Advertisements
Advertisements
प्रश्न
In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE)
Advertisements
उत्तर
AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas.
∴ Area (ΔABD) = Area (ΔACD) ... (1)
ED is the median of ΔEBC.
∴ Area (ΔEBD) = Area (ΔECD) ... (2)
On subtracting equation (2) from equation (1), we obtain
Area (ΔABD) − Area (EBD) = Area (ΔACD) − Area (ΔECD)
Area (ΔABE) = Area (ΔACE)
APPEARS IN
संबंधित प्रश्न
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
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.]
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint : From A and C, draw perpendiculars to BD.]
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).
The area of the parallelogram ABCD is 90 cm2 (see figure). Find
- ar (ΔABEF)
- ar (ΔABD)
- ar (ΔBEF)
If the medians of a ∆ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = `1/3` ar (ABC)
