 6 8 10 8 7 5 2 50

Answer:

We have,



Let the marks be represented as xi and frequency be represented as fi.

Let a = 30

 

So, x=a+fidifi=30+9050=30+1.8=31.8

Hence, the mean of the given data is 31.8.

 

Page No S.A.-2.6:

Question 27:

On a page of a telephone directory, there are 200 telephone numbers. The frequency distribution of the digits at their units place is given below:
 

Unit digit 0 1 2 3 4 5 6 7 8 9
Frequency 22 26 22 22 20 10 14 28 16 20

Without looking at the page, a number is chosen at random from the page. What is the probability that the digit at the unit's place of the number chosen is greater than 6?

Answer:

We have, the frequency distribution of the digits at their units place:

Unit digit 0 1 2 3 4 5 6 7 8 9
Frequency 22 26 22 22 20 10 14 28 16 20

 


As, the frequency of getting unit digit geater than (i.e. 7, 8 and 9) = 28 + 16 + 20 = 64


So, the required probability=Number of favourable outcomesTotal number of possible outcomes=64200=825

Hence, the probability that the digit at the unit's place of the number chosen is greater than 6 is 825.

Page No S.A.-2.6:

Question 28:

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:
 

Outcome 3 heads 2 heads 1 head No Head
Frequency 23 72 77 28

Find the experimental probability of getting

(i) 2 Heads

(ii) at least 2 Heads

Answer:

We have,

Outcome 3 heads 2 heads 1 head No Head
Frequency 23 72 77 28

As, the probability of an even E is given byPE=Number of favourable outcomesTotal number of possible outcomesSo,i P2 heads=72200=925ii Pat least 2 heads=P2 heads+P3 heads=72200+23200=95200=1940

Page No S.A.-2.6:

Question 29:

In Fig. 7, a small indoor green house is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high. What is the area of the glass? How much tape is required for all the 12 edges?

Answer:



We have,Length of the indoor green house, l=30 cm,Breadth of the incdoor green house, b=25 cm andHeight of the indoor green house, h=25 cmNow,Area of the glass=Total surface area of the cuboid=2lb+bh+hl=230×25+25×25+25×30=2750+625+750=2×2125=4250 cm2Also,The length of the tape required=4l+b+h=430+25+25=4×80=320 cm

Page No S.A.-2.6:

Question 30:

The daily earnings of 30 workers are given below:
 

Daily earning (in ₹) No. of workers
1-50 3
50-100 7
100-150 4
150-200 5
200-250 4
250-300 3
300-350 2
350-400 2

Draw a histogram and a frequency polygon to represent the above data.

Answer:

The frequency distribution table of the given data is as follows:
 

Daily earning (in ₹) No. of workers
1-50 3
50-100 7
100-150 4
150-200 5
200-250 4
250-300 3
300-350 2
350-400 2

The histogram and frequency polygon of the given data is as follows:

Page No S.A.-2.6:

Question 31:

The circumference of the base of a cone is 2207 cm and its slant height is 13 cm. Find the volume of the cone. π=227

Answer:

We have,Slant height of the cone, l=13 cmLet the radius and height of the cone be r and h, respectively.As, the circumference of the base of the cone=2207 cm2πr=22072×227×r=220744r7=2207r=220×77×44r=5 cmAlso,h=l2-r2=132-52=169-25=144=12 cmNow,Volume of the cone=13πr2h=13×227×5×5×12=22007 cm3



Page No S.A.-2.7:

Question 30:

Prove that parallelogram on the same base and between the same parallels are equal in area.

Answer:



Given: The parallelograms ABCD and ABEF be on the same base AB and between same parallels AB and FC.

​To prove: ar(ABCD) = ar(ABEF)

Proof:

As, CD=AB       Opposite sides of parallelogram ABCDAnd, FE=AB       Opposite sides of parallelogram ABEFSo, FE=CDFE+ED=CD+EDFD=CE         .....iNow,In ADF and BCE,AD=BC      Opposite sides of parallelogram ABCDAF=BE       Opposite sides of parallelogram ABEFDF=CE       From iSo, by SSS congruencyADFBCEarADF=arBCE         Congruent triangles are equal in areaarABCF-arADF=arABCF-arBCE arABCD=arABEF

Page No S.A.-2.7:

Question 31:

In Fig. 8, ABCD is a trapezium in which AB || DC. BD is a diagonal and E is the mid-point of AD. A line is drawn through E, parallel to AB, intersecting BC at F. Show that F is the mid-point of BC.

 

Answer:

Given: In trapezium ABCD, AB || DC, EF || AB and E is the mid-point of AD.

To prove: F is the mid-point of BC.

Proof:

In ΔABD,

As, EP || AB and E is the mid-point of AD      (Given: EF || AB)

Therefore, by converse of mid-point theorem − The line drawn through the mid-point of one side of a triangle and parallel to another side bisects the third side, we get

P is the mid-point of BD

Now, similarly in ΔBCD,

As, PF || DC and P is the mid-point of BD       (Since, EF || AB || CD)

Therefore, F is the mid-point of BC.

