Question

In triangle ABC, D and E are points on sides AB and AC respectively DE parallel to BC. If A(triangle ADE)=25cm square and area of trapezium DBCE=24cm square and BC=14cm, find the length of DE​

Answer 1

Answer:

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Answer 2

Answer:

Let AM be the altitude of triangle ABC

Let AM meet DE at P

Thus, AP is the altitude of triangle ADE

and, PM is the altitude of trapezium DBCE

Area triangle ABC = Area DECB + Area triangle ADE

=> 1/2 × CB × AM = 24 + 25

=> 1/2 × 14 × AM = 49

=> AM = 7 cm

Now, PM = x (say), then AP = AM - PM = 7 - x

Area triangle ADE = 25

=> 1/2 × AP × DE = 25

=> 1/2 × (7 - x) × DE = 25

x = (7DE - 50)/DE cm

Area trapezium DBCE = 24

=> 1/2 × (DE + BC) × PM = 24

=> (DE + 14) × x = 48

=> x = 48/(DE + 14)

Thus we get

(7DE - 50)/DE = 48/(DE + 14)

=> (7DE - 50)(DE + 14) = 48DE

=> 7DE² + 98DE - 50DE - 700 = 48DE

=> 7DE² = 700

=> DE² = 100

=> DE = 10 cm