In the adjoining figure mnpq and ABPQ are parallelogram and t is any point on the side bp prove that. (1) ar (mnpq) = ar (ABPQ) (2) ar (Axs) = 1/2 ar ( pqrs)
Answers
If MNPQ and ABPQ are parallelogram and t is any point on the side bp then the ar(MNPQ) = ar(ABPQ) and the ar(AQT) = 1/2 ar(MNPQ).
Hi there,
As per the data are given in the question, the second option should be prove the ar(AQT) = 1/2 ar(MNPQ). Thanks
Step-by-step explanation:
Case (1): Proving ar(MNPQ) = ar(ABPQ)
From the figure attached, we can say that
Both the parallelogram MNPQ and ABPQ are on the same base PQ and lie between the same parallel lines PQ and MB.
We know that if two parallelograms are on the same base and lie between the same parallel lines, then they have the same area.
∴ Area (parallelogram MNPQ) = Area (parallelogram ABPQ) …. (i)
Hence proved
Case 2: Proving ar(AQT) = 1/2 ar(MNPQ)
From the figure attached below, we can say that
The triangle AQT and the parallelogram ABPQ have the same base AQ and lie between the same parallel lines BP and AQ.
We know that if a triangle and a parallelogram are on the same base and lie between the same parallel lines, then the area of the triangle is equal to half the area of the parallelogram.
∴ Area (∆ AQT) = ½ * Area (parallelogram ABPQ)
But from eq. (i), we can write as
Area (∆ AQT) = ½ * Area (parallelogram MNPQ)
Hence proved
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Also View:
Theorem 9.1 parallelogram on the same base and between the same parallels are equal in area.
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