Page No S.A.-2.7:

Question 32:

In Fig. 9, O is the centre of the circle. The distance between P and Q is 4 cm. Find the ∠ROQ.

 

Answer:


Given: A circle with centre O and radius OQ = 2 cm; and PQ = 4 cm.

Construction: Join OP.

As, OQ = 2 cm and PQ = 4 cm

So, PQ is the diameter of the circle.

In OPR,

Since, OP = OR          (Radii)

So, ∠OPR = ∠ORP    (Angles opposite to equal sides are equal)

∠OPR = 35°

∠QPR = 35°

Now,

As, the subtended by an arc at the centre is double the angle subtended by the same arc at the remaining part of the circle.

So, ∠ROQ = 2∠QPR = 2 × 35° = 70°



Page No S.A.-2.8:

Question 33:

In Fig. 10, a right circular cone of diameter r cm and height 12 cm rests on the base of a right circular cylinder of radius r cm. Their bases are in the same plane and the cylinder is filled with water upto a height of 12 cm. If the cone is then removed, find the height to which water level will fall.

Answer:

We have,Radius of the cone, R=r2,Height of the cone, H=12 cm,Radius of the cylinder=r andHeight of the water level before the cone is taken out, H=12 cmLet the height of water level after the cone is taken out be h.Now,Volume of the water in the cylinder after the cone is taken out=Volume of the cylinder upto a height of 12 cm-Volume of the coneπr2h=πr2H-13πR2Hπr2h=πr2×12-13πr22×12πr2h=12πr2-13π×r24×12πr2h=12πr2-πr2πr2h=11πr2h=11πr2πr2h=11 cm The fall in water level=12-11=1 cm

So, the height to which the water level will fall is 1 cm.

Page No S.A.-2.8:

Question 34:

Draw a histogram for the following data:
 

Marks 10-15 15-20 20-25 25-30 30-40 40-60 60-80
Number of candidates 7 9 8 5 12 12 8

Answer:

We have,

As, in the above given table the classes are of unequal widths, so let us form the table with adjusted frequencies:

Now, the histogram for the given data is as follows:

Page No S.A.-2.8:

Question 10:

In Fig. 5, O is the centre of the circle. The angle by the arc BCD at the centre is 140°, BC is produced to P. Find ∠DCP.

Answer:


We have,

BOD=140°

Using the property of circles, the angle subtended by a chord at the center is twice the angle subtended by it on the remaining part of the circle, we get

BAD=12BOD=12×140°=70°

Also, in cyclic quadrilateral ABCD,

As, the opposite angles are supplementary.

So,

DCB=180°-BAD=180°-70°=110°

Now,

DCP+DCB=180°      Linear pairDCP=180°-DCBDCP=180°-110° DCP=70°

Page No S.A.-2.8:

Question 16:

In Fig. 6, ABCD is a square. If ∠PQR = 90° and PB = QC = DR, prove that ∠QPR = 45°.

Answer:



Since, ABCD is a square, then each of its sides must be equal.So, AB=BC=CD=DANow, BC=CD      Subtracting QC from both sides of the above equation, we getBC-QC=CD-QCBC-QC=CD-DR    as, PB=QC=DRBQ=RC       .....iIn PBQ and QCR,PB=QC       GivenPBQ=QCR=90°BQ=RC       From iSo, by SAS congruencyPBQQCRQP=QR   By CPCT         .....iiIn PQR,As, QP=QR        From iQRP=QPR  Angles opposite to equal sides in a  are equalAlso, PQR+QRP+QPR=180°   Angle sum property of 90°+QPR+QPR=180°   As, QRP=QPR90°+2QPR=180°2QPR=90° QPR=45°



Page No S.A.-2.9:

Question 20:

The ratio of the curved surface area to the total surface area of a right circular cylinder is 1 : 3. Find the volume of the cylinder if its total surface area is 1848 cm2.

Answer:

Let the radius of the cylinder be r and its height be h.As,CSA of the cylinderTSA of the cylinder=13CSA of the cylinder=13×TSA of the cylinderCSA of the cylinder=13×1848CSA of the cylinder=616 cm22πrh=616             .....iAlso,TSA of the cylinder=1848 cm22πrh+2πr2=1848616+2πr2=1848         Using i2πr2=1848-6162×227×r2=1232r2=1232×72×22r2=196r=196r=14 cmSubstituting r=14 cm in i, we get2π×14×h=6162×227×14×h=61688h=616h=6168h=7 cmNow,Volume of the cylinder=πr2h=227×14×14×7=4312 cm3

So, the volume of the cylinder is 4312 cm3.

 



Page No SA. 1.6:

Question 4:

In the following figure, the measure of DBC is

(a) 10°

(b) 40°

(c) 60°

(d) 30°

Answer:

Since ABC is a straight line, so
ABE+EBD+DBC=1806x+8x+4x=18018x=180x=18018=10
Now
DBC=4x=4×10=40
Hence, the correct option is (b).

Page No SA. 1.6:

Question 5:

Find the value of k, if x − 1 is a factor of x2 + x + k.

Answer:

Let f(x) = x2 + x + k.
Since x − 1 is a factor of f(x), therefore
f1=012+1+k=0k=-2
Hence, k = − 2.



