In Fig.9.29, `a r\ (B D P)\ =\ a r\ (A R C)`and `a r\ (B D P)\ =\ a r\ (A R C)`. Show that both the quadrilaterals ABCD and DCPR are trapeziums.

In Fig.9.28, A P" "||" "B Q" "||" "C R . Prove that a r" "(A Q C)" "=" "a r" "(P B R)dot

A B C D IS A PARALLELOGRAM AND O is any point in its interior. Prove that: a r\ ( A O B)+\ a r\ ( C O D)=\ a r\ ( B O C)+\ a r( A O D) a r\ ( A O B)+a r\ (C O D)=1/2\ a r(^(gm)A B C D) Given: A parallelogram A B C D\ a n d\ O is a point in its interior. To Prove: a r\ (\ A O B)+\ a r( C O D)=a r\ ( B O C)+a r( A O D)

If each diagonal of a quadrilateral separates it into two triangles of equal area then show that the quadrilateral is a parallelogram. GIVEN : A quadrilateral A B C D such that its diagonals A C and B D are such that a r( A B D)=a r( C D B ) and a r( A B C)=a r( A C D)dot TO PROVE: Quadrilateral A B C D is a parallelogram.

Diagonals A C a n d B D of a quadrilateral A B C D intersect at O in such a way that a r ( A O D)=a r ( B O C)dot Prove that A B C D is a trapezium.

Diagonals A C and B D of a quadrilateral A B C D intersect at O in such a way that a r( A O D)=a r( B O C) . Prove that A B C D is a trapezium.

If the diagonals A C ,B D of a quadrilateral A B C D , intersect at O , and seqarate the quadrilateral into four triangles of equal area, show that quadrilateral A B C D is a parallelogram. GIVEN : A quadrilateral A B C D such that its diagonals A C and B D intersect at O and separate it into four parts such that a r( A O B)=a r( B O C)=a r( C O D)=a r( A O D) TO PROVE : Quadrilateral A B C D is a parallelogram.

In a triangle, P ,\ Q\ a n d\ R are the mid-points of sides B C ,\ C A\ a n d\ A B respectively. If A C=21\ c m ,\ B C=29 c m\ a n d\ A B=30 c m , find the perimeter of the quadrilateral A R P Q

In Figure, A B C D is a trapezium in which A B||C D . Prove that: a r( A O D)=a r( B O C)dot

In Figure, A B C D is a parallelogram. Prove that: a r( B C P)=a r( D P Q) CONSTRUCTION : Join A Cdot

If A B C D is a parallelogram, then prove that a r\ (\ A B D)=\ a r\ (\ B C D)=\ a r\ (\ A B C)=\ a r\ (\ A C D)=1/2\ a r\ (|""|^(gm)A B C D)