Let the edges of the brick be \( a, b, c \) such that \( a < b < c \).
We are given:
Adding all three equations: \[ (a^2 + b^2) + (a^2 + c^2) + (b^2 + c^2) = 9 + 12 + 15 = 36 \] \[ \Rightarrow 2(a^2 + b^2 + c^2) = 36 \Rightarrow a^2 + b^2 + c^2 = 18 \quad \text{(4)} \]
Now, subtracting Equation (1) from Equation (4): \[ c^2 = 18 - (a^2 + b^2) = 18 - 9 = 9 \Rightarrow c = 3 \]
Similarly, subtracting Equation (3) from Equation (4): \[ a^2 = 18 - (b^2 + c^2) = 18 - 15 = 3 \Rightarrow a = \sqrt{3} \]
Hence, the required ratio is: \[ \frac{a}{c} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \]
Final Answer: \( \boxed{\frac{1}{\sqrt{3}}} \)
A shopkeeper sells an item at a 20 % discount on the marked price and still makes a 25 % profit. If the marked price is 500 rupees, what is the cost price of the item?
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: