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Page No 252:

Question 1:

Mark (✓) against the correct answer in each of the following:
Which of the following is a rational number?
(a) 23
(b) 23
(c) 35
(d) -35

Answer:

(d) -35

The number of the form pq, where p and q are integers and q0, are known as rational numbers.
Here -3 and 5 are integers and 50.
Therefore, -35 is a rational number.

Page No 252:

Question 2:

Mark (✓) against the correct answer in each of the following:
The value of k, for which the polynomial x3 − 4x2 + 2x + k has 3 as its zero, is
(a) 3
(b) −3
(c) 6
(d) −6

Answer:

(a) 3

Let f(x) = x3 − 4x2 + 2x + k.
3 is the zero of f(x).
Therefore, f(3) = 0
     33 -4×32 +2×3 +k = 0 27 - 36 +6+ k =0 -3 + k = 0 k = 3

Page No 252:

Question 3:

Mark (✓) against the correct answer in each of the following:
Which of the following are zeros of the polynomial x3 + 2x2 − 5x − 6?
(a) −2
(b) 2
(c) −3
(d) 3

Answer:

(b) 2 and (c) −3
Let f(x) =
x3 + 2x2 − 5x − 6.
We have to find the zeros of the polynomial.

⇒  f(x) = 0
x3 + 2x2 − 5x − 6 = 0
x3 + 3x2x2 −3x −2x −6 = 0
x2(x + 3) −x(x + 3) −2(x + 3) = 0
(x + 3)(x2x − 2)
(x + 3)(x2 − 2x + x − 2)
(x + 3){x(x − 2) +1(x − 2)} = 0
(x + 3)(x − 2)(x + 1) = 0
Either x + 3 = 0 or x − 2 = 0 or x + 1 = 0
 
x = − 3 or 2 or −1

Disclaimer: Here both 2 and − 3 are correct answers.

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Question 4:

Mark (✓) against the correct answer in each of the following:
The factorisation of x2 + 7x − 12 yields
(a) (x − 3) (x − 4)
(b) (3 + x) (4 − x)
(c) (x − 4) (3 − x)
(d) (4 − x) (3 − x)

Answer:

(c) (x − 4) (3 − x)

x2 + 7x − 12 = − (x2 − 7x + 12)
                       = − (x2 − 4x − 3x + 12)
                       = − { x(x − 4) −3 (x − 4)}
                       =  − (x − 4)(x − 3)
                       = (x − 4)(3 − x)



Page No 253:

Question 5:

Mark (✓) against the correct answer in each of the following:
In the given figure, BOC = ?
(a) 45°
(b) 60°
(c) 75°
(d) 56°
Figure

Answer:

(b) 60°

Since, OAB is a straight line, the sum of all the angles
on the same side of AOB, at point O on it, is 180o.
 3x + 5x + 4x = 180 12x = 180 x  = 15 BOC = 4×15 = 60°

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Question 6:

Mark (✓) against the correct answer in each of the following:
In the given figure, ABC is an equilateral triangle and ∆BDC is an isosceles right triangle, right-angled at D. Then ∠ACD = ?
(a) 60°
(b) 90°
(c) 120°
(d) 105°
Figure

Answer:

(d) 105°

ABC is an equilateral triangle.So, every angle of ABC is 60°.Therefore, ACB = 60°Now, BCD is an isosceles triangle and BDC = 90°.Here, BCD = CBDNow, BCD + CBD + BDC = 180°   Angle sum property        BCD + CBD+ 90° = 180°         2BCD  = 90°        BCD  = 45°From the figure,ACD =  ACB + BCD             =  60° + 45°   ACD= 105°

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Question 7:

Mark (✓) against the correct answer in each of the following:
Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is
(a) 30 cm2
(b) 45 cm2
(c) 60 cm2
(d) 78 cm2
Figure

Answer:

(c) 60 cm2

Area of an isosceles triangle = 14b4a2-b2Here, a = equal sides and b = third sideHere, a = 13 cm and b = 24 cmTherefore, area = 14×24×4×132-122                              = 6×4×132-242                              = 6×4×169-576                              = 6×676-576                              = 6×100                              = 6×10                              = 60 cm2

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Question 8:

Mark (✓) against the correct answer in each of the following:
In an isosceles right triangle, the length of the hypotenuse is 42 cm.
(a) 43 cm
(b) 6 cm
(c) 5 cm
(d) 4 cm
Figure

Answer:

(d) 4 cm

In a right-angled triangle,
Hypotenuse2 = Base2 +Perpendicular2It is given that the triangle is isosceles.Therefore, base = perpendicular = x cm sayGiven: Hypotenuse = 42 cmTherefore, 422 = x2 + x2       32 = 2x2       x2  = 16       x = 4Therefore, length of each equal side = 4 cm

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Question 9:

If x=7+43, find the value of x+1x.

