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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
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 following, AC // PS // QR and PQ // DB // SR.
Prove that: Area of quadrilateral PQRS = 2 x Area of the quad. ABCD.
In the given figure, AP is parallel to BC, BP is parallel to CQ.
Prove that the area of triangles ABC and BQP are equal.
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.
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 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.
Show that:
The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.
