Download UNIT-1 SQUARE ROOT EXERCISE 1.1.1

UNIT-1
SQUARE ROOT
EXERCISE 1.1.1
1. Find the square root of the following numbers by the factorization
method
(i)
82944
210 x 34
= (25)2 x (32)2
2 82944
2 41472
2 20736
2 10368
2
5184
2
2592
2
1296
2
648
2
324
2
162
3
81
3
27
3
9
3
= 82944 = (25 )2 × (32 )2 = 25 x 32 = 288
(ii)
155236
2 155236
2 77618
197 38809
197 197
1
= 22 X 1972
= 155236
= (2 × 197)2 = 2 X 197 = 394
(iii)
19881
3 19881
3 6627
47 2209
47
= 19881
= 32 × 472
= 3 x 47
= 141
2. Find the square root of the following numbers.
(i)
184.96
184 .96 x 100
100
=
18496
100
=
(2 3 )2 × 17 2
2 18496
2 9248
2 4624
2 2312
2 1156
2 578
289
17 289
17
=
=
=
∴ 184.96
(2 3 )2 × 17 2
10 2
8x17
10
136
10
= 13.6
10 2
(ii)
19.5364
19.5364 =
2 195364
2
97682
221
48841
221
19.5364
=
=
=
195364
10000
( 2 ×221 )2
100 2
442
100
= 4.42
195364
10000
3. Exploration: Find the squares of the numbers 9, 10, 99, 100, 1000, and
9999.Tabulate these numbers. How many digits are there in the squares
of a numbers when it has even number of digits and odd number of
digits?
92
= 81
digits-2
102
= 100
digits-3
992
= 9801
digits-4
1002
= 10000
digits-5
9902
= 998001
digits-6
10002
= 1000000
digits-7
99992
= 99980001
digits-8
They follow the rule
2n if the number has even digits
2n-1 if the number has odd digits
4. If a perfect square A has A digits, how many digits to you expect in 𝐀
If a perfect square A has n digits then, A has
n
2
Digits if n is even
(n +1)
2
Digits if n is odd.
EXERCISE 1.1.2
1. Find the square root of the following number by division method:
(i)
5329
73
7 5329
7 49
143
429
429
0
∴ 5329 = 73
(ii)
18769
137
1 18769
1 1
23
087
3
69
267
1869
1869
0
18769 = 137
(iii)
28224
168
1 2 82 24
1 1
26
182
6
156
328
2624
2624
0
28224 = 168
(iv)
186624
432
4 186624
4 16
83
266
3
249
862
1724
1724
0
186624 = 432
2. Find the least number to be added to get a perfect square.
(i)
6200
78
7 6200
7 49
148 1300
1184
116
6200 > 782
Therefore we find 792
792 = 6241
6241
6200
41
(ii)
12675
112
1 1 26 75
1 1
21
26
1
21
222
575
444
131
∴ 41 is the least number to be added to
get a perfect square.
1132 = 12769
12769
12675
94
94 is the least number to be added to 12675 to make it a
perfect square.
(iii)
88417
297
2 8 84 17
2 4
49 484
9 441
587
4317
4109
208
2982 = 88804
88804
88417
387
∴387 is the least number to be added to 88417 to make it a
perfect square.
(iv)
123456
351
3 12 34 56
3 9
65
334
5 325
701
956
701
255
123456 lies between 3512 and 3522
3512 = 123201
3522 = 123904
123904
123456
448
703 is to be added to make it a perfect square.
3. Find the least number to be subtracted from the following number to
get a perfect square:
(i)
1234
35
3 12 34
3 9
65
334
325
9
9 is the least number to be the least number to be subtracted to make
1234 a perfect square.
(ii)
4321
65
6 43 12
6 36
125 721
525
196
196 should be the least number to be subtracted to make 4321 a
perfect square.
(iii)
34567
185
1 3 45 67
1 1
28
245
8 224
365
2167
1825
342
342 should be the least number to be subtracted to make 34567 a
perfect square.
(iv)
109876
331
3 10 98 76
3 9
63
198
3
189
661
976
661
315
315 should be the least number to be subtracted to make
109876 a perfect square.
4. Find the two consecutive perfect square numbers between which the
following number occurs.
(i)
4567
67
6 45 67
6 36
127 967
889
78
The number 4567 lies between 672 and 682.
672 = 4489
682 = 4624
(ii)
56789
238
2 5 67 89
2 4
43
167
3 129
468
3889
3744
145
56644 lies between the squares of 2382 and 2392.
2382 = 56644
2392 = 57121
(iii)
88888
298
2 8 88 88
2 4
498 488
441
4788
3984
804
88888 lies between the squares of 2982 and 2992.
2982 = 88804
2992 = 89401
(iv)
123456
351
3 12 34 56
3 9
65
334
5 325
701
956
701
255
123456 lies between 3512 and 3522
3512 = 123201
3522 = 123904
5. A person has 3 rectangular plots of dimension 112m x 54m, 84m x 68m
and 140m x 87m. In different places. He wants to sell all of them and
buy a square plot of maximum possible area approximately equal to
them to the sum of these plots. What would be the dimensions of such a
square plot? How much lands he has to be lossed?
The area of the 3 rectangular plots is
(112 x 54) m = 6048 sq. m.
(84 x 68) m = 5712 sq. m.
(140 x 87) m = 12180 sq. m.
Total area = 23940 sq. m.
154
1 2 39 40
1 1
25
139
5 125
304
1440
1216
224
The area of the new square plot is 1542 = 23715 sq. m.
The person would have to loose area of
23940
23716
224 sq. m.
EXERCISE 1.1.3
1. How many digits are there after the decimal points in the following?
(i)
(3.16)2
(3.16)2 = 9.9856
(ii)
(1.234)2
(1.234)2 = 1.5227556
(iii)
Ans: 6
(0.0023)2
(0.0023)2 = 0.00000529
(iv)
Ans: 4
Ans: 8
(1.001)2
(1.001)2 = 1.002001
Ans: 6
2. Given that the numbers below are all squares of some decimal numbers,
how many digits do you expect in their square roots?
(i)
84.8241
84.8241 = 9.21
Ans: 3
(ii)
0.085849
0.085849 = 0.293
(iii)
Ans: 3
1.844164
1.844164= 1.358
(iv)
Ans: 4
0.0089510521
0.089510521= 0.09461
Ans: 5
3. Find the square root of the following number using division method.
(i)
651.7809
2 5. 5 3
2 6 51. 78 09
24
45 251
5 225
505
2678
2525
5103
15309
15309
0
651.7809 = 25.53
(ii)
0.431649
0.6 5 7
0 0.43 16 49
00
06 043
6 36
125
716
5 625
1307
9149
9149
0
1
(iii)
0.431649= 0.657
95.4529
9. 7 7
9 95. 45 29
9 81
187 1445
7 1309
1947
13629
5 13629
0
95.5849= 9.77
(iv)
73.393489
8. 5 6 7
8 73. 39 34 89
8 64
165 939
5 825
1706
11434
6 10236
17127
119889
119889
0
73.393489= 8.567
4. A square garden has an are 24686.6944m2. A trench of one meter wide
has to be dug along the boundary inside the garden. After digging the
trench, what will be the area of the out garden?
Area of the square garden is 24686.6944m2
Length of a side of the garden the is 24686.6644in 157.12m.
If a trench of one meter is dug along the boundary inside the garden the
length of a side of the garden will be (157.12-2) m. = 155.12m.
1z zxdrjhzkgfjh
1m
157.12
1222
The area of the remaining garden is (155.12) = 24062.2144m2
EXERCISE 1.1.4
1. Round off the following numbers to 3 decimal places.
1. 1.5678 ≈ 1.568
2. 2.84671 ≈ 2.847
3. 14.56789 ≈ 14.568
4. 12.987564 ≈ 12.988
5. 3.3333567 ≈ 3.333
2. Find the square root of the following numbers correct to 3 decimal
places.
(i)
12
3. 4 6 4 1
3 12. 00 00 00 00
39
64 300
4 256
686
4400
6 4115
6924
28400
4
27696
69281
80400
69281
11119
12 = 3.4641 ≈ 3.464
(ii)
1.8
1.3 4 1 6
1 1.80 00 00 00
11
23 080
3 69
264 1100
4 1056
2681
4400
1
2681
26826
171900
160956
10644
1.8 = 1.3416 ≈ 1.3416 = 1.342
(iii)
133
1 1. 5 3 2 5
1 1 33. 00 00 00 00
11
21 033
1 21
225 1200
5 1125
2303
7500
3
6909
23062
59100
2
46124
230645
1297600
1153225
144375
133 = 11.5325 ≈ 11.533
(iv)
12.34
3. 5 1 2 8
3 12.34 00 00 00
39
65 334
5 325
701 900
1 701
7022
19900
2
14044
70248
585600
561984
23616
12.34= 3.5128 ≈ 3.513
(v)
8.6666
2. 9 4 3 9
2 8.66 66 00 00
24
49 466
9 441
584 2566
3 2336
5883
23000
3
17649
58869
535100
529821
5279
8.6660 = 2.9439 ≈ 2.944
(vi)
234.234
15 . 3 0 4 7
1 2 34.23 40 00 00
11
25 134
5 125
303 923
3 909
30604
144000
4
122416
306087
2158400
2142609
15791
234.234 = 15.3047 ≈ 15.305
3. Find the square root of the following numbers correct to 4 decimal
places.
(i)
13
3. 6 0 5 5 5
3 13. 00 00 00 00 00
3 9
66 400
6 395
7205
40000
5 36025
72105
397500
5
360525
721105
3697500
3605525
91975
13 = 3.60555 ≈ 3.605
(ii)
8.12
2
2
48
8
564
4
5686
9
56985
5
56990
569906
2. 8 4 9 5 6
8. 12 00 00 00 00
4
412
384
2800
2256
54400
51201
319900
284925
3497500
3497500
3419436
78064
8.12 = 2.8956 ≈ 2.8496
(iii)
3333
5
5
107
7
1147
7
11543
3
115462
2
1154641
1
11546424
5 7. 7 3 2 1 4
33 33. 00 00 00 00 00
25
833
749
8400
8029
37100
34629
247100
230924
16171600
1154641
46295900
46185696
110504
3333 = 57.73214 ≈ 57.7321
4. Find the approximation from below to 4 decimal places to the square
root of the following numbers.
(i)
5
2
2
43
2
443
3
4466
6
447206
6
4472127
2. 2 3 6 0 6 7
5. 00 00 00 00 00 00
1
100
84
1600
1329
27100
26796
3040000
2683236
35676400
31304889
4371511
5 = 2.236067
(2.236067)2 = 4.9996<5< 5.000009 = (2.23607)2
2.236007 is the approximation from below of 5 to 5 decimal places.
(ii)
8
2
2
42
8
562
2
5648
8
56564
4
565682
2
5656847
2. 8 2 8 4 2 7
8. 00 00 00 00 00 00
4
100
384
1600
1124
47600
45184
241600
226256
1534400
1131364
40303600
39597929
705671
8 = 2.828427
(2.82842)2 = 7.999959 < 8 < 8.000016 = (2.828443) 2
2.82842 is the approximation from below of 8 , correct to 5 decimal
places.
5. A square garden has area of 900M2. Additional land measuring equal
area, surrounding it, has been added to it. If the resulting plot is also in
the form of a square, what is its side correct to 3 decimal places?
Ares of square garden is 900m2. If additional land, measuring equal area to
added to the garden, its area will be
(900+900)2 = 1800m2
Side of the new garden is 1800 m.
A = side x side = (side) 2 = 42.43m
4
4
82
2
844
4
8482
2
84846
4 2. 4 2 6
18 00.00 00 00
16
200
164
3600
3376
22400
16964
543600
509076
34524
NON-TEXTUAL QUESTIONS
1. Find the square root of the following numbers.
(i)
19.5364
19.5364 =
195364
10000
2 195364
2 97682
221 48841
221
19.5364
195364
10000
(ii)
(2x221 )2
=
10000
=
442
100
= 4.42
1.993744
1.993744
=
1993744
100000
1993744
1993744
1000000
=
(4 × 353 )2
(100 0 )2
=
1412
1000
= 1.412
2. If a perfect square A has n digits, how many digits do you expect in 𝐀?
If a perfect square A has n digits then, A has
n
2
Digits if n is even
(n + 1)
2
Digits if n are odd.
