Question

In ABC, D and E are points of AB and AC respectively such that DE||BC, if AD/DB= 3/5
AC= 4.8 the length of AE is.​

Answer 1

Answer:

The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.

It is given that AD=6 cm, DB=9 cm and AE=8 cm.

Using the basic proportionality theorem, we have

$</p><p></p><p></p><p>ABAD=ACAE=BCDE⇒ABAD=ACAE⇒156=AC8⇒6AC=15×8⇒6AC=120⇒AC=6120=20</p><p></p><p></p><p>Hence, \:  AC=20 cm.</p><p></p><p>$

Answer 2

GIVEN: In Δ ABC, D and E are points on AB and AC , DE || BC and AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BE = 5 cm.

In Δ ADE and Δ ABC,

∠ADE =∠ABC (corresponding angles)

[DE || BC, AB is transversal]

∠AED =∠ACB (corresponding angles)

[DE || BC, AC is transversal]

So, Δ ADE ~ Δ ABC (AA similarity)

Therefore, AD/AB = AE/AC = DE/BC

[In similar triangles corresponding sides are proportional]

AD/AB = DE/BC

2.4/(2.4+DB) = 2/5

2.4 × 5 = 2(2.4+ DB)

12 = 4.8 + 2DB

12 - 4.8 = 2DB

7.2 = 2DB

DB = 7.2/2

DB = 3.6 cm

Similarly, AE/AC = DE/BC

3.2/(3.2+EC) = 2/5

3.2 × 5 = 2(3.2+EC)

16 = 6.4 + 2EC

16 - 6.4 = 2EC

9.6 = 2EC

EC = 9.6/2

EC = 4.8 cm

Hence,BD = 3.6 cm and CE = 4.8 cm.

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