Advertisements
Advertisements
Question
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
Advertisements
Solution
Given: The medians BE and CF of a triangle ABC intersect at G
To prove: ar (ΔGBC) = ar (AFCE)
Proof: As median CF divides a triangle into triangle of equal area. So, ar (ΔBCF) = ar (ΔACF)
ar (ΔGBF) + ar (ΔGBC) = ar (AFGE) + ar (ΔGCE) ...(I)
Now, median BE divides a triangle into two triangle of equal area.
So, ar (ΔGBF) + ar (AFGE) = ar (ΔGCE) + ar (ΔGBC) ...(II)
Now, subtracting (II) from (I), we get
ar (ΔGBC) – ar (AFGE) = ar (ΔAFGE) – ar (ΔGBC)
ar (ΔGBC) + ar (ΔGBC) = ar (ΔAFGE) + ar (ΔAFGE)
2ar (ΔGBC) = 2ar (AFGE)
Hence, ar (ΔGBC) = ar (AFGE).
APPEARS IN
RELATED QUESTIONS
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 quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint : From A and C, draw perpendiculars to BD.]
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 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)
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 ______.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = `1/4` ar (ABC).
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).
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)
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).
