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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
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In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(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.]
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 following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]
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)
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(vii) ar (BCED) = ar(ABMN) + ar(ACFG)
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