Home
Class 9
MATHS
P and Q are any two points lying on the ...

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).

Answer

Step by step text solution for 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). by MATHS experts to help you in doubts & scoring excellent marks in Class 9 exams.

Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • AREAS

    NCERT KANNAD|Exercise EXERCISE 11.3|4 Videos

  • AREAS

    NCERT KANNAD|Exercise THINK, DISCUSS AND WRITE|3 Videos

  • CIRCLES

    NCERT KANNAD|Exercise EXERCISE 12.5|3 Videos

Similar Questions

Explore conceptually related problems

D and E are points on sides AB and AC respectively of tirangle ABC such that ar (DBC) = ar (EBC) . Prove that DE || BC

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

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

D and E are points on sides AB and AC respectively of Delta ABC such that ar (DBC) = ar EBC). Prove that DE |\| BC.

DL and BM are the ehights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470cm^2 , AB=35cm and AD=49cm, Find the length of BM and DL.

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)

If P and Q are the middle points of the sides BC and CD of the parallelogram ABCD, then AP+AQ is equal to

If D, E and F are three points on the sides BC, CA and AB, respectively, of a triangle ABC such that the lines AD, BE and CF are concurrent, then show that " "(BD)/(CD)*(CE)/(AE)*(AF)/(BF)=-1

In trapezium ABCD, AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF || AB. Show that (AE)/(ED) = (BF)/(FC)

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)

Home
Profile