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  • In Fig., X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (ABP) = ar (ACQ).

In Fig., 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 (ABP) = ar (ACQ).

Answers (1)

Solution.
Given: X and Y are the mid points of AC and AB respectively.
QP \parallel BC
CYQ, BXP are straight lines,
To prove: ar(\triangle ABP) = ar(\triangle ACQ)
Proof:
Since, X and Y are the mid-point of AC and AB.
So, XY \parallel BC
We know that triangles on the same base and between the same parallels are equal in area
Consider parallel lines BC and XY;
\triangle BYC and \triangle BXC lie on same base BC and between the same parallels BC and XY.
So, ar(\triangle BYC) = ar(\triangle BXC)
ar(\triangle BOC)+ar(\triangle BOY)=ar(\triangle BOC)+ar(\triangle XOC)
\Rightarrow ar(\triangle BOY)=ar(\triangle COX)
Adding ar(XOY) to LHS and RHS
\Rightarrow ar(BOY) + ar(XOY) = ar(\triangle COX) + ar(\triangle XOY)
\Rightarrow ar(\triangle BYX) = ar(\triangle CXY)               …(i)
We observe that the quadrilateral XYAP and XYAQ have the same base XY and these are between the same parallel lines XY and PQ.
\because ar(XYAP)=ar(YXAQ)                 …(ii)
on adding eq. (i) and (ii), we get
ar (\triangle BYX) + ar(XYAP) = ar(\triangle CXY) + ar(YXAQ)
\Rightarrow ar(\triangle ABP) = ar(\triangle ACQ)
Hence Proved.

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