Question

In the given figure DE || BC.Area of trapezium (DECB)=2x Area of triangle ADE. Find AD/AB​

Answer 1

Answer: AD/AB = (1 /√3)

Step-by-step explanation:

Given data:

DE // BC  

Area of trapezium (DECB) = 2x Area of triangle ADE ….. (i)

To find: $\frac{AD}{AB}$

Step 1:

Consider from the figure attached below, in ∆ ADE & ∆ ABC, we have

∠ADE = ∠AED ….. [corresponding angles since AD//BC]

∠AED = ∠ACB …. [corresponding angles since AD//BC]

∴ By AA similarity  

∆ ADE ~ ∆ ABC

Since sides of two similar triangles are proportional to each other

∴ $\frac{AD}{AB}$ = $\frac{AE}{AC}$ = $\frac{DE}{BC}$

Step 2:

We know, the ratio of the areas of the similar triangles is equal to the square of the ratio of their corresponding sides.

∴ [ar(∆ADE)] / [ar(∆ABC)] = (AD/AB)² = (AE/AC)² = (DE/BC)² ….. (ii)

Step 3:

Now, considering eq. (i),

Ar[trapezium (DECB)] = 2x [Ar(∆ADE)]

⇒ Ar(∆ABC) – ar(∆ADE) = 2x [Ar(∆ADE)] .... [∵ ∆ABC = trapezium (DECB)+∆ADE]

⇒ Ar(∆ABC) = [2x Ar(∆ADE)] + ar(∆ADE)

⇒ Ar(∆ABC) = [3 x Ar(∆ADE)]

⇒ [Ar(∆ADE)] / [Ar(∆ABC)] = 1/3 …. (ii)

Thus,

From (i) & (ii), we get

(AD/AB)² = 1/3

⇒ AD/AB = √(1/3) = 1 /√3