Page No SA. 1.7:

Question 6:

Rationalize the denominator of 12+3.

Answer:

The rationalising factor of 12+3 is 2-3. So
12+3=2-32+32-3              =2-322-32          a-ba+b=a2-b2              =2-34-3              =2-3
Hence, 12+3=2-3.

Page No SA. 1.7:

Question 7:

Simplify : 5+22.

Answer:

Using the identity (a + b)2 = a2 + 2ab + b2, we get
5+22=52+252+22                   =5+22×5+2                   =7+210
Hence, 5+22=7+210.

Page No SA. 1.7:

Question 8:

Show that 0.477777... can be expressed in the form pq, where p and q are integers and q ≠ 0.

Answer:

Let x=0.477777....
Multiplying it by 10, we get
10x=4.77777...                         .....i
Multiplying (i) by 10 again, we get
100x=47.7777...                      .....ii
Subtracting (i) from (ii), we get
90x=43x=4390
Hence, pq=4390.

Page No SA. 1.7:

Question 9:

Evaluate (99)3 by using suitable identity.

Answer:

Here, (99)3 = (100 − 1)3.
So, using the identity (ab)3 = a3b3 − 3ab(ab), we get
(99)3 = (100)3 − 13 − 3(100)(1)(100 − 1)
         = 1000000 − 1 − 300(99)
         = 1000000 − 1 − 29700
         = 1000000 − 29701
         = 970299
Hence, (99)3 = 970299.

Page No SA. 1.7:

Question 10:

AD is a line segment. B and C are points in the interior if AC = BD, then prove that AB = CD.

Answer:

Given: AC = BD
AB + BC = BC + CD
AB = CD                       
(Cancelling BC from both the sides)
Hence, AB = CD.

Page No SA. 1.7:

Question 11:

Find 6 rational numbers between 3 and 4.

Answer:

Let us write 3 and 4 as below:
3=3×77=2174=4×77=287
Hence, 6 rational numbers between 3 and 4 are 227, 237, 247, 257, 267 and 277.

Page No SA. 1.7:

Question 12:

If a and b are rational numbers and 2+32-3=a+b3, find the values of a and b.

Answer:

The rationalising factor for 2+32-3 is 2+3. So
2+322-32+3=a+b34+3+4322-32=a+b37+434-3=a+b37+43=a+b3
Hence, a = 7 and b = 4.

Page No SA. 1.7:

Question 13:

In the following figure, PQR = PRQ, then prove that PQS = PRT.

Answer:

Given: PQR = PRQ
Since ST is a straight line, so
PQS=180-PQR            =180-PRQ        PRQ=PQR            =PRT
Hence, PQS=PRT.

Page No SA. 1.7:

Question 14:

In the following figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130° find ∠QRS.

Answer:

Extend line PQ to B intersecting RS at A.

AT and AB are parallel lines therefore
SAB+TSA=180SAB+130=180          =130SAB=180-130=50
QAR=SAB=50        Vertically opposite angles
Now, in QRA
QRA+QAR+AQR=180QRA+50+180-110=180QRA=60
Hence, QRS=60.



Page No SA. 1.8:

Question 15:

In the following figure, if lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75°, find SQT.

Answer:

In PRT
RPT+PRT+RTP=180        Angle sum property95+40+RTP=180RTP=180-135=45
RTP and STQ are vertically opposite angles, so
STQ=RTP=45
Now, in STQ
STQ+TSQ+SQT=180        Angle sum property45+75+SQT=180SQT=180-120=60
Hence, SQT = 60°.

Page No SA. 1.8:

Question 16:

Prove that the sum of  all interior angles of a triangle is 180°.

Answer:

Let ABC be a triangle. Draw a line parallel to AC and passing hrough B as shown in the figure.

Since AC||DE, therefore
DBA=BAC    .....i     Alternative anglesEBC=ACB    .....ii    Alternative angles
 Here,  DBE is a straight line, so
DBA+ABC+EBC=180BAC+ABC+ACB=180       From i and ii
Hence proved.

Page No SA. 1.8:

Question 17:

In the following figure, D and E are points on the base BC of a triangle ABC such that BD = CE and AD = AE. Prove that Δ ABE  Δ ACD.

Answer:

Since AD = AE, so ADE is a an isosceles triangle and therefore
ADE=AED180-ADE=180-AEDBDA=AEC                                .....i
Now, in ABD and ACE, we have
BD=EC                             GivenAD=AE                             GivenBDA=AEC                From iABDACE            By SAS
AB=AC                       BY c.p.c.t
In ABE and ACD
BD=EC                              GivenBD+ED=EC+EDBE=CD
AB = AC
AE = AD

ABE=ACD            By SSS
Hence, Δ ABE  Δ ACD.

Page No SA. 1.8:

Question 18:

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BAD=ABE and EPA=DPB
. Show that

(i) Δ DAP Δ EBP

(ii) AD = BE

Answer:

(i)

In Δ DAP and Δ EBP, we have
EPA=DPBEPA+EPD=DPB+EPD       Adding EPD on both sidesAPD=EPB      
DAP=EBP               GivenAP=BP                           GivenDAPEBP            BY ASA
Hence, DAPEBP.

