Mathematics
If the medians of a ΔABC intersect at G, show that :
ar (ΔAGB) = ar (ΔAGC) = ar (ΔBGC) = ar (ΔABC).
Theorems on Area
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Answer
In ΔABC,
Median AD divides ΔABC into two Δs of equal area.
∴ ar (ΔABD) = ar (ΔACD) ….(1)
In ΔGBC,
Median GD divides ΔGBC into two Δs of equal area.
∴ ar (ΔGBD) = ar (ΔGCD) ….(2)
Subtracting equation (2) from equation (1), we get :
ar (ΔABD) − ar (ΔGBD) = ar (ΔACD) − ar (ΔGCD)
ar (ΔAGB) = ar (ΔAGC) ………..(3)
In ΔABC,
Median BE divides ΔABC into two Δs of equal area.
∴ ar (ΔABE) = ar (ΔBCE) ….(4)
In ΔGAC,
Median GE divides ΔGAC into two Δs of equal area.
∴ ar (ΔGEA) = ar (ΔGEC) ….(5)
Subtracting equation (5) from equation (4), we get :
ar (ΔABE) − ar (ΔGEA) = ar (ΔBCE) − ar (ΔGEC)
ar (ΔAGB) = ar (ΔBGC) ………..(6)
From equation (3) and (6), we get :
∴ ar (ΔAGB) = ar (ΔAGC) = ar (ΔBGC)
From figure,
ar (ΔAGB) + ar (ΔAGC) + ar (ΔBGC) = ar (ΔABC)
3ar (ΔAGB) = ar (ΔABC)
ar (ΔAGB) = × ar (ΔABC)
Hence, proved that ar (ΔAGB) = ar (ΔAGC) = ar (ΔBGC) = ar (ΔABC).
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Related Questions
In the given figure, D is the mid-point of BC and E is the mid-point of AD. Prove that :
ar (ΔABE) = ar (ΔABC).
In the given figure, a point D is taken on side BC of ΔABC and AD is produced to E, making DE = AD. Show that :
ar (ΔBEC) = ar (ΔABC).
D is a point on base BC of a ΔABC such that 2BD = DC. Prove that :
ar (ΔABD) = ar (ΔABC).
In the given figure, AD is a median of ΔABC and P is a point on AC such that :
ar (ΔADP) : ar (ΔABD) = 2 : 3.
Find :
(i) AP : PC
(ii) ar (ΔPDC) : ar (ΔABC)