3. Find the square root
(i)
378225
6 1 5
6 378225
6 36
121 182
1 121
1225
6125
6125
0
378225 = 615
(ii)
923521
9 6 1
9 92 35 21
9 81
186 1135
6 1116
1921 1921
1921
0
923521= 961
4. Find the least number to be added to get the perfect square.
(i)
456123
6 7 5
6 45 61 23
6 36
127 961
7 889
1345 7223
6725
498
456123 > 6752
Therefore we find 6752
6752 = 456123
456976
456123
853
853 should be added to 456123 to make it a perfect
square.
(ii)
7891011
2 8 0 9
2 7 891011
4
48x8 389
384
560x0
510
0
5609x9
51011
50481
530
28102 = 7896100
7896100
7891011
5089
5089 should be added to 7891011 to make it a perfect square.
ADDITIONAL QUESTIONS
1. There are 500 students in a school. For the sports day display they have
to arrange themselves so as to have number of rows equal to number of
columns. How many children would be left out in this arrangement?
Ans: since number of rows should be equal to number of column rows xxx ≈ 500
x2 = 500
22 row and column are possible
22 x 22 = 484 students can perform the drill.
Remaining students 500-484 = 16students
2. Students of class IX wanted to rise the bund to help the needy student of
their school. Each student donated as many rupees as the number of
students in the class and the amount collected was 6044 find the number
of students.
Ans: Let the number of students = x
Then each students will donates = rupees x
Amount collected xxx = x2
x2 = 6044
x = 6044 = 78
No. of students in that class = 78
3. Find the smallest square number that is divisible by each of the number,
8, 15 and 20.
Ans: The smallest number divisible by 8, 15, and 20 is their LCM.
L.C.M of 8, 15, and 20 is 120.
But 120 is not a square number
120 = 2 x 2 x 2 x 3 x 5
22
So to make it perfect square we have to multiply by 2 x 3 x 5
Then
= 2x2 x 2x3x5 x (2x3x5)
= 2x2 x 2x2 x 3x3 x 5x5
= 22
22
32
52
= 3600
3600 is a perfect number that is divisible by 8, 15, and 20.
4. Express 13 as sum of two consecutive integer we have n = 13.
Ans:
n 2 −1
2
n 2 +1
2
=
=
13 2 − 1
2
13 2 + 1
2
=
=
169 − 1
2
169 + 1
2
=
=
168
2
168
2
= 84
= 85
132 = 84+85
169 = 84+85
132 can be expressed as sum of 84 and 85
UNIT-1
MULTIPLICTION OF POLYNOMIALS
EXERCISE 3.1.2
1. Evaluate the following products:
(i)
ax2 ( bx + c)
= ax2 (bx) + ax2 (c)
= abx3 + acx2
(ii)
ab (a+b)
= ab (a) + ab (b)
= a2b + ab2
(iii)
a2b2 (ab2+a2b)
= a2b2 (ab2) + a2b2 (a2b)
= a3b4 + a4b3
(iv)
b4(b6 + b8)
= b4 (b6) + b4 (b8)
= b10 + b12
2. Evaluate the following products:
(i)
(x+3) (x+2)
= (x+3) x + (x+2) x
= x2 + 3x + 2x + 6
= x2 + 5x + 6
(ii)
(x+5) (x–2)
= (x+5) x + (x+5) (–2)
= x2 +5x – 2x –10
= x2 + 3x –10
(iii)
( y–4 ) ( y+6 )
= (y – 4) y +(y – 4)6
= y2 – 4y + 6y – 24
= y2 + 2y – 24
(iv)
(a–5) (a–6)
= (a–5) a + (a–6) (–6)
= a2 – 5a – 6a + 30
= a2 – 11a +30
(v)
(2x+1) (2x–3)
= (2x+1)2x + (2x+1) (–3)
= 4x2 + 2x – 6x – 3x
= 4x2 – 4x –3
(vi)
(a +b)(c +d)
= (a + b) c + (a+b) d
= ac + bc + ad + bd
(vii) ( 2x – 3y ) ( x – y )
= (2x – 3y) x + (2x – 3y) (–y)
= 2x2 – 3xy – 2xy +3y2
= 2x2 – 5xy + 3y2
(viii) ( 𝟕𝐱 + 𝟓 ) ( 𝟓𝐱 + 𝟕 )
= ( 𝟕𝐱 + 𝟓 ) ( 𝟓𝐱) + ( 𝟕𝐱 + 𝟓 ) ( 𝟕)
= 𝟑𝟓 x2 + 5x + 7x + 𝟑𝟓
= 𝟑𝟓x2 + 12x +
𝟑𝟓
(xi)
(2a+3b) (2a–3b)
= (2a+3b) 2a + (2a+3b) (–3b)
= 4a2+ 6ab – 6ab – 9b2
= 4a2 – 9b2
(xii) (6xy–5) (6xy+5)
= (6xy – 5) (6xy) + (6xy – 5) 5
= 36x2y2 – 30xy + 30xy – 25
= 36x2y2 – 25
2
(xiii)
x
2
=
=
=
2
+3
x
4
x2
4
x2
x
2
+3
x
6
14
x
x
+ –
8
– –21
x
–7
+
– 21
2
x
+ 3 (– 7 )
3. Expand the following using appropriate identity:
(i)
(a +5)2
Using (a + b)2 = a2 +2ab +b 2 we get
a=a
(a +5)2
b=5
= a2 +2.a.5 +b 2
= a2 +10a +25
(ii)
(2a +3)2
Using (a + b)2 = a2 +2ab +b 2 we get
a = 2a
b=3
(2a +3)2 = (2a)2 +2.2a.3 +b 2
= 4a2 + 12a + 9
(iii)
𝟏
( x + )2
𝐱
Using (a + b)2 = a2 +2ab +b 2 we get
a = 2a
b=
1
x
1
1
1
x
x
x
(X + )2 = x2 + 2.x. + (
= x2 + 2 +
𝟏
𝐱𝟐
)2
(iv)
( 12a + 6b )2
Using (a + b)2 = a2 +2ab +b 2 we get
a = 12a and b = 6b
12a + 6b )2= ( 12a)2 + 2. 12 a + 6b + ( 6b)2
= 12a2 +2 72 ab +6b 2
= 12a2 +2 36 × 2 ab +6b 2
= 12a 2 + 12 𝟐 ab +6b2
(v)
(𝛑 +
𝟐𝟐 2
)
𝟕
Using (a + b)2 = a2 +2ab +b 2 we get
a = 𝛑 and b =
(𝛑 +
22 2
)
7
= 𝛑2 + 2. 𝛑
= 𝛑2 +
44 π
= 𝛑2 +
𝟒𝟒𝛑
7
𝟕
22
7
22
7
22
+ ( )2
7
22
+ ( )2
7
+
𝟒𝟖𝟒
𝟒𝟗
(vi)
(y – 3)2
Using (a – b)2 = a2 – 2ab + b 2
a = y and b = –3
(y – 3)2 = y2 – 2. y.3 + 32
= y2 – 6y + 9
(vii) (3a – 2b)2
Using (a – b)2 = a2 – 2ab + b 2
a = 3a and b = –2b
(3a – 2b)2 = (3a)2 – 2.3a.2b + (2b)2
= 9a2 – 12ab + 4b2
𝟏
(viii) ( y – )2
𝐲
Using (a – b)2 = a2 – 2ab + b 2
a = y and b =
1
y
1
1
y
y
(y – )2 = y2 – 2.y. + (
= y2 – 2 +
𝟏
𝐲𝟐
1 2
)
y
(ix)
( 10x – 5y)2
Using (a – b)2 = a2 – 2ab + b 2
a = 10xand b = 5y)
( 10x – 5y)2 = ( 10x)2 – 2. 10x. 5y)+ ( 5y))2
= 10x2 – 2 50 xy + 5y2
= 10x2 – 2.5 2 xy + 5y2
= 10x2 – 10 2 xy + 5y2
(x)
(𝛑 –
𝟐𝟐 2
)
𝟕
Using (a + b)2 = a2 +2ab +b 2 we get
a = 𝛑 and b =
(𝛑 –
22 2
)
7
22
7
= 𝛑2 – 2. 𝛑
22
= 𝛑2 –
44 π
22
= 𝛑2 –
𝟒𝟒𝛑
7
𝟕
7
+(
+ ( )2
7
+
𝟒𝟖𝟒
𝟒𝟗
22
7
)2
(xi)
(2x+3) (2x+5)
Using (x + a) (x + b) = x2 + x (a + b) ab we get
x = 2x, a = 3 and b = 5
(2x+3) (2x+5) = (2x)2 +2x (3 + 5) + 3.5
= 4x2 +16x +15
(xii) (3x – 3) (3x + 4)
Using (x + a) (x + b) = x2 + x (a + b) ab we get
x = 3x, a = –3 and b = 4
(3x – 3) (3x + 4) = (3x)2 + 3x [(–3)+(4)] + (-3)4
= 9x2 + 3x –12
= 9x2 + 3x –12
4. Expand :
(i)
(x + 3 ) (x – 3)
Using (a + b) (a – b) = a2 – b2 we get
a = x, b = 3
(x + 3) (x – 3) = x2 – 32
= x2 – 9
(3x – 5y) (3x + 5y)
(ii)
Using (a + b) (a – b) = a2 – b2 we get
a = 3x, b = 5y
(3x – 5y) (3x + 5y) = (3x)2 – (5y)2
= 9x2 – 25y2
(iii)
x
3
+
y
x
2
3
+
y
2
Using (a + b) (a – b) = a2 – b2 we get
x
y
3
2
a= ,b=
x
3
+
y
x
2
3
+
y
2
=
(iv)
=(
𝐱𝟐
𝟗
x 2
)
3
+
-(
y
2
)2
𝐲𝟐
𝟒
(x2 + y2) (x2 – y2)
Using (a + b) (a – b) = a2 – b2 we get
a = x2, b = y2
(x2 + y2) (x2 – y2) = (x2)2 – (y2)2
= x4 – y4
(v)
(a2 + 4b2) (a + 2b) (a - 2b)
Using (a + b) (a – b) = a2 – b2 for 2nd and 3rd term we get
(a2 + 4b2) (a + 2b) (a - 2b) = (a2 + 4b2) [a2 – (2b)2]
= (a2 + 4b2) (a2 – 4b2)
Using the above identity once again we get
= (a2)2 – (4b2)2
= a4 – 16b4
(vi)
(x – 4) (x + 4) (x – 3) (x + 4)
Using (a + b) (a – b) = a2 – b2 we get
(x – 4) (x + 4) (x – 3) (x + 4) = (x2 – 42) (x2 – 32)
= (x2 – 16) (x2 – 9)
Using (a + b) (a – b) = x2 – x (a + b) + ab
= (x2)2 – x (16+9) +16.9
= x4 – 25x2 +144
𝟏
(vii) (x – a) (x + a)
1
(x2 – a2)
x2 x
1–
1+
1
x2
x2
a2
𝐱𝟐
𝐚𝟐
x2
+
– x2 x
−
−
a2
x2
𝐱
−
𝟏
𝟏
𝐚
𝐱
+
𝟏
𝐚
1
a2
1
a2
– a2 x
1
x2
+ a2 x
+1
𝐚𝟐
𝐱𝟐
5. Simplify the following:
(i)
(2x – 3y)2 + 12xy
= (2x)2 + (3y)2 – 2.2x.3y + 12xy
= 4x2 + 9y2 -12xy +12xy
= 4x2 + 9y2
(ii)
(3m + 5n)2 – (2n)2
= (3m)2 + (5n)2 + 2.3m.5n – 4n2
= 9m2 + 25n2 + 30mn – 4n2
= 9m2 + 30 mn +21n2
1
a2
(iii)
(4a – 7b)2 – (3a)2
= (4a)2 – 2.4a.7b + (7b)2 – (3a)2
= 16a2 – 56ab + 49b 2 - 9a2
= 7a2 -56ab + 49b2
(iv) (x +
𝟏 2
)
𝐱
(m +
𝟏
𝐦
1
= (x2 + 2. x. +
x
= x2 + 2 +
= x2 + 2 +
1
x2
1
x2
= x2 – m2+
(v)
)2
1 2
)
x2
– (m2 + 2. m.
