Home
Class 9
MATHS
In P is a point in the interior of a par...

In P is a point in the interior of a parallelogram ABCD. Show that`
(ii) `ar (APD) + ar (PBC) = ar (APB) + ar (PCD)`

Promotional Banner

Topper's Solved these Questions

  • AREA OF PARALLELOGRAMS AND TRIANGLES

    SUBHASH PUBLICATION|Exercise EXERCISE 11.3|8 Videos
  • AREA OF PARALLELOGRAMS AND TRIANGLES

    SUBHASH PUBLICATION|Exercise EXERCISE 11.4|3 Videos
  • AREA OF PARALLELOGRAMS AND TRIANGLES

    SUBHASH PUBLICATION|Exercise EXERCISE 11.4|3 Videos
  • APPENDIX A

    SUBHASH PUBLICATION|Exercise EXERCISE A 1.4|10 Videos
  • CIRCLES

    SUBHASH PUBLICATION|Exercise EXERCISE 12.6|7 Videos

Similar Questions

Explore conceptually related problems

IN the figure, 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)

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 E,F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = (1)/(2) ar (ABCD)

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)

In the figure 11.23, E is any point on median AD of a Delta ABC. Show that ar (ABE) = ar (ACE).

Diagonals AC and BD of a quadrilateral ABCD each other at P. Show that ar (APB) xx ar (CPD) =ar (APD) xx ar (BPC)

In the following figure, ABCD, DCFR and ABFE are parallelograms. Show that ar (ADE) = ar (BCF)

In the Given figure, E is any point on median AD of a triangle ABC Show that ar (ABE) = ar (ACE)

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) xx ar (CPD) = ar (APD) xx ar (BPD). [Hint : From A and C, draw perpendiculars to BD.

In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar (ACB) = ar (ACF) (ii) ar (AEDF) = ar (ABCDE)