(ii)

DAPEBPAD=BE                        By c.p.c.t

Page No SA. 1.8:

Question 19:

In which quadrant or on which axis do each of the points (−2, 4), (3, −1),(−1, 0), (−3, −5) and (0, −1) lie?

Answer:

The given points A(−2, 4), B(3, −1), C(−1, 0), D(−3, −5) and E(0, −1) are plotted as shown below:


From the figure, the points (−2, 4), (3, −1), and (−3, −5) lie in second, third and fourth quadrants respectively.
And the points (−1, 0) and (0, −1) lie on x-axis and y-axis respectively.



Page No SA. 1.9:

Question 20:

Plot the points E(4, 2), I (0, 2), L (−1, 3) and N (2, 0) on the Cartesian plane. Join these points in order and name the shape thus obtained.

Answer:

The points E(4, 2), I (0, 2), L (−1, 3) and N (2, 0) are plotted in the xy-plane and joined as shown below:


Hence, the shape obtained is a triangle.

Page No SA. 1.9:

Question 21:

Factorise : x3 + 13x2 + 32x + 20

Answer:

Let us split the middle terms as shown below:
x3 + 13x2 + 32x + 20 = x3 + 10x2 +  3x2 + 30x + 2x + 20
                                  = x2(x + 10) + 3x(x + 10) + 2(x + 10)
                                  = (x + 10) (x2 + 3x + 2)
                                  = (x + 10) (x2 + 2x + x + 2)
                                  = (x + 10) [x(x + 2) + (x + 2)]
                                  = (x + 10)(x + 2) (x + 1)

Hence, x3 + 13x2 + 32x + 20 = (x + 1) (x + 2)(x + 10).

Page No SA. 1.9:

Question 22:

Without actually calculating the cubes, find the value of the following:

(−12)3 + (7)3 + (5)3

Answer:

Let a = −12, b = 7 and c = 5.
Here, a + b + c = −12 + 7 + 5 = 0.
If a + b + c = 0, then a3 + b3 + c3 = 3abc. Therefore
(−12)3 + (7)3 + (5)3 = 3(−12)(7)(5) = −1260
Hence, (−12)3 + (7)3 + (5)3 = −1260.

Page No SA. 1.9:

Question 23:

If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz.

Answer:

Putting x + y + z = 0 in the identity x3 + y3 + z3 − 3xyz = (x + y + z) (x2 + y2 + z2xyyzzx), we get
x3 + y3 + z3 − 3xyz = (0) (x2 + y2 + z2xyyzzx)
x3+y3+z3-3xyz=0x3+y3+z3=3xyz
Thus, if x + y + z = 0, then x3 + y3 + z3 = 3 xyz.

Page No SA. 1.9:

Question 24:

Without actual division, prove that 2x4 − 6x3 + 3x2 + 3x − 2 is exactly divisible by x2 − 3x + 2.

Answer:

Let fx=2x4-6x3+3x2+3x-2 and gx=x2-3x+2.
gx=x2-3x+2       =x2-2x-x+2       =xx-2-1x-2       =x-1x-2
Thus, f(x) is divisible by g(x) if f(1) = 0 and f(2) = 0. Now
f1=214-613+312+31-2=2-6+3+3-2=0f2=224-623+322+32-2=32-48+12+6-2=0
Hence, 2x4 − 6x3 + 3x2 + 3x − 2 is exactly divisible by x2 − 3x + 2.

Page No SA. 1.9:

Question 25:

Factorise: a7ab6

Answer:

Factorise the expression a7ab6 as shown below:
a7-ab6=aa6-b6            =aa32-b32            =aa3+b3a3-b3          x-yx+y=x2-y2
Now, using the identities x3+y3=x+yx2+y2-xy and x3-y3=x-yx2+y2+xy, we get
a7-ab6=aa3+b3a3-b3             =aa+ba2+b2-aba-ba2+b2+ab             =aa-ba+ba2+b2-aba2+b2+ab
Hence, a7-ab6=aa-ba+ba2+b2-aba2+b2+ab.

Page No SA. 1.9:

Question 26:

In the following figure the side QR of Δ PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at a point T, then prove that ∠QTR = 12QPR.

Answer:

From the figure,
PQT=TQR=12PQR         .....iPRT=TRS=12PRS          .....ii
Here, PRS is an external angle for PQR. So
PRS=PQR+QPR2TRS=2TQR+QPR       .....iii      From (i) and (ii) 
TRS is an external angle for TQR. So
TRS=QTR+TQR
Substituting TRS=QTR+TQR in (iii), we get
2QTR+TQR=2TQR+QPR2QTR+2TQR=2TQR+QPR2QTR=QPRQTR=12QPR
Hence, QTR=12QPR.

Page No SA. 1.9:

Question 27:

Three bus-stops situated at A, B and C in the following figure are operated by handicapped person. These three bus − stops are equidistant from each other. OB is the bisector of ∠ABC and OC is the bisector of ∠ACB.



(i) Find ∠BOC.

(ii) Do you think employment provided to handicapped persons is important for the development of the society? Express your views with relevant points.

