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In figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that, Area of ∆ABP + area of ∆DPC = Area of ∆APD.

In figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that, Area of ∆ABP + area of ∆DPC = Area of ∆APD. Theorems on Area, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Theorems on Area

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Answer

We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallel lines.

∆APD and || gm ABCD are on the same base AD and between the same || lines AD and BC,

∴ Area of ∆APD = 12\dfrac{1}{2} Area of || gm ABCD …….(i)

From figure,

Area of ||gm ABCD = Area of ∆APD + Area of ∆ABP + Area of ∆DPC

Dividing the above equation by 2 we get,

Area of ||gm ABCD2=Area of ∆APD2+Area of ∆ABP2+Area of ∆DPC2\dfrac{\text{Area of ||gm ABCD}}{2} = \dfrac{\text{Area of ∆APD}}{2} + \dfrac{\text{Area of ∆ABP}}{2} + \dfrac{\text{Area of ∆DPC}}{2}

From (i),

Area of ∆APD=Area of ∆APD2+Area of ∆ABP2+Area of ∆DPC2Area of ∆APD−Area of ∆APD2=Area of ∆ABP2+Area of ∆DPC2Area of ∆APD2=Area of ∆ABP2+Area of ∆DPC2Area of ∆APD=Area of ∆ABP+Area of ∆DPC\text{Area of ∆APD} = \dfrac{\text{Area of ∆APD}}{2} + \dfrac{\text{Area of ∆ABP}}{2} + \dfrac{\text{Area of ∆DPC}}{2} \\[1em] \text{Area of ∆APD} - \dfrac{\text{Area of ∆APD}}{2} = \dfrac{\text{Area of ∆ABP}}{2} + \dfrac{\text{Area of ∆DPC}}{2} \\[1em] \dfrac{\text{Area of ∆APD}}{2} = \dfrac{\text{Area of ∆ABP}}{2} + \dfrac{\text{Area of ∆DPC}}{2} \\[1em] \text{Area of ∆APD} = \text{Area of ∆ABP} + \text{Area of ∆DPC}

Hence, proved that Area of ∆APD = Area of ∆ABP + Area of ∆DPC.

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