Question

1. $\triangle klm \sim \triangle def$

(image of two similar triangles: triangle def with sides ed=5, df=8; triangle klm with side km=8x, kl=25)

Explanation:

Step1: Identify Corresponding Sides

Since \(\triangle KLM \sim \triangle DEF\), the corresponding sides are proportional. The right angles are at \(D\) and \(K\), so \(DE\) corresponds to \(KM\), and \(DF\) corresponds to \(KL\). So \(DE = 5\), \(DF = 8\), \(KM = 8x\), \(KL = 25\).

Step2: Set Up Proportion

The proportion from similar triangles is \(\frac{DE}{KM}=\frac{DF}{KL}\). Substituting the values, we get \(\frac{5}{8x}=\frac{8}{25}\). Wait, no, actually, let's check the correspondence again. Wait, maybe \(DE\) corresponds to \(KL\) and \(DF\) corresponds to \(KM\)? Wait, the right angle: in \(\triangle DEF\), right angle at \(D\), sides \(DE = 5\), \(DF = 8\). In \(\triangle KLM\), right angle at \(K\), sides \(KL = 25\), \(KM = 8x\). So the correct correspondence: \(\triangle DEF \sim \triangle KLM\) (order matters), so \(DE\) corresponds to \(KL\), \(DF\) corresponds to \(KM\), \(EF\) corresponds to \(LM\). So \(\frac{DE}{KL}=\frac{DF}{KM}\). So \(DE = 5\), \(KL = 25\), \(DF = 8\), \(KM = 8x\). So \(\frac{5}{25}=\frac{8}{8x}\).

Step3: Solve the Proportion

Simplify \(\frac{5}{25}=\frac{1}{5}\), so \(\frac{1}{5}=\frac{8}{8x}\). The \(8\) in numerator and denominator: \(\frac{1}{5}=\frac{1}{x}\) (since \(\frac{8}{8x}=\frac{1}{x}\)). So \(x = 5\)? Wait, no, wait: \(\frac{5}{25}=\frac{8}{8x}\) simplifies to \(\frac{1}{5}=\frac{1}{x}\) (because \(\frac{8}{8x}=\frac{1}{x}\)). So cross - multiply: \(1\times x=5\times1\), so \(x = 5\). Wait, let's do it step by step. \(\frac{5}{25}=\frac{8}{8x}\). Simplify left side: \(\frac{1}{5}\). Right side: \(\frac{8}{8x}=\frac{1}{x}\) (divide numerator and denominator by 8). So \(\frac{1}{5}=\frac{1}{x}\), so \(x = 5\).

Wait, maybe I mixed up the correspondence. Let's check the order: \(\triangle KLM \sim \triangle DEF\), so \(K\) corresponds to \(D\), \(L\) corresponds to \(E\), \(M\) corresponds to \(F\). So angle \(K\) (right angle) corresponds to angle \(D\) (right angle). So side \(KL\) (adjacent to right angle at \(K\)) corresponds to side \(DE\) (adjacent to right angle at \(D\)), and side \(KM\) (adjacent to right angle at \(K\)) corresponds to side \(DF\) (adjacent to right angle at \(D\))? Wait, no, in \(\triangle DEF\), legs: \(DE = 5\) (vertical), \(DF = 8\) (horizontal). In \(\triangle KLM\), \(KL = 25\) (vertical), \(KM = 8x\) (horizontal). So the ratio of vertical sides: \(DE/KL = 5/25 = 1/5\), ratio of horizontal sides: \(DF/KM = 8/(8x)=1/x\). Since they are similar, these ratios should be equal: \(1/5 = 1/x\), so \(x = 5\).

Answer:

\(x = 5\)