Question

solve it's help me guys ​

Answer 1

Answer:

here,

DE//BC

so,

angleD=angleB

angleA is common

therefore triangleADE is similar to triangle ABC

(by AA criteria)

so, (AE/AC)^2 or (AD/DB)^2=area of triangle ADE/area of triangle ABC

i.e, 4/9=area of triangle ADE /36

area of triangle ADE =16cm^2

now the area of quad. BCED will be 36-16

i.e, 20cm^2

Answer 2

Given:

AD/DB = AE/EC = 2,

Then by converse of BPT,

DE || BC.

∴ ∠ADE = ∠ABC (corresponding angles)

∠AEC = ∠ACB (corresponding angles).

Therefore, ∆ADE ~ ∆ABC. (By AA criteria.)

As per data, ar(∆ABC) = 36 cm².

We know,

ar(∆ADE)/ar(∆ABC) = AD²/AB² = DE²/BC² = AE²/AC². __(A)

& AD/DB = 2

=> AD/(AB - AD) = 2

{Just getting the value in terms of AB which will help in getting the answer.}

=> 2AB - 2AD = AD

=> 2AB = 3AD

=> AD/AB = 2/3

=> AD²/AB² = (2/3)² = 4/9

=> ar(∆ADE)/ar(∆ABC) = 4/9

[From A]

=> ar(∆ADE) = 4/9 • ar(∆ABC) = 4/9 × 36 cm² = 16 cm²

Clearly, now ar(□BCED) = ar(∆ABC) - ar(∆ADE) = 36 cm² - 16 cm² = 20 cm². (D)