Question

Let a be the area of the triangle ABC. Then the value of (64a)² is

Answer 1

Correct Answer: 1007.99 - 1008.01

Let the sides of the triangle be \( a - d \), \( a \), and \( a + d \).

We are given that \( A - C = \frac{\pi}{2} \) and the circumradius \( R = 1 \).

Now, we start with the following relation:

\(\frac{a + d}{\sin A} = \frac{a}{\sin B} = \frac{a - d}{\sin C} = 2\)

From this, we can conclude that:

\(A = \frac{\pi}{2} + C\)

Using the sine law:

\(\sin A = \sin \left( \frac{\pi}{2} + C \right) \Rightarrow \sin A = \cos C\)

Now, we substitute this into the equation:

\(\frac{a + d}{2} = \sqrt{1 - \sin^2 C } \Rightarrow \left( \frac{a + d}{2} \right)^2 = 1 - \left( \frac{a - d}{2} \right)^2\)

Expanding both sides:

\(2 \left( a^2 + d^2 \right) / 4 = 1 \Rightarrow a^2 + d^2 = 2 \quad \text{(Equation 1)}\)

Now, using the cosine law:

\(\cos B = \frac{(a - d)^2 + (a + d)^2 - a^2}{2(a^2 - d^2)}\)

By simplifying further:

\(\sqrt{1 - \sin^2 B} = \frac{2(a^2 + d^2) - a^2}{2(a^2 - d^2)} \Rightarrow a^2 = 7/4, d^2 = 1/4\)

Now, using the area formula:

\(\Delta = \frac{a(a^2 - d^2)}{4}\)

Substitute the values of \( a^2 \) and \( d^2 \):

\(\alpha = \frac{\sqrt{7}}{2} \times \frac{6}{4 \times 4}\)

Finally, we calculate \( (64 \alpha)^2 \):

\((64 \alpha)^2 = 1008\)

Thus, the final result is \( (64 \alpha)^2 = 1008 \).

Answer 2

The formula for the ratio \( r \) is given by:

\(r = \frac{\Delta}{S} = \frac{\sqrt{7}}{2} \times \frac{6}{16} = \frac{4}{16} = \frac{1}{4} = 0.25\)

Thus, the value of the ratio \( r \) is 0.25.