Question

70 In △ABC, the line segment DE connects sides AB and BC so that DE ∥ AC . The lengths of the sides of △DBE are one third of the lengths of the sides of △ABC. What is the area of trapezoid ADEC if the area of △ABC is 27 cm2?

Answer 1

Answer:

REF.image

To prove : AB×EF=AD×EC

⇒

EC

AB

=

EF

AD

Proof :

AB=AC (∵ ABC is isosceles)

∴∠B=∠C (angles opposite to equal sides are equal) - (1)

In ΔABD and ΔECF

∠ABD=∠ECF (from (1))

∠ADB=∠EFC (Both are 90

∘

)

Using AA similarity

ΔADB∼ΔECF

⇒

EC

AB

=

EF

AD

⇒AB×EF=AD×EC

∴ Hence proved.

Answer 2

Given :-

• DE || AC.
• The lengths of the sides of △DBE are one third of the lengths of the sides of △ABC.
• The area of △ABC is 27 cm² .

To Find :-

• Area of trapezoid ADEC = ?

Solution :-

in ∆DBE and ∆ABC , we have ,

→ ∠DBE = ∠ABC (common)

→ ∠BDE = ∠BAC ( DE || AC.)

→ ∠BED = ∠BCA ( DE || AC.)

then,

→ ∆DBE ~ ∆ABC .

now, we know that, If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

therefore,

→ (Area ∆DBE) / (Area ∆ABC) = (BD/AB)²

→ (Area ∆DBE) / (Area ∆ABC) = (1/3)²

→ (Area ∆DBE) / (Area ∆ABC) = (1/9)

→ (Area ∆DBE) / 27 = (1/9)

→ Area ∆DBE = 3 cm².

hence,

→ Area of trapezoid ADEC = Area of ∆ABC - Area of ∆DBE

→ Area of trapezoid ADEC = 27 - 3

→ Area of trapezoid ADEC = 24 cm². (Ans.)

Learn more :-

In ABC, AD is angle bisector,

angle BAC = 111 and AB+BD=AC find the value of angle ACB=?

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