Question

In TRiangle ABC , DE || BC, DE = 3cm, BC = 9cm and ar(ADE) = 30cm2. Find ar (trap. BCED

ANS-240cm2

Answer 1

In TRiangle ABC , DE || BC, DE = 3cm, BC = 9cm and ar(ADE) = 30cm2. Find ar (trap. BCED.

Good question,

Here is your perfect answer!

Since DE||BC, triangle ADE & triangle ABC are similar.

By similar triangle's area theorem,

=) ar(ADE) / ar(ABC) = (DE/BC)²

=) 30/ ar(ABC) = (3/9)²

=) 30/ar(ABC) = 1/9

=) 30*9 cm² = ar(ABC)

=) 270 cm² = ar(ABC),

Area of trapezium = ar(ABC) - ar(ADE)

= 270 - 30

= 240 cm².

Answer 2

since, DE || BC and DB is transversal,

angle ADE ~ angle ABC( AA Test) ___(1)

also,

EC is transversal,

angle AEC~ angle ACB____(2)

IN ∆ ABC and ∆ADE

angle ABC~angle ADE(from1)

angle ACB~angle AED(from 2)

∆ABC~∆ADE(AA test)

area(∆ABC)/ area (∆ADE)=BC²/DE²

area(∆ABC)/30=9²/3²

area(∆ABC)/30=81/9

area(∆ABC)=270

area(∆ADE)+ area( quad BCED)=270

30+area(quad BCED)= 270

area(quad BCED) =270-30

area (quad BCED) = 240