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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
D and E are points on sides AB and AC respectively of ΔABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
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, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-
(i) ΔMBC ≅ ΔABD
(ii) ar (BYXD) = 2 ar(MBC)
(iii) ar (BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar (CYXE) = ar(ACFG)
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
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).
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PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.
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).
The area of the parallelogram ABCD is 90 cm2 (see figure). Find
- ar (ΔABEF)
- ar (ΔABD)
- ar (ΔBEF)
