Question:
In \(△ABC\), \(DE || BC\), \(\frac {AD}{DB}=\frac 35\) and \(AC=5.6\) cm, then \(AE =\)\(?\)
In \(△ABC\), \(DE || BC\), \(\frac {AD}{DB}=\frac 35\) and \(AC=5.6\) cm, then \(AE =\)\(?\)
Updated On: Apr 29, 2025
- 3 cm
- 5 cm
- 2.1 cm
- 7 cm
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The Correct Option is C
Solution and Explanation
In $\triangle ABC$, $DE \parallel BC$. We are given that $\frac{AD}{DB} = \frac{3}{5}$ and $AC = 5.6$ cm.
By the Basic Proportionality Theorem (also known as Thales' Theorem), we have $$ \frac{AD}{DB} = \frac{AE}{EC} $$ Also, we have $\frac{AD}{AB} = \frac{AE}{AC}$. We are given $\frac{AD}{DB} = \frac{3}{5}$. Then $$ \frac{AD}{AD + DB} = \frac{AD}{AB} = \frac{3}{3+5} = \frac{3}{8} $$ Since $\frac{AD}{AB} = \frac{AE}{AC}$, we have $$ \frac{3}{8} = \frac{AE}{5.6} $$ So, $$ AE = \frac{3}{8} \times 5.6 = \frac{3}{8} \times \frac{56}{10} = \frac{3 \times 7}{10} = \frac{21}{10} = 2.1 $$ Therefore, $AE = 2.1$ cm.
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