In TRiangle ABC , DE || BC, DE = 3cm, BC = 9cm and ar(ADE) = 30cm2. Find ar (trap. BCED
ANS-240cm2
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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.
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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
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
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