Question:
In triangle ABC, ‘D’ and ‘E’ are two points on sides AB and AC, respectively such that DE is parallel to BC. If AD=16 cm, BD=(5x-16) cm, AE=2x cm and EC=(25-2x) cm, then find the value of ‘x’.
In triangle ABC, ‘D’ and ‘E’ are two points on sides AB and AC, respectively such that DE is parallel to BC. If AD=16 cm, BD=(5x-16) cm, AE=2x cm and EC=(25-2x) cm, then find the value of ‘x’.
Updated On: Sep 13, 2024
- \(2\sqrt10\)
- \(3\sqrt5\)
- \(4\sqrt5\)
- \(3\sqrt10\)
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The Correct Option is A
Solution and Explanation
The correct option is (A): \(2\sqrt10\).
In triangle ABC, DE is parallel to BC
Therefore, triangle ADE is similar to triangle ABC
Therefore,
\((\frac{AD}{AB}) = (\frac{AE}{AC})\)
Or, {\(\frac{16}{(5x – 16 + 16)}\)} = {\(\frac{2x}{(25 – 2x + 2x)}\)}
Or, 5x × 2x = 16 × 25
Or, 10x2 = 400
Or, x2 = 40
Or, x = \(2\sqrt{10}\) (Since, length cannot be negative).
In triangle ABC, DE is parallel to BC
Therefore, triangle ADE is similar to triangle ABC
Therefore,
\((\frac{AD}{AB}) = (\frac{AE}{AC})\)
Or, {\(\frac{16}{(5x – 16 + 16)}\)} = {\(\frac{2x}{(25 – 2x + 2x)}\)}
Or, 5x × 2x = 16 × 25
Or, 10x2 = 400
Or, x2 = 40
Or, x = \(2\sqrt{10}\) (Since, length cannot be negative).
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