Advertisements
Advertisements
Question
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).
Options
True
False
Advertisements
Solution
This statement is False.
Explanation:
In the given figure, join PG.
Since, G is the mid-point of CD.
Thus, PG is a median of ΔDPC and it divides the triangle into parts of equal areas.
Then, ar (ΔDPG) = ar (ΔGPC) = `1/2` ar (ΔDPC) ...(i)
Also, we know that, if a parallelogram and a triangle lie on the same base and between the same parallels, then area of triangle is equal to half of the area of parallelogram.
Here, parallelogram EFGD and ΔDPG lie on the same base DG and between the same parallels DG and EF.
So, ar (ΔDPG) = `1/2` ar (EFGD) ...(ii)
From equations (i) and (ii),
`1/2` ar (ΔDPG) = `1/2` ar (EFGD)
⇒ ar (ΔDPC) = ar (EFGD)
APPEARS IN
RELATED QUESTIONS
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that
ar (ABE) = ar (ACF)
In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
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 the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).
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)
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = `1/4` ar (ABC).