– (m2 + 2 +
– m2 + 2 –
𝟏
𝐱𝟐
–
𝟏
𝐦𝟐
1
m2
1
m2
+4
(m2 + 2n2)2 – 4m2n2
= m4 +2m4.2n2 +4n4 – 4m2 n2
= m4+ 4m2n2 + 4n2 – 4m2n2
= m4 – 4n2
)
1
m
+
1 2
)
m2
(vi)
(3a – 2)2 – (2a -3)2
= (9a2 – 2.3a.2 + 22) – (4a2 – 2.3a.2 + 92)
= 9a2 –12a + 4 – 4a2 + 12a – 9
= 5a2 – 5
=5(a2 – 1)
EXERCISE 3.1.3
1. Find the following products:
(i)
(x + 4) (x + 5) (x + 2)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
a = 4, b = 5 and c =2
(x + 4) (x + 5) (x + 2) = x3 + x2(4 + 5 + 2) + x (4.5 + 5.2 + 2.4) + 4.5.2
= x3 + 11x2 + x (20 + 10 + 8) +40
= x3 + 11x2 + 38x +40
(ii)
(y + 3) (y + 2) ( y – 1)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = y, a = 3, b = 2 and c = -1
(y + 3) (y + 2) (y – 1) = y3 + y2 (3 + 2 – 1) + y (3.2 + 2(-1) + (-1)3 + 3.2(-1)
= y3 + 4y2 + y (6 – 2 – 3) – 6
= y3 + 4y2 + y – 6
(iii)
(a + 2) (a – 3) (a + 4)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = a, a = 2, b = –3 and c = 4
(a + 2) (a – 3) (a – 4) = a3 + a2 (2 – 3 + 4) + a [2(–3) + (–3)4 + 4.2) + 2 (–3) 4
= a3 + 3a2 + a (–6 – 12 + 8) – 24
= a3 + 3a2 – 10a – 24
(iv)
(m – 1) (m – 2) (m – 3)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = m, a = –1, b = –2 and c = –3
(m – 1) (m – 2) (m – 3) = m3 + m2 (–1 – 2 – 3) + m [(–1) (–2) + (–2) (–3) +
(– 3)(–1)] + (–1) (–2) (–3)
= m3 + m2 (–6) + m [2 + 6 + 3] – 6
= m3 - 6m2 + 11m – 6
(v)
( 𝟐 + 𝟑) ( 𝟐+ 𝟓) ( 𝟐+ 𝟕 )
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = 2, a = 3, b = 5 and c = 7
( 2 + 3) ( 2+ 5) ( 2+ 7 ) = ( 2)3 + ( 2)2
3. 5 +
=2 2+2
3) +
7 + 2
5. 7 + 7. 3 + 3. 5. 7
3) +
5 +
15 + 35 +
(vi)
5+
7 + 2
21 + 105
105 x 101 x 102
We can write this as
(100 + 5) (100 + 1) (100 + 2)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = 100, a = 5, b = 1 and c = 2
(100 + 5) (100 + 1) (100 + 2) = 1003 + 1002 (5 + 1 + 2) + 100 (5.1 + 1.2
+ 2.5) + 5.1.2
= 1000000 + 10000 (8) + 100(5 + 2 +10) + 10
= 1000000 + 80000 + 1700 +10
= 1081710
(vii) 95 x 98 x 103
We can write this as
(100 - 5) (100 - 2) (100 + 3)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = 100, a = -5, b = -2 and c = 3
(100 – 5) (100 – 2) (100 + 3) = 1003 + 1002 (–5 – 2 +3) + 100(–5) (–2) + (–2) 3
+ 3 (-5) + (-5) (-2) 3
= 1000000 + 10000(–4) +100 (10 – 6 –16) + 30
= 1000000 – 40000 – 1100 +30
= 958930
(viii) 1.01 x 1.02 x 1.03
We can write this as
(1 + 0.01) (1 + 0.02) (1 + 0.03)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = 1, a = 0.01, b = 0.02 and c = 0.03
(1 + 0.01) (1 + 0.02) (1 + 0.03) = 13 + 12 (0.01 + 0.02 + 0.03) + 1 [(0.01) (0.02) +
(0.02) (0.03) + (0.03) (0.01)] + (0.01) (0.02) (0.03)
= 1 + 0.06 + (0.0002 + 0.0006 + 0.0003) + 0.000006
= 1.061106
2. Find the coefficients of x2 and x in the following:
(i)
(x + 4) (x + 1) (x + 2)
= x3 + x2 (4 + 1 + 2) + x (4.1 + 1.2 + 2.4) + 4.1.2
= x3 + 7x2 + 14x + 8
Coefficient of x2 is 7
Coefficient of x is 14
(ii)
(x – 5) (x – 6) (x – 1)
= x3 + x2 (–5 – 6 –1) + x [(–5) (–6) + (–6) (–1) – 1(–5)] + (–5) (–6) (–1)
= x3– 12x2 + x (30 + 6 + 5) – 30
= x3– 12x2 + 41x – 30
Coefficient of x2 is –12 and x is 41
(iii)
(2x + 1) (2x – 2) (2x – 5)
= (2x)3 + (2x)2 [1–2–5] + 2x [(1)(–2) + (–2)(–5) + (–5)(1)] + 1 (–2) (–5)
= 8x3 + 4x2 (–6) + 2x [–2 +10–5] + 10
= 8x3 – 24x2 + 6x + 10
Coefficient of x2 is –24 and x is 6
𝐱
𝐱
𝐱
𝟐
𝟐
𝟐
(iv) ( + 1) ( + 2) ( + 3)
x
x
x
2
2
2
= ( )3 + ( )2 [1 + 2 + 3] +
=
=
x3
8
x3
8
+
x2
4
[1.2 + 2.3 + 3.1] + 1.2.3
x
(6) + (2 + 6 + 3) + 6
2
3
11
2
2
+ x2 +
x+6
𝟑
𝟏𝟏
𝟐
𝟐
Coefficient of x2 is and x is
3. The length, breadth and height of a cuboid are (x +3), (x - 2) and (x -1)
respectively. Find its volume.
Volume of a cuboid = length x breadth x height
V = (x +3) (x – 2) (x – 1)
Using the identity (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca)
+ abc we get
V = x3 + x2 (3 – 2 – 1) + x [3(–2) + (–2) (–1) + (–1)3] + 3(–2) (–1)
= x3 - 0x2 + X (–6 + 2 – 3) + 6
= x3 - 7x2 + 6
4. The length, breadth and height of a metal box are cuboid are (x +5), (x –
2) and (x – 1) respectively. What is its volume?
Volume of the metal box = length x breadth x height
V = (x +5) (x – 2) (x – 1)
Using the identity (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca)
+ abc we get
V = x3 + x2 + (–5 – 2 –1) + x [5(–2) + (–2) (–1) + (–1)5] + 5 (–2) (–1)
= x3 + 2x2 + x [–10 + 2 – 5] + 10
= x3 + 2x2 – 13x 10
5. Prove that
(a + b) (b + c) (c + a) = (a + b + c) (ab + bc + ca) – abc
[Hint: write a + b = a + b + c – c, b + c = a + b + c – a, c + a = a + b + c – d]
x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get
L. H.S. = (a + b + c)3 + (a + b + c)2 (–c – a – b) + (a + b + c) [(–c) (–a) +
(–a) (–b) + (–b) (–c)] – (–c) (–a) (–b)
= (a + b + c)3-(a + b + c)2[(a + b + c)] +(a + b + c) (ac + ab + bc) – abc
= (a + b + c)3- (a + b + c)3 + (a + b + c) (ac + ab + bc) – abc
= (a + b + c) (ac + ab + bc) – abc
= R. H. S.
6. Find the cubes of the following:
(i)
(2x +y)3
Using (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get
a = 2x, b = y
(2x +y)3 = (2x)3 + 3(2x)2 y + 3 (2x) y2 +y3
= 8x3 + 12 x2y + 6 xy2 +y3
(ii)
(2x + 3y)3
Using (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get
a = 2x, b = 3y
(2x + 3y)3 = (2x)3 + 3(2x)2 (3y) + 3 (2x) (3y)2 + (3y)3
= 8x3 + 36 x2y + 54 xy2 + 27y3
(iii)
(4a + 5b)3
Using (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get
a = 4a, b = 5b
(4a + 5b)3 = (4a)3 + 3(4a)2(5b) + 3(4a)(5b)2 + (5b)3
= 64a3 + 240a2b + 300ab 2 + 125b 3
𝐱
(iv) ( x + )3
𝟏
Using (a + b)3 = a3 + 3a2 b + 3ab2 + b3 we get
a = x, b =
x
1
x
x
1
1
(x + )3 = x3 + 3x2
= x3 + 3x +
= x3 + 3x +
(v)
x
x
1
1
+ 3x ( )3 + ( )3
3x
1
x
x3
3
x
2 +
+
1
x3
233
We write this as (20 + 3)3
Using identity (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get
a = 20, b = 3
(20 + 3)3 = (20)3 + 3 (20)2(3) + 3(20) 32 + 33
= 8000 + 3600 + 540 + 27
= 12167
(vi)
513
We write this as (50 + 1)3
Using identity (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get
a = 50, b = 1
(50 + 1)3 = (50)3 – 3 x (50)2 x 1 + 3(50) (1)2 + 13
= 125000 + 7500 + 150 +1
= 132651
(vii) 1013
We write 101 as (100 + 1)3
Using identity (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get
a = 100, b = 1
(100 + 1)3 = 1003 + 3. 1002 + 3. 100.12 + 13
= 1000000 + 30000 + 300 + 1
= 1030301
(viii) 2.13
We write 2.1 (2 + 0.1)3
Using identity (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get
a = 2, b = 0.1
(2 + 0.1)3 = 23 + 3 x 22(0.1) + 3 x 2 x (0.1)2 + (0.1)3
= 8 + 1.2 + 0.06 + 0.001
= 9.261
7. Find the cubes of the following:
(i)
(2a – 3b)3
Using (a – b)3 = a3 – 3a2b + 3ab 2 – b3 we get
a = 2a, b = 3b
(2a – 3b)3 = (2a)3 – 3 (2a)2(3b) + 3 (2a)(3b)2 – (3b)3
= 8a3 – 36a2b + 54ab2 -27b
(ii)
𝟏
( x – )3
𝐱
Using (a – b)3 = a3 – 3a2b + 3ab 2 – b3 we get
a = x, b =
1
x
1
1
x
x
(x – )3 = x3 – 3x2
(iii)
1
1
x
x
+ 3x( )3 – ( )3
= x3 – 3x +
3x
= x3 – 3x +
3
x2
x
–
–
1
x3
1
x3
(√3 x – 2)2
Using (a – b)3 = a3 – 3a2b + 3ab 2 – b3 we get
a = √3 x, b = 2
(√3 x – 2)2 = (√3x)3 – 3 (√3x)2 .2 + 3. √3x x 22 – 23
= 3√3 x3 – 6. 3 x2 + 12√3 x – 8
=3√3 x3 – 18x2 + 12√3 x – 8
(iv)
(2x – √5)3
Using (a – b)3 = a3 – 3a2b + 3ab 2 – b3 we get
a = 2x, b = √5
(2x – √5)3 = (2x)3 – 3(2x)2 √5 + 3. 2x. √5)2 – (√5)3
= 8x3 – 12√5x2 + 30x – 5√5
(v)
493
We can write 49 = 50 – 1
Using (a – b)3 = a3 – 3a2b + 3ab 2 – b3 we get
a = 50, b = 1
50 – 1)3 = 503 – 3.502.1 + 3.50.12 – 13
= 125000 – 3 x 2500 + 150 – 1
= 125000 – 7500 +149
= 117649
(vi)
183
We wrote 18 = 20 – 2
Using (a – b)3 = a3 – 3a2b + 3ab 2 – b3 we get
a = 20, b = 2
(20 – 2)3 = 203 – 3.202.2 + 3.20.22 – 23
= 8000 – 6x400 + 60x4 - 8
= 8000 – 2400 +240 – 8
= 5832
(vii) 953
We write 95 = 100 – 5
Using (a – b)3 = a3 – 3a2b + 3ab 2 – b3 we get
a = 100, b = 5
(100 – 5)3 = 1003 – 3.1002.5 + 3.100.52 – 53
= 1000000 – 150000 + 7500 – 125
= 857375
(viii) 1083
We write 1083 = (110 – 2)
Using (a – b)3 = a3 – 3a2b + 3ab 2 – b3 we get
a = 110, b = -2
(110 – 2)3 = 1103 – 3. (110)2.2 + 3.110x22 – 23
= 1331000 – 72600 + 1320 -8
= 1259712
𝟏
𝟏
𝐱
𝐱𝟑
8. If x + = 3, prove that x3 +
= 18.
