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प्रश्न
Show that:
The ratio of the areas of two triangles of the same height is equal to the ratio of their bases.
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उत्तर
Consider the following figure:
Here AP ⊥ BC
Since Ar. ( ΔABD ) = `1/2` BD x AP
And, Ar. ( ΔADC ) =`1/2` DC x AP
`["Area"( ΔABD)]/["Area"(Δ ADC )] = [1/2 BD xx AP]/[1/2 DC xx AP]= (BD)/(DC)`
Hence proved.
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संबंधित प्रश्न
The given figure shows the parallelograms ABCD and APQR.
Show that these parallelograms are equal in the area.
[ Join B and R ]
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.
ABCD is a trapezium with AB // DC. A line parallel to AC intersects AB at point M and BC at point N.
Prove that: area of Δ ADM = area of Δ ACN.
ABCD and BCFE are parallelograms. If area of triangle EBC = 480 cm2; AB = 30 cm and BC = 40 cm.
Calculate :
(i) Area of parallelogram ABCD;
(ii) Area of the parallelogram BCFE;
(iii) Length of altitude from A on CD;
(iv) Area of triangle ECF.
In the given figure, M and N are the mid-points of the sides DC and AB respectively of the parallelogram ABCD.
If the area of parallelogram ABCD is 48 cm2;
(i) State the area of the triangle BEC.
(ii) Name the parallelogram which is equal in area to the triangle BEC.
In the following figure, CE is drawn parallel to diagonals DB of the quadrilateral ABCD which meets AB produced at point E.
Prove that ΔADE and quadrilateral ABCD are equal in area.
ABCD is a parallelogram a line through A cuts DC at point P and BC produced at Q. Prove that triangle BCP is equal in area to triangle DPQ.
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 )
In parallelogram ABCD, E is a point in AB and DE meets diagonal AC at point F. If DF: FE = 5:3 and area of ΔADF is 60 cm2; find
(i) area of ΔADE.
(ii) if AE: EB = 4:5, find the area of ΔADB.
(iii) also, find the area of parallelogram ABCD.
Show that:
The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.
