QUESTION IMAGE
Question
in the diagram, \\( \triangle a b c \sim \triangle d b e \\).
what is the length of \\( a b \\)?
Step1: Find length of DB
In right triangle \( \triangle DBE \), \( DE = 4 \), \( BE = 3 \). By Pythagorean theorem, \( DB=\sqrt{DE^{2}+BE^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{16 + 9}=\sqrt{25}=5 \).
Step2: Set up proportion for similar triangles
Since \( \triangle ABC\sim\triangle DBE \), the ratios of corresponding sides are equal. So \( \frac{AB}{DB}=\frac{AC}{DE} \). We know \( AC = 10 \), \( DE = 4 \), \( DB = 5 \). Substitute values: \( \frac{AB}{5}=\frac{10}{4} \).
Step3: Solve for AB
Cross - multiply: \( 4\times AB=5\times10 \), so \( 4AB = 50 \), then \( AB=\frac{50}{4}=\frac{25}{2}=12.5 \). Wait, correction: Wait, actually, the corresponding sides: \( AC \) corresponds to \( DE \), \( BC \) corresponds to \( BE \), and \( AB \) corresponds to \( DB \)? Wait, no, let's re - check the similar triangles. \( \angle C=\angle E = 90^{\circ} \), \( \angle B=\angle B \), so \( \triangle ABC\sim\triangle DBE \) by AA similarity. So the ratio of \( AC \) to \( DE \) is equal to the ratio of \( AB \) to \( DB \)? Wait, no, \( AC \) is corresponding to \( DE \), \( BC \) corresponding to \( BE \), and \( AB \) corresponding to \( DB \)? Wait, \( AC = 10 \), \( DE = 4 \), \( BE = 3 \), \( BC=BE + EC \)? Wait, no, maybe I made a mistake in corresponding sides. Wait, actually, \( DE\parallel AC \) (since both are perpendicular to \( BC \)), so \( \triangle DBE\sim\triangle ABC \) with \( DE \) corresponding to \( AC \), \( BE \) corresponding to \( BC \), and \( DB \) corresponding to \( AB \). Wait, no, \( BE \) is a part of \( BC \). Wait, let's re - define the similar triangles. \( \angle B \) is common, \( \angle BED=\angle BCA = 90^{\circ} \), so \( \triangle DBE\sim\triangle ABC \). Therefore, \( \frac{DB}{AB}=\frac{DE}{AC}=\frac{BE}{BC} \). We found \( DB = 5 \), \( DE = 4 \), \( AC = 10 \). So \( \frac{5}{AB}=\frac{4}{10} \). Cross - multiply: \( 4AB=5\times10 = 50 \), so \( AB=\frac{50}{4}=12.5 \)? Wait, no, that can't be. Wait, no, the correct correspondence: \( AC \) is corresponding to \( DE \), \( AB \) is corresponding to \( DB \)? No, wait, \( AC = 10 \), \( DE = 4 \), \( BE = 3 \), and \( DB = 5 \). Wait, maybe the ratio is \( \frac{AB}{DB}=\frac{AC}{DE} \), so \( AB=\frac{AC\times DB}{DE}=\frac{10\times5}{4}=\frac{50}{4}=12.5 \). But let's check again. Wait, in \( \triangle DBE \), \( DE = 4 \), \( BE = 3 \), so \( DB = 5 \). In \( \triangle ABC \), \( AC = 10 \), and since they are similar, \( \frac{AC}{DE}=\frac{AB}{DB} \), so \( AB=\frac{AC\times DB}{DE}=\frac{10\times5}{4}=12.5 \). But wait, maybe the problem is that \( BC \) is equal to \( BE \) plus something? No, the diagram shows \( BE = 3 \), \( AC = 10 \), \( DE = 4 \). Wait, another way: Let \( AB=x \), \( DB = 5 \) (from 3 - 4 - 5 triangle). Since \( \triangle ABC\sim\triangle DBE \), \( \frac{AC}{DE}=\frac{AB}{DB} \), so \( \frac{10}{4}=\frac{x}{5} \), so \( x=\frac{10\times5}{4}=\frac{50}{4}=12.5 \). But 12.5 is \( \frac{25}{2} \). Wait, but maybe the corresponding sides are \( AC \) to \( DE \), \( BC \) to \( BE \), so \( \frac{AC}{DE}=\frac{BC}{BE} \), so \( \frac{10}{4}=\frac{BC}{3} \), so \( BC=\frac{30}{4}=7.5 \), then in \( \triangle ABC \), \( AC = 10 \), \( BC = 7.5 \), so \( AB=\sqrt{10^{2}+7.5^{2}}=\sqrt{100 + 56.25}=\sqrt{156.25}=12.5 \). Yes, that's correct. So the length of \( AB \) is \( 12.5 \) or \( \frac{25}{2} \). Wait, but let's do it again.
First, in \( \triangle DBE \), \( DE = 4 \), \( BE = 3 \), right - angled at \( E \), so \( DB=\sqrt{4^{2}+3^{2}} = 5 \) (3 - 4 - 5 right triangle).
Since \( \triangle ABC\sim\triangle DBE \) (AA:…
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\( 12.5 \) (or \( \frac{25}{2} \))