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Question
In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
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Solution
It is given that
Area (ΔDRC) = Area (ΔDPC)
As ΔDRC and ΔDPC lie on the same base DC and have equal areas, therefore, they must lie between the same parallel lines.
∴ DC || RP
Therefore, DCPR is a trapezium.
It is also given that
Area (ΔBDP) = Area (ΔARC)
⇒ Area (BDP) − Area (ΔDPC) = Area (ΔARC) − Area (ΔDRC)
⇒ Area (ΔBDC) = Area (ΔADC)
Since ΔBDC and ΔADC are on the same base CD and have equal areas, they must lie between the same parallel lines.
∴ AB || CD
Therefore, ABCD is a trapezium.
RELATED QUESTIONS
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).
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]
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 Fig. below, ABC and BDE are two equilateral triangles such that D is the mid-point of
BC. AE intersects BC in F. Prove that
(1) ar (Δ BDE) = `1/2` ar (ΔABC)
(2) Area ( ΔBDE) `= 1/2 ` ar (ΔBAE)
(3) ar (BEF) = ar (ΔAFD)
(4) area (Δ ABC) = 2 area (ΔBEC)
(5) ar (ΔFED) `= 1/8` ar (ΔAFC)
(6) ar (Δ BFE) = 2 ar (ΔEFD)
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = `1/4` ar (ABC).
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)
O is any point on the diagonal PR of a parallelogram PQRS (Figure). Prove that ar (PSO) = ar (PQO).
A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
