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P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
If a triangle and a parallelogram are on the same base and between the same parallel lines, the area of the triangle will be half of the area of the parallelogram.
It can be observed that ΔBQC and parallelogram ABCD lie on the same base BC and these are between the same set of parallel lines AD and BC.
∴ Area (ΔBQC) = 1/2 Area (ABCD).....(1)
Similarly, ΔAPB and parallelogram ABCD lie on the same base AB and between the same set of parallel lines AB and DC.
∴ Area (ΔAPB) = 1/2 Area (ABCD).....(2)
From Equations (1) and (2), we see that,
Area (ΔBQC) = Area (ΔAPB)
☛ Check: Class 9 Maths NCERT Solutions Chapter 9
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC)
NCERT Maths Solutions Class 9 Chapter 9 Exercise 9.2 Question 3
If ABCD is parallelogram such that P and Q are any two points lying on the sides DC and AD respectively, then we have proved that area of (APB) = area of (BQC).
☛ Related Questions:
- In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that (i) ar (APB) + ar (PCD) = 1/2 ar (ABCD) (ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD) [Hint: Through P, draw a line parallel to AB.]
- In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that (i) ar (PQRS) = ar (ABRS) (ii) ar (AXS) = 1/2 ar (PQRS)
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- In Fig.9.23, E is any point on median AD of a ∆ ABC. Show that ar (ABE) = ar (ACE).