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In the figure, $X$ and $Y$ are the mid-points of $AC$ and $AB$ respectively, $QP \parallel BC$ and $CYQ$ and $BXP$ are straight lines. Prove that $ar(\triangle ABP) = ar(\triangle ACQ)$.
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Academic Mathematics NCERT Class 9

Given:

$X$ and $Y$ are the mid-points of $AC$ and $AB$ respectively, $QP \parallel BC$ and $CYQ$ and $BXP$ are straight lines.

To do:

We have to prove that $ar(\triangle ABP) = ar(\triangle ACQ)$.

Solution:

$X$ and $Y$ are mid points of sides $AC$ and $AB$.

This implies,

$XY \parallel BC$

Similarly,

$XY \parallel PQ$

$\triangle BXY$ and $\triangle CXY$ are on the same base $XY$ and between the same parallels.

This implies,

$ar(\triangle BXY) = ar(\triangle CXY)$...…(i)

Trapezium $XYAP$ and trapezium $XYAQ$ are on the same base $XY$ and between the same parallels

This implies,

$ar(XYAP) = ar(XYAQ)$......…(ii)

Adding (i) and (ii), we get,

$ar(\triangle BXY) + ar(XYAP) = ar(\triangle CXY) + ar(XYAQ)$

$ar(\triangle ABP) = ar(\triangle ACQ)$.

Hence proved.

Updated on: 10-Oct-2022

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