Advertisements
Advertisements
Question
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).
Advertisements
Solution
AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas.
∴ Area (ΔABD) = Area (ΔACD)
⇒ Area (ΔABD) = 1/2Area (ΔABC)... (1)
In ΔABD, E is the mid-point of AD. Therefore, BE is the median.
∴ Area (ΔBED) = Area (ΔABE)
⇒ Area (ΔBED) = 1/2Area (ΔABD)
⇒ Area (ΔBED) = 1/2*1/2Area (ΔABC) [From equation (1)]
⇒ Area (ΔBED) = 1/4Area (ΔABC)
RELATED QUESTIONS
In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE)
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that
(i) BDEF is a parallelogram.
(ii) ar (DEF) = 1/4ar (ABC)
(iii) ar (BDEF) = 1/2ar (ABC)
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 given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)
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 a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :
(1) ar (Δ PBQ) = ar (Δ ARC)
(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)
(3) ar (Δ RQC) =`3/8` ar (Δ ABC) .
In the following figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = `1/2` ar (EFGD).
O is any point on the diagonal PR of a parallelogram PQRS (Figure). Prove that ar (PSO) = ar (PQO).
In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)
