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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
In the given figure, AD // BE // CF.
Prove that area (ΔAEC) = area (ΔDBF)
In the following, AC // PS // QR and PQ // DB // SR.
Prove that: Area of quadrilateral PQRS = 2 x Area of the quad. ABCD.
In parallelogram ABCD, P is a point on side AB and Q is a point on side BC.
Prove that:
(i) ΔCPD and ΔAQD are equal in the area.
(ii) Area (ΔAQD) = Area (ΔAPD) + Area (ΔCPB)
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.
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.
E, F, G, and H are the midpoints of the sides of a parallelogram ABCD.
Show that the area of quadrilateral EFGH is half 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 )
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.
