Question

ncert 9 ex 9.2 Q no. 4​

Answer 1

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Question ---> In figure, P is a point in the interior of a parallelogram ABCD.

Show that:

(i) ar (APB) + ar (PCD) = ar (ABCD)

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

Ans. (i) Draw a line passing through point P and parallel to AB which intersects AD at Q and BC at R respectively.

Now APB and parallelogram ABRQ are on the same base AB and between same parallels AB and QR.

ar (APB) = ar (gm ABRQ) ……….(i)

Also PCD and parallelogram DCRQ are on the same base AB and between same parallels AB and QR.

ar (PCD) = ar (gm DCRQ) ……….(ii)

Adding eq. (i) and (ii),

ar (APB) + ar (PCD)

= ar (gm ABRQ) + ar (gm DCRQ)

ar (APB) = ar (gm ABCD) ……….(iii)

(ii) Draw a line through P and parallel to AD which intersects AB at M and DC at N. Now APD and parallelogram AMND are on the same base AD and between same parallels AD and MN.

ar (APD) = ar (gm AMND) ……….(iv)

Also PBC and parallelogram MNCB are on the same base BC and between same parallels BC and MN.

ar (PBC) = ar (gm MNCB) ……….(v)

Adding eq. (i) and (ii),

ar (APD) + ar (PBC)

= ar (gm AMND) + ar (gm MNCB)

ar (APD) = ar (gm ABCD) ……….(vi)

From eq. (iii) and (vi), we get,

ar (APB) + ar (PCD) = ar (APD) + ar (PBC)

or ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

Hence proved.

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Answer 2

Answer:

(i) Draw a line passing through point P and parallel to AB which intersects AD at Q and BC at R respectively.

Now APB and parallelogram ABRQ are on the same base AB and between same parallels AB and QR.

ar (APB) = ar (gm ABRQ) ……….(i)

Also PCD and parallelogram DCRQ are on the same base AB and between same parallels AB and QR.

ar (PCD) = ar (gm DCRQ) ……….(ii)

Adding eq. (i) and (ii),

ar (APB) + ar (PCD)

= ar (gm ABRQ) + ar (gm DCRQ)

ar (APB) = ar (gm ABCD) ……….(iii)

(ii) Draw a line through P and parallel to AD which intersects AB at M and DC at N. Now APD and parallelogram AMND are on the same base AD and between same parallels AD and MN.

ar (APD) = ar (gm AMND) ……….(iv)

Also PBC and parallelogram MNCB are on the same base BC and between same parallels BC and MN.

ar (PBC) = ar (gm MNCB) ……….(v)

Adding eq. (i) and (ii),

ar (APD) + ar (PBC)

= ar (gm AMND) + ar (gm MNCB)

ar (APD) = ar (gm ABCD) ……….(vi)

From eq. (iii) and (vi), we get,

ar (APB) + ar (PCD) = ar (APD) + ar (PBC)

or ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

Hence proved.