Answer:

Given,  x = 7 + 43 x = 4 + 43 + 3 x = 22 + 2×2×3 + 32 x = 2+32 x =± 2+3When x = 2+3, 1 x = 12+3 1 x = 12+3×2-32-3 1 x = 2-322-32 1 x = 2-34-3 1 x = 2-3 Therefore, when x =- 2+3, 1 x =- 2-3 = -2+3 Therefore, x +1 x=2+3 +  2-3    or  - 2+3 + -2+3                                            = 4 or -4  

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Question 10:

Factorise: (7a3+56b3).

Answer:

   7a3 + 56b3= 7a3 + 8b3=7a3 + 2b3= 7a+2ba2 + 2b2 -a×2b= 7a+2ba2 + 4b2 - 2ab

Page No 253:

Question 11:

Find the value of a for which (x 1) is a factor of the polynomial (a2x3 − 4ax + 4a − 1).

Answer:

Let f(x) = a2x3 − 4ax + 4a − 1.
(x 1) is a factor of f(x)
Therefore, f(1) = 0
       a2×13 - 4×a×1 + 4a - 1 = 0 a2 - 1 = 0 a2  =1Therefore, a = ±1

Page No 253:

Question 12:

In the given figure, if AC = BD, show that AB = CD. State Euclid's axiom used for it.
Figure

Answer:

From the given figure,
                           AC  = AB + BC
                            BD  =  BC + CD
It is given that AC = BD.
Putting the values of AC and BD, we get:
                               AB + BC =  BC + CD..................(1)
According to Euclid's axiom, when equals are subtracted from equals, the remainders are also equal.
Subtracting BC from both sides in equation (1), we get:
                            AB + BC - BC = BC + CD - BC
                           AB = CD

Hence, proved.                                

Page No 253:

Question 13:

In ABC, if 2∠A = 3∠B = 6∠C, calculate the measure of ∠B.

Answer:


In a ABC, 2A = 3B = 6C = xo (say)

Therefore, A = x°2                    B = x°3                    C = x°6By angle sum property in ABC,A + B + C = 180°x°2+x°3+x°6 = 180°x° = 180° Therefore, B = x°3 = 180°3 = 60°.

Page No 253:

Question 14:

In the given figure, BAC = 30°, ∠ABC = 50° and ∠CDE = 40°. Find ∠AED.
Figure

Answer:

Side BC of ABC is produced to D.
ACD=CAB+ABC  [Exterior angle property]ACD=30+50°ACD=80°

Side CE of CED is produced to A.
AED=ECD+EDC [Exterior angle property]AED=80+40°AED=120°



Page No 254:

Question 15:

If x=5+35-3 and y=5-35+3, find the value of (x2 + y2).
or
Simplify: (7+35)(3+5)-(7-35)(3-5)

Answer:

Given,x=5+35-3,y=5-35+3Therefore,x2=5+35-32=5+3+2155+3-215=8+2158-215Andy2=5-35+32=5+3-2155+3+215=8-2158+215

Now,x2 + y2 = 8+2158-215 + 8-2158+215               =8+2152 + 8-21528-2158+215               =82+2152+2×8×215+82+2152-2×8×21582-2152               =64 + 60 + 64 + 6064 - 60               =2484                = 62

                                                      OR

  (7+35)(3+5)-(7-35)(3-5)= (7+35)(3-5) - (7-35)(3+5)(3+5)(3-5)=21 - 75 + 95 -15 - 21 -75 +95 + 1532 - 52=459 - 5=454 = 5

Page No 254:

Question 16:

If 2 and -13 are the zeros of the polynomial 3x3 − 2x2 − 7x − 2, find the third zero of the polynomial.

Answer:

To find the zeros of a polynomial, we equate the polynomial to 0.  3x3 - 2x2 -7x -2 = 03x3 - 6x2 + 4x2-8x +x -2 =03x2x-2 + 4xx-2 +1x-2=0x-23x2 + 4x +1=0x-23x2 + 3x +x +1=0x-23xx+1+1x+1=0x-2x+13x+1=0 x-2 = 0 or x+1 = 0 or 3x+1 = 0 x =2 or -1 or -13As 2 and -13 are the given zeros, the third zero is -1.

Page No 254:

Question 17:

Find the remainder when the polynomial f(x) = 4x3 − 12x2 + 14x − 3 is divided by (2x − 1).

Answer:

Let px = 4x3 - 12x2 + 14x -3.

