if triangle ABC DE is parallel to BC and AD/DB=2/3 calculate the value of area of trapezium DECB/AREA of the triangle of ABC
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In △ABC, we have
DE∣∣BC
⇒
AD
AB
=
AE
AC
Thus, in triangles ABC and ADE, we have
AD
AB
=
AE
AC
and, ∠A=∠A
Therefore, by SAS-criterion of similarity, we have
△ABC∼△ADE
⇒
AD
AB
=
DE
BC
.......(i)
It is given that
AD
AB
=
3
2
⇒
AD
DB
=
2
3
⇒
AD
DB
+1=
2
3
+1
⇒
AD
DB+AD
=
2
5
⇒
AD
AB
=
2
5
.............(ii)
From (i) and (ii), we get
DE
BC
=
2
5
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