Math, asked by kuhusingh8322, 9 months ago

In Fig. 15.89, ABCD is a ||ᵍᵐ. O is any point on AC. PQ||AB and LM||AD. Prove that:
ar(||ᵍᵐ DLOP) = ar(||ᵍᵐ BMOQ)

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Answered by nikitasingh79
0

Given :  ABCD is a ||ᵍᵐ. O is any point on AC. PQ||AB and LM||AD.

 

To Prove  : ar(||ᵍᵐ DLOP) = ar(||ᵍᵐ BMOQ)

 

Proof :

Since, a diagonal of a parallelogram divides it into two triangles of equal area :  

∴ ar (∆ADC) = ar (∆ABC)

ar (∆APO) + ar (|| gm DLOP) + ar (∆OLC)

ar (∆AOM) + ar (|| gm DLOP) + ar (∆OQC) …………….(1)

 

Since,  AO and CO are diagonals of || gms AMOP and OQCL.

∴  ar (∆APO) = ar (∆AMO) ……………..(2)

ar(∆OLC) = ar(∆OQC) ………………..(3)

 

On Subtracting eq (2) from (3), we get

ar (|| gm DLOP)  = ar (|| gm BMOQ).

Hence  ar(||ᵍᵐ DLOP) = ar(||ᵍᵐ BMOQ) .

HOPE THIS ANSWER WILL HELP YOU…..

 

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Answered by Anonymous
0

Answer:

Step-by-step explanation:

ar (∆ADC) = ar (∆ABC)

ar (∆APO) + ar (|| gm DLOP) + ar (∆OLC)

ar (∆AOM) + ar (|| gm DLOP) + ar (∆OQC) …………….(1)

 

Since,  AO and CO are diagonals of || gms AMOP and OQCL.

∴  ar (∆APO) = ar (∆AMO) ……………..(2)

ar(∆OLC) = ar(∆OQC) ………………..(3)

 

On Subtracting eq (2) from (3), we get

ar (|| gm DLOP)  = ar (|| gm BMOQ)

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