By the remainder theorem, we know that when p(x) is divided by (2x - 1), the remainder is p12.
Now,
p12=4×123-12×122+14×12-3      =12-3+7-3      =32

Page No 254:

Question 18:

Factorise: (p q)3 + (q r)3 + (r p)3

Answer:

(p q)3 + (q r)3 + (r p)
Let p q = x
q r = y
r p = z
Therefore, x + y+ z =
p q + q r + r p = 0
So,
(p q)3 + (q r)3 + (r p)3 = x3 + y3 + z3
                                             = 3xyz   (when x + y + z = 0, then x3 + y3 + z3 = 3xyz )
                                             = 3(p q)(q r)(r p)

Page No 254:

Question 19:

In the given figure, in ABC, it is given that ∠B = 40° and ∠C = 50°. DE || BC and EF || AB. Find: (i) ∠ADE + ∠MEN (ii) ∠BDE and (iii) ∠BFE.
Figure

Answer:

i DEBCTherefore,ADE = ABC = 40°        Alternate anglesEFAB and MF is the produced line of EF.Therefore, MFAB. MEN = ADE = 40°        Alternate anglesADE + MEN = 40° + 40° = 80°ii DEBC and AB is the transversal. Interior angles on the same side of a transversal are supplementary. BDE = 180° - ABC                         = 180° - 40°                  = 140°iii EFAB and BC is the transversal. Interior angles on the same side of a transversal are supplementary. BFE = 180° - ABC                  = 180° - 40°                  = 140°

Page No 254:

Question 20:

In the given figure, ABC and ∆ABD are such that AD = BC, ∠1 = ∠2 and ∠3 = ∠4. Prove that BD = AC.
Figure

Answer:

In ABC and ABD,AD= BC  GivenAB = AB  Common sideGiven:     1 = 2and 3 = 4Therefore, 1 +3 = 2 + 4      DAB = ABC By SAS, DAB  CBATherefore, BD = AC Corresponding sides of congruent triangles are equal

Page No 254:

Question 21:

In the given figure, C is the mid-point of AB. If DCA = ∠ECB and ∠DBC = ∠EAC, prove that DC = EC.
Figure

Answer:

In ECA and DCB,EAC=DBC    given AC = BC  C is the mid point of ABNow,DCA= ECBDCA + DCE= ECB + DCEACE= BCDTherefore, by ASA congruency criteria, ECA DCB. DC = EC    Corresponding sides of congruent triangles are equal 

Page No 254:

Question 22:

In ABC, ALBC and AM is the bisector of ∠A. Show that LAM=12(B-C).
Figure

Answer:

In ABC,A = 180° - B+Ci.e. BAC = 180° - B+CNow, AM is the angle bisector of A. BAM = 12180° - B+C BAM= 90° - 12B+C -----iIn ABL,ALB = 90° BAL = 180°-B + ALB  BAL = 180°-B + 90° BAL = 90°-B        ----------iiNow,LAM = BAM - BALUsing i  and ii,  we have:LAM =90° - 12B+C - 90°-B             =90° - 12B- 12C - 90°+BLAM =12B - 12CHence, proved.



Page No 255:

Question 23:

In the given figure, AB || CD, BAE = 100° and ∠AEC = 30°. Find ∠DCE.
Figure

Answer:

Construction:Draw GHFC as shown in the book by dotted line GH,which meets AB at A and CD at H.Now,GAE = AEC = 30°   alternate angles
And GAB = BAE - GAE                     = 100° - 30°                     = 70°Now, ABCD and GHis the transversal.DHA= GAB = 70°         corresponding anglesAgain GHFC and DC is the transversal.  DCE = DHA=70°      corresponding angles

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Question 24:

Factorise: a3 b3 + 1 + 3ab

Answer:

a3 - b3 + 1 + 3ab= a3+-b3 + 13 -3a-b1=a+ -b+1a2+-b2+12-a×-b--b×1-a×1=a-b+1a2 + b2 +1 +ab + b -a

Page No 255:

Question 25:

If x=1(2-3), show that the value of (x3 − 2x2 − 7x + 5) is 3.
Or
Simplify: 11+2+12+3+13+4+......+18+9

Answer:

x=12-3By rationalising we get:x=12-3×2+32+3x=2+34-3x=2+3x-2=3Squaring both sides, we get:x-22=3x2+4-4x=3x2-4x+1=0So,we can write x3-2x2-7x+5 as x3-4x2+x+2x2-8x+2+3.Hence, x3-2x2-7x+5=x3-4x2+x+2x2-8x+2+3=xx2-4x+1+2x2-4x+1+3=x0+20+3=3

                                                                 OR

11+2+12+3+13+4+......+18+9By rationalising we get:11+2 = 12+1×2-12-1= 2-12-1 = 2-112+3=13+2×3-23-2 = 3-23-2


=3-2Similarly, 13+4=4-314+5=5-415+6=6-516+7=7-617+8=8-718+9=9-8 Adding all the terms of the left side and right side, we get: 11+2+12+3+13+4+.....+18+9=9-1=3-1=2

Page No 255:

Question 26:

If x=a+2b+a-2ba+2b-a-2b , then show that bx2-ax+b=0.

