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Question
The given figure shows a parallelogram ABCD with area 324 sq. cm. P is a point in AB such that AP: PB = 1:2
Find The area of Δ APD.
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Solution
(i) The ratio of the area of triangles with the same vertex and bases along the same line is equal to the ratio of their respective bases.
So, we have
`["Area of "Δ"APD"]/["Area of" Δ"BPD" ] = "AP"/"BP"` = `1/2`
Area of parallelogram ABCD = 324 sq.cm
The area of the triangles with the same base and between the same parallels are equal.
We know that the area of the triangle is half the area of the parallelogram if they lie on the same base and between the parallels.
Therefore, we have,
Area ( ΔABD ) = `1/2` x Area [ || gm ABCD ]
=`324/2`
= 162 sq.cm
From the diagram it is clear that,
Area (Δ ABD ) = Area ( ΔAPD ) + Area ( ΔBPD )
⇒ 162 = Area ( ΔAPD ) + 2Area ( ΔAPD )
⇒ 162 = 3Area ( ΔAPD )
⇒ Area ( ΔAPD ) =`162/3`
⇒ Area ( ΔAPD ) = 54 sq.cm
(ii) Consider the triangle ΔAOP and ΔCOD
∠AOP = ∠COD ....[ vertically opposite angles ]
∠CDO = ∠APD .....[ AB and DC are parallel and DP is the transversal, alternate interior angles are equal ]
Thus, by Angle-Angle similarly,
ΔAOP ∼ ΔCOD.
Hence the corresponding sides are proportional.
`"AP"/"CD"= "OP"/"OD" = "AP"/"AB"`
= `"AP"/"AP + PB"`
= `"AP"/"3AP"`
=`1/3`
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