A line segment DE is drawn parallel to base BC of - ABC which cuts AB at point D and AC at point E- If AB - 5 BD and EC - 3-2 cm- Find the length of AE-

The correct option is A 12.8 cm AB = 5 BD AB BD = 4 BD AD = 4 BD or AD/AB = nbsp;4/5 In ADE and ABC , ∠ DAE = ∠ BAC (Com ...

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Byju's Answer

Standard X

Mathematics

Construction of a Similar Triangle - When Vertices Are Distinct

A line segmen...

Question

A line segment DE is drawn parallel to base BC of $Δ A B C$ which cuts AB at point D and AC at point E. If AB = 5 BD and EC = 3.2 cm. Find the length of AE.

12.8 cm

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1.28 cm

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2.8 cm

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12.6 cm

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Solution

The correct option is A 12.8 cm
$A B = 5 B D$
$A B - B D = 4 B D$
$A D = 4 B D$
or $\frac{A D}{A B} = \frac{4}{5}$
In $△ A D E$ and $△ A B C$ ,
$∠ D A E = ∠ B A C$ (Common)
$∠ A D E = ∠ A B C$ (Corresponding angles)
$∠ A E D = ∠ A C B$ (Corresponding angles)
Thus, $△ A D E ≅ △ A B C$ (AAA rule)
Thus, $\frac{A D}{A B} = \frac{A E}{A C}$
$\frac{4}{5} = \frac{A E}{A E + E C}$
$5 A E = 4 A E + 3.2 \times 4$
$A E = 12.8$ cm

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A line segment DE is drawn parallel to base BC of ΔABC which cuts AB at point D and AC at point E. If AB = 5 BD and EC = 3.2 cm. Find the length of AE.

The correct option is A 12.8 cmAB=5BDAB−BD=4BDAD=4BDor ADAB=45In △ADE and △ABC,∠DAE=∠BAC (Common)∠ADE=∠ABC (Corresponding angles)∠AED=∠ACB (Corresponding angles)Thus, △ADE≅△ABC (AAA rule)Thus, ADAB=AEAC45=AEAE+EC5AE=4AE+3.2×4AE=12.8 cm

A line segment DE is drawn parallel to base BC of $Δ A B C$ which cuts AB at point D and AC at point E. If AB = 5 BD and EC = 3.2 cm. Find the length of AE.