Advertisements
Advertisements
प्रश्न
In the following figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)

Advertisements
उत्तर
Given: ABCDE is a pentagon.
BP || AC and EQ|| AD.
To prove: ar (ABCDE) = ar (APQ)
Proof: We know that, triangles on the same base and between the same parallels are equal in area.
Here, ΔADQ and ΔADE lie on the same base AD and between the same parallels AD and EQ.
So, ar (ΔADQ) = ar (ΔADE) ...(i)
Similarly, ΔACP and ΔACB lie on the same base AC and between the same parallels AC and BP.
So, ar (ΔACP) = ar (ΔACB) ...(ii)
On adding equations (i) and (ii), we get
ar (ΔADQ) + ar (ΔACP) = ar (ΔADE) + ar (ΔACB)
On adding ar (ΔACD) both sides, we get
ar (ΔADQ) + ar (ΔACP) + ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD)
⇒ ar (ΔAPQ) = ar (ABCDE)
Hence proved.
APPEARS IN
संबंधित प्रश्न
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).

D and E are points on sides AB and AC respectively of ΔABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

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.]
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)
If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is ______.
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.
In ∆ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Figure), then prove that ar (BPQ) = `1/2` ar (∆ABC).

