KSEEB SSLC Solutions for Class 10 Maths – Surds (English Medium)
Exercise 7.1:
Question I(1):
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Question I(2):
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Question I(3):
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Question I(4):
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Question I(5):
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Question I(6):
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Question I(7):
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Question I(8):
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Question I(9):
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Question I(10):
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Question II(1):
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Question II(2):
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Question II(3):
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Question II(4):
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Question II(5):
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Question II(6):
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Question III(1):
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Question III(2):
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Question III(3):
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Question III(4):
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Exercise 7.2:
Question I(1):
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Question I(2):
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Question I(3):
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Question I(4):
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Question I(5):
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Question I(6):
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Question I(7):
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Question I(8):
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Question II(1):
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Question II(2):
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Question II(3):
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Question II(4):
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Question II(5):
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Question II(6):
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Question II(7):
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Question II(8):
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Question III(1):
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Question III(2):
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Question III(3):
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Question III(4):
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Question III(5):
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Question III(6):
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Exercise 7.3:
Question I:
Solution :
Question II:
If the product of two binomial surds is a rational number, then each surd is called the conjugate of the other.
Solution :
Question III(a):
Find the rationalizing factor of the following binomial surd:
Solution :
Question III(b):
Find the rationalizing factor of the following binomial surd:
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Question III(c):
Find the rationalizing factor of the following binomial surd:
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Question III(d):
Find the rationalizing factor of the following binomial surd:
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Exercise 7.4:
Question I(A)(1):
Rationalise the denominator and simplify.
Solution :
*Note: Answer given in the book is incorrect.
Question I(A)(2):
Rationalise the denominator and simplify.
Solution :
Question I(A)(3):
Rationalise the denominator and simplify.
Solution :
Question I(A)(4):
Rationalise the denominator and simplify.
Solution :
Question I(A)(5):
Rationalise the denominator and simplify.
Solution :
Question I(B)(1):
Rationalise the denominator and simplify.
Solution :
Question I(B)(2):
Rationalise the denominator and simplify.
Solution :
Question I(B)(3):
Rationalise the denominator and simplify.
Solution :
Question I(B)(4):
Rationalise the denominator and simplify.
Solution :
Question I(B)(5):
Rationalise the denominator and simplify.
Solution :
Question I(C)(1):
Rationalise the denominator and simplify.
Solution :
\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
⇒ \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\) (\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\))
⇒ \(\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}\) = \(\frac{(\sqrt{3}+\sqrt{2})^{2}}{3-2}\)
∵ (a+b) (a-b) = a² – b²
⇒ \(\frac{3+2+2(\sqrt{3})(\sqrt{2})}{3-2}\)
⇒ \(\frac{5+2 \sqrt{6}}{1}\)
⇒ 5 + 2√6
Question I(C)(2)
Rationalise the denominator and simplify.
Solution :
Question I(C)(3):
Rationalise the denominator and simplify.
Solution :
*Note : Answer given in the book is incorrect.
Question I(C)(4):
Rationalise the denominator and simplify.
Solution :
Question II(1):
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Question II(2):
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Question II(3):
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Question II(4):
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Question II(5):
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Question II(6):
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Question II(7):
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