QUESTION IMAGE
Question
if $\triangle gyk \sim \triangle bak$, then the value of $x$ is.
10
15
20
25
Step1: Recall Similar Triangles Property
For similar triangles, corresponding sides are proportional. So, \(\frac{GY}{BA}=\frac{GK}{BK}=\frac{YK}{AK}\).
Step2: Identify Corresponding Sides
From \(\triangle GYK \sim \triangle BAK\), we have \(\frac{GY}{BA}=\frac{YK}{AK}\). Given \(GY = x\), \(BA = 12\), \(YK = 25\), \(AK = 15\). Wait, alternatively, check another pair: \(\frac{GK}{BK}=\frac{GY}{BA}\). \(GK = 10\), \(BK = 6\)? No, wait, \(GK = 10\), \(BK = 6\)? Wait, no, \(GK\) is 10, \(BK\) is 6? Wait, no, let's check the sides. Wait, \(GK = 10\), \(BK = 6\)? Wait, no, the segments: \(GK = 10\), \(BK = 6\)? Wait, no, actually, \(GK\) is 10, \(BK\) is 6? Wait, no, let's re - check. The triangles are \(\triangle GYK\) and \(\triangle BAK\). So the sides: \(GY\) corresponds to \(BA\), \(YK\) corresponds to \(AK\), \(GK\) corresponds to \(BK\)? Wait, no, maybe \(\frac{GY}{BA}=\frac{YK}{AK}=\frac{GK}{BK}\). Wait, \(GK = 10\), \(BK = 6\)? No, \(GK\) is 10, \(BK\) is 6? Wait, no, the length of \(GK\) is 10, \(BK\) is 6? Wait, no, the vertical segment: \(GK = 10\), \(BK = 6\)? Wait, no, the other way: \(GK\) is 10, \(BK\) is 6? Wait, maybe I mixed up. Wait, let's take the sides: \(YK = 25\), \(AK = 15\), \(GY = x\), \(BA = 12\), \(GK = 10\), \(BK = 6\). Wait, the ratio of \(YK\) to \(AK\) is \(\frac{25}{15}=\frac{5}{3}\). The ratio of \(GK\) to \(BK\) is \(\frac{10}{6}=\frac{5}{3}\). So that's consistent. Now, the ratio of \(GY\) to \(BA\) should also be \(\frac{5}{3}\). So \(\frac{x}{12}=\frac{5}{3}\). Wait, no, wait, \(\frac{GY}{BA}=\frac{YK}{AK}\), so \(\frac{x}{12}=\frac{25}{15}\). Simplify \(\frac{25}{15}=\frac{5}{3}\). Then \(x=\frac{5}{3}\times12 = 20\). Wait, let's check with another ratio. \(\frac{GK}{BK}=\frac{10}{6}=\frac{5}{3}\), and \(\frac{YK}{AK}=\frac{25}{15}=\frac{5}{3}\), so the ratio is \(\frac{5}{3}\). Then \(\frac{GY}{BA}=\frac{5}{3}\), so \(GY=\frac{5}{3}\times BA=\frac{5}{3}\times12 = 20\).
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