Answer:

Given,x = a+2b + a-2ba+2b -a-2bRationalising the denominator, we get:x = a+2b + a-2ba+2b -a-2b×a+2b + a-2ba+2b +a-2b x = a+2b + a-2b2a+2b2 - a-2b2 x = a+2b2+a-2b2+2a+2ba-2ba+2b - a-2b x = a+2b+a-2b + 2a2-2b24bx = 2a + 2 a2-4b24bx = a +a2-4b22b 2bx = a + a2-4b2 2bx - a = a2-4b2Squaring both sides we get:2bx - a2 = a2-4b22 4b2x2 + a2 - 4abx = a2 - 4b2 4b2x2 - 4abx+ 4b2 = 0 4bbx2 - ax+ b = 0Therefore, bx2 - ax+ b = 0Hence, proved.

Page No 255:

Question 27:

If (x3 + mx2 x + 6) has (x − 2) as a factor and leaves a remainder r when divided by (x − 3), find the values of m and r.

Answer:

Let fx = x3 + mx2 -x + 6.Here, x-2 is a factor of fx.Therefore, f2 = 0 23 + m×22 -2 + 6 = 0 8 + 4m -2 + 6 = 0  4m + 12 = 0  4m  = - 12  m = -3Now, using m = -3 in fx, we get:fx = x3 -3x2 -x + 6When fx is divided by x-3, the remainder will be f3,and the remainder is given as r.Now, f3 = 33 -3×32 -3 + 6      or,  r  = 27-27 -3 + 6      or, r   = 3Therefore, m = -3 and  r = 3.

Page No 255:

Question 28:

If r and s be the remainders when the polynomials (x3 + 2x2 − 5ax − 7) and (x3 + ax2 − 12ax + 6) are divided by (x + 1) and (x − 2), respectively and 2r + s = 6, find the value of a.

Answer:

Let fx = x3 + 2x2 -5ax -7.When fx is divided by x+1, the remainder is r.Therefore,  f-1 = r -13 + 2-12 -5a-1 -7 = r -1 + 2 + 5a - 7 = r 5a -6 = r r = 5a - 6 -------1Again, let  gx = x3 + ax2 -12ax +6.When gx is divided by x-2, the remainder is s.Therefore,  g2 = s 23 + a22 -12a2 +6 = s 8+ 4a -24a+6 = s  s = -20a + 14  -----2From 1×2 + 2, we have:2r + s = 25a - 6 + -20a + 14 6 = 10a - 12 - 20a + 14 6 = -10a + 2-10a = 4a = -25

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Question 29:

Prove that:
(a+b)3(b+c)3+(c+a)3-3(a+b)(b+c)(c+a)=2(a3+b3+c3-3abc).

Answer:

L.H.S. = a+b3 + b+c3+c+a3-3a+bb+cc+a             = a+b+b+c+c+c+aa+b2+b+c2+c+a2-a+bb+c-b+cc+a-c+aa+b             =2a+2b+2ca2 + b2 + 2ab + b2 + c2 +2bc + c2 + a2 + 2ca - ab -ac - b2 - bc - bc -ab -c2 -ca - ca-bc-a2-ab             =2a+b+ca2 + b2+ c2-ab-bc-ca             =2a3 + b3 +  c3 -3abc             =R.H.S.Hence, proved.

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Question 30:

On a graph paper plot the following points:
A(3,3), B(2, 4), C(5, 5), D(0, 2), E(3, −3) and F(−5,−5).
Which of these points are the mirror images on (i) x-axis (ii) y-axis?

Answer:

Points A(3,3), B(2,4), C(5,5), D(0,2), E(3,-3) and F(-5,-5)
are shown on the graph paper below.



(3,-3) is the mirror image of (3,3) on the x-axis .

Page No 255:

Question 31:

In the given figure, in a ABC, BEAC, ∠EBC = 40° and ∠DAC = 30°. Find the values of x, y and z.
Figure

Answer:



In BCE, CBE + CEB + BCE = 180°       Angle sum property 40°+90°+x° = 180° x° + 130° = 180° x° = 180° - 130°  x° = 50°Therefore,x = 50In  ADC,ADC + ACD + CAD = 180°       Angle sum property ADC + 50° + 30° = 180° ADC =180° - 80° ADC =100°Now,ADB + ADC = 180°          (Linear pair) y° + 100° = 180° y° =180°-100° y° =80°Therefore,y = 80

Let, line segments AD and BE intersect at F.Therefore, side EF of AEF is produced to B.AFB = FAE + AEF   (Exterior angle property)    z°   = 30°+90°             z°   = 120°Therefore, z = 120.



Page No 256:

Question 32:

In the given figure, ABC is a triangle in which AB = AC. D is a point in the interior of ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC.
Figure

Answer:

InBCD,DBC = DCBTherefore, DBC is an isosceles triangle and the opposite sides of equal angles are equal. DB = DC ----1
AB = ACSo, ABC is isosceles triangle and the opposite angles of the equal sides are equal. ABC = ACBand  DBC = DCBABC -DBC= ACB -DCB ABD = ACD ---------2
AndAB = AC ------(3)                  given
From (1), (2) and (3), we have:

ADBADC           [By SAS congruency criteria ] BAD = CADTherefore, AD bisects BAC.

