Advertisements
Advertisements
प्रश्न
If the medians of a ∆ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = `1/3` ar (ABC)
Advertisements
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.
APPEARS IN
संबंधित प्रश्न
In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE)
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).
D and E are points on sides AB and AC respectively of ΔABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
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 given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
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 below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).
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 ______.
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)