1
Given x + = 3
x
Cubing both sides we get
1
(x + )3 = 33
x
Using (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get
a=x
b=
1
x
1
1
1
1
x
x
x
x
(x + )3 = (x)3 + ( )3 + 3x. (x + )
= 27 = x3 +
= x3 +
= x3 +
1
x3
1
x3
1
x3
+ 3 (3)
= 27 – 9
= 18
9. If p + q = 5 and pq = 6, find p3 + q3
(p + q)3 = p3 + 3pq (p + q) + q 3
53 = p3 + 3.6 (5) + q 3
125 = p 3 + 90 + q 3
p3 + q3 = 125 – 90
p3 + q3 = 35
10. If a – b = 3 and ab = 10, find a 3 – b3
Given a – b = 3 and ab = 10
(a – b)3 = a3 – b3 - 3ab (a – b)
33 = a3 – b3 – 3.10 (3)
27 = a3 – b3 – 90
a3 – b3 = 27 + 90
a3 – b3 = 117
11. If a2 +
𝟏
𝐚𝟐
= 20 and a 3 +
a3 +
1
a3
𝟏
𝐚𝟑
= 30, find a +
1
1
a
a2
= (a + ) (a2 +
1
30 = (a + ) = (20 – 1)
a
1
30 = (a + ) x 19
a
30
19
= a+
1
a
1
𝟑𝟎
a
𝟏𝟗
a+ =
1
-ax )
a
𝟏
𝐚
EXERCISE 3.1.4
1. Expand the following:
(i)
(a + b + 2c)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
a = a, b = b and c = 2c
(a + b + 2c)2 = a2 + b2 + (2c)2 + 2ab + 2b (2c) + 2 (2c)a
= a2 + b2 + 4c 2 + 2ab + 4bc + 4ca
(ii)
(x + y + 3z)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
a = x, b = y and c = 3z
(x + y + 3z)2 = x2 + y2 + (3z)2 + 2.x.y + 2y(3z) + 2.(3z)x
= x2 + y2 + 9z2 + 2xy + 6yz + 6zx
(iii)
(p + q - 2r)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
a = p, b = q and c = -2r
(p + q - 2r)2 = p2 + q2 + (-2r)2 + 2.p.q + 2q(-2r) +2(-2r)p
= p2 + q2 + 4r2 + 2pq – 4pr – 4pr
(iv)
a
b
2
2
( +
c
+ )2
2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
a
b
2
2
a= ,b=
a
b
2
2
( +
c
and c =
2
c
a
b
c
a
b
2
2
2
2
2
2
+ )2 =( )2 + ( )2 + ( )2 + 2 ( ) (
c
a
2
2
2( )(
=
(v)
a2
4
+
b2
4
+
c2
4
b
c
2
2
)+2( )( )+
)
+
ab
2
+
bc
2
+
ca
2
(x2 + y2 + z)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
a = x2, b = y2 and c = z
(x2 + y2 + z)2 = (x2)2 + (y2)2 + (z)2 + 2x2y2 + 2y2z +2zx2
= x4 + y4 + z2 + 2x2y2 + 2y2z +2zx2
(vi)
(m – 3 -
1
m
)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
a = m, b = -3 and c = (m – 3 -
1
m
)2 = m2 + (-3)2 + (
= m2 + 9 +
1
m2
1
m
1
m
)2 + 2.m(-3) + 2(-3)(-
- 6m +
6
m
-2
1
m
) + 2 (-
1
m
)m
= m2 +
1
m2
+
6
m
– 6m + 7
(vii) (-a + b – c)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
a = –a
c = –c
b=b
(-a + b – c)2 = (–a)2 + b2 + (–c)2 + 2(–a)b + 2b(–c) +2(–c)a
= a2 + b2 + c 2 – 2ab – 2bc +2ca
(viii) (x + 5 +
1
2x
)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
a=x
(x + 5 +
1
2x
b=5
c=
1
2x
1
1
2x
2x
)2 = x2 + 52 + ( )2 + 2.x.5 + 2.5.
= x2 + 25 +
= x2 +
1
4x 2
1
4x 2
5
+10x + + 1
x
5
+10x + + 26
x
1
+ 2( )x
2x
2. Simplify the following:
(i)
(a – b + c)2 – (a – b – c)2
Using (a + b + c)2 = a2 + b2 + c 2 + 2ab + 2bc +2ca we get
(a – b + c)2 – (a – b – c)2
= [ a2 + (-b)2 + c 2 + 2a(-b) + 2(-b)c + 2ca ] – [a + (-b)2 + (–c)2 + 2a(–b)
+ 2(–b)(–c) + 2(–c)a]
= a2 + b2 + c 2 – 2ab – 2bc +2ca – [a2 + b2 + c2 – 2ab + 2bc –2ca]
= a2 + b2 + c 2 – 2ab – 2bc +2ca – a2 – b2 – c2 + 2ab – 2bc +2ca
= 4ac – 4bc
= 4c (a – b)
(ii)
(3x + 4y + 5)2 – (x + 5y – 4)2
Using (a + b + c)2 = a2 + b2 + c 2 + 2ab + 2bc +2ca we get
(3x + 4y + 5)2 – (x + 5y – 4)2
= [(3x)2 + (4y)2 + 52 + 2.3x.4y + 2.4y.5 + 2.5(3x)] – [x2 + (5y)2 + (-4)2 +
2. X.5y + 2.5y (-4) + 2 (-4). x]
= 9x2 + 16y2 + 25 + 24xy + 40y + 30x – [x2 + 25y2 + 16 + 10xy – 40y-8x
= 9x2 + 16y2 + 25 + 24xy + 40y + 30x - x2 - 25y2 - 16 - 10xy + 40y + 8x
= 8x2 – 9y2 – 14xy + 80y + 38x + 9
(iii)
(2m – n - 3p)2 + 4mn - 6np + 12pm
Using (a + b + c)2 = a2 + b2 + c 2 + 2ab + 2bc +2ca we get
(2m – n - 3p)2 + 4mn - 6np + 12pm
= (2m)2 + (–n)2 + (-3p)2 + 2.2m(–n) + 2(n)(–3p) + 2 (-3p)(2m) + 4mn –
6np + 12pm
= 4m2 + n2 + 9p2 – 4mn – 6np – 12pm + 4mn – 6np + 12pm
= 4m2 + n2 + 9p2
(iv)
(x + 2y + 3z + r)2 + (x + 2y + 3z – r)2
Using (a + b + c)2 = a2 + b2 + c 2 + 2ab + 2bc +2ca we get
a = (x + 2y)
b = 3z
c=r
(x + 2y + 3z + r)2 + (x + 2y + 3z – r)2
= (x + 2y)2 + (3z)2 + r2 + 2 (x + 2y)3z + 2.3z.r + 2.r.(x + 2y) + (x + 2y)2 +
(3z)2 – (r)2 + 2 (x + 2y) 3z + 2.3z (-r) + 2 (-r)(x + 2y)
= 2(x + 2y)2 + 9z2 + r2 + 6 (x + 2y)z + 6zr + 2r (x + 2y) + 9z2 + r2 +
6 (x + 2y) z - 6zr – 2r (x + 2y)
= 2(x2 + 2.x.2y +4y2) + 18z2 + 2r2 + 12 (x + 2y)z
= 2x2 + 8xy + 8y2 + 18z2 + 2r2 +12xz + 24 yz
= 2x2 + 8y2 + 18z2 + 2r2 + 8xy +12xy + 24 yz
3. If a + b + c = 12 and a 2 + b2 + c 2 = 50, find ab + bc + ca.
Given a + b + c = 12 squaring both sides
(a + b + c)2 = 122
a2 + b2 + c 2 + 2ab + 2bc + 2ca = 144
Given a2 + b2 + c 2 = 50
50 + 2ab + 2bc + 2ca = 144
2(ab + bc + ca) = 144 – 50
2(ab + bc + ca) = 94
ab + bc + ca =
94
2
ab + bc + ca = 47
4. If a2 + b2 + c 2 = 35 and ab + bc + ca = 23, find all possible values of
a + b +c.
Given a2 + b2 + c 2 = 35 and ab + bc + ca = 23,
(a + b +c)2 = a2 + b2 + c 2 + 2ab + 2bc + 2ca
= a2 + b2 + c2 + 2(ab + bc + ca)
= 35 + 2 (23)
= 35 + 46
(a + b + c)2 = 81
(a + b +c) = ± 81
a + b +c = ± 9
5. Express 4x + 9y + 16z + 12xy – 24yz – 16zx as the square of a trinomial
Using and comparing the coefficient of
(a +b+c)2 = a2 + b2 + c 2 + 2ab + 2bc +2ca we get
4x + 9y + 16z + 12xy – 24yz – 16zx
= (2x)2 + (3y)2 + (–4z2) + 2.2x.3y.(–4z) + 2.(–4z) + 2.(–4z).2x
= (2x + 3y – 4z)2
6. If x and y are real numbers and satisfy the equation
(2x + 3y – 4z)2 + (5x – y - 4)2 = 0 find x, y.
[Hint: If a, b are real numbers such that a2 + b2 = 0, then a = b = 0.]
Given if a2 + b2 = 0, then a = b = 0
(2x + 3y – 4z)2 = 0 and (5x – y - 4)2 = 0
2x + 3y = 5 x 1
5x – y = 4 x 3
2x + 3y = 5
2x + 3y = 5
21 + 3y = 5
15x – 3y = 12
2 + 3y = 5
17x
3y = 5 – 2
= 17
x=
17
3y = 3
17
3
x=1
y= =1
3
x=1
y=1
EXERCISE 3.1.5
I.
If a + b + c = 0, prove the following:
(i)
(b + c) (b – c) + a (a + 2b) = 0
Given a + b + c = 0 the a + b = -c, b + c = -a, c + a = -b we have
L.H.S = (b + c) (b – c) + a (a + 2b)
= (-a) (b – c) + a (a + b + b)
= -ab + ac + a (-c + b)
= -ab + ab - ac + ac
= 0 = R.H.S
(ii)
a (a2 – bc) + b (b2 – c) + c (c2 – ab) = 0
L.H.S = a (a2 – bc) + b (b 2 – c) + c (c2 – ab)
= a3 – abc + b 3 - abc + c 3 – abc
= a3 + b3 + c 3 – 3abc
We know that if a + b + c = 0 then
a3 + b3 + c 3 = 3abc
Hence we have
= 3abc – 3abc
= 0 = R.H.S
(iii)
a (b2 + c 2) + b (c 2 + a2) + c (a 2 + b2) = –3abc
L.H.S = a (b 2 + c 2) + b (c 2 + a2) + c (a2 + b2)
= ab2 + ac 2+ bc 2 + ba2 + ca2 +cb 2
= ab2 + ba2 + b2c + bc 2 + ac 2 + a2c
= ab (a + b) + bc (b + c) + ac (a + c)
= ab (–c) + bc (–a) + ac (–b)
= –abc – abc – abc
= –3abc = R.H.S
[a + b + c =0, a + b = -c, b + c = -a, c + a = -b]
(iv)
(ab + bc + ca)2 = a2b2 + b2c2 + c 2a2
L.H.S = (ab + bc + ca)2
= (ab)2 + (bc)2 + (ca)2 +2ab.bc + 2bc. ca + 2ca.ab
= a2b2 + b2c2 + c2a2 + 2ab 2c + 2bc 2a + 2ca2b
= a2b2 + b2c2 + c2a2 + 2abc + (0)
= a2b2 + b2c2 + c 2a2 = R.H.S
(v)
a2 – bc = b2 – ca = c 2 – ab = - (ab + bc + ca)
a. a – bc
b2 – ca
c2 – ab
a (–b – c) – bc
b. b – ca
c. c – ab
b(–c – a) – ca
c(–a – b) – ab
–ab – ac – ba
–bc – ab – ca
–ac – bc – ab
– (ab + bc + ca)
– (ab + bc + ca)
– (ab + bc + ca)
……… (1)
……. (2)
From equation (1) (2) and (3)
a2 – bc = b 2 – ca = c 2 – ab = - (ab + bc + ca)
(vi)
2a2 + bc = (a – b) (a –c)
L.H.S = 2a2 + bc
= a2 + a2 + bc
= a2 + a x a + bc
= a2 + a (- b – c) + bc
= a2 – ab – ac + bc
= a (a – b) – c (a - b)
= (a – b) (a –c) = R.H.S
………(3)
(vii) (a + b) (a – b) + ca - cb = 0
We have
a+b+c=0
a+b=c
L.H.S = (a + b) (a – b) + ac – cb
= –c (a – b) + ac – cd
= – ca + bc + ac – cb
= 0 = R.H.S
(viii) a2 + b2 + c 2 = -2(ab + bc + ca)
We have
a+b+c=0
Squaring we get
(a + b + c)2 = 0
a2 + b2 + c 2 + 2ab + 2bc +2ca = 0
a2 + b2 + c 2 = – 2ab – 2bc – 2ca
a2 + b2 + c 2 = – 2(ab + bc + ca)
Hence the proof
2.