Page No 256:

Question 33:

In the given figure, ABCD is a square and EF is parallel to diagonal DB and EM = FM. Prove that: (i) BF = DE (ii) AM bisects BAD.
Figure

Answer:

i  We have,CD = BC-------1                                   Sides of a square are equal and BDC = DBC = 45°Given: EFDB BDC = FEC         Corresponding angles FEC =  45°Similarly,  EFC =  45°CEF is an isosceles triangle.and CE = CF  -------2From 1 - 2, we get: CD - CE = BC -  CF DE = BFBF = DE

ii In ABF and  ADE,AB= AD        All sides of a square are equalABF = ADE      Both angles are equal to 90°BF = DE        (Proved above)   ABF   ADE       by SAS congruency criteriaBAF = DAE ----3And AF = AE ------4Now, In AME and  AMF,AM = AM      Common sideEM = FM      GivenAE = AF          from 4    AME   AMF      By SSS congruency criteriaEAM = FAM   FAM = EAM -----5From 3 + 5, we get:BAF+FAM = DAE+EAMBAM = DAMTherefore, AM bisects BAD.

Page No 256:

Question 34:

In the given figure, AB || CD. If BAE = 100° and ∠ECD = 120° then x = ?
Figure

Answer:





Draw EFABCD.
Then, AEF+CEF=x°
Now, EFAB and AE is the transversal.
 AEF+BAE=180°    Sum of consecutive interior angles is supplementary AEF+100°=180°AEF=80°
Again, EFCD and CE is the transversal.
CEF+ECD=180°   Sum of consecutive interior angles is supplementaryCEF+120°=180°CEF=60°
Therefore,
x°=AEF+CEF  =80+60°  =140°Therefore, x = 140



Page No 257:

Question 1:

Mark (✓) against the correct answer in each of the following:
An irrational number between 2 and 2.5 is
(a) 3
(b) 2.3
(c) 5
(d) 2.34

Answer:

(c) 5

An irrational number between a and b is ab.

Page No 257:

Question 2:

Mark (✓) against the correct answer in each of the following:
Which of the following is a polynomial in one variable?
(a) x2 + x2
(b) 3x+9
(c) x2+2x-x+3
(d) 3+2x-x2

Answer:

(d) 3+2x-x2

Here, the powers of x in all the terms are non-negative integers.
Therefore, it is a polynomial in one variable.

Page No 257:

Question 3:

Mark (✓) against the correct answer in each of the following:
1(18-32)=?
(a) 2
(b) 12
(c) -2
(d) -12

Answer:

(d) -12

  118 - 32=132-42=1-2= -12

Page No 257:

Question 4:

Mark (✓) against the correct answer in each of the following:
If p(x)=(x4-x2+x), then p12=?
(a) 116
(b) 316
(c) 516
(d) 716

Answer:

(c) 516

px = x4 - x2 +xTherefore,p12 = 124 - 122 +12 p12 =116 - 14 +12  p12 =1 - 4 + 816  p12 =516



Page No 258:

Question 5:

Mark (✓) against the correct answer in each of the following:
If p(x) = x3 + x2 + ax + 115 is exactly divisible by (x + 5) then a = ?
(a) 8
(b) 6
(c) 5
(d) 3

Answer:

(d) 3

p(x) = x3 + x2 + ax + 115 is exactly divisible by (x + 5).
Therefore, p(-5) = 0
 -53 + -52 + a×-5 + 115 = 0 -125 + 25 - 5a + 115 = 0 -5a + 15 = 0 5a = 15 a = 3

Page No 258:

Question 6:

Mark (✓) against the correct answer in each of the following:
The equation of y-axis is
(a) y = 0
(b) x = 0
(c) y = 0
(d) y = constant

Answer:

(b) x = 0

The equation of y-axis is x = 0.

Page No 258:

Question 7:

Mark (✓) against the correct answer in each of the following:
In the given figure, the value of x is
(a) 10
(b) 12
(c) 15
(d) 20
Figure

Answer:

(d) 20

   ACD + BCD = 180°        Linear pair 4x° + 5x° = 180° 9x°= 180° x°= 20°Therefore, x = 20

Page No 258:

Question 8:

Mark (✓) against the correct answer.
In the given figure, CE || BA and EF || CD. If BAC = 40°, ∠ACB = 65° and ∠CEF = x° then the value of x is
(a) 40
(b) 65
(c) 75
(d) 105
Figure

Answer:

(d) 105

CEAB  ACE = BAC = 40°          Alternate anglesBCE = BCA+ ACE = 65° + 40° = 105°Again EFCD and line segment EC intersects them. CEF = BCE           alternate angle x° = 105°Therefore, x = 105.

Page No 258:

Question 9:

Factorise: 2x2+3x+2.

Answer:

2x2 + 3x + 2 = 2x2 +2x + x + 2                             =2xx + 2 +1x+2                             =x + 22 x + 1

Page No 258:

Question 10:

Prove that 5 is an irrational number.

