Advertisements
Advertisements
प्रश्न
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).
विकल्प
True
False
Advertisements
उत्तर
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
संबंधित प्रश्न
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that
(i) BDEF is a parallelogram.
(ii) ar (DEF) = 1/4ar (ABC)
(iii) ar (BDEF) = 1/2ar (ABC)
D and E are points on sides AB and AC respectively of ΔABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
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 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.
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.
O is any point on the diagonal PR of a parallelogram PQRS (Figure). Prove that ar (PSO) = ar (PQO).
In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)
If the medians of a ∆ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = `1/3` ar (ABC)
