C-11-Q062mediumsingle_mcq

From $\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}$ , the underlying given is—

a $a:b = c:d$
b $a + b = c + d$
c $a = c, b = d$
d $abcd = 1$

ব্যাখ্যা

This identity is obtained by applying componendo-dividendo to a proportion: starting from $a:b=c:d$ , squaring gives $a^2:b^2=c^2:d^2$ , and combining cross terms produces $\dfrac{a^2+b^2}{a^2-b^2}=\dfrac{ac+bd}{ac-bd}$ . Hence the underlying given is $a:b=c:d$ .

এই অধ্যায়ের আরও প্রশ্ন

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Algebraic Ratio and Proportion — সব প্রশ্ন দেখুন (180টি) →

From $\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}$ , the underlying given is—

a:b = c:d

a + b = c + d

a = c, b = d

abcd = 1

From a2+b2a2−b2=ac+bdac−bd\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}a2−b2a2+b2​=ac−bdac+bd​, the underlying given is—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

From a2+b2a2−b2=ac+bdac−bd\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}a2−b2a2+b2​=ac−bdac+bd​, the underlying given is—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

From $\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}$ , the underlying given is—

From $\dfrac{a^2+b^2}{a^2-b^2} = \dfrac{ac+bd}{ac-bd}$ , the underlying given is—