5. Express the following as the sum of consecutive
odd numbers.
b. 122 c. 132
d. 152 e.
6. Find the square of the following numbers.
b. 126
d. 67
a. 102
e. 182
a. 35
c. 128
e. 48
Answers
Answer:
ok
6) answer (i)
\left(32\right)^2=\left(30+2\right)^2=\left(30\right)^2+2\times30\times2+\left(2\right)^2(32)
2
=(30+2)
2
=(30)
2
+2×30×2+(2)
2
[\because\left(a+b\right)^2=a^2+2ab+b^2∵(a+b)
2
=a
2
+2ab+b
2
]
= 900 + 120 + 4 = 1024
(ii) \left(35\right)^2=\left(30+5\right)^2=\left(30\right)^2+2\times30\times5+\left(5\right)^2(35)
2
=(30+5)
2
=(30)
2
+2×30×5+(5)
2
[\because\left(a+b\right)^2=a^2+2ab+b^2∵(a+b)
2
=a
2
+2ab+b
2
= 900 + 300 + 25 = 1225
(iii) \left(86\right)^2=\left(80+6\right)^2=\left(80\right)^2+2\times80\times6+\left(6\right)^2(86)
2
=(80+6)
2
=(80)
2
+2×80×6+(6)
2
[\because\left(a+b\right)^2=a^2+2ab+b^2∵(a+b)
2
=a
2
+2ab+b
2
= 8100 + 540 + 9 = 8649
(iv) \left(93\right)^2=\left(90+3\right)^2=\left(90\right)^2+2\times90\times3+\left(3\right)^2(93)
2
=(90+3)
2
=(90)
2
+2×90×3+(3)
2
[\because\left(a+b\right)^2=a^2+2ab+b^2∵(a+b)
2
=a
2
+2ab+b
2
]
= 8100 + 540 + 9 = 8649
(v) \left(71\right)^2=\left(70+1\right)^2=\left(70\right)^2+2\times70\times1+\left(1\right)^2(71)
2
=(70+1)
2
=(70)
2
+2×70×1+(1)
2
[\because\left(a+b\right)^2=a^2+2ab+b^2∵(a+b)
2
=a
2
+2ab+b
2
= 4900 + 140 + 1 = 5041
(vi) \left(46\right)^2=\left(40+6\right)^2=\left(40\right)^2+2\times40\times6+\left(6\right)^2(46)
2
=(40+6)
2
=(40)
2
+2×40×6+(6)
2
[\because\left(a+b\right)^2=a^2+2ab+b^2∵(a+b)
2
=a
2
+2ab+b
2
]
= 1600 + 480 + 36 = 2116
