Correct option is (A) (a + b) (a – b) = a2 – b2
\(101\times99\) = (100+1) (100-1)
Let a = 100, b = 1
Then, \(101\times99\) \(=(a+b)(a-b)=a^2-b^2=100^2-1^2\) = 10000 - 1 = 9999
Thus, the identity used...
Correct option is (C) (a – b) (a + b) = a2 – b2
\(103\times97\) = (100 + 3) (100 - 3)
Let a = 100, b = 3
Then \(103\times97\) \(=(a+b)(a-b)\) \(=a^2-b^2\)
\(=100^2-3^2\)
= 10000 - 9 =...
Correct option is (A) ΔABC ≅ ΔDEF
In \(\triangle ABC\;and\;\triangle EFD,\)
AB = EF (Both hypotenuse are equal)
\(\angle C=\angle D\) \(=90^\circ\) (Angles opposite to hypotenuse)
AC = ED (Given)
\(\therefore\) \(\triangle ABC\cong\triangle EFD\) ...
