Advertisements
Advertisements
प्रश्न
Show that:
The ratio of the areas of two triangles of the same height is equal to the ratio of their bases.
Advertisements
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.
APPEARS IN
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 the parallelograms ABCD and APQR.
Show that these parallelograms are equal in the area.
[ Join B and R ]
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.
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 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 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.
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 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.
Show that:
The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.
