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प्रश्न
In ΔABC, E and F are mid-points of sides AB and AC respectively. If BF and CE intersect each other at point O,
prove that the ΔOBC and quadrilateral AEOF are equal in area.
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उत्तर
E and F are the midpoints of the sides AB and AC.
Consider the following figure.
Therefore, by midpoint theorem, we have, EF || BC
Triangles BEF and CEF lie on the common base EF and between the parallels, EF and BC
Therefore, Ar.( ΔBEF ) = Ar.( ΔCOF )
⇒ Ar.( ΔBOE ) + Ar.( ΔEOF ) = Ar.( ΔEOF ) + Ar.( ΔCOF )
⇒ Ar.(ΔBOE ) = Ar.( ΔCOF )
Now BF and CE are the medians of the triangle ABC
Medians of the triangle divide it into two equal areas of triangles.
Thus, we have, Ar. (ΔABF) = Ar. (ΔCBF)
Subtracting Ar. ΔBOE on both the sides, we have
Ar. (ΔABF) - Ar. (ΔBOE) = Ar. (ΔCBF) - Ar. (ΔBOE)
Since, Ar. ( ΔBOE ) = Ar. ( ΔCOF ),
Ar. (ΔABF) - Ar. (ΔBOE) = Ar. (ΔCBF) - Ar. (ΔCOF)
Ar. ( quad. AEOF ) = Ar. ( ΔOBC ) , hence proved
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संबंधित प्रश्न
In the given figure, if the area of triangle ADE is 60 cm2, state, given reason, the area of :
(i) Parallelogram ABED;
(ii) Rectangle ABCF;
(iii) Triangle ABE.
The given figure shows a rectangle ABDC and a parallelogram ABEF; drawn on opposite sides of AB.
Prove that:
(i) Quadrilateral CDEF is a parallelogram;
(ii) Area of the quad. CDEF
= Area of rect. ABDC + Area of // gm. ABEF.
In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS. Show that:
(i) 2 Area (POS) = Area (// gm PMLS)
(ii) Area (POS) + Area (QOR) = Area (// gm PQRS)
(iii) Area (POS) + Area (QOR) = Area (POQ) + Area (SOR).
In the following figure, DE is parallel to BC.
Show that:
(i) Area ( ΔADC ) = Area( ΔAEB ).
(ii) Area ( ΔBOD ) = Area( ΔCOE ).
In the figure given alongside, squares ABDE and AFGC are drawn on the side AB and the hypotenuse AC of the right triangle ABC.
If BH is perpendicular to FG
prove that:
- ΔEAC ≅ ΔBAF
- Area of the square ABDE
- Area of the rectangle ARHF.
ABCD is a trapezium with AB parallel to DC. A line parallel to AC intersects AB at X and BC at Y.
Prove that the area of ∆ADX = area of ∆ACY.
In the following figure, BD is parallel to CA, E is mid-point of CA and BD = `1/2`CA
Prove that: ar. ( ΔABC ) = 2 x ar.( ΔDBC )
The given figure shows a parallelogram ABCD with area 324 sq. cm. P is a point in AB such that AP: PB = 1:2
Find The area of Δ APD.
In parallelogram ABCD, P is the mid-point of AB. CP and BD intersect each other at point O. If the area of ΔPOB = 40 cm2, and OP: OC = 1:2, find:
(i) Areas of ΔBOC and ΔPBC
(ii) Areas of ΔABC and parallelogram ABCD.
Show that:
The ratio of the areas of two triangles of the same height is equal to the ratio of their bases.