Suppose a, b, c are non-zero real numbers such that a + b + c = 0,
Prove the following:
(i)
𝐚𝟐
𝐛𝐜
+
𝐛𝟐
𝐜𝐚
𝐜𝟐
+
𝐚𝐛
L.H.S =
=
=
=3
a2
+
bc
b2
ca
+
c2
ab
a 2 .a+b 2 .b +c 2 .c
abc
a 3 +b 3 +c 3
abc
………. (1)
We have a + b + c = 0, a + b = –c
Cubing we get
(a + b)3 = (–c)3
a3 + b3 + 3ab + (a + b) = –c3
a3 + b3 – 3ab = –c 3
a3 + b3 + c 3 = 3abc
Substituting (2) in (1)
L.H.S =
3abc
abc
=3
………. (2)
(ii)
(
𝐚+𝐛
𝐜
+
𝐛+𝐜
𝐚
+
𝐜+𝐚
𝐛
)(
𝐛
𝐜+𝐚
+
𝐜
+
𝐚+𝐛
𝐚
)
𝐛+𝐜
Whenever b + c ≠ 0, c + a ≠ 0, a + b ≠ 0
We have a + b + c =0
a + b = –c
b + c = –a
c + a = –b
L.H.S = (
=(
a +b
c
−c
+
c
b+c
+
a
−a
a
+
c+a
+
b
−a
b
)(
)(
b
−b
b
c+a
+
+
c
−c
c
a +b
+
a
−a
+
a
b +c
)
)
= (-1-1-1) (-1-1-1)
= (-3) (-3)
= 9 = R.H.S
(iii)
𝐚𝟐
𝟐𝐚𝟐 +𝐛𝐜
+
𝐛𝟐
𝟐𝐛𝟐 𝐜𝐚
+
𝐜𝟐
𝟐𝐜 𝟐 𝐚𝐛
= 1, provided the denominators do not become
0.
L.H.S =
=
=
a2
2a 2 +bc
+
b2
2b 2 ca
a2
a−b (a−c)
a2
a −b (a−c)
+
-
+
c2
2c 2 ab
b2
b −c (b −a)
b2
a −b (b −c)
+
+
c2
c−a (c−b)
c2
a−c (b−c)
=
=
=
a 2 b−c − b 2 a−c + c 2 a−b
a−b (b−c)(a−c)
a 2 b− a 2 c − b 2 a + b 2 c + c 2 a− c 2 b
ab − b 2 − ac +bc (a −c)
a 2 b− a 2 c − b 2 a + b 2 c + c 2 a− c 2 b
a 2 b − ab 2 − a 2 c + abc − abc + b 2 c + ac 2
= 1 = R.H.S
3.
If a + b + c = 0, prove that b2 – 4ac is a square.
We have a + b + c = 0
b = - (a + c)
Squaring on both sides
b2 = [- (a + c)]2
= (a + c)2
b2 = a2 + c 2 + 2ac
Subtracting 4ac on both sides
b2 – 4ac = a2 + c2 + 2ac – 4ac
= a2 - 2ac + c 2
b2 – 4ac = (a - c)2
We find that b 2 – 4ac is the square of (a – c)
4.
If a, b, c are real numbers such that a + b + c = 2s, prove the following:
(i)
s (s – a) + s (s - b) + s (s – c) = s 2
L.H.S. = s (s – a) + s (s - b) + s (s – c)
= s 2 – as + s 2 – bs + s 2 – cs
= 3s 2 – as – bs – cs
= 3s 2 – s (a + b + c)
= 3s 2 – s (2s)
(a + b + c = 2s)
= 3s 2 – 2s 2
= s 2 = R.H.S.
(ii)
s 2 (s – a)2 + s (s - b)2 + s (s – c)2 = a2 + b2 + c 2
L.H.S.
= s 2 (s – a)2 + s (s - b)2 + s (s – c)2
= s 2 + s 2 + a2 – 2sa + s 2 + b2 + s 2 + c 2 – 2as – 2bs – 2cs
= 4s 2 + a2 + b2 + c 2 – 2s – 2bs – 2cs
= 4s 2 + a2 + b2 + c 2 – 2s (a + b + c)
= 4s 2 + a2 + b2 + c 2 – 2as (2s)
= 4s 2 + a2 + b2 + c 2 – 4s 2
= a2 + b2 + c 2
= R.H.S.
(a + b + c = 2s)
(iii)
(s – a) (s – b) + (s – b) (s – c) + (s – c) (s – a) + s 2 = ab + bc + ca
L.H.S. = (s – a) (s – b) + (s – b) (s – c) + (s – c) (s – a) + s 2
= s 2 – as – bs + ab + s 2 – bs – cs + bc + s 2 – cs – as + ac + s 2
= 4s 2 – 2 as – 2bs – 2cs + ab + bc +ca
= 4s 2 – 2s (a + b + c) + ab + bc +ca
= 4s 2 – 2s (2s) + ab + bc +ca
= 4s 2 – 4s 2 + ab + bc +ca
= ab + bc +ca
= R.H.S.
(iv)
a2 – b2 – c2 + 2bc = 4 (s – b) ( s – c)
L.H.S = a2 – b2 – c2 + 2bc
= a2 – (b2 + c2 – 2bc)
= a2 – (b – c)2
= (a + b – c) [a – (b – c)]
= (a + b – c) (a + b – c)
= (2s – c – c) (2s – b – b)
= (2s – 2c) (2s – 2b)
= 2(s – c) (2) (s – b)
= 4 (s – c) (s – b)
= R.H.S.
(a + b + c = 2s)
5. If a, b, c are real numbers, a + b + c =2s and s – a ≠ 0, s – b ≠ 0, s – c ≠ 0,
Prove that
L.H.S =
=
=
=
=
=
=
=
=
𝐚
(𝐬−𝐚)
a
(s−a )
+
+
𝐛
(𝐬−𝐛)
b
(s −b)
+
+
𝐜
(𝐬−𝐜)
c
(s−c)
+2=
𝐚𝐛𝐜
𝐬−𝐜
𝐬−𝐛 (𝐬−𝐜)
+2
a s −b s−c + b s−a s −c + c s −a s−b + 2(s −a)(s −b)(s−c)
s−a s −b (S−c)
[a s 2 −bs −cs +bc +b s 2 −as −cs +ac +c s 2 −as −bs +ab +2 s 2 −as −bs +ab s −c ]
s −a s −b (S−c)
as 2 −abs −acs +abc +b s 2 −abc −bcs +abc +cs 2 −acs −bcs +abc +2(s 3 −as 2 −b s 2 +abs
−cs 2 +acs +bcs −abc ]
s −a s −b (S −c )
[s 2 a+b+c − 2abc −2acs − 2bcs − 3abc + 2s 3 −2 as 2 − 2bs 2 + 2abs −2cs 2 + 2acs +
2bcs −2abc ]
s −a s −b (S −c)
[s 2 2s + abc +2s 3 −2s 2 a+b+c ]
(s −a) s−b (S−c)
[2s 3 + abc +2s 3 −2s 2 2s ]
(s −a) s −b (S −c)
4s 3 + abc −4s 3
(s−a ) s −b (S −c)
abc
(s−a ) s −b (S −c)
= R.H.S.
6. If a + b + c = 0, prove that a 2 – bc = b2 – ca = c 2 – ab =
(𝐚𝟐 + 𝐛𝟐 +𝐜 𝟐 )
𝟐
We have a + b + c = 0
Squaring will get
(a + b + c)2 = 0
a2 + b2 + c 2 + 2ab + 2bc + 2ca = 0
a2 + b2 + c 2 + 2b (a + c) + 2ca = 0
a2 + b2 + c 2 + 2b (-b) + 2ca = 0
a2 + b2 + c 2 = 2b2 – 2ca
(a + c = -b)
a2 + b2 + c 2 = 2 (b2 – ca)
a 2 + b 2 +c 2
2
IIIly
= b2 – ca
……… (1)
(a + b + c)2 = 0
a2 + b2 + c 2 + 2ab + 2bc + 2ca = 0
a2 + b2 + c 2 + 2ab + 2c (b + a) = 0
a2 + b2 + c 2 + 2ab + 2c (-c) = 0
(a + c = -c)
a2 + b2 + c 2 = 2 (c 2 – ab)
a 2 + b 2 +c 2
2
= (c 2 – ab)
……… (2)
Also a2 + b2 + c 2 + 2ab + 2bc + 2ca = 0
a2 + b2 + c 2 + 2a (b + c) + 2bc = 0
a2 + b2 + c 2 + 2a (-a) + 2bc = 0
a2 + b2 + c 2 = 2 (a2 – bc)
……..(3)
From (1), (2) and (3) we get,
2
2
2
a – bc = b – ca = c – ab =
(a 2 + b 2 +c 2 )
2
7. If 2(a 2 + b2) = (a + b)2, prove that a = b.
2a2 + 2b 2 = a2 + b2 + 2ab
2a2 + 2b 2 – a2 – b2 – 2ab = 0
a2 + b2 – 2ab = 0
(a – b)2 = 0
a – b =0
a =b
8. If x2 – 3x + 1 = 0, prove that x2 +
𝟏
𝐱𝟐
= 7.
Given x2 – 3x + 1 = 0
x2 + 1 = 3x
1
x + = 3 (dividing both sides by x)
x
Squaring both sides we get
1
1
x
x2
(x + )2 = 32 = x2 +
x2 +
1
x2
x2 +
1
x2
1
+ 2x. = 9
x
+ 2= 9
=9–2=7
ADDITIONAL PROBLEMS ON
“MULTIPLICATION OF POLYNOMIALS”
I.
Find the following products:
(i)
(2a + 3b) (4a 2 – 6ab + 9b2)
= 8a3 + 12a2b – 12ab2 – 18ab2 + 18ab 2 + 27b 3
= 8a3 + 27b3
(ii)
(3x + 4y) (9x2 – 12xy +16y2)
= 27x3 + 36x2y - 36x2y - 48xy2 + 48xy2 + 64y3
= 27x3 + 64y3
(iii)
(5x – 2y) (25x2 + 10xy + 4y2)
= 125x3 – 50x2y + 50x2y – 20xy2 + 20xy2 – 8y3
= 125x3 – 8y3
(iv)
(a3 – 2) (a 3 + 2a3 + 4)
= a9 – 2a6 + 2a6 – 4a3 + 4a3 – 8
= a9 – 8
2.
By which factor should the following get multiplied to be in the form
a3 + b3.
(i)
(2x + 1)
We have a3 + b3 = (a + b) (a2 – ab +b2)
a = 2x
b=1
(2x)3 + 1 = (2x + 1) [(2x)2 – 2x.1 + 12]
8x + 1 = (2x + 1) (4x2 – 2x + 1)
We have to multiply (2x + 1) with (4x2 – 2x + 1)
(ii)
4x2 – 6x + 9
We have a3 + b3 = (a + b) (a2 – ab +b2)
a = 2x
b=3
(2x)3 + 33 = (2x + 3) (4x2 – 6x + 9)
(2x + 3) should be multiplied
(iii)
9a2 – 15a +25
We have a3 + b3 = (a + b) (a2 – ab +b2)
a = 3a
b=5
(3a)3 + 5 = (3a + 5) (9a – 15a + 25)
Hence (3a + 5) should be multiplied.
(iv)
4a + 3
We have a3 + b3 = (a + b) (a2 – ab +b2)
a = 4a
b=3
(4a)3 + 3 = (4a + 3) [(4a)2 – 4a.3 + 32)
= (4a + 3) (16a2 – 12a + 9)
Hence (16a2 – 12a + 9) should be multiplied.
