Question

If A: B=C: D, then the value of \(\frac{A^2+B^2}{C^2+D^2}\) is

• A. \(\frac{1}{2}\)
• B. \(\frac{A+ B}{C +D}\)
• C. \(\frac{A- B}{C -D}\)
• D. \(\frac{A\times B}{C \times D}\)

Answer 1

The Correct Option is D

To solve the problem, we need to find the value of \(\frac{A^2 + B^2}{C^2 + D^2}\) given \(A : B = C : D\). Let's go through the solution step-by-step:

1. We are provided with the ratio \(A : B = C : D\). This can also be written as \(\frac{A}{B} = \frac{C}{D}\).
2. Cross multiplying gives us \(A \times D = B \times C\).
3. Now, we want to find \(\frac{A^2 + B^2}{C^2 + D^2}\). To simplify this, we'll express both numerator and denominator in terms of \(A \times B\) and \(C \times D\).
4. Using the identity \(A = \frac{C \times B}{D}\) from the proportional relationship \(A \times D = B \times C\), and similarly, \(B = \frac{A \times D}{C}\), we can substitute these into the expression:
5. Recalling that for any terms \(X\) and \(Y\) where \(X : Y = X_1 : Y_1\), we have: \[ \frac{X^2 + Y^2}{X_1^2 + Y_1^2} = \frac{\left(\frac{X}{Y}\right)^2 + 1}{\left(\frac{X}{Y}\right)^2 + 1} = 1 \] Applying this to our scenario with \(A, B, C, D\), things simplify further:
6. \[ \frac{A^2 + B^2}{C^2 + D^2} = \frac{\left(\frac{A}{B}\right)^2 + 1}{\left(\frac{C}{D}\right)^2 + 1} = \text{\(A \times B\) : \(C \times D \)} \] Therefore, it simplifies to: \[ \frac{A \times B}{C \times D} \]

Hence, the correct answer is \(\frac{A \times B}{C \times D}\).