Answer:

(i) In the figure, AB = BC = CA.

∴ ∆ABC is an equilateral triangle.

⇒ ∠ABC = ∠ACB = ∠BAC = 60º 

It is given that OB is the bisector of ∠ABC.

 ∴ ∠ABO = ∠OBC = 12ABC=12×60°=30°     .....(1)

Also, OC is the bisector of ∠ACB. 

 ∴ ∠ACO = ∠OCB = 12ACB=12×60°=30°     .....(2)

In ∆OBC,

∠OBC + ∠OCB + ∠BOC = 180º         (Angle sum property)

⇒ 30º + 30º + ∠BOC = 180º

⇒ ∠BOC = 180º − 60º = 120º

(ii)  Yes, the employment provided to handicapped persons is important for the development of the society as employment to handicapped persons provide them with better opportunities of success and prosperity. It helps in their social and economic growth which in turns helps in the development of society.



Page No SA.1.10:

Question 28:

In the following figure, S is any point on the side QR of Δ PQR. Prove that PQ + QR + RP > 2 PS.

Answer:

The sum of any two sides of a triangle is grater than the third side.
In PQS and PSR, we have
PQ + QS > PS                  ..... (i)
PR + SR > PS                  ..... (ii)
Adding (i) and (ii), we get
PQ + PR + (QS + SR) > 2PS
Since QS + SR = QR, therefore
PQ + PR + QR > 2PS
Hence, PQ + QR + RP > 2PS.

Page No SA.1.10:

Question 29:

Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Answer:

Let a, b and c be the sides of the given triangle. Then
a = 18 cm, b = 10 cm and a + b + c = 42 cm
Substituting a = 18, b = 10 in a + b + c = 42, we get
18 + 10 + c = 42 c=42-28=14 cm
Now
s=a+b+c2=18+10+142=21 cm
Using Heron's formula, we get
Area of triangle=ss-as-bs-c                       =2121-1821-1021-14                       =213117                       =2111 cm2
Hence, the area of the triangle is 2111 cm2.

Page No SA.1.10:

Question 30:

Δ ABC is an isosceles triangle in which AB= AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.

Answer:

Since AB= AC and AD = AB, therefore AC = AD. Thus
ABC=ACB                .....i ADC=ACD                .....ii
Now, in BDC
BDC+DBC+BCD=180                        Angle sum propertyACD+ACB+ACD+ACB=180         From (i) and (ii)2ACD+2ACB=180    ACD+ACB=1802BCD=90
Hence, BCD is a right angle.

Page No SA.1.10:

Question 31:

Simplify:

5+25-2-5-25+2

Answer:

The rationalising factor for 5+25-2 and 5-25+2 are 5+2 and 5-2 respectively. So
5+25-2-5-25+2=5+25+25-25+2-5-25-25+25-2                               =5+2252-22-5-2252-22               a-ba+b=a2-b2                               =5+2+455-4-5+2-455-4                               =7+45-7+45=85
Hence, 5+25-2-5-25+2=85.



Page No SA1.2:

Question 1:

Decimal expansion of a rational number cannot be
(a) non-terminating
(b) non-terminating and recurring
(c) terminating
(d) non-terminating and non-recurring

Answer:

The decimal expansion of a rational number is either terminating or non-terminating recurring.
So, a rational number cannot be non-terminating and non-recurring.
Hence, the correct option is (d).

Page No SA1.2:

Question 2:

One of the factors of (9x2 − 1) − (1 + 3x)2, is

(a) 3 + x

(b) 3 − x

(c) 3x − 1

(d) 3x + 1

Answer:

To find all the factors of (9x2 − 1) − (1 + 3x)2, let us factorise it.
9x2-1-1+3x2=9x2-1-1+6x+9x2       a+b2=a2+2ab+b2                              =9x2-1-1-6x-9x2                              =-2-6x                              =-21+3x
Thus, the factors of (9x2 − 1) − (1 + 3x)2 are –2 and 3x + 1.
Hence, the correct option is (d).

Page No SA1.2:

Question 3:

Simplify: 164-63433+18×2435-196

Answer:

Let us simplify the expression by simplifying each term separately.
164=2×2×2×24=23433=7×7×73=72435=3×3×3×3×35=3196=14×4=14
Therefore
164-63433+18×2435-196=2-6×7+18×3-14=2-42+54-14=56-56=0
Hence, 164-63433+18×2435-196=0.

Page No SA1.2:

Question 4:

An exterior angle of a triangle is 110° and the two interior opposite angles are equal.
Each of these angles is

(a) 70°

(b) 55°

(c) 35°

(d) 110°

Answer:

Let ABC be the triangle in which BCD=110 is the exterior triangle as shown in the figure.

Given: ABC=BAC
An exterior angle of a triangle is equal to the sum of the opposite interior angles, therefore

ABC+BAC=BCDABC+ABC=110          ABC=BAC2ABC=110ABC=55
Thus, the measure of the equal angles is 55.
Hence, the correct option is (b).

