Advertisements
Advertisements
Question
The base BC of triangle ABC is divided at D so that BD = `1/2`DC.
Prove that area of ΔABD = `1/3` of the area of ΔABC.
Solution
In ΔABC, ∵ BD = `1/2"DC" ⇒ "BD"/"DC" = 1/2`
∴ Ar.( ΔABD ) : Ar.( ΔADC ) = 1:2
But Ar.( ΔABD ) + Ar.( ΔADC ) = Ar.( ΔABC )
Ar.( ΔABD ) + 2Ar.( ΔABD ) = Ar.( ΔABC )
3 Ar.( ΔABD ) = Ar.( ΔABC )
Ar.( ΔABD ) = `1/3`Ar.( ΔABC )
APPEARS IN
RELATED QUESTIONS
The following figure shows a triangle ABC in which P, Q, and R are mid-points of sides AB, BC and CA respectively. S is mid-point of PQ:
Prove that: ar. ( ∆ ABC ) = 8 × ar. ( ∆ QSB )
In the given figure; AD is median of ΔABC and E is any point on median AD.
Prove that Area (ΔABE) = Area (ΔACE).
In the figure of question 2, if E is the mid-point of median AD, then
prove that:
Area (ΔABE) = `1/4` Area (ΔABC).
In the following figure, OAB is a triangle and AB || DC.
If the area of ∆ CAD = 140 cm2 and the area of ∆ ODC = 172 cm2,
find : (i) the area of ∆ DBC
(ii) the area of ∆ OAC
(iii) the area of ∆ ODB.
