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प्रश्न
Show that:
The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.
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उत्तर
Consider the following figure :
Here
Ar. ( ΔABC ) = `1/2` BM x AC
and, Ar. ( ΔADC ) = `1/2` DN x AC
`["Area"( Δ"ABD")]/["Area(Δ ADC )"] = [1/2 "BM" xx "AC"]/[1/2 "DN" xx "AC"]= "BM"/"DN"`
hence proved
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संबंधित प्रश्न
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, AD // BE // CF.
Prove that area (ΔAEC) = area (ΔDBF)
In the following figure, DE is parallel to BC.
Show that:
(i) Area ( ΔADC ) = Area( ΔAEB ).
(ii) Area ( ΔBOD ) = Area( ΔCOE ).
Show that:
A diagonal divides a parallelogram into two triangles of equal area.
ABCD is a parallelogram in which BC is produced to E such that CE = BC and AE intersects CD at F.
If ar.(∆DFB) = 30 cm2; find the area of parallelogram.
In a parallelogram ABCD, point P lies in DC such that DP: PC = 3:2. If the area of ΔDPB = 30 sq. cm.
find the area of the parallelogram ABCD.
ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively.
Prove that area of triangle APQ = `1/8` of the area of parallelogram ABCD.
In the given figure, the diagonals AC and BD intersect at point O. If OB = OD and AB//DC,
show that:
(i) Area (Δ DOC) = Area (Δ AOB).
(ii) Area (Δ DCB) = Area (Δ ACB).
(iii) ABCD is a parallelogram.
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.
Show that:
The ratio of the areas of two triangles of the same height is equal to the ratio of their bases.