Page No SA1.2:

Question 5:

In ΔPQR, if R>Q, then

(a) QR > PR

(b) PQ > PR

(c) PQ < PR

(d) QR < PR

Answer:

In a given triangle, the side opposite to the greater angle is greater.
The side opposite to angles R and Q are PQ and PR respectively.
In ΔPQR, R>Q, therefore
PQ > PR
Hence, the correct option is (b).

Page No SA1.2:

Question 6:

Evaluate the product without multiplying directly : 104×97.

Answer:

Let us find the product 104 × 97 using the identity x+ax-b=x2+a-bx-ab.
104×97=100+4100-3              =1002+4-3×100-4×3      x+ax-b=x2+a-bx-ab              =10000+100-12              =10088
Hence, 104 × 97 = 10088.



Page No SA1.3:

Question 7:

"Lines are parallel if they do not intersect," prove the above with suitable diagram.

Answer:

Let AB and CD be two parallel lines.

If possible, let AB and CD intersect at a point P.

But, two lines in the plane are intersecting lines if they have a common point.

This is a contradiction, because AB and CD are two parallel lines.



Hence, the lines are parallel if they do not intersect.

Page No SA1.3:

Question 8:

Prove that two distinct lines cannot have more than one point in common.

Answer:

Let AB and CD be two distinct lines. We need to prove that AB and CD cannot have more than one point in common.



Let us assume that AB and CD have two points, say P and Q, in common.

Then, AB contains both points P and Q.

Also, CD contains both points P and Q.

Now, incidence axiom states that there is one and only one line passing through two distinct points P and Q.

So, AB and CD are same lines.

But, this is a contradiction because AB and CD are two distinct lines.

Hence, two distinct lines cannot have more than one point in common.

Page No SA1.3:

Question 9:

Plot the points (3, 4), (−3,−3), (−7, 6) and (0, −6) on Cartesian plane and give their positions in quadrants / axes.

Answer:

Take 1 cm = 1 unit and draw the x-axis and y-axis.
Now plot the points (3, 4), (−3,−3), (−7, 6) and (0, −6) as shown below:


From the graph, the points (3, 4), (−7, 6) and (−3,−3) lie in the first, second and third quadrants respectively
and (0, −6) lies on the y-axis.

Page No SA1.3:

Question 10:

Find the area of the triangle with sides 35 cm, 54 cm and 61 cm.

Answer:

We will use Heron's formula to find the area of the triangle.
Here, a = 35 cm, b = 54 cm and c = 61 cm. Now
s=a+b+c2=35+54+612=1502=75
Therefore
Area of triangle=ss-as-bs-c                        =7575-3575-5475-61                        =75×40×21×14                        =4×3×5×75                        =4205
Hence, the area of the required triangle is 4205 sq. units.

Page No SA1.3:

Question 11:

If x = 2 + 3, find x-1x.

Answer:

Given: x=2+3
1x=12+3     =2-32+32-3     =2-322-32                             a+ba-b=a2-b2     =2-34-3=2-3
Now
x-1x=2+3-2-3         =2+3-2+3         =23                       
Hence, x-1x=23.

Page No SA1.3:

Question 12:

Simplify x4 x3 x2x345.

Answer:

Here, we need to simplify the inner most radicals first.
x4 x3 x2x345=x4 x3 x2×x12345=x4 x3 x2+12345=x4 x3 ×x52×345=x4 x3+5645=x4 x23645=x4 ×x23245=x4+23245=x119245=x11924×5=x119120
Hence, x4 x3 x2x345=x119120.

Page No SA1.3:

Question 13:

Expand -x2+y+142

Answer:

Using the identity ( a + b + c )2 = a2 + b2 + c2 + 2(ab + bc + ac), we get
-x2+y+142=-x22+y2+142+2-x2×y+y×14+-x2×14                        =x24+y2+116+2-xy2+y4-x8                        =x24+y2-xy+y2-x4+116
Hence, -x2+y+142=x24+y2-xy+y2-x4+116.

Page No SA1.3:

Question 14:

Evaluate the following : 32x+13 -27x24-9x2. 

Answer:

Using the identity (a + b)3 = a3 + b3 + 3ab(a + b), we get
32x+13 -27x24-9x2=3x23+13+3×3x2×13x2+1 -27x24-9x2=27x38+1+27x24+9x2 -27x24-9x2=27x38+1
Hence, 32x+13 -27x24-9x2=27x38+1.

Page No SA1.3:

Question 15:

In Δ ABC, if AB is the greatest side, then prove that C>60.

Answer:

It is given that, AB is the longest side, so AB > BC and AB > AC.
Since, the angle opposite to the longer side is greater, therefore
C>A                                                      .....iC>B                                                      .....ii
Adding (i) and (ii), we get
2C>A+B
Adding C on both sides, we have
2C+C>A+B+C3C>180            A+B+CC>60
Hence, C>60.



Page No SA1.4:

Question 16:

In the following figure, AE = AD, BAE=CAD. Prove that AB = AC.

Answer:

Since AE = AD, so AED=ADE. Now
180-AED=180-ADEAEB=ADC
In ABE and ACD
AE=AD                                                     (Given) 
BAE=CAD                                          (Given)
AEB=ADC                                       (Prove above)
 ABEACD                                (By ASA)
Thus, by corresponding parts of congruent triangles
AB = AC

Page No SA1.4:

Question 17:

Sides AB and CD of a quadrilateral ABCD are produced as shown in the following figure. Show that x+y=a+b.