Answer:

We will show that 5 is irrational by contradiction.
Let us assume that 5 is rational.

Therefore we can represent it in the form of ab.
5 = ab
Let us assume that ab is in its lowest form, i.e. a and b are co-prime.
5 = ab a = 5ba2 = 5b2 -----1

Therefore, 5 is the factor of a2 . If 5 is a factor of a2, 5 will also be the factor of a.

So ,we can write a = 5c.

5c2 = 5b225c2 = 5b2b2 = 5c2

Therefore b2 is divisible by 5.
Since, it is divisible by 5, b will also be divisible by 5.

Therefore, a and b have at least 5 as a common factor.

But this contradicts that a and b are co-prime.

This contradiction has arisen because of our incorrect assumption that 5 is rational.

Therefore, 5 is irrational.

Page No 258:

Question 11:

Draw the graph of the equation y = 2x + 3.

Answer:

To draw the graph of y = 2x + 3, make a table of values of x and y satisfying the given equation.
Now, we have


 

   x       y
    0            3
    -32       0

Page No 258:

Question 12:

If x=(3+8), find the value of x2+1x2.

Answer:

   x=3+8 1x=13+8 1x=13+8×3-83-8 1x =3-89-8 1x=3-8
and x+1x=3+8+3-8=6So, x+1x=6Squaring both sides, we get:x+1x2=62x2+1x2+2×x×1x=36x2+1x2+2=36x2+1x2=36-2 x2+1x2=34

Page No 258:

Question 13:

Find the area of the triangle whose sides measure 52 cm, 56 cm and 60 cm.

Answer:

Let the sides of the triangle bea = 52 cmb = 56 cmc = 60 cm s = a + b+ c2 = 52 +56 + 602 = 84 cm

Area of triangle = ss-as-bs-c                             =8484-5284-5684-60                             =84×32×28×24                             =3×28×4×4×2×28×2×2×2×3                             =2×2×2×2×3×3×4×4×28×28                             =2×2×3×4×28                             =1344 cm2

Page No 258:

Question 14:

In the given figure, AB || CD. Find the value of x.
Figure

Answer:



Draw EFABCD.
Now, EFAB and BF is the transversal.
ABF+EFB=180°   Sum of consecutive interior angles is supplementary40°+EFB=180°EFB=140°
Also,
EFCD and FD is the transversal.
CDF+EFD=180°   Sum of consecutive interior angles is supplementary35°+EFD=180°EFD=145°
Therefore,
      x°=EFB+EFD x°=140+145° x°=285°Therefore, x = 285

Page No 258:

Question 15:

Find the values of a and b so that the polynomial (x4 + ax3 − 7x2 + 8x + b) is exactly divisible by (x + 2) as well as (x + 3).

Answer:

Let  px = x4 + ax3 - 7x2 + 8x + bNow as px is exactly divisible by x+2 and x+3, p-2 = 0            and p-3 = 0For p-2 = 0 -24 + a-23 - 7-22 + 8-2 + b = 0 16 -8a -28 -16 + b = 0 -8a + b = 28  ---------1For p-3 = 0 -34 + a-33 - 7-32 + 8-3 + b = 0 81 -27a -63 -24 + b = 0 -27a + b = 6  ---------2Subtracting (2) from (1), we get:  19a = 22Therefore, a = 2219Putting a = 2219 in 1, we get: -8×2219+b = 28 b = 28 + 17619 = 70819

Page No 258:

Question 16:

Using remainder theorem, find the remainder when p(x) = x3 − 3x2 + 4x + 50 is divided by (x + 3).

Answer:

px = x3 - 3x2 + 4x + 50According to remainder theorem, when px is divided by x+3 the remainder is p-3.Now,p-3 =  -33 - 3-32 + 4-3 + 50           = -27 -27 -12 + 50            = -16Therefore, remainder = -16



Page No 259:

Question 17:

Factorise: (2x3 + 54).

Answer:

 2x3 + 54= 2x3 + 27= 2x3 + 33= 2x+3x2 - x×3+ 32=2x+3x2 -3x+ 9

Page No 259:

Question 18:

Find the product (a bc) (a2 + b2 + c2 + ab + ac bc).

Answer:

  a-b-ca2 + b2 + c2 + ab + ac - bc= a+-b+-ca2 + -b2 + -c2 - a-b - a-c - -b-c= a3 + -b3 + -c3 -3a-b-c= a3 - b3 - c3 -3abc

Page No 259:

Question 19:

In ABC, if ∠A − ∠B = 33° and ∠B − ∠C = 18°, find the measure of each angle of the triangle.

