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In the given figure DE is parallel to the base BC of the triangle ABC and AD:DB = 4:3. Find the ratio of DE:BC
and area of triangle DEF: area of triangle BFC
Answers
Answer:
Step-by-step explanation:
In ∆ABC , DE||BC
AD/DB= AE/EC
[ By B.P.T]
AD/DB= AE/(AC-AE)
3/5 = AE / (4.8 - AE)
3(4.8 - AE) = 5 AE
14.4 - 3AE = 5AE
14.4 = 5AE+3AE
14.4 = 8AE
AE = 14.4/8
AE= 1.8
Hence, AE is = 1.8 cm
Answer:
In △ABC, we have
DE||BC
⇒ ∠ADE=∠ABC and ∠AED=∠ACB [Corresponding angles]
Thus, in triangles ADE and ABC, we have
∠A=∠A [Common]
∠ADE=∠ABC
and, ∠AED=∠ACB
∴ △AED∼△ABC [By AAA similarity]
⇒
AB
AD
=
BC
DE
We have,
DB
AD
=
4
5
⇒
AD
DB
=
5
4
⇒
AD
DB
+1=
5
4
+1
⇒
AD
DB+AD
=
5
9
⇒
AD
AB
=
5
9
⇒
AB
AD
=
9
5
∴
BC
DE
=
9
5
In △DFE and △CFB, we have
∠1=∠3 [Alternate interior angles]
∠2=∠4 [Vertically opposite angles]
Therefore, by AA-similarity criterion, we have
△DFE∼△CFB
⇒
Area(△CFB)
Area(△DFE)
=
BC
2
DE
2
⇒
Area(△CFB)
Area(△DFE)
=(
9
5
)
2
=
81
25
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