Answer:

In quadrilateral ABCD
CDA=180-yABC=180-x
Since, the sum of the angles of a quadrilateral is 180, therefore
CDA+DAB+ABC+BCD=360180-y+a+180-x+b=360a+b=x+y
Hence, a+b=x+y.

Page No SA1.4:

Question 18:

AB is a line segment and line l is its perpendicular bisector. Show that every point on line l is equidistant from A and B.

Answer:

In the following figure, AB is the given line segment and l is its perpendicular bisector.
A any point on the given line l.


Now, in ∆AOP and∆BOP
AO=BOAOP=BOP=90     l is perpendicular bisector of ABOP=OP                             Common
AOPBOPAP=BP                     By c.p.c.t

Hence, every point on line l is equidistant from A and B.

Page No SA1.4:

Question 19:

Find the area of a triangle whose sides are 5 cm, 12 cm and 13 cm. Also, find its shortest altitude.

Answer:

Here,
52 + 122
= 25 + 144
= 169
=  132.
Thus, the triangle is a right angled triangle.
So, the length of hypotenuse is 13 and the length of the perpendicular sides are 5 and 12.
Therefore, the length of its shortest altitude is 5.
Now
Area of triangle=12×base×height                         =12×12×5                         =30 sq. units
Hence, the area of the required triangle is 30 sq. units.

Page No SA1.4:

Question 20:

The perimeter of an isosceles triangle is 42 cm and its base is 32 times each of the equal sides. Find the length of each side and area of the triangle.

Answer:

Let a, b and c be the sides of the triangle. Here, a = b and cbase=32a.
Since the perimeter of the triangle is 42 cm, so
a+b+c=42a+a+32a=42a=12 cmc=32×a=32×12=18 cm
Therefore, the lengths of the sides of the triangle are 12 cm, 12 cm and 18 cm.
Now
s=a+b+c2=12+12+182=21 cmArea of triangle=ss-as-bs-c                       =2121-1221-1221-18                       =21993=277 cm2
Hence, the area of the triangle is 277 cm2.

Page No SA1.4:

Question 21:

Varun was facing some difficulty in simplyfying 17 -3. His classmate Priya gave him a clue to rationalise the denominator for simplification. Varun Simplified the expression and thanked Priya for this good will. How Varun simplified 17 -3.? What value does it indicate?

Answer:

The rationalising factor for 17 -3 is 7+3. So
17-3=7+37-37+3                 =7+372-32                 =7+37-3                 =147+3
Varun simplified 17 -3 by multiplying the numerator and denominator by 7+3.
It indicates the value of co-operation and help.



Page No SA1.5:

Question 22:

Prove that xaxb1ab .xbxc1bc .xcxa1ca =1

Answer:

Using the laws of exponents xmxn=xm-n and xmn=xmn, we get
xaxb1ab =xa-b1ab =xa-bab=x1b-1axbxb1bc =xb-c1bc =xb-cbc=x1c-1bxcxa1ca =xc-a1ca =xc-aca=x1a-1c
Now
xaxb1ab .xbxc1bc .xcxa1ca =x1b-1a·x1c-1b·x1a-1c=x1b-1a+1c-1b+1a-1c           xm·xn=xm+n=x0=1
Hence, xaxb1ab .xbxc1bc .xcxa1ca =1.

Page No SA1.5:

Question 23:

If the polynomials az3 + 4z2 + 3z − 4 and z3 − 4z + a leave the same remainder when divided by z − 3, find a.

Answer:

Let pz=az3+4z2+3z-4 and qz=z3-4z+a.
Since, p(z) and q(z) leave the same remainder when divided by z − 3, so
p3=q3a33+432+33-4=33-43+a27a+36+9-4=27-12+a26a=-26a=-1
Hence, a = −1.

Page No SA1.5:

Question 24:

Factorize 2x256x + 112.

Answer:

Splitting middle term of 2x2-56x+112, we get
2x2-56x+112=2x2-12x-13x+112      12+13=56, 12×13=16                          =2xx-14-13x-14                         =x-142x-13
Hence, the required factorisation is x-142x-13.

Page No SA1.5:

Question 25:

Find the quotient when f(x) = x3 + 3x2 + 3x + 5 is divided by g(x) = x + 2. Also, find the remainder.

Answer:

Here, f(x) = x3 + 3x2 + 3x + 5 and g(x) = x + 2.
           x2+x+1x+2  x3+3x2+3x+5            x3+2x2        -     -                            x2+3x+5                     x2+2x                 -    -                                        x+5                                x+2                            -   -                                          3
Hence, the quotient is x2 + x + 1 and the remainder is 3.

Page No SA1.5:

Question 26:

Prove that:

(x + y)3 + (y + z)3 + (z + x)3 − 3(x + y) (y + z) (z + x) = (x3 + y3 + z3 − 3xyz).

