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Question
In parallelogram ABCD, P is the mid-point of AB. CP and BD intersect each other at point O. If the area of ΔPOB = 40 cm2, and OP: OC = 1:2, find:
(i) Areas of ΔBOC and ΔPBC
(ii) Areas of ΔABC and parallelogram ABCD.
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Solution
(i) Joining AC we have the following figure
Consider the triangles ΔPOB and ΔCOD
∠POB = ∠DOC ...[ vertically opposite angles ]
∠OPB = ∠ODC ...[ AB and DC are parallel, CP and BD are the transversals, alternate interior angles are equal ]
Therefore, by Angle-Angle similarly criterion of congruence, ΔPOB ∼ ΔCOD
Since P is the mid-point AP = BP, and AB = CD, we have CD = 2 BP
Therefore, We have,
`"BP"/"CD" = "OP"/"OC"= "OB"/"OD" = 1/2`
⇒ OP : OC = 1: 2
(ii) Since from part ( i ), we have
`"BP"/"CD" = "OP"/"OC"= "OB"/"OD" = 1/2` ,
The ratio between the areas of two similar triangles is equal to the ratio between the square of the corresponding sides.
Here, ΔDOC and ΔPOB are similar triangles.
Thus, we have ,
`" Ar.( ΔDOC)"/"Ar.( ΔPOB )" ="DC"^2/"PB"^2"`
⇒ `"Ar.(ΔDOC )"/"Ar.( ΔPOB )" ="(2PB)"^2/"PB"^2"`
⇒ `"Ar.( ΔDOC )"/"Ar.( ΔPOB )" ="4PB"^2/"PB"^2"`
⇒ `"Ar.( ΔDOC )"/"Ar.( ΔPOB )"` = 4
⇒ Ar.( ΔDOC ) = 4Ar, ( ΔPOB )
= 4 x 40
= 160 cm2
Now consider Ar. ( ΔDBC ) = Ar. ( ΔDOC ) + Ar. (Δ BOC )
= 160 + 80
= 160 cm2
Two triangles are equal in the area if they are on equal bases and between the same parallels.
Therefore, Ar. ( ΔDBC ) = Ar. ( ΔABC ) = 240 cm2
The median divides the triangles into areas of two equal triangles.
Thus, CP is the median of the triangle ABC.
Hence, Ar. ( ΔABC ) = 2 Ar. ( ΔPBC )
Ar. ( ΔPBC ) = `"Ar.( ΔABC )"/2`
Ar. ( ΔPBC ) = 120 cm2
( iii ) From part (ii) we have,
Ar. ( ΔABC ) = 2Ar. ( PBC ) = 240 cm2
The area of a triangle is half the area of the parallelogram if both are on equal bases and between the same parallels.
Thus, Ar. ( ΔABC ) = `1/2` Ar. [ || gm ABCD ]
AR. [ || gm ABCD ] = 2 Ar. ( ΔABC )
AR. [ || gm ABCD ] = 2 x 240
AR. [ || gm ABCD ] = 480 cm2
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