C-11-Q099mediumsingle_mcqIf a:b=c:da:b = c:da:b=c:d, then (2a+b)2(2c+d)2\dfrac{(2a + b)^2}{(2c + d)^2}(2c+d)2(2a+b)2 equals—a(2a−b)2(2c−d)2\dfrac{(2a-b)^2}{(2c-d)^2}(2c−d)2(2a−b)2ba2c2\dfrac{a^2}{c^2}c2a2cb2d2\dfrac{b^2}{d^2}d2b2dall of the aboveব্যাখ্যাWith ac=bd=k\frac{a}{c}=\frac{b}{d}=kca=db=k, we have 2a+b=k(2c+d)2a+b=k(2c+d)2a+b=k(2c+d) and 2a−b=k(2c−d)2a-b=k(2c-d)2a−b=k(2c−d), so (2a+b)2(2c+d)2=k2=a2c2=b2d2=(2a−b)2(2c−d)2\frac{(2a+b)^2}{(2c+d)^2}=k^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{(2a-b)^2}{(2c-d)^2}(2c+d)2(2a+b)2=k2=c2a2=d2b2=(2c−d)2(2a−b)2. All three listed expressions are equal.