Answer:

Using the identities a3+b3+c3-3abc=a+b+ca2+b2+c2-ab-bc-ca and a+b2=a2+2ab+b2, we get
x+y3+y+z3+z+x3-3x+yy+zz+x=x+y+y+z+z+xx+y2+y+z2+z+x2-x+yy+z-y+zz+x-z+xx+y=2x+y+z2x2+y2+z2+2xy+2yz+2zx-xy-xz-y2-yz-yz-xy-z2-zx-xz-yz-x2-xy=2x+y+zx2+y2+z2+2xy+2yz+2zx-xy-xz-yz=2x+y+zx2+y2+z2-xy-xz-yz
=2x3+y3+z3-3xyz
Hence, (x + y)3 + (y + z)3 + (z + x)3 − 3(x + y) (y + z) (z + x) = 2(x3 + y3 + z3 − 3xyz).

Page No SA1.5:

Question 27:

In a ΔPQR, if PQ = QR and mid-points of three sides PQ, QR and RP are L, M and N respectively. Prove that LNMN.

Answer:

Given: In PQR, PQ=QR and mid-points of three sides PQ, QR and RP are L, M and N respectively.

Now, in PLN and RMN
P=R                         (∵ PQ = QR)
PN = RN                          (N is mid-point of PR)
PL = RM                          (PQ = QR and L and M are mid-points of PQ and RQ respectively )
PLNRMN        (By SAS)
Thus, by corresponding parts of congruent triangles, we have
LN = MN

Page No SA1.5:

Question 28:

In the following figure, if OA = OD and 1=2. Prove that ΔOCB is an isosceles triangle.

Answer:

Given: OA = OD and 1=2
In OCA and OBD
OA = OD                                (Given)
AOC=DOB                      (Vertically opposite angles)
1=2                                 Given180-1=2-180CAO=BDO
CAOBDO                  (By ASA)
Thus, by corresponding parts of congruent triangles, we have
OC = OB
Hence, ΔOCB is an isosceles triangle.

Page No SA1.5:

Question 29:

In the following figure, ray OS stands on a line POQ, ray OR and ray OT are angle bisectors of POS and QOS respectively.
If POS = x, find ROT.

 

Answer:

Given: POQ is a straight line and ray OR and ray OT are angle bisectors of POS and QOS respectively.
Angles POS and QOS form a linear pair, so
POS+SOQ=180
Since OR and OT are angle bisectors of POS and QOS respectively, therefore
2ROS+SOQ=1802ROS+2SOT=180ROS+SOT=90ROT=90
Hence, ROT=90 which is independent of x.



Page No SA1.6:

Question 30:

If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.

Answer:

Let m and l be two lines and p be the transversal line.

Let ATR and CST be the corresponding angles and CM and TN respectively be their angle bisectors. Therefore
ATN=NTP=xCSM=MST=y
Here, SM and TN are parallel. Therefore
NTR=MSTx=y
Now
CST+STA=2y+180-2x                         =180+2y-2x                         =180+2x-2x         x=y                         =180
Thus, the sum of the interior angles on the same side of the line p and between the lines l and m is 180.
Therefore, l and m are parallel.
Hence, if a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then the two lines are parallel.

Page No SA1.6:

Question 31:

PQRS is a quadrilateral in which diagonals PR and QS intersect at O. Prove that PQ + QR + RS + SP < 2 (PR + QS).

Answer:

Let PQRS be the quadrilateral in which diagonals PR and QS intersect at O.

The sum of any two sides of a triangle is greater than the third side.
So, in POQ, POS, ROS and QOR, we have
OP + OQ > PQ                                    .....(i)
OP + OS > PS                                     .....(ii)
OS + OR > RS                                    .....(iii)
OR + OQ > QR                                  .....(iv)
Adding (i), (ii), (iii) and (iv), we get
2(OP + OQ + OS + OR) > PQ + PS + RS + QR
⇒  
2(OP + OR + OQ + OS ) > PQ + PS + RS + QR
⇒  
2(PR + QS ) > PQ + PS + RS + QR
⇒  PQ + PS + RS + QR < 2(PR + QS)

Page No SA1.6:

Question 32:

The value of 271/3. 41/2 is

(a) 6

(b) 12

(c) 18

(d) − 6

Answer:

Let us first simplify 2712 and 412 separately.
2713=3313=33×13=31=3412=2212=22×12=21=2
Therefore
2713·412=3×2=6
Hence, the correct option is (a).

Page No SA1.6:

Question 33:

Which of the following is an irrational number?

(a) 927

(b) −64

(c) 1.44

(d) 14

Answer:

Let us simplify each of the numbers 927, −64, 1.44 and 14 separately.
927=13=13-64=-8×8=-81.44=144100=1210=1.214=14=12
Here, −64, 1.44 and 14 are rational numbers while 927 is an irrational number.
Hence, the correct option is (a).
 

Page No SA1.6:

Question 34:

If the perimeter of an equilateral triangle is 60 m, then its area is

(a) 103 m2

(b)  153 m2

(c) 203 m2

(d) 1003 m2

Answer:

Let a be the side of the equilateral triangle. Then
Perimeter = 60 m
a+a+a=603a=60a=20 m
Now
Area=a234=20234=40034=1003 m2
Hence, the correct option is (d).



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