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प्रश्न
If the medians of a ∆ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = `1/3` ar (ABC)
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उत्तर
Given: In ΔABC, AD, BE and CF are medians and intersect at G.
To prove: ar (ΔAGB) = ar (ΔAGC) = ar (ΔBGC) = `1/3` ar (ΔABC)
Proof: We know that, a median of a triangle divides it into two triangles of equal area.
In ΔABC, AD is a median
∴ ar (ΔABD) = ar (ΔACD) ...(i)
In ΔBGC, GD is a median
∴ ar (ΔGBD) = ar (ΔGCD) ...(ii)
On subtracting equation (ii) from equation (i), we get
ar (ΔABD) – ar (ΔGBD) = ar (ΔACD) – ar (ΔGCD)
⇒ ar (ΔAGB) = ar (ΔAGC) ...(iii)
Similarly, ar (ΔAGB) = ar (ΔBGC) ...(iv)
From equations (iii) and (iv),
ar (ΔAGB) = ar (ΔBGC) = ar (ΔAGC) ...(v)
Now, ar (ΔABC) = ar (ΔAGB) + ar (ΔBGC) + ar (ΔAGC)
⇒ ar (ΔABC) = ar (ΔAGB) + ar (ΔAGB) + ar (ΔAGB) ...[From equation (v)]
⇒ ar (ΔABC) = 3 ar (ΔAGB)
⇒ ar (ΔAGB) = `1/3` ar (ΔABC) ...(vi)
From eqattions (v) and (vi),
ar (ΔBGC) = `1/3` ar (ΔABC)
And ar (ΔAGC) = `1/3` ar (ΔABC)
Hence proved.
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संबंधित प्रश्न
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).
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(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]
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.]
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(4) area (Δ ABC) = 2 area (ΔBEC)
(5) ar (ΔFED) `= 1/8` ar (ΔAFC)
(6) ar (Δ BFE) = 2 ar (ΔEFD)
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