3.
By which factor should the following get multiplied to be in the form
a3 - b3.
(i)
(5a – 3)
We have a3 - b3 = (a - b) (a2 + ab +b 2)
a = 5a
b=3
(5a)3 – 33 = (5a – 3 ) [(5a2) + (5a) 3 + 32]
125a3 – 27 = (5a – 3) (25a2 + 15a + 9)
(25a2 + 15a + 9) should be multiplied.
(ii)
`16a2 + 20a + 25
We have a3 - b3 = (a - b) (a2 + ab +b 2)
a = 4a
b=5
(4a)3 – 53 = (4a – 5 ) (4a2 + 20a + 25)
(4a – 5) should be multiplied.
(iii)
3x – 2y
We have a3 - b3 = (a - b) (a2 + ab +b 2)
a = 3x
b = 2y
(3x)3 – (2y)3 = (3x – 2y) [(3x)2 + 3x.2y + (2y)2]
27x3 – 8y3 = (3x – 2y) (9x2 + 6xy + 4y2)
(9x2 + 6xy + 4y2) should be multiplied.
(iv)
16x2 – 20x + 25
We have a3 - b3 = (a - b) (a2 + ab +b 2)
a = 4x
b=5
(4x)3 – 53 = (4x – 5 ) (4x2 + 20x + 25)
(4x – 5) should be multiplied.
4.
Use the appropriate identity to compute the following:
(i)
(103)2
We write 103 = 100 + 3
Using (a + b)2 = a2 + 2ab + b 2
We get
a = 100
b=3
(100 + 3)2 = 1002 + 2.100.3 +32
= 10000 + 600 + 9
= 10609
(ii)
(107)2
We write 107 = 100 + 7
Using (a + b)2 = a2 + 2ab + b 2
We get
a = 100
b=7
(100 + 7)2 = 1002 + 2.100.7 +72
= 10000 + 1400 + 49
= 11449
(iii)
𝟏
( 50 )2
𝟐
We write 50
1
2
= 50.5 = 50. +0.5
Using (a + b)2 = a2 + 2ab + b 2
We get
a = 50
b = 0.5
(50 + 0.5)2 = 502 + 2 x 50(0.5) + (0.5)2
= 2500 + 100 (0.5) +0.25
= 2500 + 50 +0.25
= 2550.25
(iv)
(998)2
We write 998 = 1000 - 2
Using (a + b)2 = a2 + 2ab + b 2
We get
a = 1000
b = -2
(1000 - 2)2 = 10002 - 2.1000.2 +22
= 1000000 – 4000 + 4
= 996004
(v)
107 x 93
We write 107 = 100 + 7
and
93 = 100 -7
Using (a + b) (a – b) = a2 + b2
We get
a = 100
b=7
(100 + 7) (100 – 7)
= 1002 – 72
= 10000 – 49
= 9951
(vi)
1008 x 992
We write 1008 = 1008 + 8
and
992 = 1000 - 8
Using (a + b) (a – b) = a2 + b2
We get
a = 1000
(1000 + 8) (1000 – 8)
= 10002 – 82
= 100000 – 64
= 99936
b=8
(vii) (101)3
We write 101 = 100 + 1
Using (a + b)3 = a3 + 3a2b + 3ab 2 + b3
We get
a = 100
b=1
(100 + 1)3 = 1003 + 3(100)2.1 + 3(100).12 + 13
= 1000000 + 30000 + 300 +1
= 10303001
(viii) (1002)3
We write 1002 = 1000 + 2
Using (a + b)3 = a3 + 3a2b + 3ab 2 + b3
We get
a = 1000
b=2
(1000 + 2)3 = 10003 + 3(1000)2.2 + 3(1000).22 + 23
= 1000000000 + 6000000 + 12000 + 8
= 1006012008
5. If x + y = 7 and xy = 12, find x2 + y2 and x2 + y2.
Given x + y = 7 and xy = 12
We take
x+y=7
Squaring both sides
(x + y)2 = 72
x2 + y2 + 2xy = 49
x2 + y2 + 2.12 = 49
x2 + y2 = 49 – 24
x2 + y2 = 25
IIIly x + y = 7
Cubing both sides
(x + y)3 = 73
x3 + y3 + 3xy(x + y) = 343
x3 + y3 + (3 x 12 x 7) = 343
x3 + y3 = 343 – 252
x3 + y3 = 91
6.
𝟏
𝟏
𝐱
𝐱𝟐
If x + = 3, find x2 +
and x4 +
𝟏
𝐱𝟒
1
x+ =3
x
Squaring both sides
(x +
x2 +
x2 +
x2 +
1 2
) =
x
1
x2
1
x2
1
x2
32
1
+ 2.x. = 9
x
= 9– 2
=7
Squaring again
(x2 +
x4 +
x4 +
x4 +
1
x2
1
x4
1
x4
1
x4
)2 = 72
+ 2.x2.
1
x2
= 49
= 49 – 2
= 74
7. If a – b = 2 and ab = 15, find a 3 – b3.
a–b =2
Cubing both sides
(a – b)3 = 23
a3 – b3 – 3ab (a – b) = 8
a3 – b3 – 3 x 15(2) = 8
a3 – b3 = 8 + 90
a3 – b3 = 98
8. If a + b + c = 2s then prove that
(s – a)3 + (s – b)3 + 3c (s – a) (s – b) = c 3
Let
s – a = x and s – b = y
Now L.H.S. will be
x+y
x3 + y3 + 3cxy
s – a + s –b
x3 + y3 + 3(x + y) xy
(x + y = c)
(x + y)3
c3
2s – a - b
a+b+c–a–b
c
9. If a + b + c = 2s, prove that
16s (s – a) (s – b) (s – c) = 2a 2b2 + 2b2c2 + 2c 2a2 – a2 – a4 – b4 – c4
Given a + b + c = 2s
L.H.S. = 16s (s – a) (s – b) (s – c)
= 2s 2(s – a) 2(s – b) 2(s – c)
= 2s (2s – 2a) (2s – 2b) (2s – 2c)
= (a + b + c) (a + b + c – 2a) (a + b + c – 2b) (a + b + c – 2c)
= (a + b + c) (b + c – a) (a + c – b) (a + b – c)
= (ab + b2 + bc + ac + bc + c 2 – a2 – ab – ac)
= (a2 + ac – ab – ab + bc – b2 – ac – c2 + bc)
= (b2 + c2 – a2 + bc ) (a2 – b2 – c + 2bc)
= a2b2 + a2c2 – a4 + 2a2 bc – b4 – b2c 2 + a2b2 – 2b3c – c 2b2 – c4 + a2c2
- 2bc 3 + 2b3c + 2bc 3 – 2a3bc + 4 b2c2
= 2a2b2 + 2 b2c2 + 2a2c2 – a4 – b4 – c4
= R.H.S.
10. If a2 + b2 = c 2, prove that
(a + b + c) (b + c – a) (c + a – b) (a + b – c) = 4a2b2
L.H.S = (a + b + c) (b + c – a) (c + a – b) (a + b – c)
Re arranging
(a + b + c) (a + b – c)
(c + b – a) c – (b – a)
(a + b)2 – c2
c2 – (b – a2)
a2 + b2 + 2ab – c 2
c2 – (a2 + b2 – 2ab)
(a2 + b2) + (2ab – c2)
c2 – (a2 + b2) + 2ab
(c 2+ 2ab – c 2)
c2 – c2 + 2ab
(2ab)
(2ab)
4a2b2
11. If x + y = a and xy = b, prove that (1 + x2) (1 + y2) = a2 + (1 + b2).
` L.H.S =(1 + x2) (1 + y2)
= 1 + x2 + y2 + x2y2
Add and subs tract 2xy
= 1 + x2 + y2 + 2xy – 2xy + x2y2
= 1 + (x + y)2 – 2xy + (xy)2
= 1 + a2 – 2b + b 2
[x + y = a, xy = b]
= a2 + (1 – 2b + b 2)
= a2 + (1 – b)2
= R.H.S
[(a - b)2 = a2 – 2ab + b 2]
𝟏
𝟔
𝐱
𝐱𝟐
12.If x – = 4, show that x3 + 6x2 +
-
𝟏
𝐱𝟑
= 184.
1
Given x – = 4
x
Squaring both sides
1
(x – )2 = 42
x
x2 +
x2 +
x2 +
1
x2
1
x2
1
x2
1
- 2x. = 16
x
= 16 + 2
= 18
………..(1)
1
Again take x – = 4
x
Cubing on both sides
1
(x – )3 = 43
x
x3 –
x3 –
x3 –
x3 –
1
1
x3
1
x3
1
x3
1
1
x
x
– 3x. (x – ) = 64
x3
– 3(4) = 64
= 64 + 12
…………..(2)
= 76
Now consider
L.H.S x3 + 6x2 +
= x3 –
1
x3
6
x2
-
1
x3
+ 6 (x2 +
= 76 + 6 x 18
= 76 + 108
= 184 R.H.S
1
x2
)
13. If xy(x + y) = 1, prove that
𝟏
𝐱 𝟑 𝐲𝟑
– x3 – y3 = 3
Given xy(x + y) = 1
By dividing the equation by xy
x+y=
1
xy
Cubing the both sides
1
(x + y)3 = ( )3
xy
x3 + y3 + 3xy(x + y) =
x3 + y3 + 3(1) =
1
x 3 y3
1
x 3 y3
– x3 + y3 = 3
1
x 3 y3
[xy (x + y) = 1]
14.Suppose a, b, are the sides of a triangle such that 2s = a + b + c.
Prove that
L.H.S =
=
=
=
=
=
=
=
=
𝐚𝟐 − 𝐛𝟐 + 𝟐𝐛𝐜 − 𝐜 𝟐
𝟐 =
𝐛𝟐 + 𝟐𝐛𝐜 + 𝐜 𝟐 − 𝐚
𝐬−𝐛 (𝐬−𝐜)
𝐬( 𝐬−𝐚)
a 2 − b 2 + 2bc − c 2
b 2 + 2bc + c 2 − a 2
a 2 − ( b 2 − 2bc + c 2 )
(b 2 + 2bc + c 2 ) − a 2
a 2 − (b − c)2
( b + c 2 ) − a2
a + b−c
[a− b − c ]
[ b + c + a] [ b + c − a]
[a2 – b2 = (a + b) (a – b)]
a + b – c (a – b+c)
a + b + c (b +c −a)
a + b + c – c −c (a + b + c – b −b )
a + b + c (a + b + c – a −a )
2s − 2c (2s − 2b)
2s (2s − 2a)
4 s − c (s − b )
4s (s − a)
s − b (s − c)
s (s − a )
= R.H.S
15.Suppose a, b, are the sides of a triangle such that 2s = a + b + c.
Prove that
`L.H.S. =
=
=
=
=
𝟏
+
𝐬 −𝐚
𝟏
𝐬 −𝐚
+
𝟏
𝐬 −𝐛
𝟏
𝐬 −𝐛
+
+
𝟏
𝐬− 𝐜
𝟏
𝐬 −𝐜
−
−
𝟏
𝐬
=
𝐚𝐛𝐜
𝐬 𝐬 − 𝐚 𝐬 − 𝐛 (𝐬 − 𝐜)
𝟏
𝐬
s s −b s −c + s s −a s −c + s s −a s −b − [ s −a s−b s −c ]
s s −a s −b (S −c)
s s 2 −bs −cs +bc +s s 2 −as −cs +ac +s s 2 −as −bs +ab
s s −a s −b (S −c)
s 3 − bs 2 −cs 2 + bcs + s 3 − a s 2 − cs 2 + acs + s 3 − cs 2 − bs 2 + abs
s s −a s −b (S−c)
3s 3 −2as 2 −2b s 2 −2bc + abs + bcs + acs – s 3 + a s 2 + bs 2 − abs + cs 2 −
acs – bcs +abc
s s−a s −b (S−c)
= [2s 3 − as 2 − bs 2 − cs 2 + abc ]
= [2s 3 − s 2 a + b + c + abc ]
= [2s 3 − s 2 2s + abc ]
=
=
2 s 3 − 2s 3 + abc
s (s −a) s −b (S −c)
abc
s (s −a) s −b (S −c)
= RHS
1
s(s−a ) s −b (S −c)
1
s (s −a) s −b (S −c)
1
s (s−a) s −b (S−c)
(a + b + c = 2s)
EXTRA QUESTIONS
I.
