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?
Answers
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.
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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