Math, asked by ayushi1012, 11 months ago

23.
In the given figure, DE is parallel to the base BC of triangle ABC
Area of ADEF
and AD: DB = 3: 5. Find the ratio
Area of ACFB
(A) 25:9
(B) 9:25
B
C) 9:64
(D) 25:64

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Answers

Answered by bhagyashreechowdhury
5

The ratio of the area of ∆DEF and area of ∆CFB is option (C): 9:64.

Step-by-step explanation:

It is given that,

In ∆ABC, DE // BC

AD:DB = 3:5

Step 1:

Consider ∆ADE and ∆ABC, we have

∠A = ∠A ….. [common angle]

∠ADE = ∠ABC ……. [corresponding angles]

By AA similarity, ∆ADE ~ ∆ABC

Since the corresponding sides of similar triangles are proportional to each other.

AD/AB = DE/BC

⇒ AD/(AD+DB) = DE/BC

⇒ 3/(3+5) = DE/BC

DE/BC = 3/8 ……. (i)

Step 2:

Consider ∆DFE and ∆CFB, we have

∠DFE = ∠BFC …… [vertically opposite angles]

∠EDF = ∠BCF ……. [alternate angles]

By AA similarity, ∆DFE ~ ∆CFB

We know that the ratio of the two similar triangles is equal to the ratio of the square of their corresponding sides.

[Area (∆DFE)] / [Area (∆CFB)] = [DE²]/[BC²]

Substituting the value from (i)

[Area (∆DFE)] / [Area (∆CFB)] = [3²]/[8²] = [9/64]option (C)

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