Expand using appropriate identity:
(i)
(3a + 5b)2
Using (a + b)2 = a2 + 2ab + b 2 we get
a = 3a
b = 5b
(3a + 5b)2 = (3a)2 + 2.3a.5b + (5b)2
= 9a2 + 30ab + 25b2
(ii)
(
𝟏
𝟐
𝐱+
𝟐
𝟑
𝐲)2
Using (a + b)2 = a2 + 2ab + b 2 we get
a=
(
1
2
1
2
2
x
x+
b= y
3
2
3
=
(iii)
1
x
2y
2
2
3
y)2 = ( x)2 + 2. .
x2
4
2
4y 2
3
3
+ xy +
2y
+ ( )2
3
(2a + 3b + 4c)2
Using (a + b + c)2 = a2 + b2 + c 2 + 2ab + 2bc + 2ca we get
a = 2a
b = 3b
c = 4c
(2a + 3b + 4c)2 = (2a)2 + (3b)2 + (4c)2 + 2.2a.3b + 2.3b.4c + 2.4c2a
= 4a2 + 9b2 + 16c 2 + 12ab + 24bc + 16ca
(4a – 3b)2
(iv)
Using (a – b)2 = a2 – 2ab + b2 we get
a = 4a
b = 3b
(4a – 3b)2 = (4a)2 – 2.4a.3b + (3b)2
= 16a2 – 24ab + 9b2
(3a – 2b + 5c)2
(v)
Using (a + b + c)2 = a2 + b2 + c 2 + 2ab + 2bc + 2ca we get
a = 3a
b = -2b
c = 5c
(3a – 2b + 5c)2 = (3a)2 + (–2b)2 + (5c)2 + 2(3a) (–2b) + 2(–2b) (5c)
+ 2(5c) (3a)
= 9a2 + 4b2 + 25c 2 – 12ab – 20bc + 30c
II.
If (x –
𝟏
𝟐𝐱
𝟏
(i) x2 +
(i)
(x –
1
2x
) = 3, find the valve of
(ii) x4 +
𝟒𝐱 𝟐
𝟏
𝟏𝟔𝐱 𝟒
)=3
Squaring on both sides
(x –
x2 +
x2 +
1
2x
1
4x
)2 = 32
– 2.x.
2
1
4x 2
1
2x
=9
= 9 + 1 = 10
(iii) x3 –
𝟏
𝟖𝐱 𝟑
(ii)
x2 +
1
= 10
4x 2
Squaring on both sides
(x2 +
x4 +
x4 +
x4 +
(iii)
(x –
1
4x 2
1
16 x 4
1
1
16 x 4
2x
+ 2.x2 x
1
= 100
4x 2
1
16 x 4
1
)2 = 102
+ = 100
2
= 100 -
1
2
= 99
1
2
)=3
Cubing on both sides
(x –
1
2x
(x3 –
1
8x 3
= x3 –
x3 –
)3 = 33
- 3.x x
1
8x 3
1
8x 3
1
2x
= (x –
1
2x
) = 27
3
= (3) = 27
2
9
54 + 9
2
2
= 27 + =
=
63
2
𝟏
If a2 +
3.
(i) a +
(i)
= 14, find the values of
𝐚𝟐
𝟏
(ii) a –
𝐚
a2 +
1
a2
𝟏
(iii) a2 –
𝐚
= 14
Additing 2 on both sides
(a2 +
1
a2
+ 2) = 14 + 2
1
(a2 + 2 )2 = 16
a
1
a + = 12 = ±4
a
(ii)
a2 +
1
a2
= 14
Subtracting 2 on both sides
1
(a2 +
a2
– 2) = 14 – 2
1
(a – )2 = 12
a
1
(a – ) = 12 = ±2 3
a
(iii)
a2 –
1
a2
1
1
a
a
= (a – ) (a – )
= (±4) (±2 3)
= (±8 3)
𝟏
𝐚𝟐
4. If (3a + 4b) = 16 and ab = 4 find the value of 9a 2 + 16b2
We have (3a + 4b) = 16
Squaring on both sides
(3a + 4b)2 = 162
9a2 + 2.3a.4b + 16b 2 =256
9a2 + 16b 2 = 256 – 24ab
= 256 – 24 x 4
= 256 -96
= 160
5. If a2 – 4a – 1 = 0 2a ≠ 0, find the values of
𝟏
(ii) a –
(i) (a + )
𝐚
We have a2 – 4a – 1 = 0
a2 – 4a = 1
Adding 4 on both sides
a2 – 4a + 4 = 1 + 4
(a + 2)2 = 5
a+2=
5
a= 5–2
1
a
=
1
5−2
𝟏
𝐚
(i)
1
a+ = 5–2+
a
=
=
=
=
(ii)
5−2
( 5−2)2 + 1
5−2
5 + 4 − 2.2 5 + 1
5−2
10 − 4 5
5−2
1
a
=
=
=
=
5−2
( 5−2)2 + 1
a– = 5–2=
1
1
5−2
( 5−2)2 − 1
5−2
5+4–2
5− 2 − 1
5−2
9 – 4 5−1
5−2
9–4 5
5−2
4(2 – 5 )
−(2 – 5 )
=–4
6. If a + 2b + 3c = 0, show that a 3 + 8b3 +27c 3 – 18abc
We have a + 2b + 3c = 0
a + 2b = –3c
Cubing on both sides
(a + 2b)3 = (–3c)3
a3 + 8b3 + 3a (2b) (a+2b) = -27c 3
a3 + 8b3 + 6ab (-3c) = -27c 3
a3 + 8b3 + 27c 3 = 18abc
𝐚
𝐛
𝐛
𝐜
7. If = prove that (a + b + c) (a – b + c) = a 2 + b2 + c 2
a
b
b
c
[Hint: Let = = k. then b = ck, a = bk, b = ck and a = ck2]
L.H.S = (a + b + c) (a – b + c)
= (ck2 + ck + c) (ck2 - ck + c)
= c (k2 + k + 1) c (k2 - k + 1)
= c 2 (k2 + 1 + k) (k2 +1 - k)
= c 2 [(k2 – 1)2 – k]
= c 2 [k4 + 2k2 – 1 – k2]
= c 2 [k4 + k2 + 1]
………. (1)
R.H.S = a2 + b2 + c 2
= (ck2)2 + (ck)2 + c 2
= c 2k4 + c 2k2 + c2
= c 2 (k4 + k2 + 1)
……….. (2)
From (1) and (2) we get
LHS = RHS
8. Evaluate (i) (968)2 – (32)2
(i)
(ii) (98.7)2 – (1.3)2
(968)2 – (32)2
Using (a2 – b2) = (a + b) (a – b) we get
a = 968 and
b = 32
(968)2 – (32)2 = (968 + 32) (968 – b)
= 1000 x 936
= 936000
(ii)
(98.7)2 – (1.3)2
Using (a2 – b2) = (a + b) (a – b) we get
a = 98.7 and
b = 1.3
(98.7)2 – (1.3)2 = (98.7 + 1.3) (98.7 – 1.3)
= 100 x 97.4
= 9740
9. If x + y = 8 and xy = 12, find x4 + y4.
Take x + y = 8
Squares we get
(x + y)2 = 82
x2 + y2 + 2xy = 64
x2 + y2 + 2 x 12 = 64
x2 + y2 = 64 – 24
x2 + y2 = 40
Squaring both sides we get
(x2 + y2)2 = 402
x4 + y4 + 2x2y2 = 1600
x4 - y4 + 2(xy)2 = 1600
x4 + y4 + 2(12)2 = 1600
x4 + y4 + 2 x 144 = 1600
x4 + y4 = 1600 – 288
x4 + y4 = 1312
10. If a + b = 8 and ab = 15, find a3 + b3.
Take a + b = 8
Cubing both sides
(a + b)3 = 83
a3 + b3 + 3ab (a + b) = 512
a3 + b3 + 3 x 15(8) = 512
a3 + b3 + 360 = 512
a3 + b3 = 512 – 360
a3 + b3 = 152
𝟏
𝟏
𝐱
𝐱𝟑
11. If x – = 3, find x3 –
1
Take x – = 3
x
Cubing on both sides
1
(x – )3 = 33
x
1
1
1
x
x
x
x3 + ( )3 + 3x + (x – ) = 27
1
x3 +
x3
1
x3 +
x3
1
x3 +
x3
+ 3(3) = 27
= 27 – 9
= 18
𝟏
𝟏
𝐱
𝐱𝟑
12. If x – = 5, find x3 –
1
Take x – = 5
x
Cubing on both sides
1
(x – )3 = 53
x
1
1
1
x
x
x
x3 – ( )3 – 3x + (x – ) = 125
x3 –
x3 –
x3 –
x3 –
1
– 3(5) = 125
x3
1
x3
– 15 = 125
1
= 125 + 15
x3
1
= 140
x3
13. Suppose x, y, z are non-zero real numbers such that x + y + z = 0
(𝐱 + 𝐲)𝟐
𝐱𝐲
(𝐲 + 𝐳) 𝟐
+
(𝐳 + 𝐱)𝟐
+
𝐲𝐳
= 3.
𝐳𝐱
We have x + y = -z, y + z = -x, and z +x = -y
LHS =
=
=
=
=
(x + y) 2
xy
(−z)2
xy
z2
xy
+
+
x2
xy
+
(y + z) 2
(−x) 2
zy
+
z 3 + x3 + y 3
xyz
3 xyz
xyz
= 3 RHS
yz
+
+
(z + x) 2
zx
(−y)2
zx
y2
xy
(if x + y + z = 0 then x3 + y3 + z3 = 3xyz)
14. If a + b + c = 2s prove that s(s – a) + (s – b) + (s – c) = s 2
LHS = s(s – a) + (s – b) + (s – c)
s 2 – sa + s 2 – sb + s 2 - sc
3s 2 – s (a + b + c)
3s 2 – s (2s)
s 2 = RHS
𝟏
𝟏
𝐱
𝐱𝟐
15. If x + = 3, find x2 +
= 3√5
1
x+ =3
x
Squaring on both sides
1
(x + )2 = 32
x
x2 +
x2 +
x2 +
1
x2
1
x2
1
x2
1
+ 2 x =9
x
= 9– 2
………..(1)
=7
1
1
1
x
x
x
Now (x2 – )2 = (x – )2 – 4x x
1
(x – )2 = 32 – 4
x
1
(x – )2 = 5
x
1
x– = 5
x
………… (2)
Now x2 –
12
x
1
1
x
x
= (x + ) (x – )
From (1) and (2)
x2 +
𝟏
𝐱𝟐
= 3 5 = R.H.S
Important formulae
1. (a + b)2 = a2 + 2ab + b 2
2. (a – b)2 = a2 – 2ab + b 2
3. (a + b) (a – b) = a2 – b2
4. (x + a) (x + b) = x2 + x (a + b) + ab
5. (x + a) (x+ b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
6. (a + b)3 = a3 + b3 + 3ab (a + b)
7. (a + b)3 = a3 – b3 – 3ab (a – b)
8. a3 + b3 = (a + b) (a2 – ab + b 2)
9. a3 – b3 = ( a – b) (a2 + ab + b 2)
10. (a + b + c)2 = a2 + b2 + c 2 + 2ab + 2bc + 2ca
UNIT-1
POLYGONS
EXERCISE 4.1.3
1. In each of the following polygons find in degrees the sum of the interior
angles and the sum of exterior angles.
