Question

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Explanation:

Step1: Check for similar - triangles or ratio

Assume the two triangles are similar.

Step2: Find the ratio of corresponding sides

For the right - hand triangle, sides are \(35:56:63 = 5:8:9\) (dividing by 7).

Step3: Set up proportion for left - hand triangle

Let the missing side of the left - hand triangle be \(x\). If the triangles are similar, and considering the ratio \(5:8:9\), we can assume the sides of the left - hand triangle follow the same ratio. If we assume the side corresponding to 35 in the first triangle is 48, and the side corresponding to 56 is \(x\).
We set up the proportion \(\frac{35}{48}=\frac{56}{x}\). Cross - multiply: \(35x = 48\times56\), then \(x=\frac{48\times56}{35}=\frac{48\times8}{5}=\frac{384}{5} = 76.8\) (this is wrong assumption).
Let's use the Pythagorean theorem. For the right - hand triangle \(35^{2}+56^{2}=1225 + 3136=4361\) and \(63^{2}=3969\), it's not a right - triangle.
Let's assume the triangles are related by a linear scaling factor.
If we consider the sides of the second triangle \(35,56,63\) and the sides of the first triangle \(48,54\).
We use the law of cosines in the second triangle to find an angle, say \(\cos C=\frac{35^{2}+56^{2}-63^{2}}{2\times35\times56}=\frac{1225 + 3136-3969}{3920}=\frac{4361 - 3969}{3920}=\frac{392}{3920}=0.1\)
In the first triangle, using the law of cosines \(c^{2}=48^{2}+54^{2}-2\times48\times54\times0.1\)
\[

$$\begin{align*} c^{2}&=2304+2916-(2\times48\times54\times0.1)\\ &=2304 + 2916-(518.4)\\ &=5220-518.4\\ &=4701.6 \end{align*}$$

\]
\(c=\sqrt{4701.6}\approx68.6\) (wrong)
Let's assume the triangles are similar in a non - standard way. If we consider the ratio of the sides pairwise.
Let's try another approach. If we assume the sides of the two triangles are in proportion.
We know that if the triangles are similar, we can consider the ratios of the sides.
Let's assume the sides of the first triangle \(a = 48\), \(b = 54\), and the sides of the second triangle \(A=35\), \(B = 56\), \(C = 63\)
We note that \(35:48
eq56:54\)
Let's use the fact that if we consider the triangles' side - length relationships in terms of a common factor.
If we assume the first triangle's sides are \(a = 48\), \(b = 54\), and we want to find \(c\)
We use the law of cosines \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\)
In the second triangle, we can find an angle. Let's assume we use the fact that the triangles might be related by a simple arithmetic or geometric relationship.
If we consider the following:
We know that \(35 = 5\times7\), \(56=7\times8\), \(63 = 7\times9\)
For the first triangle, \(48=6\times8\), \(54 = 6\times9\)
The missing side should be \(6\times5=30\)

Answer:

30