Question

in triangle ABC seg DE parallel side BC. If 2 area of triangle ADE = area of quadrilateral DBCE find AB : AD show that BC = square root 3 × DE​

Answer 1

Solution :-

given ,

→ 2 * Area (∆ADE) = Area (Quad.DBCE)

→ Area (∆ADE) / Area (Quad.DBCE) = 1/2

so,

→ Area (∆ADE) / Area (∆ADE) + Area (Quad.DBCE) = 1/(1 + 2) = 1/3

→ Area (∆ADE) / Area (∆ABC) = 1/3

now, in ∆ADE and ∆ABC, we have,

→ ∠ADE = ∠ABC (given that, DE || BC, so , corresponding angles .)

→ ∠AED = ∠ACB (corresponding angles .)

then,

→ ∆ADE ~ ∆ABC (By AA similarity.)

now, we know that,

• Ratio of areas of two similar ∆'s = Ratio of square of their corresponding sides.

therefore,

→ Area (∆ADE) / Area (∆ABC) = DE²/BC²

→ (1/3) = (DE/BC)²

square root both sides,

→ DE / BC = 1/√3

hence,

→ BC = √3DE . (Proved).

Learn more :-

*जर △ ABC ~ △ DEF असून AB = 12 सेमी and DE = 14 सेमी. तर △ ABC आणि △ DEF यांच्या क्षेत्रफळाचे गुणोत्तर किती?.*

1️⃣ 49/9...

brainly.in/question/37096819

*In triangle ABC, DE || AB. If CD = 3 cm, EC = 4 cm, BE = 6 cm, then DA is equal to …………….*

1️⃣ 7.5 cm

2️⃣ 3 cm

3️⃣ 4.5...

brainly.in/question/37020783

In the figure , Line PQ || Side BC AP = 6 , PB = 8 , AQ = x and QC = 12 , then write the value of x.

https://brainly.in/question/36586072

Answer 2

Step-by-step explanation:

This is ur answer just try to make a suitable diagram if u want to be a geometry champ just try to think which theorems we can use here bcs diagrams and appropriate required theorems are heart and soul of geometry