(i)
Hexagon (ii) Octagon (iii) pentagon (iv) nonagon (v) decagon
Solution:
(i)
Hexagon
Numbers of sides, n = 6
Sum of the interior angles = (2n – 4) 90°
= (2 x 6 – 4) x 90° = (12 – 4) 90 = 8 x 90 = 720°
Sum of the exterior angles = 360°
(ii)
Octagon
Numbers of sides, n = 8
Sum of the interior angles = (2n – 4) 90°
= (2 x 8 – 4) x 90° = (16 – 4) 90 = 12 x 90 = 1080°
Sum of the exterior angles = 360°
(iii)
Pentagon
Numbers of sides, n = 5
Sum of the interior angles = (2n – 4) 90°
= (2 x 5 – 4) x 90° = (10 – 4) 90 = 6 x 90 = 540°
Sum of the exterior angles = 360°
(iv)
Nonagon
Numbers of sides, n = 9
Sum of the interior angles = (2n – 4) 90°
= (2 x 9 – 4) x 90° = (18 – 4) 90 = 14 x 90 = 1260°
Sum of the exterior angles = 360°
(v)
Decagon
Numbers of sides, n = 10
Sum of the interior angles = (2n – 4) 90°
= (2 x 10 – 4) x 90° = (20 – 4) 90 = 16 x 90 = 1440°
Sum of the exterior angles = 360°
2. How many sides dose a polygon have if the sum of the interior angle is
(i) 540°
(ii) 900°
(iv) 7 straight angles
(iii) 1440°
(v) 8 straight angles
Solution:
(i)
Sum of the interior angles = (2n – 4) 90° = 540°
→ 2n – 4 =
540
90
=6
→ 2n = 6 + 4 = 10
n=
10
2
=5
The polygon has 5 sides
(ii)
Sum of the interior angles = (2n – 4) 90° = 900°
→ 2n – 4 =
900
90
= 10
→ 2n = 10 + 4 = 14
→
n=
14
2
=7
The polygon has 7 sides
(iii)
The Sum of the interior angles = (2n – 4) 90° = 1440°
→ 2n – 4 =
1440
90
= 16
→ 2n = 16 + 4 = 20
→
n=
20
2
= 10
The polygon has 10 sides
(iv)
Sum of the interior angles = (2n – 4) 90° = 7 straight angles = 7 x 180
→ (2n – 4) =
→ 2n – 4 =
7 x 180
90
7 x 180
90
= 7 x 2 = 14
→
2n = 14 + 4 = 18
→
n=
18
2
=9
The polygon has 9 sides
(v)
Sum of the interior angles = (2n – 4) 90° = 8 right angles
→ (2n – 4) 90° = 8 x 90°
→ 2n – 4 =
8 x 90
90
=8
→
2n = 8 + 4 = 12
→
n=
12
2
=6
The polygon has 6 sides
3. Find the measure of each exterior angle of a regular polygon with sides:
(i) 40
(ii) 30
(iii) 20
(iv) 18
(v) 16
(vi) 2x
(vii) (2a +4b)
Solutions:
(i)
Number of sides n = 40
Measure of each exterior angle = (
(ii)
40
= 9°
360
n
)°=
360
30
= 12°
360
n
)°=
360
20
= 18°
Number of sides n = 18
Measure of each exterior angle = (
(v)
360
Number of sides n = 20
Measure of each exterior angle = (
(iv)
n
)°=
Number of sides n = 30
Measure of each exterior angle = (
(iii)
360
360
n
)°=
360
18
= 20°
Number of sides n = 16
Measure of each exterior angle = (
360
n
)°=
360
16
= 22.5°
(vi)
Number of sides n = 2x
Measure of each exterior angle = (
360
)°=
n
360
2x
=(
𝟏𝟖𝟎
𝐱
)°
(vii) Number of sides n = 2a + 4b
Measure of each exterior angle = (
360
)°=
n
360
2a + 4b
=(
𝟏𝟖𝟎
𝐚 + 𝟐𝐛
)°
4. Find the numbers of sides of regular polygon. If each exterior angle
measures.
(i)
10°
(ii) 20°
(v) 45°
(iii) 30°
(vi) 60°
(iv) 40°
(vii) 72°
(viii) 120°
Solution:
(i)
Measure of each exterior angle = x° = 10°
Number of sides =
(ii)
360
x
=
360
10
= 35
Measure of each exterior angle = x° = 20°
Number of sides =
360
x
=
360
20
= 18
(iii)
Measure of each exterior angle = x° = 30°
Number of sides =
(iv)
360
30
= 12
360
x
=
360
40
=9
Measure of each exterior angle = x° = 45°
Number of sides =
(vi)
x
=
Measure of each exterior angle = x° = 40°
Number of sides =
(v)
360
360
x
=
360
45
=8
Measure of each exterior angle = x° = 60°
Number of sides =
360
x
=
360
60
=6
(vii) Measure of each exterior angle = x° = 72°
Number of sides =
360
x
=
360
72
=5
(viii) Measure of each exterior angle = x° = 120°
Number of sides =
360
x
=
360
120
=3
5. Find the numbers of sides of a regular polygon it each exterior is equal
to
(i)
Its adjacent interior angle.
(ii)
Twice its adjacent interior angle.
(iii)
Half its adjacent interior angle.
(iv)
One-third of its adjacent interior angle.
Solution:
(i)
Exterior angle + its adjacent interior = 180°
e + i = 180°
e=i
e + e = 180°
2e = 180°
e=
180
2
= 90°
Number of sides =
(ii)
360
e
e + i = 180°
e = 2i
→ i=
→e+
e
2
e
2
= 180°
→
2e+e
→
3e = 180 x 2 = 360°
→
e=
2
= 180°
360
3
= 120°
=
360
90
=4
Number of sides =
(iii)
360
e
=
360
120
=3
e + i = 180°
e=
i
2
→ i = 2e
→ e + 2e= 180°
→
3e = 180°
→
e=
180
3
= 60°
Number of sides =
(iv)
360
e
=
360
60
=6
e + i = 180°
e=
i
3
→ i = 3e
→ e + 3e= 180°
→
4e = 180°
→
e=
180
4
= 45°
Number of sides =
360
e
=
360
45
=8
6. find the number of sides in a regular polygon if each interior angle is
(i)
twice its adjacent exterior angle
(ii)
Four times the adjacent exterior angle.
(iii)
Eight times the adjacent exterior angle.
(iv)
Seventeen times the adjacent exterior angle.
Solution:
(i)
e + i = 180°
i = 2e
→ e + 2e = 180°
→ 3e = 180°
→e=
180
3
= 60°
Number of sides =
(ii)
360
e
=
360
60
=6
e + i = 180°
i = 4e
→ e + 4e = 180°
→ 5e = 180°
→e=
180
5
= 36°
Number of sides =
360
e
=
360
36
= 10
(iii)
e + i = 180°
i = 8e
→ e + 8e = 180°
→ 9e = 180°
180
→e=
9
= 20°
Number of sides =
(iv)
360
e
=
360
20
= 18
Seventeen times the adjacent exterior angle
e + i = 180°
i = 17e
→ e + 17e = 180°
→ 18e = 180°
→e=
180
18
= 10°
Number of sides =
360
e
=
360
10
= 36
7. The angles of a convex polygon are in the ratio 2 : 3 : 5 : 9 :11. Find the
measure of each angle.
Solution:
Ratio of the angle = 2 : 3 : 5 : 9 :11.
Let the angle be 2x, 3x, 5x, 9x and 11x.
The polygon has five angles and therefore five sides.
Sum of angle = (2n - 4)90°
= (2 x 5 – 4)90°
= (10 – 4)90
= 6 x 90°
2x + 3x + 5x + 9x +11x = 6 x 90°
30x = 6 x 90°
x=
6 x 90
30
= 6 x 3 = 18°
The angles are 2x = 2 x 18 = 36°
3x = 3 x 18 = 54°
5x = 5 x 18 = 90°
9x = 9 x 18 = 162°
11x = 11 x 18 = 198°
8. Prove that the opposite sides of a regular hexagon are parallel.
Solution:
E
D
F
C
A
B
Construction: in the regular hexagon ABCDEF, join FC. In the quadrilateral
ABCF, AF = BC and FAB = CBA = 120°
ABCF is an isosceles trapezium
FC | | AB
Similarly ED | | FC
……. (1)
……. (2)
From (1) and (2) AB | | ED
Similarly BC | | FE and CD | | FA
ADDITIONAL PROBLEMS ON POLYGONS
1. In an equiangular polygon, the measure of each exterior angle is 25% of
the measure of each interior angle. How many sides it has?
Solution:
e + i = 180°
e= ix
25
100
1
= i
4
→ i = 4e
→ e + 4e = 180°
→ 5e = 180°
e=
180
5
= 36°
Number of sides =
360
e
=
360
36
= 10
2. In a heptagon, two of the angles 130° each and the remaining angles are
equal. Find the equal angles.
Solution:
A heptagon has 7 sides
Sum of interior angles = (2n – 4)90°
= (2 x 7 – 4)90
= (14 – 4) 90
= 10 x 90 = 900°
Measure of two equal angles = 130° each their sum = 130 x 2 = 260
Sum of the remaining 5 equal angles = 900 – 260 = 640
Measure of each of the 5 angles =
640
5
= 128°
3. A polygon has n sides two of its angles are right angles and each of the
remaining angles is 144 degrees find the value of n.
Solution:
Numbers of sides = n
Sum of two right angles = 90° x 2 = 180°
Sum of the remaining (n - 2) angles,
Each equal to 144° = (n - 2)144°
Sum of interior angles = (2n – 4)90°
= 180 + (n – 2)144
180n - 360 = 180 + 144n – 288
180n – 144n = 180 + 360 – 288
36n = 540 – 288 = 252°
n=
252
36
=7
4. Is there a polygon which has only two types of interior angles 120° and
60°? If so how many sides such a polygon has?
[Hint: if n is the number of sides and if k angles are of 60° then (n – k)
angles are of 120° and (n - 2) 180 = 60k + 120(n – k)
Solution:
180n – 360 = 60k + 120n – 120k
180n – 120k - 60k + 120k = 360
60n + 60k = 360
n+k=6
When k = 1 n = 5 and
When k = 2 n = 4
Thus the polygon can be quadrilateral with two 60°angles and two 120°
Angles. or a pentagon with one 60° angle and four 120° angles.
5. In the adjacent figure, the pentagon is such that AB = BC = CD and
AE = ED. Moreover ABC = BCD = AED = 130°.
Find the measures of BAE and EDC
B
C
130°
130°
A
D
130°
E
Solution:
Construction: Join AD.
ABC = BCD = 130°
AB = CD
ABCD In an isosceles trapezium
ABC + BAD = 180°
130° + BAD = 180
BAD = 180 – 130 = 50°
ADC = BAD = 60
In
EAD, EA = ED
EAD = EDA
But EAD + EDA = 180 – 130 = 50
EAD = EDA =
50
2
= 25°
BAE = BAD + DAE = 50 + 25 = 75°
Similarly EDC = 75°
6. In the adjacent figure you have a star ABCDEFGHIKL. All the sides of
the star are equal. What is the measure of angles at the vertices A, C, E,
G and k?
K
E
F
H
K
D
L
B
A
Solution:
All
s
are isosceles.
x+ x + A + x + x + C + ……… = 180 x 5 = 900
10x + A + C + E + G + K = 900
10x + 5y = 900
5(2x + y = 180)
C
7. Prove that the perpendicular drawn from the vertex of a regular
pentagon to the opposite side bisects that side,
Solution:
D
E
C
A
F
B
Join DA and DB
In
DEA and
DCB
DE = DC
EA = CB
DEA = DCB
DEA =
In
DCB (SAS)
DAF and
DBF
Hypotenuse DA = Hypotenuse DB
DF = DF
DAF ≈
DBF (RHS)
AF = FB (CPCT)
DF bisects AB
8. For what values of m and n is it possible for the external angle of a
regular m-gon to be equal to the internal angle of a regular n-gon.
Check your answer.
𝛑
𝛑
𝐧
𝟐𝐦
[Hint: (n – 2) =
]
Solution:
The sum of the exterior angles of a m-gon. = 2𝛑
The measure of each exterior angle =
π
2m
The sum of the interior angles of a n-gon. = (n – 2) 𝛑
The measure of each exterior angle =
Now,
π
2m
2
m
=
=
(n – 2)π
n
(n – 2)π
n
n–2
n
2n = mn – 2m
2n = m (n – 2)
m=
2n
n–2
m and n can have all positive integral values satisfying the
above condition.
Such that m ≥ 3 and n ≥ 3
9. In a convex polygon, sum of all the angles except one angle is 2280°.
How many sides does the polygon have? What is the measure of this
exceptional angle?
[Hint: if x is the exceptional angle, then (n – 2) 180 = 2280 + x, and
0 < x < 180]
Solution:
If the polygon has n sides, the
Sum of the interior angle = (n – 2)
If the exceptional angle measures x°
→ (n – 2)180 = 2280 + x
→ 2280 + x is a multiple of 180
→ 120 + x = 180
Or x = 180 – 120 = 60
Thus (n – 2)180 = 2280 + 60 = 2340
n -2 =
2340
180
= 13
n = 13 + 2 = 15
And the exceptional angle = 60
12
180 2280
180
480
360
120