Answer:

  A - B = 33° ------1and  B - C = 18° ------2From 1 - 2,  we get: A - B -  B - C  = 33° - 18°A - B - B + C  = 15°A  - 2B + C  = 15°-----3By the angle sum property, we know thatA + B + C = 180°------4From 4 - 3,  we get: A + B + C - A  - 2B + C = 180° -15°A + B + C -A  + 2B -C = 165° 3B = 165° B = 55°Putting  B = 55° in 1, we get:A - 55° = 33°A  = 88°Putting  B = 55° in 2, we get: 55° - C = 18° C = 55° - 18°= 37°Therefore, A  = 88°, B = 55° and  C = 37°

Page No 259:

Question 20:

In the figure, in ABC, the angle bisectors of ∠B and ∠C meet at a point O. Find the measure of ∠BOC.
Figure

Answer:

In  ABC, using angle sum property, we have: CAB + ABC + BCA = 180°70° + ABC + BCA = 180° ABC + BCA = 110° -----1 Now,BO is the angle bisector of ABC. OBC = 12ABCandCO is the angle bisector of BCA. OCB = 12BCAOBC + OCB = 12ABC + BCA-----2In OBC, using angle sum property, we get: BOC + OBC + OCB = 180°Using 2 we get: BOC + 12ABC + BCA = 180°Now, using 1 we get: BOC + 12110° = 180°BOC + 55° = 180°BOC = 180°-55°= 125°Therefore, BOC = 125°

Page No 259:

Question 21:

In the given figure, AB || CD. If BAO = 110°, ∠AOC = 20° and ∠OCD = x°, find the value of x.
Figure

Answer:



Draw OEABCD.
Now, OEAB and OA is the transversal.
OAB+AOE=180°   Sum of consecutive interior angles is supplementaryOAB+AOC+COE=180°110°+20°+COE=180°COE=50°
Also, 
OECD and OC is the transversal.
OCD+COE=180°   Sum of consecutive interior angles is supplementaryx°+50°=180°x°=130°Therefore, x = 130

Page No 259:

Question 22:

In a right-angled triangle, prove that the hypotenuse is the longest side.

Answer:



Let ABC be a right-angled triangle, in which C is the right angle.InABC,ACB = 90°ACB + CBA + BAC = 180°      angle sum property 90° +CBA + BAC = 180° CBA + BAC = 180°-90°  CBA + BAC = 90° So,CBA and BAC are acute angles.CBA <90° and  BAC <90°CBA <ACB and  BAC <ACB AB> AC and AB > BC side opposite to greater angle is greaterTherefore, hypotenuse AB is the longest side.Hence, proved.

Page No 259:

Question 23:

In the given figure, prove that:
x = α + β + γ.
Figure

Answer:



Join BD to produce it to E.
Side BD of triangle ABD is produced to E.
ADE=BAD + ABD----1    (Exterior angle property)
Side BD of triangle BCD is produced to E.
CDE =CBD+BCD----2         (Exterior angle property)
Adding 1 and 2, we get:ADE + CDE =BAD + ABD + CBD +BCD CDA = BAD + BCD+ABD + CBD x° = BAD + BCD+CBA x° = α° + γ°+β° x = α +β+ γ.  

Page No 259:

Question 24:

Find six rational numbers between 3 and 4.

Answer:

Let x=3,y=4,n=6d=y-xn+1=4-36+1=17Then, six rational numbers are : x+d,x+2d,x+3d,x+4d,x+5d,x+6d= 3+17, 3+27, 3+37, 3+47, 3+57, 3+67=227, 237, 247, 257, 267, 277

Page No 259:

Question 25:

If 5+35-3=a+15b, find the values of a and b.
OR
Factorise: (5a-7b)3+(9c-5a)3+(7b-9c)3.

Answer:

5+35-3=a+15bRationalising the denomiantor of the left side, we get: 5+35-3×5+35+3=a+15b 5+3252-32=a+15b52+32 + 25×35 - 3=a+15b5+3+2152=a+15b8+2152=a+15b24+152=a+15b 4+15 =a+15b Comparing the left and right sides, we geta=4, b=1.
                                                                                                                
                                                                                                                OR

5a - 7b3 + 9c-5a3 + 7b-9c3Letx = 5a - 7by = 9c-5az = 7b-9c x+y+z = 0If x+y+z = 0,then x3 + y3 + z3 = 3xyzTherefore,5a - 7b3 + 9c-5a3 + 7b-9c3 =3 5a - 7b9c-5a7b-9c                                                                   

Page No 259:

Question 26:

Factorise: 12(x2+7x)2-8(x2+7x)(2x-1)-15(2x-1)2.

Answer:

12x2 + 7x2 -8x2 + 7x2x-1 -152x-12= 12x2 + 7x2 -18x2 + 7x2x-1 +10x2 + 7x2x-1 -152x-12=6x2 + 7x2x2 + 7x-32x-1 + 5 2x-12x2 + 7x-32x-1=2x2 + 7x-32x-16x2 + 7x+ 5 2x-1=2x2 + 14x -6x + 36x2 + 42x + 10x -5=2x2 + 8x + 36x2 + 52x -5

Page No 259:

Question 27:

If (x3 + ax2 + bx + 6) has (x − 2) as a factor and leaves a remainder 3 when divided by (x − 3), find the values of a and b.

