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प्रश्न
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
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उत्तर
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).
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संबंधित प्रश्न
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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).
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
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, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-
(i) ΔMBC ≅ ΔABD
(ii) ar (BYXD) = 2 ar(MBC)
(iii) ar (BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar (CYXE) = ar(ACFG)
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
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 the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = `1/4` ar (ABC).
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)
If the medians of a ∆ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = `1/3` ar (ABC)
