Question

answer the questions​

Answer 1

Answer:

tell me which standard , I will complete it accordingly..

Answer 2

Answer:

1

Given that ABCD is a trapezium with AB || DC and Diagonal AC and BD intersect each other at O.

To prove: Area (AOD) = Area (BOC)

Proof: ΔADC and ΔBDC are on the same base DC and between same parallel AB and DC.

∴Area (ΔADC) = Area (ΔBDC) [triangles on the same base and between same parallel are equal in area]

Subtract Area (ΔDOC) from both side

Area (ΔADC) – Area (ΔDOC) = Area (ΔBDC) – Area (ΔDOC)

Area (ΔAOD) = Area (ΔBOC)

Hence proved.

2

Required to prove that:

Ar( tri BQP) = 1/2 Ar( tri ABC)

PROOF: Since PD// AQ ( given)

=> ar( triPDQ)= ar( tri PDA) ( as triangles on the same base PD, & between the same parallels PD & AQ , will be equal in area) . . .(1)

Ar( tri BQP) = ar( tri PBD) + ar(tri PDQ)

=>Ar( tri BQP)= ar( triPBD)+ ar( tri PDA)…By (1)

=> ar( triBQP) = Ar( tri BDA)

=> ar( tri BQP) = 1/2 (ar tri ABC) ( since AD is median , which divides the triangle ABC into 2 triangles with equal area)

Hence proved that : ar(tri BQP)= 1/2 ( Ar tri ABC)