Answer:

Let fx = x3 + ax2 +bx + 6Here, x-2 is a factor of fxTherefore, f2 = 0 23 + a×22 +2b + 6 = 0 8 + 4a +2b + 6 = 0  4a + 2b = -14  2a + b  = - 7 -----1Therefore, when fx is divided by x-3the remainder will be f3 , which is given to be 3. f3 =3  33 +a×32 +3b + 6  =3 27 + 9a + 3b + 6 = 39a + 3b + 33 = 39a + 3b = -30 3a + b = -10Subtracting (1) from (2) , we get: a =  -3Putting a = -3 in  1, we get:2×-3 + b = -7 -6 + b = -7 b = -1Therefore, a = -3 and b = -1.

Page No 259:

Question 28:

Without actual division, show that (x3 − 3x2 − 13x + 15) is exactly divisible by (x2 + 2x − 3).

Answer:

x3 - 3x2 -13x + 15= x3 -5 x2 +2x2 - 10x -3x + 15= x2x-5 +2xx-5 -3x-5=x-5x2 +2x -3Here, x2 +2x -3 is a factor of x3 - 3x2 -13x + 15.Therefore, x3 - 3x2 -13x + 15 is exactly divisible by x2 +2x -3.



Page No 260:

Question 29:

Factorise: a3 b3 + 1 + 3ab.

Answer:

 a3 - b3 + 1+ 3ab= a3 + -b3 +13 -3a-b1=a + -b + 1a2 + -b2+12 -a-b - -b1 - 1a= a-b+1a2+b2 +1+ab+b-a

Page No 260:

Question 30:

In the given figure, AB || CD, ECD = 100°, ∠EAB = 50° and ∠AEC = x°. Find the value of x.
Figure

Answer:



Draw EFABCD.
Now, EFAB and EA is the transversal.
EAB+AEF=180°  Sum of consecutive interior angles is supplementary 50° + AEF=180°  AEF=180° - 50° = 130°
Also,
EFCD and EC is the transversal.
ECD+CEF=180°   Sum of consecutive interior angles is supplementary100°+CEF=180°CEF=80°
Now,
AEC+CEF=AEFx° + 80° = 130°x° =  130° -  80°  = 50°Therefore, x = 50

Page No 260:

Question 31:

Prove that the bisectors of the angles of a linear pair are at right angles.

Answer:




In the figure,ACD and BCD form a linear pair. ACD + BCD = 180°CE and CF bisect ACD and BCD, respectively.  ACD + BCD = 180° ACD2 + BCD 2= 90° ECD + DCF = 90°     CE and CF bisect ACD and BCD respectively ECF = 90° ECF is the angle between CE and CF, which bisect the linear pair of ACD and BCD.Hence, it is proved that the angle bisectors of a linear  pair are at right angles to each other.

Page No 260:

Question 32:

In the given figure, AD bisects BAC in the ratio 1 : 3 and AD = DB. Determine the value of x.
Figure

Answer:

In the figure,

 BAC+108°=180°  [Linear pair ]BAC=72°

Now, divide 72° in the ratio 1 : 3.
a+3a=72°a=18°a=18° and 3a=54°
Hence, the angles are 18o and 54o.
BAD=18° and DAC=54°

Given,
AD=DBDAB=DBA=18°

In ABC, we have:
BAC+ABC+ACB=180°   angle sum property72°+18°+x°=180°x°=90°Therefore, x = 90

Page No 260:

Question 33:

In the given figure, AM BC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, find ∠MAN.
Figure

Answer:

In ABC, using angle-sum property, we have:ABC + BCA + CAB = 180°70° + 20° + CAB = 180°90° + CAB = 180°CAB = 180° - 90°CAB = 90° Now, AN is the angle bisector of A. BAN = 12CAB                  = 12×90°                   = 45°In ABM, using angle-sum property, we have:ABM + BMA + MAB = 180°70° + 90°+ MAB = 180°160°+ MAB = 180° MAB = 180°-160° = 20°From the figure,MAN = BAN - MAB             =  45° - 20°             =  25° Therefore, MAN= 25°

Page No 260:

Question 34:

If the bisector of the vertical angle of a triangle bisects the base, prove that the triangle is isosceles.

Answer:



In ABC,BAD = CADand BD = CDExtend AD to point E so that CE is parallel to AB.Join CE. Taking transversal AE intersecting AB and CE,BAD =DEC    interior alternate angleWe know that BAD = CAD = DECIn ACE,CAD= DECAC= CEIn ABD and ECD,ADB = EDC    vertically-opposite angleBAD = CED     interior alternate angleBD = CD     givenBy AAS, ABDECDTherefore,AB = EC   by cpctWe know EC = AC   already provenTherefore,AB = AC and ABC is isosceles.



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