Advertisements
Advertisements
प्रश्न
In the following figure, 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).

Advertisements
उत्तर
Given: X and Y are the mid-points of AC and AB respectively. Also, QP || BC and CYQ, BXP are straight lines.
To prove: ar (ΔABP) = ar (ΔACQ)
Proof: Since, X and Y are the mid-points of AC and AB respectively.
So, XY || BC
We know that, triangles on the same base and between the same parallels are equal in area.
Here, ΔBYC and ΔBXC lie on same base BC and between the same parallels BC and XY.
So, ar (ΔBYC) = ar (ΔBXC)
On subtracting ar (ΔBOC) from both sides, we get
ar (ΔBYC) – ar (ΔBOC) = ar (ΔBXC) – ar (ΔBOC)
=» ar (ΔBOY) = ar (ΔCOX)
On adding ar (ΔXOY) both sides, we get
ar (ΔSOY) + ar (ΔXOY) = ar (ΔCOX) + ar (ΔXOY)
⇒ ar (ΔBYX) = ar (ΔCXY) ...(i)
Hence, we observe that quadrilaterals XYAP and YXAQ are on the same base XY and between the same parallels XY and PQ.
ar (XYAP) = ar (YXAQ) ...(ii)
On adding equations (i) and (ii), we get
ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ)
⇒ ar (ΔABP) = ar (ΔACQ)
Hence proved.
APPEARS IN
संबंधित प्रश्न
In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE)

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that
ar (ABCD) = ar (PBQR).
[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)]

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar(PRQ) = 1/2 ar(ARC)
(ii) ar(RQC) = 3/8 ar(ABC)
(iii) ar(PBQ) = ar(ARC)
In a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :
(1) ar (Δ PBQ) = ar (Δ ARC)
(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)
(3) ar (Δ RQC) =`3/8` ar (Δ ABC) .
In a ΔABC, if L and M are points on AB and AC respectively such that LM || BC. Prove
that:
(1) ar (ΔLCM ) = ar (ΔLBM )
(2) ar (ΔLBC) = ar (ΔMBC)
(3) ar (ΔABM) ar (ΔACL)
(4) ar (ΔLOB) ar (ΔMOC)
In the following figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = `1/2` ar (EFGD).

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See figure). Prove that ar (LZY) = ar (MZYX)

O is any point on the diagonal PR of a parallelogram PQRS (Figure). Prove that ar (PSO) = ar (PQO).

