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प्रश्न
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)
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उत्तर
(i) In ΔABC,
E and F are the mid-points of side AC and AB respectively.
Therefore, EF || BC and EF = 1/2BC (Mid-point theorem)
However, BD = 1/2BC (D is the mid-point of BC)
Therefore, BD = EF and BD || EF
Therefore, BDEF is a parallelogram.
(ii) Using the result obtained above, it can be said that quadrilaterals BDEF, DCEF, AFDE are parallelograms.
We know that diagonal of a parallelogram divides it into two triangles of equal area.
∴Area (ΔBFD) = Area (ΔDEF) (For parallelogram BD)
Area (ΔCDE) = Area (ΔDEF) (For parallelogram DCEF)
Area (ΔAFE) = Area (ΔDEF) (For parallelogram AFDE)
∴Area (ΔAFE) = Area (ΔBFD) = Area (ΔCDE) = Area (ΔDEF)
Also,
Area (ΔAFE) + Area (ΔBDF) + Area (ΔCDE) + Area (ΔDEF) = Area (ΔABC)
⇒ Area (ΔDEF) + Area (ΔDEF) + Area (ΔDEF) + Area (ΔDEF) = Area (ΔABC)
⇒ 4 Area (ΔDEF) = Area (ΔABC)
⇒ Area (ΔDEF) = 1/4Area (ΔABC)
(iii) Area (parallelogram BDEF) = Area (ΔDEF) + Area (ΔBDF)
⇒ Area (parallelogram BDEF) = Area (ΔDEF) + Area (ΔDEF)
⇒ Area (parallelogram BDEF) = 2 Area (ΔDEF)
⇒ Area (parallelogram BDEF)`= 2xx1/4"Area "(ΔABC)`
⇒ Area (parallelogram BDEF) = 1/2Area (ΔABC)
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).
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)
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
(i) ar (BDE) = 1/4 ar (ABC)
(ii) ar (BDE) = 1/2 ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = 1/8 ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
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)
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 (ΔBEF)
O is any point on the diagonal PR of a parallelogram PQRS (Figure). Prove that ar (PSO) = ar (PQO).
