Since CQ || PD, triangles DPQ and DPC share base PD and parallel lines, giving Area(ΔDPQ) = Area(ΔDPC). Thus, Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPQ) = Area(ΔBPD) + Area(ΔDPC) = Area(ΔBDC) = (1/2) x Area(ΔABC).
In ΔABC, D is the midpoint of AB. P is any point on BC and Q is a point on AB such that CQ || PD. PQ is joined (Fig. 6.34). Prove that Area (ΔBPQ) = 1/2 Area (ΔABC).
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Given CQ is parallel to PD.
Triangles DPQ and DPC lie on the same base PD and between the same parallel lines PD and CQ.
Therefore, their areas are equal:
Area(ΔDPQ) = Area(ΔDPC).
Now, consider triangle BPQ:
Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPQ)
Substitute Area(ΔDPC) for Area(ΔDPQ):
Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPC) = Area(ΔBDC).
Since D is the midpoint of AB, CD is a median of triangle ABC.
A median divides the triangle into two equal halves:
Area(ΔBDC) = (1/2) x Area(ΔABC).
Hence, Area(ΔBPQ) = (1/2) x Area(ΔABC).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/