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In ΔABC, D and E are points on AB and AC respectively such that DE ∥ BC. If AE = 2 cm, EC = 3 cm and BC = 10 cm, then DE is equal to:

  1. 4 cm

  2. 5 cm

  3. 203\dfrac{20}{3} cm

  4. 15 cm

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Answer

In ΔABC, D and E are points on AB and AC respectively such that DE ∥ BC. If AE = 2 cm, EC = 3 cm and BC = 10 cm, then DE is equal to: Similarity of Triangles, RSA Mathematics Solutions ICSE Class 10.

From figure,

AC = AE + EC = 2 + 3 = 5 cm.

In ΔABC and ΔADE,

∠A = ∠A [Common angle]

∠ABC = ∠ADE [Corresponding angles are equal]

∴ ΔABC ∼ ΔADE (By A.A. axiom).

In similar triangles, ratios of corresponding sides are equal.

∴AEAC=DEBC⇒25=DE10⇒DE=2×105⇒DE=205⇒DE=4 cm.\therefore \dfrac{AE}{AC} = \dfrac{DE}{BC} \\[1em] \Rightarrow \dfrac{2}{5} = \dfrac{DE}{10} \\[1em] \Rightarrow DE = \dfrac{2 \times 10}{5} \\[1em] \Rightarrow DE = \dfrac{20}{5} \\[1em] \Rightarrow DE = 4 \text{ cm.}

Hence, option 1 is the correct option.

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